{
  "cells": [
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "dxOT-q6TqcAw"
      },
      "source": [
        "To run this, press \"*Runtime*\" and press \"*Run all*\" on a **free** Tesla T4 Google Colab instance!\n",
        "<div class=\"align-center\">\n",
        "<a href=\"https://unsloth.ai/\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/unsloth%20new%20logo.png\" width=\"115\"></a>\n",
        "<a href=\"https://discord.gg/unsloth\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/Discord button.png\" width=\"145\"></a>\n",
        "<a href=\"https://docs.unsloth.ai/\"><img src=\"https://github.com/unslothai/unsloth/blob/main/images/documentation%20green%20button.png?raw=true\" width=\"125\"></a></a> Join Discord if you need help + ⭐ <i>Star us on <a href=\"https://github.com/unslothai/unsloth\">Github</a> </i> ⭐\n",
        "</div>\n",
        "\n",
        "To install Unsloth on your own computer, follow the installation instructions on our Github page [here](https://docs.unsloth.ai/get-started/installing-+-updating).\n",
        "\n",
        "You will learn how to do [data prep](#Data), how to [train](#Train), how to [run the model](#Inference), & [how to save it](#Save)\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "gmzFABkdqcAx"
      },
      "source": [
        "### News"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "ZnfTyJwdqcAx"
      },
      "source": [
        "Unsloth now supports Text-to-Speech (TTS) models. Read our [guide here](https://docs.unsloth.ai/basics/text-to-speech-tts-fine-tuning).\n",
        "\n",
        "Read our **[Qwen3 Guide](https://docs.unsloth.ai/basics/qwen3-how-to-run-and-fine-tune)** and check out our new **[Dynamic 2.0](https://docs.unsloth.ai/basics/unsloth-dynamic-2.0-ggufs)** quants which outperforms other quantization methods!\n",
        "\n",
        "Visit our docs for all our [model uploads](https://docs.unsloth.ai/get-started/all-our-models) and [notebooks](https://docs.unsloth.ai/get-started/unsloth-notebooks).\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "AE3eYvpBqcAx"
      },
      "source": [
        "### Installation"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 2,
      "metadata": {
        "id": "nRVlVpIPqcAx"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "Collecting unsloth\n",
            "  Using cached unsloth-2025.5.9-py3-none-any.whl.metadata (47 kB)\n",
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            "^C\n",
            "\u001b[31mERROR: Operation cancelled by user\u001b[0m\u001b[31m\n",
            "\u001b[0m"
          ]
        }
      ],
      "source": [
        "# %%capture\n",
        "import os\n",
        "if \"COLAB_\" not in \"\".join(os.environ.keys()):\n",
        "    !pip install unsloth vllm\n",
        "else:\n",
        "    # [NOTE] Do the below ONLY in Colab! Use [[pip install unsloth vllm]]\n",
        "    !pip install --no-deps unsloth vllm==0.8.5.post1"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "IJ_lZtZhqcAy"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
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            "  Using cached xformers-0.0.30.tar.gz (10.2 MB)\n",
            "  Installing build dependencies ... \u001b[?25ldone\n",
            "\u001b[?25h  Getting requirements to build wheel ... \u001b[?25lerror\n",
            "  \u001b[1;31merror\u001b[0m: \u001b[1msubprocess-exited-with-error\u001b[0m\n",
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            "  \u001b[31m×\u001b[0m \u001b[32mGetting requirements to build wheel\u001b[0m did not run successfully.\n",
            "  \u001b[31m│\u001b[0m exit code: \u001b[1;36m1\u001b[0m\n",
            "  \u001b[31m╰─>\u001b[0m \u001b[31m[23 lines of output]\u001b[0m\n",
            "  \u001b[31m   \u001b[0m Traceback (most recent call last):\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/home/user/.pyenv/versions/3.13.2/lib/python3.13/site-packages/pip/_vendor/pyproject_hooks/_in_process/_in_process.py\"\u001b[0m, line \u001b[35m353\u001b[0m, in \u001b[35m<module>\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31mmain\u001b[0m\u001b[1;31m()\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31m~~~~\u001b[0m\u001b[1;31m^^\u001b[0m\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/home/user/.pyenv/versions/3.13.2/lib/python3.13/site-packages/pip/_vendor/pyproject_hooks/_in_process/_in_process.py\"\u001b[0m, line \u001b[35m335\u001b[0m, in \u001b[35mmain\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     json_out['return_val'] = \u001b[31mhook\u001b[0m\u001b[1;31m(**hook_input['kwargs'])\u001b[0m\n",
            "  \u001b[31m   \u001b[0m                              \u001b[31m~~~~\u001b[0m\u001b[1;31m^^^^^^^^^^^^^^^^^^^^^^^^\u001b[0m\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/home/user/.pyenv/versions/3.13.2/lib/python3.13/site-packages/pip/_vendor/pyproject_hooks/_in_process/_in_process.py\"\u001b[0m, line \u001b[35m118\u001b[0m, in \u001b[35mget_requires_for_build_wheel\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     return hook(config_settings)\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/tmp/pip-build-env-2d8ax1lo/overlay/lib/python3.13/site-packages/setuptools/build_meta.py\"\u001b[0m, line \u001b[35m331\u001b[0m, in \u001b[35mget_requires_for_build_wheel\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     return \u001b[31mself._get_build_requires\u001b[0m\u001b[1;31m(config_settings, requirements=[])\u001b[0m\n",
            "  \u001b[31m   \u001b[0m            \u001b[31m~~~~~~~~~~~~~~~~~~~~~~~~\u001b[0m\u001b[1;31m^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\u001b[0m\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/tmp/pip-build-env-2d8ax1lo/overlay/lib/python3.13/site-packages/setuptools/build_meta.py\"\u001b[0m, line \u001b[35m301\u001b[0m, in \u001b[35m_get_build_requires\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31mself.run_setup\u001b[0m\u001b[1;31m()\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31m~~~~~~~~~~~~~~\u001b[0m\u001b[1;31m^^\u001b[0m\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/tmp/pip-build-env-2d8ax1lo/overlay/lib/python3.13/site-packages/setuptools/build_meta.py\"\u001b[0m, line \u001b[35m512\u001b[0m, in \u001b[35mrun_setup\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31msuper().run_setup\u001b[0m\u001b[1;31m(setup_script=setup_script)\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31m~~~~~~~~~~~~~~~~~\u001b[0m\u001b[1;31m^^^^^^^^^^^^^^^^^^^^^^^^^^^\u001b[0m\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"/tmp/pip-build-env-2d8ax1lo/overlay/lib/python3.13/site-packages/setuptools/build_meta.py\"\u001b[0m, line \u001b[35m317\u001b[0m, in \u001b[35mrun_setup\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31mexec\u001b[0m\u001b[1;31m(code, locals())\u001b[0m\n",
            "  \u001b[31m   \u001b[0m     \u001b[31m~~~~\u001b[0m\u001b[1;31m^^^^^^^^^^^^^^^^\u001b[0m\n",
            "  \u001b[31m   \u001b[0m   File \u001b[35m\"<string>\"\u001b[0m, line \u001b[35m24\u001b[0m, in \u001b[35m<module>\u001b[0m\n",
            "  \u001b[31m   \u001b[0m \u001b[1;35mModuleNotFoundError\u001b[0m: \u001b[35mNo module named 'torch'\u001b[0m\n",
            "  \u001b[31m   \u001b[0m \u001b[31m[end of output]\u001b[0m\n",
            "  \n",
            "  \u001b[1;35mnote\u001b[0m: This error originates from a subprocess, and is likely not a problem with pip.\n",
            "\u001b[?25h\n",
            "\u001b[1m[\u001b[0m\u001b[34;49mnotice\u001b[0m\u001b[1;39;49m]\u001b[0m\u001b[39;49m A new release of pip is available: \u001b[0m\u001b[31;49m24.3.1\u001b[0m\u001b[39;49m -> \u001b[0m\u001b[32;49m25.1.1\u001b[0m\n",
            "\u001b[1m[\u001b[0m\u001b[34;49mnotice\u001b[0m\u001b[1;39;49m]\u001b[0m\u001b[39;49m To update, run: \u001b[0m\u001b[32;49mpip install --upgrade pip\u001b[0m\n",
            "\u001b[1;31merror\u001b[0m: \u001b[1msubprocess-exited-with-error\u001b[0m\n",
            "\n",
            "\u001b[31m×\u001b[0m \u001b[32mGetting requirements to build wheel\u001b[0m did not run successfully.\n",
            "\u001b[31m│\u001b[0m exit code: \u001b[1;36m1\u001b[0m\n",
            "\u001b[31m╰─>\u001b[0m See above for output.\n",
            "\n",
            "\u001b[1;35mnote\u001b[0m: This error originates from a subprocess, and is likely not a problem with pip.\n"
          ]
        }
      ],
      "source": [
        "# %%capture\n",
        "#@title Colab Extra Install { display-mode: \"form\" }\n",
        "import os\n",
        "if \"COLAB_\" not in \"\".join(os.environ.keys()):\n",
        "    !pip install unsloth vllm\n",
        "else:\n",
        "    !pip install --no-deps unsloth vllm==0.8.5.post1\n",
        "    # [NOTE] Do the below ONLY in Colab! Use [[pip install unsloth vllm]]\n",
        "    # Skip restarting message in Colab\n",
        "    import sys, re, requests; modules = list(sys.modules.keys())\n",
        "    for x in modules: sys.modules.pop(x) if \"PIL\" in x or \"google\" in x else None\n",
        "    !pip install --no-deps bitsandbytes accelerate xformers==0.0.29.post3 peft \"trl==0.15.2\" triton cut_cross_entropy unsloth_zoo\n",
        "    !pip install sentencepiece protobuf \"datasets>=3.4.1\" huggingface_hub hf_transfer\n",
        "    !pip install transformers==4.51.3\n",
        "\n",
        "    # vLLM requirements - vLLM breaks Colab due to reinstalling numpy\n",
        "    f = requests.get(\"https://raw.githubusercontent.com/vllm-project/vllm/refs/heads/main/requirements/common.txt\").content\n",
        "    with open(\"vllm_requirements.txt\", \"wb\") as file:\n",
        "        file.write(re.sub(rb\"(transformers|numpy|xformers)[^\\n]{1,}\\n\", b\"\", f))\n",
        "    !pip install -r vllm_requirements.txt"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "ZkH_y8UC9lvv"
      },
      "source": [
        "### Unsloth"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "jN75nmdx9lvw"
      },
      "source": [
        "Goal: To convert `Qwen3-4B-Base` into a reasoning model via GRPO by using OpenR1's Math dataset.\n",
        "\n",
        "We first pre fine-tune the model to make GRPO skip trying to match formatting - this speeds GRPO up."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 5,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 1000,
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        },
        "id": "DkIvEkIIkEyB",
        "outputId": "b20d048a-450d-4c0a-c889-cee88da7d090"
      },
      "outputs": [
        {
          "name": "stderr",
          "output_type": "stream",
          "text": [
            "<unknown>:476: SyntaxWarning: invalid escape sequence '\\.'\n",
            "<unknown>:839: SyntaxWarning: invalid escape sequence '\\)'\n",
            "<unknown>:924: SyntaxWarning: invalid escape sequence '\\)'\n",
            "<unknown>:476: SyntaxWarning: invalid escape sequence '\\.'\n",
            "<unknown>:839: SyntaxWarning: invalid escape sequence '\\)'\n",
            "<unknown>:924: SyntaxWarning: invalid escape sequence '\\)'\n",
            "<unknown>:476: SyntaxWarning: invalid escape sequence '\\.'\n",
            "<unknown>:839: SyntaxWarning: invalid escape sequence '\\)'\n",
            "<unknown>:924: SyntaxWarning: invalid escape sequence '\\)'\n"
          ]
        },
        {
          "ename": "ImportError",
          "evalue": "Unsloth: If you are in Colab, we updated the top cell install instructions - please change it to below then press Disconnect Runtime and then Restart it.\n\n%%capture\n# Installs Unsloth, Xformers (Flash Attention) and all other packages!\n!pip install \"unsloth[colab-new] @ git+https://github.com/unslothai/unsloth.git\"\n!pip install --no-deps \"xformers<0.0.27\" \"trl<0.9.0\" peft accelerate bitsandbytes\n\nOtherwise in local machines, your xformers version of 0.0.30 is too new.\nPlease downgrade xformers via `pip install --force-reinstall \"xformers<0.0.27\"",
          "output_type": "error",
          "traceback": [
            "\u001b[31m---------------------------------------------------------------------------\u001b[39m",
            "\u001b[31mImportError\u001b[39m                               Traceback (most recent call last)",
            "\u001b[36mCell\u001b[39m\u001b[36m \u001b[39m\u001b[32mIn[5]\u001b[39m\u001b[32m, line 1\u001b[39m\n\u001b[32m----> \u001b[39m\u001b[32m1\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01munsloth\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m FastLanguageModel\n\u001b[32m      2\u001b[39m \u001b[38;5;28;01mimport\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01mtorch\u001b[39;00m\n\u001b[32m      3\u001b[39m max_seq_length = \u001b[32m2048\u001b[39m \u001b[38;5;66;03m# Can increase for longer reasoning traces\u001b[39;00m\n",
            "\u001b[36mFile \u001b[39m\u001b[32m~/projects/ai_stack/notebooks/.venv/lib/python3.12/site-packages/unsloth/__init__.py:158\u001b[39m\n\u001b[32m    148\u001b[39m         warnings.warn(\n\u001b[32m    149\u001b[39m             \u001b[33m\"\u001b[39m\u001b[33mUnsloth: CUDA is not linked properly.\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\\\n\u001b[32m    150\u001b[39m             \u001b[33m\"\u001b[39m\u001b[33mTry running `python -m bitsandbytes` then `python -m xformers.info`\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\\\n\u001b[32m   (...)\u001b[39m\u001b[32m    154\u001b[39m             \u001b[33m\"\u001b[39m\u001b[33mUnsloth will still run for now, but maybe it might crash - let\u001b[39m\u001b[33m'\u001b[39m\u001b[33ms hope it works!\u001b[39m\u001b[33m\"\u001b[39m\n\u001b[32m    155\u001b[39m         )\n\u001b[32m    156\u001b[39m \u001b[38;5;28;01mpass\u001b[39;00m\n\u001b[32m--> \u001b[39m\u001b[32m158\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mmodels\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m *\n\u001b[32m    159\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01msave\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m *\n\u001b[32m    160\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mchat_templates\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m *\n",
            "\u001b[36mFile \u001b[39m\u001b[32m~/projects/ai_stack/notebooks/.venv/lib/python3.12/site-packages/unsloth/models/__init__.py:15\u001b[39m\n\u001b[32m      1\u001b[39m \u001b[38;5;66;03m# Copyright 2023-present Daniel Han-Chen & the Unsloth team. All rights reserved.\u001b[39;00m\n\u001b[32m      2\u001b[39m \u001b[38;5;66;03m#\u001b[39;00m\n\u001b[32m      3\u001b[39m \u001b[38;5;66;03m# Licensed under the Apache License, Version 2.0 (the \"License\");\u001b[39;00m\n\u001b[32m   (...)\u001b[39m\u001b[32m     12\u001b[39m \u001b[38;5;66;03m# See the License for the specific language governing permissions and\u001b[39;00m\n\u001b[32m     13\u001b[39m \u001b[38;5;66;03m# limitations under the License.\u001b[39;00m\n\u001b[32m---> \u001b[39m\u001b[32m15\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mloader\u001b[39;00m\u001b[38;5;250m  \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m FastLanguageModel\n\u001b[32m     16\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mllama\u001b[39;00m\u001b[38;5;250m   \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m FastLlamaModel\n\u001b[32m     17\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mmistral\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m FastMistralModel\n",
            "\u001b[36mFile \u001b[39m\u001b[32m~/projects/ai_stack/notebooks/.venv/lib/python3.12/site-packages/unsloth/models/loader.py:15\u001b[39m\n\u001b[32m      1\u001b[39m \u001b[38;5;66;03m# Copyright 2023-present Daniel Han-Chen & the Unsloth team. All rights reserved.\u001b[39;00m\n\u001b[32m      2\u001b[39m \u001b[38;5;66;03m#\u001b[39;00m\n\u001b[32m      3\u001b[39m \u001b[38;5;66;03m# Licensed under the Apache License, Version 2.0 (the \"License\");\u001b[39;00m\n\u001b[32m   (...)\u001b[39m\u001b[32m     12\u001b[39m \u001b[38;5;66;03m# See the License for the specific language governing permissions and\u001b[39;00m\n\u001b[32m     13\u001b[39m \u001b[38;5;66;03m# limitations under the License.\u001b[39;00m\n\u001b[32m---> \u001b[39m\u001b[32m15\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01m_utils\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m is_bfloat16_supported, HAS_FLASH_ATTENTION, HAS_FLASH_ATTENTION_SOFTCAPPING\n\u001b[32m     16\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mllama\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m FastLlamaModel, logger\n\u001b[32m     17\u001b[39m \u001b[38;5;28;01mfrom\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[34;01m.\u001b[39;00m\u001b[34;01mmistral\u001b[39;00m\u001b[38;5;250m \u001b[39m\u001b[38;5;28;01mimport\u001b[39;00m FastMistralModel\n",
            "\u001b[36mFile \u001b[39m\u001b[32m~/projects/ai_stack/notebooks/.venv/lib/python3.12/site-packages/unsloth/models/_utils.py:196\u001b[39m\n\u001b[32m    194\u001b[39m \u001b[38;5;66;03m# Temporarily disable 0.0.27 and higher - inference issues\u001b[39;00m\n\u001b[32m    195\u001b[39m \u001b[38;5;28;01mif\u001b[39;00m Version(xformers_version) >= Version(\u001b[33m\"\u001b[39m\u001b[33m0.0.27\u001b[39m\u001b[33m\"\u001b[39m):\n\u001b[32m--> \u001b[39m\u001b[32m196\u001b[39m     \u001b[38;5;28;01mraise\u001b[39;00m \u001b[38;5;167;01mImportError\u001b[39;00m(\n\u001b[32m    197\u001b[39m         \u001b[33m\"\u001b[39m\u001b[33mUnsloth: If you are in Colab, we updated the top cell install instructions - please change it to below \u001b[39m\u001b[33m\"\u001b[39m\\\n\u001b[32m    198\u001b[39m         \u001b[33m\"\u001b[39m\u001b[33mthen press Disconnect Runtime and then Restart it.\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\\\n\u001b[32m    199\u001b[39m         \u001b[33m\"\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\\\n\u001b[32m    200\u001b[39m         \u001b[33m\"\u001b[39m\u001b[38;5;132;01m%%\u001b[39;00m\u001b[33mcapture\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\n\u001b[32m    201\u001b[39m         \u001b[33m\"\u001b[39m\u001b[33m# Installs Unsloth, Xformers (Flash Attention) and all other packages!\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\n\u001b[32m    202\u001b[39m         \u001b[33m'\u001b[39m\u001b[33m!pip install \u001b[39m\u001b[33m\"\u001b[39m\u001b[33munsloth[colab-new] @ git+https://github.com/unslothai/unsloth.git\u001b[39m\u001b[33m\"\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m'\u001b[39m\n\u001b[32m    203\u001b[39m         \u001b[33m'\u001b[39m\u001b[33m!pip install --no-deps \u001b[39m\u001b[33m\"\u001b[39m\u001b[33mxformers<0.0.27\u001b[39m\u001b[33m\"\u001b[39m\u001b[33m \u001b[39m\u001b[33m\"\u001b[39m\u001b[33mtrl<0.9.0\u001b[39m\u001b[33m\"\u001b[39m\u001b[33m peft accelerate bitsandbytes\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m'\u001b[39m\\\n\u001b[32m    204\u001b[39m         \u001b[33m'\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m'\u001b[39m\\\n\u001b[32m    205\u001b[39m         \u001b[33mf\u001b[39m\u001b[33m\"\u001b[39m\u001b[33mOtherwise in local machines, your xformers version of \u001b[39m\u001b[38;5;132;01m{\u001b[39;00mxformers_version\u001b[38;5;132;01m}\u001b[39;00m\u001b[33m is too new.\u001b[39m\u001b[38;5;130;01m\\n\u001b[39;00m\u001b[33m\"\u001b[39m\\\n\u001b[32m    206\u001b[39m         \u001b[33m'\u001b[39m\u001b[33mPlease downgrade xformers via `pip install --force-reinstall \u001b[39m\u001b[33m\"\u001b[39m\u001b[33mxformers<0.0.27\u001b[39m\u001b[33m\"\u001b[39m\u001b[33m'\u001b[39m\n\u001b[32m    207\u001b[39m     )\n\u001b[32m    208\u001b[39m \u001b[38;5;28;01mpass\u001b[39;00m\n\u001b[32m    210\u001b[39m \u001b[38;5;28;01mif\u001b[39;00m   Version(torch_version) < Version(\u001b[33m\"\u001b[39m\u001b[33m2.2.0\u001b[39m\u001b[33m\"\u001b[39m) \u001b[38;5;129;01mand\u001b[39;00m Version(xformers_version) >= Version(\u001b[33m\"\u001b[39m\u001b[33m0.0.24\u001b[39m\u001b[33m\"\u001b[39m):\n",
            "\u001b[31mImportError\u001b[39m: Unsloth: If you are in Colab, we updated the top cell install instructions - please change it to below then press Disconnect Runtime and then Restart it.\n\n%%capture\n# Installs Unsloth, Xformers (Flash Attention) and all other packages!\n!pip install \"unsloth[colab-new] @ git+https://github.com/unslothai/unsloth.git\"\n!pip install --no-deps \"xformers<0.0.27\" \"trl<0.9.0\" peft accelerate bitsandbytes\n\nOtherwise in local machines, your xformers version of 0.0.30 is too new.\nPlease downgrade xformers via `pip install --force-reinstall \"xformers<0.0.27\""
          ]
        }
      ],
      "source": [
        "from unsloth import FastLanguageModel\n",
        "import torch\n",
        "max_seq_length = 2048 # Can increase for longer reasoning traces\n",
        "lora_rank = 32 # Larger rank = smarter, but slower\n",
        "\n",
        "model, tokenizer = FastLanguageModel.from_pretrained(\n",
        "    model_name = \"unsloth/Qwen3-4B-Base\",\n",
        "    max_seq_length = max_seq_length,\n",
        "    load_in_4bit = False, # False for LoRA 16bit\n",
        "    fast_inference = True, # Enable vLLM fast inference\n",
        "    max_lora_rank = lora_rank,\n",
        "    gpu_memory_utilization = 0.7, # Reduce if out of memory\n",
        ")\n",
        "\n",
        "model = FastLanguageModel.get_peft_model(\n",
        "    model,\n",
        "    r = lora_rank, # Choose any number > 0 ! Suggested 8, 16, 32, 64, 128\n",
        "    target_modules = [\n",
        "        \"q_proj\", \"k_proj\", \"v_proj\", \"o_proj\",\n",
        "        \"gate_proj\", \"up_proj\", \"down_proj\",\n",
        "    ],\n",
        "    lora_alpha = lora_rank*2, # *2 speeds up training\n",
        "    use_gradient_checkpointing = \"unsloth\", # Reduces memory usage\n",
        "    random_state = 3407,\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "W9DuiVRLhMco"
      },
      "source": [
        "### GRPO chat template\n",
        "Since we're using a base model, we should set a chat template. You can make your own chat template as well!\n",
        "1. DeepSeek uses `<think>` and `</think>`, but this is **not** necessary - you can customize it however you like!\n",
        "2. A `system_prompt` is recommended to at least guide the model's responses."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 52
        },
        "id": "6UjowCbT-cFz",
        "outputId": "9390bd61-7864-4f23-c40c-4f69c8752cd3"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "'You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION>'"
            ]
          },
          "execution_count": 4,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "reasoning_start = \"<start_working_out>\" # Acts as <think>\n",
        "reasoning_end   = \"<end_working_out>\"   # Acts as </think>\n",
        "solution_start  = \"<SOLUTION>\"\n",
        "solution_end    = \"</SOLUTION>\"\n",
        "\n",
        "system_prompt = \\\n",
        "f\"\"\"You are given a problem.\n",
        "Think about the problem and provide your working out.\n",
        "Place it between {reasoning_start} and {reasoning_end}.\n",
        "Then, provide your solution between {solution_start}{solution_end}\"\"\"\n",
        "system_prompt"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "zGgs0MJkDkYL"
      },
      "source": [
        "We create a simple chat template below. Notice `add_generation_prompt` includes prepending `<start_working_out>` to guide the model to start its reasoning process."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "Y3fF9gMujY02"
      },
      "outputs": [],
      "source": [
        "chat_template = \\\n",
        "    \"{% if messages[0]['role'] == 'system' %}\"\\\n",
        "        \"{{ messages[0]['content'] + eos_token }}\"\\\n",
        "        \"{% set loop_messages = messages[1:] %}\"\\\n",
        "    \"{% else %}\"\\\n",
        "        \"{{ '{system_prompt}' + eos_token }}\"\\\n",
        "        \"{% set loop_messages = messages %}\"\\\n",
        "    \"{% endif %}\"\\\n",
        "    \"{% for message in loop_messages %}\"\\\n",
        "        \"{% if message['role'] == 'user' %}\"\\\n",
        "            \"{{ message['content'] }}\"\\\n",
        "        \"{% elif message['role'] == 'assistant' %}\"\\\n",
        "            \"{{ message['content'] + eos_token }}\"\\\n",
        "        \"{% endif %}\"\\\n",
        "    \"{% endfor %}\"\\\n",
        "    \"{% if add_generation_prompt %}{{ '{reasoning_start}' }}\"\\\n",
        "    \"{% endif %}\"\n",
        "\n",
        "# Replace with out specific template:\n",
        "chat_template = chat_template\\\n",
        "    .replace(\"'{system_prompt}'\",   f\"'{system_prompt}'\")\\\n",
        "    .replace(\"'{reasoning_start}'\", f\"'{reasoning_start}'\")\n",
        "tokenizer.chat_template = chat_template"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "vEcLdymBEHdk"
      },
      "source": [
        "Let's see how our chat template behaves on an example:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 87
        },
        "id": "BciEDYSSYFNj",
        "outputId": "730f6a22-ebdc-4c25-ef57-b80d92d9c516"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "\"You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION><|endoftext|>What is 1+1?<start_working_out>I think it's 2.<end_working_out><SOLUTION>2</SOLUTION><|endoftext|>What is 2+2?<start_working_out>\""
            ]
          },
          "execution_count": 6,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "tokenizer.apply_chat_template([\n",
        "    {\"role\" : \"user\", \"content\" : \"What is 1+1?\"},\n",
        "    {\"role\" : \"assistant\", \"content\" : f\"{reasoning_start}I think it's 2.{reasoning_end}{solution_start}2{solution_end}\"},\n",
        "    {\"role\" : \"user\", \"content\" : \"What is 2+2?\"},\n",
        "], tokenize = False, add_generation_prompt = True)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "_mdsuGjxHrjT"
      },
      "source": [
        "### Pre fine-tuning for formatting\n",
        "We now use a subset of NVIDIA's [Open Math Reasoning dataset](https://huggingface.co/datasets/nvidia/OpenMathReasoning) which was filtered to only include high quality DeepSeek R1 traces.\n",
        "\n",
        "We'll only filter ~59 or so examples to first \"prime\" / pre fine-tune the model to understand our custom GRPO formatting."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 693,
          "referenced_widgets": [
            "b57476ddcf71441a9be7a7adb5344c97",
            "cab74710156848319ca1bd9f49956341",
            "6eb3e586a7944970b38740e5f66ac216",
            "9f4b810c28484866aee40c2178b54d94",
            "fd3c03604a4446c59d9b9512a5ddbec6",
            "cc44e1389c764206a2cda64169a94db3",
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              "summary": "{\n  \"name\": \"dataset\",\n  \"rows\": 7507,\n  \"fields\": [\n    {\n      \"column\": \"expected_answer\",\n      \"properties\": {\n        \"dtype\": \"category\",\n        \"num_unique_values\": 868,\n        \"samples\": [\n          \"672\",\n          \"335\",\n          \"575757\"\n        ],\n        \"semantic_type\": \"\",\n        \"description\": \"\"\n      }\n    },\n    {\n      \"column\": \"problem\",\n      \"properties\": {\n        \"dtype\": \"string\",\n        \"num_unique_values\": 3895,\n        \"samples\": [\n          \"A club with 7 members forms three-person committees, but no two committees can have more than one member in common. What is the maximum number of committees that can be formed?\",\n          \"Find the smallest integer \\\\( a > 2 \\\\) such that \\\\( 2 \\\\mid a \\\\), \\\\( 3 \\\\mid (a+1) \\\\), \\\\( 4 \\\\mid (a+2) \\\\), \\\\( 5 \\\\mid (a+3) \\\\), and \\\\( 6 \\\\mid (a+4) \\\\).\",\n          \"Given the polynomial equation \\\\(x^3 - x = -1\\\\) with roots \\\\(a\\\\), \\\\(b\\\\), and \\\\(c\\\\), find the value of \\\\(\\\\frac{1}{1+a} + \\\\frac{1}{1+b} + \\\\frac{1}{1+c}\\\\).\"\n        ],\n        \"semantic_type\": \"\",\n        \"description\": \"\"\n      }\n    },\n    {\n      \"column\": \"generated_solution\",\n      \"properties\": {\n        \"dtype\": \"string\",\n        \"num_unique_values\": 7507,\n        \"samples\": [\n          \"<think>\\nOkay, let's see. I need to solve this problem where there are three prime numbers p, q, and r. The equations given are pq + qr + rp = 191 and p + q = r - 1. The goal is to find p + q + r. Hmm, primes, so they must be 2 or odd primes. Let me start by analyzing the problem step by step.\\n\\nFirst, the equation p + q = r - 1. If I can express r in terms of p and q, maybe I can substitute that into the first equation. Let's try that. So, from the second equation, r = p + q + 1. That seems straightforward. Now, substituting this into the first equation, we get pq + qr + rp = 191. Let's replace r with (p + q + 1).\\n\\nSo, substituting, the first equation becomes:\\n\\npq + q(p + q + 1) + p(p + q + 1) = 191.\\n\\nLet me expand each term:\\n\\nFirst term: pq.\\n\\nSecond term: q*(p + q + 1) = pq + q\\u00b2 + q.\\n\\nThird term: p*(p + q + 1) = p\\u00b2 + pq + p.\\n\\nNow, adding all these together:\\n\\npq + (pq + q\\u00b2 + q) + (p\\u00b2 + pq + p) = 191.\\n\\nCombine like terms:\\n\\npq + pq + pq = 3pq.\\n\\nq\\u00b2 + p\\u00b2.\\n\\nq + p.\\n\\nSo altogether, the equation becomes:\\n\\np\\u00b2 + q\\u00b2 + 3pq + p + q = 191.\\n\\nHmm. Let me note that.\\n\\nAlternatively, maybe I can factor this expression. Let me see. Let's try to group terms. Wait, another approach: since r = p + q + 1, then p + q + r = (p + q) + (p + q + 1) = 2(p + q) + 1. So if I can find p + q, then multiplying by 2 and adding 1 gives me the answer. That's useful. So perhaps instead of dealing with the first equation directly, I can express it in terms of p + q.\\n\\nBut maybe let's work with the equation we derived: p\\u00b2 + q\\u00b2 + 3pq + p + q = 191.\\n\\nWait, another thought: p\\u00b2 + q\\u00b2 + 3pq. Let's recall that (p + q)^2 = p\\u00b2 + 2pq + q\\u00b2, so p\\u00b2 + q\\u00b2 + 3pq = (p + q)^2 + pq.\\n\\nTherefore, the equation can be rewritten as:\\n\\n(p + q)^2 + pq + p + q = 191.\\n\\nLet me set S = p + q, and P = pq. Then the equation becomes:\\n\\nS\\u00b2 + P + S = 191.\\n\\nBut from the second equation, we know that r = S + 1, so since we need to find S + r = S + (S + 1) = 2S + 1. So our target is 2S + 1. So if we can find S, we can find the answer.\\n\\nNow, the equation is S\\u00b2 + P + S = 191. But S and P are related as S = p + q, P = pq. For two primes p and q, their sum and product. Since p and q are primes, maybe we can list possible primes that add up to S and multiply to P. However, since S and P are variables here, maybe we can express P in terms of S from the equation.\\n\\nFrom S\\u00b2 + P + S = 191, so P = 191 - S\\u00b2 - S.\\n\\nTherefore, P = -S\\u00b2 - S + 191. But P = pq must be positive, so -S\\u00b2 - S + 191 > 0. Therefore, S\\u00b2 + S < 191. Let's see the possible values of S. Since S is the sum of two primes, which are at least 2 each, so S is at least 2 + 2 = 4. Also, since S\\u00b2 + S < 191, let's solve S\\u00b2 + S - 191 < 0. Let's find the roots of S\\u00b2 + S - 191 = 0.\\n\\nUsing quadratic formula: S = [-1 \\u00b1 sqrt(1 + 4*191)] / 2 = [-1 \\u00b1 sqrt(765)] / 2. sqrt(765) is approx 27.66, so S \\u2248 (-1 + 27.66)/2 \\u2248 13.33. So the positive root is approximately 13.33, so S must be less than 13.33. Therefore, S can be integers from 4 up to 13.\\n\\nSo possible S values: 4,5,6,7,8,9,10,11,12,13.\\n\\nBut since p and q are primes, their sum S must be even or odd. Since except for 2, all primes are odd. So if both p and q are odd primes, their sum is even. If one is 2 and the other is odd, then their sum is odd. So S can be even or odd. So possible values of S (from 4 to 13) can be checked.\\n\\nBut maybe instead of all possible S, let's check possible S values from 4 to 13 and see if for each S, P = 191 - S\\u00b2 - S, and check if P can be expressed as the product of two primes that add up to S.\\n\\nAlternatively, maybe first check which S gives P as a product of two primes.\\n\\nLet's start with S=4. Then P = 191 - 16 -4 = 171. Then check if 171 can be written as product of two primes that add up to 4. But 4 is the sum. The primes could be 2 and 2 (since 2+2=4). Then 2*2=4, but P here is 171. 4 \\u2260 171, so S=4 is invalid.\\n\\nNext S=5: P=191 -25 -5=161. 161 factors into 7*23. Check if 7 + 23 = 30, which is not 5. So that's not possible. Alternatively, are there primes adding to 5? 2 and 3, since 2+3=5. Then P=2*3=6. But 6 \\u2260 161. So S=5 invalid.\\n\\nS=6: P=191 -36 -6=149. 149 is a prime number, so can't be expressed as product of two primes. So invalid.\\n\\nS=7: P=191 -49 -7=135. 135 factors into 5*27 (but 27 not prime), 3*45, 9*15, none primes. So no. So invalid.\\n\\nS=8: P=191 -64 -8=119. 119 factors into 7*17. Check if 7+17=24\\u22608. Primes adding to 8 are 3+5=8 or 5+3. Then P=15. But 15\\u2260119. So invalid.\\n\\nS=9: P=191 -81 -9=101. 101 is prime, so no.\\n\\nS=10: P=191 -100 -10=81. 81=9*9, but 9 not prime. So no.\\n\\nS=11: P=191 -121 -11=59. 59 is prime. So no.\\n\\nS=12: P=191 -144 -12=35. 35=5*7. Check if 5 +7=12? Yes! 5 +7=12. So here, S=12, which is p + q=12, and pq=35. 5 and 7 are primes. So this works.\\n\\nSo then p=5 and q=7, or p=7 and q=5. Then r = S + 1 =12 +1=13. Check if r is prime: 13 is prime. So yes. So then p, q, r are 5,7,13 or 7,5,13. Then p + q + r =5 +7 +13=25. So 25 would be the answer.\\n\\nWait, let me check S=13 as well just to be thorough. S=13: P=191 -169 -13=9. 9=3*3. Check if 3 +3=6\\u226013. So no. So no.\\n\\nTherefore, only S=12 gives valid primes. So the answer is 25.\\n\\nLet me verify the original equations. pq + qr + rp. Let p=5, q=7, r=13. Then 5*7 +7*13 +13*5 =35 +91 +65=35+91=126+65=191. Which matches. And p + q =5 +7=12. r -1=13 -1=12. So that also matches. So it's correct. Therefore, the answer is 25.\\n</think>To solve the problem where three prime numbers \\\\( p, q, \\\\) and \\\\( r \\\\) satisfy the equations \\\\( pq + qr + rp = 191 \\\\) and \\\\( p + q = r - 1 \\\\), we proceed as follows:\\n\\n1. **Express \\\\( r \\\\) in terms of \\\\( p \\\\) and \\\\( q \\\\):**\\n   From the equation \\\\( p + q = r - 1 \\\\), we can solve for \\\\( r \\\\):\\n   \\\\[\\n   r = p + q + 1\\n   \\\\]\\n\\n2. **Substitute \\\\( r \\\\) into the first equation:**\\n   Substitute \\\\( r = p + q + 1 \\\\) into the equation \\\\( pq + qr + rp = 191 \\\\):\\n   \\\\[\\n   pq + q(p + q + 1) + p(p + q + 1) = 191\\n   \\\\]\\n   Expanding and combining like terms:\\n   \\\\[\\n   pq + pq + q^2 + q + p^2 + pq + p = 191\\n   \\\\]\\n   Simplify:\\n   \\\\[\\n   p^2 + q^2 + 3pq + p + q = 191\\n   \\\\]\\n\\n3. **Introduce new variables:**\\n   Let \\\\( S = p + q \\\\) and \\\\( P = pq \\\\). The equation becomes:\\n   \\\\[\\n   S^2 + P + S = 191\\n   \\\\]\\n\\n4. **Express \\\\( r \\\\) in terms of \\\\( S \\\\):**\\n   Since \\\\( r = p + q + 1 = S + 1 \\\\), we need to find \\\\( S \\\\) such that \\\\( S^2 + P + S = 191 \\\\) and \\\\( P = pq \\\\) is the product of two primes \\\\( p \\\\) and \\\\( q \\\\) that sum to \\\\( S \\\\).\\n\\n5. **Determine possible values for \\\\( S \\\\):**\\n   Solve the inequality \\\\( S^2 + S < 191 \\\\):\\n   \\\\[\\n   S^2 + S - 191 < 0\\n   \\\\]\\n   Using the quadratic formula \\\\( S = \\\\frac{-1 \\\\pm \\\\sqrt{1 + 4 \\\\cdot 191}}{2} \\\\):\\n   \\\\[\\n   S = \\\\frac{-1 \\\\pm \\\\sqrt{765}}{2}\\n   \\\\]\\n   Since \\\\( \\\\sqrt{765} \\\\approx 27.66 \\\\), we have:\\n   \\\\[\\n   S \\\\approx \\\\frac{-1 + 27.66}{2} \\\\approx 13.33\\n   \\\\]\\n   Therefore, \\\\( S \\\\) must be an integer between 4 and 13.\\n\\n6. **Check possible values of \\\\( S \\\\):**\\n   - For \\\\( S = 12 \\\\):\\n     \\\\[\\n     P = 191 - 12^2 - 12 = 191 - 144 - 12 = 35\\n     \\\\]\\n     Check if \\\\( 35 \\\\) can be written as the product of two primes that sum to 12:\\n     \\\\[\\n     35 = 5 \\\\times 7 \\\\quad \\\\text{and} \\\\quad 5 + 7 = 12\\n     \\\\]\\n     This works. So \\\\( p = 5 \\\\) and \\\\( q = 7 \\\\).\\n\\n7. **Find \\\\( r \\\\):**\\n   \\\\[\\n   r = S + 1 = 12 + 1 = 13\\n   \\\\]\\n\\n8. **Verify the solution:**\\n   - Check \\\\( pq + qr + rp = 191 \\\\):\\n     \\\\[\\n     5 \\\\times 7 + 7 \\\\times 13 + 13 \\\\times 5 = 35 + 91 + 65 = 191\\n     \\\\]\\n   - Check \\\\( p + q = r - 1 \\\\):\\n     \\\\[\\n     5 + 7 = 12 \\\\quad \\\\text{and} \\\\quad 13 - 1 = 12\\n     \\\\]\\n\\nSince all conditions are satisfied, the final answer is:\\n\\\\[\\n\\\\boxed{25}\\n\\\\]\",\n          \"<think>\\nOkay, let's see. I need to solve this problem where x and y are positive integers satisfying 2(x + y) = gcd(x, y) + lcm(x, y). And I have to find the ratio of the lcm to the gcd of x and y. Hmm, okay, let's break this down.\\n\\nFirst, I remember that for any two positive integers, the product of the lcm and gcd of those numbers is equal to the product of the numbers themselves. So, lcm(x, y) * gcd(x, y) = x * y. That might come in handy here. Let me note that down: lcm(x,y)*gcd(x,y) = x*y.\\n\\nGiven the equation 2(x + y) = gcd(x, y) + lcm(x, y), maybe I can express everything in terms of gcd and the ratio of x and y. Since gcd and lcm are involved, it might be helpful to let d = gcd(x, y), and then express x and y as x = d*a and y = d*b, where a and b are coprime integers (their gcd is 1). That's a standard approach for problems involving gcd and lcm.\\n\\nSo let me set d = gcd(x, y). Then x = d*a, y = d*b, with gcd(a, b) = 1. Then, the lcm(x, y) would be d*a*b, because lcm(x, y) = x*y / gcd(x, y) = (d*a*d*b)/d = d*a*b. Right, so lcm(x,y) = d*a*b.\\n\\nSubstituting these into the original equation: 2(x + y) = gcd(x, y) + lcm(x, y).\\n\\nSubstituting x = d*a, y = d*b, gcd = d, lcm = d*a*b. Then:\\n\\n2(d*a + d*b) = d + d*a*b.\\n\\nFactor out d from the left side: 2d(a + b) = d(1 + a*b).\\n\\nSince d is a positive integer, we can divide both sides by d, yielding:\\n\\n2(a + b) = 1 + a*b.\\n\\nSo now the equation simplifies to 2(a + b) = a*b + 1, where a and b are coprime positive integers. Hmm, okay. Now we have a simpler equation to solve: a*b - 2a - 2b + 1 = 0. Let me rearrange that:\\n\\na*b - 2a - 2b + 1 = 0.\\n\\nHmm, maybe factor this equation? Let me see. Adding 4 to both sides might help in factoring. Let's try:\\n\\na*b - 2a - 2b + 1 + 4 - 4 = 0\\n\\nSo, a*b - 2a - 2b + 4 = 3.\\n\\nWait, not sure. Alternatively, perhaps rearrange the terms:\\n\\na*b - 2a - 2b = -1.\\n\\nThen, add 4 to both sides:\\n\\na*b - 2a - 2b + 4 = 3.\\n\\nNow, left side can be factored as (a - 2)(b - 2) = 3. Because expanding (a - 2)(b - 2) gives a*b - 2a - 2b + 4. Yes, that's right. So:\\n\\n(a - 2)(b - 2) = 3.\\n\\nSince a and b are positive integers and coprime, we need to find pairs (a, b) such that their product is 3 when each is reduced by 2. Also, since a and b are coprime, (a - 2) and (b - 2) must be divisors of 3, which is prime. The positive divisors of 3 are 1 and 3.\\n\\nSo possible pairs (since a and b are positive integers, a - 2 and b - 2 must be at least such that a and b are positive. Let's see:\\n\\nCase 1: (a - 2) = 1 and (b - 2) = 3. Then, a = 3, b = 5. Check if gcd(a, b) = 1. gcd(3, 5) = 1, which is good.\\n\\nCase 2: (a - 2) = 3 and (b - 2) = 1. Then, a = 5, b = 3. Similarly, gcd(5, 3) = 1. So this is also valid.\\n\\nBut also, since 3 is prime, the only positive divisors are 1 and 3. But since we're considering positive integers, we could also consider if one of them is negative? But since a and b are positive, a - 2 and b - 2 must be positive or zero? Wait, but 3 is positive, so the factors must both be positive. Because if one of (a - 2) or (b - 2) were negative, their product would be negative, but 3 is positive. So both (a - 2) and (b - 2) must be positive. Thus, only the two cases above.\\n\\nAlternatively, maybe (a - 2) and (b - 2) could be 3 and 1 in some order, which gives the two cases. So the possible (a, b) are (3, 5) and (5, 3). Since a and b are interchangeable (since x and y are symmetric in the problem), these two cases would yield the same results.\\n\\nSo now, let's see. For (a, b) = (3, 5), then x = d*3, y = d*5. Similarly, for (a, b) = (5, 3), x = d*5, y = d*3. But since the problem is symmetric in x and y, both cases are equivalent.\\n\\nNow, since we need to find the ratio lcm(x, y)/gcd(x, y), let's compute that.\\n\\nRecall that lcm(x, y)/gcd(x, y) = (d*a*b)/d = a*b. So it's simply a*b. Since in both cases, a and b are 3 and 5, the product is 15. Therefore, the ratio is 15.\\n\\nWait, that seems too straightforward. Let me check.\\n\\nIf the ratio is a*b, then yes. Because lcm(x, y) is d*a*b and gcd(x, y) is d, so their ratio is (d*a*b)/d = a*b. Since a and b are 3 and 5, 3*5=15. Therefore, the answer is 15. So the answer is 15.\\n\\nBut let me verify with an example. Let's take d=1. Then x=3, y=5. Then gcd(3,5)=1, lcm=15. Then 2(x + y) = 2*(8) = 16. The right side is 1 + 15=16. So that works. If d=1, then 2(3+5)=16=1 +15.\\n\\nWhat if d=2? Then x=6, y=10. gcd(6,10)=2, lcm=30. Then 2(6 + 10)=2*16=32. The right side is 2 +30=32. So that also works. Then the ratio lcm/gcd is 30/2=15. So regardless of d, the ratio is always 15. Wait, because if x = d*a, y = d*b, then lcm(x,y)/gcd(x,y) = (d*a*b)/d = a*b, which is 15 as in the first case. So regardless of d, the ratio is always 15. Therefore, the answer is 15.\\n\\nSo even if d is some other positive integer, the ratio remains a*b =15. Therefore, the required ratio is 15.\\n\\nTherefore, the answer is 15. So \\\\boxed{15}.\\n\\n**Final Answer**\\n\\\\boxed{15}\\n</think>Given \\\\( x \\\\) and \\\\( y \\\\) are positive integers such that \\\\( 2(x + y) = \\\\gcd(x, y) + \\\\text{lcm}(x, y) \\\\), we need to find \\\\( \\\\frac{\\\\text{lcm}(x, y)}{\\\\gcd(x, y)} \\\\).\\n\\nFirst, let \\\\( d = \\\\gcd(x, y) \\\\). Then, we can express \\\\( x \\\\) and \\\\( y \\\\) as \\\\( x = d \\\\cdot a \\\\) and \\\\( y = d \\\\cdot b \\\\), where \\\\( \\\\gcd(a, b) = 1 \\\\). The least common multiple (lcm) of \\\\( x \\\\) and \\\\( y \\\\) is given by \\\\( \\\\text{lcm}(x, y) = d \\\\cdot a \\\\cdot b \\\\).\\n\\nSubstituting these into the given equation:\\n\\\\[\\n2(d \\\\cdot a + d \\\\cdot b) = d + d \\\\cdot a \\\\cdot b\\n\\\\]\\nDividing both sides by \\\\( d \\\\):\\n\\\\[\\n2(a + b) = 1 + a \\\\cdot b\\n\\\\]\\nRearranging terms, we get:\\n\\\\[\\na \\\\cdot b - 2a - 2b + 1 = 0\\n\\\\]\\nAdding 4 to both sides to factorize:\\n\\\\[\\na \\\\cdot b - 2a - 2b + 4 = 3\\n\\\\]\\nThis can be factored as:\\n\\\\[\\n(a - 2)(b - 2) = 3\\n\\\\]\\nThe positive integer solutions for \\\\((a - 2)\\\\) and \\\\((b - 2)\\\\) are 1 and 3. Thus, the possible pairs \\\\((a, b)\\\\) are \\\\((3, 5)\\\\) and \\\\((5, 3)\\\\). Since \\\\( a \\\\) and \\\\( b \\\\) are coprime, both pairs are valid.\\n\\nThe ratio \\\\( \\\\frac{\\\\text{lcm}(x, y)}{\\\\gcd(x, y)} \\\\) is given by:\\n\\\\[\\n\\\\frac{\\\\text{lcm}(x, y)}{\\\\gcd(x, y)} = \\\\frac{d \\\\cdot a \\\\cdot b}{d} = a \\\\cdot b\\n\\\\]\\nFor both pairs \\\\((3, 5)\\\\) and \\\\((5, 3)\\\\), the product \\\\( a \\\\cdot b = 15 \\\\).\\n\\nThus, the final answer is:\\n\\\\[\\n\\\\boxed{15}\\n\\\\]\",\n          \"<think>\\nOkay, so I need to find the remainder when the product of all odd numbers from 1 to 2005 is divided by 1000. Hmm, let's think about how to approach this. \\n\\nFirst, the product is 1 \\u00d7 3 \\u00d7 5 \\u00d7 ... \\u00d7 2005. That's a lot of numbers! Since we're dealing with division by 1000, maybe modular arithmetic can help here. The remainder when divided by 1000 is equivalent to the product modulo 1000. But calculating such a huge product directly seems impossible. There has to be a smarter way.\\n\\nI remember that when dealing with factorials and remainders, factors of 2 and 5 can create trailing zeros. However, here we're only multiplying odd numbers, so there are no factors of 2. But there might still be factors of 5. Wait, 1000 is 8\\u00d7125, which is 2^3 \\u00d7 5^3. Since the product is all odd numbers, it won't have factors of 2, but it can have factors of 5. Therefore, the product will be divisible by 5^3, but since there are no 2s, the product modulo 1000 might not be zero. Hmm, maybe I need to compute the product modulo 1000, but adjusting for the factors of 5?\\n\\nAlternatively, maybe split the problem into modulo 8 and modulo 125, then use the Chinese Remainder Theorem (CRT) to combine the results. Since 1000 = 8 \\u00d7 125, and 8 and 125 are coprime, CRT says that if I can find the remainder modulo 8 and modulo 125, then I can combine them to find the remainder modulo 1000. That might be a good approach.\\n\\nLet me start with modulo 8. The product is 1\\u00d73\\u00d75\\u00d77\\u00d79\\u00d7...\\u00d72005. But modulo 8, odd numbers repeat every 8 numbers. Let's see, the residues modulo 8 of odd numbers are 1,3,5,7,1,3,5,7,... So the pattern repeats every 4 terms. Wait, no. Wait, the numbers go 1,3,5,7,9\\u22611,11\\u22613,13\\u22615,15\\u22617, etc. So every 8 numbers, the cycle of residues 1,3,5,7 repeats twice. Wait, actually, modulo 8, the odd residues cycle every 4 numbers. Let's confirm:\\n\\n1 mod 8 =1\\n\\n3 mod8=3\\n\\n5 mod8=5\\n\\n7 mod8=7\\n\\n9 mod8=1\\n\\n11 mod8=3\\n\\n13 mod8=5\\n\\n15 mod8=7\\n\\nYes, every 4 terms, the cycle repeats. So how many terms are in the product 1\\u00d73\\u00d75\\u00d7...\\u00d72005? Let's find the number of terms first. The nth odd number is 2n-1. So 2n-1=2005 => n=(2005+1)/2=2006/2=1003. So there are 1003 terms.\\n\\nSo 1003 terms, each group of 4 terms (mod8) is 1\\u00d73\\u00d75\\u00d77=105. Then 105 mod8= 105 - 13\\u00d78=105-104=1. So each group of 4 terms multiplies to 1 mod8. Then how many full groups of 4 are there in 1003 terms? Let's divide 1003 by 4. 1003 \\u00f74=250.75. So 250 full groups, each contributing 1 mod8, and then a remainder of 3 terms. \\n\\nSo the total product modulo8 is (1^250) \\u00d7 (last three terms). The last three terms would be the terms after the 250th group. The 250th group ends at term 250\\u00d74=1000. So the 1001st term is 2\\u00d71001 -1=2001. Wait, no: the first term is 1=2\\u00d71-1, second term 3=2\\u00d72-1, so term k is 2k-1. Therefore, term 1001 is 2\\u00d71001 -1=2002-1=2001. Then the 1002nd term is 2003, 1003rd term is 2005. So the last three terms are 2001, 2003, 2005. Let's compute each mod8:\\n\\n2001 \\u00f78: 8\\u00d7250=2000, so 2001 mod8=1\\n\\n2003 mod8=3\\n\\n2005 mod8=5\\n\\nSo the last three terms modulo8 are 1\\u00d73\\u00d75=15 mod8=7.\\n\\nTherefore, total product mod8 is (1^250) \\u00d77=1\\u00d77=7 mod8.\\n\\nSo the remainder modulo8 is7.\\n\\nNow, we need to compute the product modulo125. This seems more complicated. Let's think.\\n\\nThe product is the product of all odd numbers from1 to2005. Wait, 2005=5\\u00d7401. So we can write the product as (1\\u00d73\\u00d75\\u00d77\\u00d7...\\u00d72005). Let's note that there are a lot of factors of 5 in this product, which would make the product divisible by 5 multiple times. However, modulo125 is 5^3, so if the product has at least three factors of 5, then modulo125 would be 0. Wait, but maybe even if it's divisible by 5^3, but we need to compute the actual remainder. Wait, but perhaps the product is divisible by 5^3, but when divided by 5^3, the remaining product modulo8 or something else. Wait, maybe not. Let me check how many factors of5 are in the product.\\n\\nThe number of factors of5 in the product:\\n\\nEach multiple of5 contributes at least one factor of5. Since we're dealing with odd numbers, the multiples of5 that are odd. So numbers divisible by5 but not by2. So numbers like5,15,25,...,2005. Let's count how many multiples of5 are in the product. The first term is5, which is5\\u00d71, then15=5\\u00d73,..., up to2005=5\\u00d7401. So the multiples of5 are5\\u00d7(1,3,5,...,401). Wait, 5\\u00d7k, where k is odd from1 to401. Because 5\\u00d7401=2005. So how many terms are there?\\n\\nThe number of terms k from1 to401 where k is odd. Since401 is odd, the number is (401 +1)/2=201. So there are201 multiples of5 in the product. Each contributes at least one factor of5. Additionally, multiples of25 contribute an extra factor of5. Similarly, multiples of125, 625, etc., contribute more factors.\\n\\nSo let's compute the total number of factors of5 in the product.\\n\\nNumber of multiples of5:201 (as above)\\n\\nNumber of multiples of25: These are numbers in the product divisible by25. Since the product includes numbers of the form5\\u00d7(odd numbers). So multiples of25 are numbers divisible by25, which are 25,75,125,...,2000. But 2005 is not divisible by25. Wait, wait, in the original product (all odd numbers up to2005), the multiples of25 must be odd multiples. So 25\\u00d71,25\\u00d73,..., up to the largest odd multiple less than or equal to2005.\\n\\n25\\u00d7k \\u22642005, where k is odd. Let's compute k_max:\\n\\n25k \\u22642005 => k \\u22642005/25=80.2. So the largest integer k is80, but since k has to be odd, the largest odd k is79. So 25\\u00d779=1975. Then 25\\u00d781=2025, which is over. So the multiples of25 in the product are25\\u00d71,25\\u00d73,...,25\\u00d779. Number of terms: (79-1)/2 +1=39 +1=40. Wait, from1 to79 odd numbers: number is (79+1)/2=40. So there are40 multiples of25.\\n\\nSimilarly, multiples of125:125\\u00d71,125\\u00d73,..., up to125\\u00d7k\\u22642005. 125\\u00d7k \\u22642005 =>k\\u226416.04. So k_max=15 (odd). So 125\\u00d715=1875. So multiples are125\\u00d71,125\\u00d73,...,125\\u00d715. Number of terms: (15-1)/2 +1=7+1=8.\\n\\nMultiples of625:625\\u00d71=625, next is625\\u00d73=1875, next is625\\u00d75=3125>2005. So only two multiples:625 and1875. But1875 is already counted as a multiple of125. So factors of625 contribute an extra factor of5 each. So number of multiples of625 is 2. 625 and1875.\\n\\nMultiples of3125:3125>2005, so none.\\n\\nSo total number of factors of5:\\n\\nFrom multiples of5:201\\n\\nFrom multiples of25:40 (each contributes an extra)\\n\\nFrom multiples of125:8 (each contributes another extra)\\n\\nFrom multiples of625:2 (each contributes another extra)\\n\\nTotal:201 +40 +8 +2=251.\\n\\nSo total factors of5 in the product:251.\\n\\nSimilarly, factors of2: since all numbers are odd, there are none. So the product is divisible by5^251 but not by2. So when dividing by5^3, since 251\\u22653, the product is divisible by5^3. Therefore, the product modulo125 is0? Wait, no, wait. Wait, 125 is5^3. If the product has at least three factors of5, then when divided by5^3, the quotient is an integer, but the remainder when divided by125 is0. So if the product is divisible by125, then the remainder is0. But the question is, when we divide the product by1000, which is8\\u00d7125, the remainder is to be found. However, we already considered that modulo8 is7, and modulo125 is... Hmm, but if modulo125 is0, then using CRT, we can say the remainder is a number congruent to7 mod8 and0 mod125. So we need to solve for x\\u22610 mod125 andx\\u22617 mod8.\\n\\nBut wait, let me confirm if modulo125 is indeed0. Let's check: since the product has at least three factors of5, then yes, the product is divisible by5^3, so product \\u22610 mod125. Therefore, modulo125 is0. Therefore, we have:\\n\\nx \\u22617 mod8\\n\\nx \\u22610 mod125\\n\\nWe need to find x such that x \\u22610 mod125 andx \\u22617 mod8. Let\\u2019s solve this system.\\n\\nLet x=125k. Then 125k\\u22617 mod8. Since125 mod8=5, so 5k\\u22617 mod8.\\n\\nWe solve 5k\\u22617 mod8.\\n\\nMultiply both sides by inverse of5 mod8. The inverse of5 mod8 is5, since5\\u00d75=25\\u22611 mod8. So multiply both sides by5:\\n\\nk\\u22617\\u00d75 mod8 =>k\\u226135 mod8 =>35\\u00f78=4*8=32, 35-32=3. So k\\u22613 mod8.\\n\\nTherefore, k=8m +3 for some integer m. Therefore, x=125(8m +3)=1000m +375. Therefore, the smallest non-negative solution is375. Therefore, the remainder is375 when divided by1000. So the answer is375.\\n\\nWait, but before accepting that, let me verify my steps again because this is tricky.\\n\\nFirst, confirming that the product is divisible by5^3: yes, since there are 251 factors of5, so 5^251 divides the product. Therefore, the product is divisible by5^3, so product\\u22610 mod125. That's correct.\\n\\nThen, solving x\\u22617 mod8 and x\\u22610 mod125. So x=125k. Then 125k\\u22615k mod8. So 5k\\u22617 mod8. Multiply both sides by inverse of5 mod8, which is5, since5\\u00d75=25\\u22611 mod8. So k\\u226135 mod8\\u22613 mod8. Therefore, k=8m +3, so x=125\\u00d73 +1000m=375 +1000m. So the minimal positive solution is375. So remainder is375 when divided by1000. That seems correct.\\n\\nBut wait, to make sure, let me check with an example. Let's take x=375. 375 \\u00f78=46*8=368, remainder7. So 375 mod8=7. 375 mod125=0. So yes, 375 satisfies both conditions. Therefore, the remainder is375. Therefore, the answer is\\\\boxed{375}.\\n\\nWait, but I need to make sure that my calculation for modulo8 and modulo125 are correct. Let me double-check the modulo8 calculation.\\n\\nEarlier, I considered that the product modulo8 is7. Let me recast that.\\n\\nThe product is1\\u00d73\\u00d75\\u00d77\\u00d7\\u2026\\u00d72005. When taking modulo8, each cycle of4 terms (1,3,5,7) multiplies to1\\u00d73\\u00d75\\u00d77=105\\u22611 mod8. Since 105/8=13\\u00d78=104, 105-104=1. Then, since there are1003 terms, how many cycles of4 are there? 1003 divided by4 is250 cycles with a remainder of3 terms. Then the product is (1)^250 \\u00d7 last3 terms.\\n\\nThe last3 terms are2001,2003,2005. 2001 mod8=1 (2001-8\\u00d7250=2001-2000=1), 2003=2001+2\\u21921+2=3 mod8, 2005=2001+4\\u21921+4=5 mod8. So last3 terms are1\\u00d73\\u00d75=15\\u22617 mod8. Therefore, total product\\u22611^250 \\u00d77\\u22617 mod8. Correct.\\n\\nTherefore, the logic holds. Therefore, the remainder is375. So I think that's the correct answer.\\n\\n**Final Answer**\\n\\\\boxed{375}\\n</think>To find the remainder when the product \\\\(1 \\\\times 3 \\\\times 5 \\\\times \\\\cdots \\\\times 2005\\\\) is divided by 1000, we use modular arithmetic and the Chinese Remainder Theorem (CRT).\\n\\n### Step 1: Calculate the product modulo 8\\n\\nThe sequence of odd numbers modulo 8 repeats every 4 terms: \\\\(1, 3, 5, 7\\\\). The product of each cycle is:\\n\\\\[\\n1 \\\\times 3 \\\\times 5 \\\\times 7 = 105 \\\\equiv 1 \\\\mod 8\\n\\\\]\\n\\nThere are 1003 terms in the product. Dividing 1003 by 4 gives 250 full cycles and a remainder of 3 terms. The remaining terms are 2001, 2003, and 2005. We calculate these modulo 8:\\n\\\\[\\n2001 \\\\equiv 1 \\\\mod 8, \\\\quad 2003 \\\\equiv 3 \\\\mod 8, \\\\quad 2005 \\\\equiv 5 \\\\mod 8\\n\\\\]\\nThe product of these remaining terms is:\\n\\\\[\\n1 \\\\times 3 \\\\times 5 = 15 \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\nThus, the product modulo 8 is:\\n\\\\[\\n1^{250} \\\\times 7 \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\n### Step 2: Calculate the product modulo 125\\n\\nWe need to determine the number of factors of 5 in the product. The sequence of odd numbers includes multiples of 5, 25, 125, and 625.\\n\\n- Multiples of 5: \\\\(5, 15, 25, \\\\ldots, 2005\\\\)\\n  - These are of the form \\\\(5 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1, 2, \\\\ldots, 200\\\\)\\n  - Number of such terms: \\\\(\\\\frac{2005}{5} = 401\\\\), and half of these are odd, so \\\\(201\\\\) multiples of 5.\\n\\n- Multiples of 25: \\\\(25, 75, 125, \\\\ldots, 1975\\\\)\\n  - These are of the form \\\\(25 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1, 2, \\\\ldots, 39\\\\)\\n  - Number of such terms: \\\\(\\\\frac{1975}{25} = 79\\\\), and half of these are odd, so \\\\(40\\\\) multiples of 25.\\n\\n- Multiples of 125: \\\\(125, 375, 625, 875, 1125, 1375, 1625, 1875\\\\)\\n  - These are of the form \\\\(125 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1, 2, \\\\ldots, 15\\\\)\\n  - Number of such terms: \\\\(\\\\frac{1875}{125} = 15\\\\), and half of these are odd, so \\\\(8\\\\) multiples of 125.\\n\\n- Multiples of 625: \\\\(625, 1875\\\\)\\n  - These are of the form \\\\(625 \\\\times (2k+1)\\\\) for \\\\(k = 0, 1\\\\)\\n  - Number of such terms: \\\\(\\\\frac{1875}{625} = 3\\\\), and half of these are odd, so \\\\(2\\\\) multiples of 625.\\n\\nTotal factors of 5:\\n\\\\[\\n201 + 40 + 8 + 2 = 251\\n\\\\]\\n\\nSince \\\\(251 \\\\geq 3\\\\), the product is divisible by \\\\(5^3 = 125\\\\). Therefore, the product modulo 125 is:\\n\\\\[\\n0 \\\\mod 125\\n\\\\]\\n\\n### Step 3: Combine results using the Chinese Remainder Theorem\\n\\nWe have:\\n\\\\[\\nx \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\\\[\\nx \\\\equiv 0 \\\\mod 125\\n\\\\]\\n\\nLet \\\\(x = 125k\\\\). Then:\\n\\\\[\\n125k \\\\equiv 7 \\\\mod 8\\n\\\\]\\nSince \\\\(125 \\\\equiv 5 \\\\mod 8\\\\), we have:\\n\\\\[\\n5k \\\\equiv 7 \\\\mod 8\\n\\\\]\\n\\nThe multiplicative inverse of 5 modulo 8 is 5, so:\\n\\\\[\\nk \\\\equiv 7 \\\\times 5 \\\\equiv 35 \\\\equiv 3 \\\\mod 8\\n\\\\]\\n\\nThus, \\\\(k = 8m + 3\\\\) for some integer \\\\(m\\\\). Therefore:\\n\\\\[\\nx = 125(8m + 3) = 1000m + 375\\n\\\\]\\n\\nThe smallest non-negative solution is:\\n\\\\[\\nx = 375\\n\\\\]\\n\\nTherefore, the remainder when the product is divided by 1000 is:\\n\\\\[\\n\\\\boxed{375}\\n\\\\]\"\n        ],\n        \"semantic_type\": \"\",\n        \"description\": \"\"\n      }\n    }\n  ]\n}",
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              "      <td>&lt;think&gt;\\nOkay, let's see. I need to solve the ...</td>\n",
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              "      <th>19248</th>\n",
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            ],
            "text/plain": [
              "      expected_answer                                            problem  \\\n",
              "0                  14  Given $\\sqrt{x^2+165}-\\sqrt{x^2-52}=7$ and $x$...   \n",
              "6                  -2  Find the value of the parameter $a$ for which ...   \n",
              "9                  18  What is the sum of all real numbers $x$ for wh...   \n",
              "13                  2  Evaluate the sum \\(\\sum_{n=1}^\\infty \\frac{\\ph...   \n",
              "17                 30  What is the largest positive integer that divi...   \n",
              "...               ...                                                ...   \n",
              "19243             244  Let \\( p \\), \\( q \\), and \\( r \\) be the disti...   \n",
              "19245               1  A bug is on the $0$ of a number line. At any p...   \n",
              "19247               4  A bus left point X for point Y. Two hours late...   \n",
              "19248              18  Each interior angle of a regular n-gon measure...   \n",
              "19250          0.8960  Find the probability that the second blue resu...   \n",
              "\n",
              "                                      generated_solution  \n",
              "0      <think>\\nOkay, let's see. I need to solve the ...  \n",
              "6      <think>\\nOkay, so I need to find the value of ...  \n",
              "9      <think>\\nOkay, so I need to solve the equation...  \n",
              "13     <think>\\nOkay, so I need to evaluate the infin...  \n",
              "17     <think>\\nAlright, so I need to find the larges...  \n",
              "...                                                  ...  \n",
              "19243  <think>\\nOkay, so I need to find the value of ...  \n",
              "19245  <think>\\nOkay, so I have this problem where a ...  \n",
              "19247  <think>\\nOkay, let's tackle this problem step ...  \n",
              "19248  <think>\\nOkay, let's see. I need to find the n...  \n",
              "19250  <think>\\nOkay, so I need to find the probabili...  \n",
              "\n",
              "[7507 rows x 3 columns]"
            ]
          },
          "execution_count": 7,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "from datasets import load_dataset\n",
        "import pandas as pd\n",
        "import numpy as np\n",
        "\n",
        "dataset = load_dataset(\"unsloth/OpenMathReasoning-mini\", split = \"cot\")\n",
        "dataset = dataset.to_pandas()[\n",
        "    [\"expected_answer\", \"problem\", \"generated_solution\"]\n",
        "]\n",
        "\n",
        "# Try converting to number - if not, replace with NaN\n",
        "is_number = pd.to_numeric(pd.Series(dataset[\"expected_answer\"]), errors = \"coerce\").notnull()\n",
        "# Select only numbers\n",
        "dataset = dataset.iloc[np.where(is_number)[0]]\n",
        "\n",
        "dataset"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "JVRFqoSdIEVK"
      },
      "source": [
        "We have to format the dataset to follow our GRPO style formatting:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "Z9ydcV_Abfi6"
      },
      "outputs": [],
      "source": [
        "def format_dataset(x):\n",
        "    expected_answer = x[\"expected_answer\"]\n",
        "    problem = x[\"problem\"]\n",
        "\n",
        "    # Remove generated <think> and </think>\n",
        "    thoughts = x[\"generated_solution\"]\n",
        "    thoughts = thoughts.replace(\"<think>\", \"\").replace(\"</think>\", \"\")\n",
        "\n",
        "    # Strip newlines on left and right\n",
        "    thoughts = thoughts.strip()\n",
        "    # Add our custom formatting\n",
        "    final_prompt = \\\n",
        "        reasoning_start + thoughts + reasoning_end + \\\n",
        "        solution_start + expected_answer + solution_end\n",
        "    return [\n",
        "        {\"role\" : \"system\",    \"content\" : system_prompt},\n",
        "        {\"role\" : \"user\",      \"content\" : problem},\n",
        "        {\"role\" : \"assistant\", \"content\" : final_prompt},\n",
        "    ]\n",
        "\n",
        "dataset[\"Messages\"] = dataset.apply(format_dataset, axis = 1)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "X5NI47rOIRP2"
      },
      "source": [
        "Check to see if it worked:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 193
        },
        "id": "LTdXBKcslhRH",
        "outputId": "7985177a-63b7-4af8-b5f1-dfb66470d643"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "\"You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION><|endoftext|>Given $\\\\sqrt{x^2+165}-\\\\sqrt{x^2-52}=7$ and $x$ is positive, find all possible values of $x$.<start_working_out>Okay, let's see. I need to solve the equation √(x² + 165) - √(x² - 52) = 7, and find all positive values of x. Hmm, radicals can be tricky, but maybe if I can eliminate the square roots by squaring both sides. Let me try that.\\n\\nFirst, let me write down the equation again to make sure I have it right:\\n\\n√(x² + 165) - √(x² - 52) = 7.\\n\\nOkay, so the idea is to isolate one of the radicals and then square both sides. Let me try moving the second radical to the other side:\\n\\n√(x² + 165) = 7 + √(x² - 52).\\n\\nNow, if I square both sides, maybe I can get rid of the square roots. Let's do that:\\n\\n(√(x² + 165))² = (7 + √(x² - 52))².\\n\\nSimplifying the left side:\\n\\nx² + 165 = 49 + 14√(x² - 52) + (√(x² - 52))².\\n\\nThe right side is expanded using the formula (a + b)² = a² + 2ab + b². So the right side becomes 7² + 2*7*√(x² - 52) + (√(x² - 52))², which is 49 + 14√(x² - 52) + (x² - 52).\\n\\nSo putting it all together:\\n\\nx² + 165 = 49 + 14√(x² - 52) + x² - 52.\\n\\nHmm, let's simplify the right side. The x² terms will cancel out, right? Let's subtract x² from both sides:\\n\\n165 = 49 + 14√(x² - 52) - 52.\\n\\nSimplify the constants on the right:\\n\\n49 - 52 is -3, so:\\n\\n165 = -3 + 14√(x² - 52).\\n\\nNow, add 3 to both sides to isolate the radical term:\\n\\n165 + 3 = 14√(x² - 52).\\n\\nSo 168 = 14√(x² - 52).\\n\\nDivide both sides by 14:\\n\\n168 / 14 = √(x² - 52).\\n\\n12 = √(x² - 52).\\n\\nNow, square both sides again to eliminate the square root:\\n\\n12² = x² - 52.\\n\\n144 = x² - 52.\\n\\nAdd 52 to both sides:\\n\\n144 + 52 = x².\\n\\n196 = x².\\n\\nSo x = √196 = 14.\\n\\nBut wait, since the problem states that x is positive, we only take the positive root. So x = 14.\\n\\nBut hold on, when dealing with squaring equations, sometimes extraneous solutions can come up. I should check if this solution actually satisfies the original equation.\\n\\nLet's plug x = 14 back into the original equation:\\n\\n√(14² + 165) - √(14² - 52) = ?\\n\\nCalculate each term:\\n\\n14² is 196.\\n\\nSo first radical: √(196 + 165) = √361 = 19.\\n\\nSecond radical: √(196 - 52) = √144 = 12.\\n\\nSo 19 - 12 = 7, which is exactly the right-hand side. So yes, it checks out.\\n\\nTherefore, the only solution is x = 14. Since the problem says x is positive, we don't have to consider negative roots. So I think that's the answer.\\nTo solve the equation \\\\(\\\\sqrt{x^2 + 165} - \\\\sqrt{x^2 - 52} = 7\\\\) for positive \\\\(x\\\\), we proceed as follows:\\n\\n1. Start with the given equation:\\n   \\\\[\\n   \\\\sqrt{x^2 + 165} - \\\\sqrt{x^2 - 52} = 7\\n   \\\\]\\n\\n2. Isolate one of the square roots by moving \\\\(\\\\sqrt{x^2 - 52}\\\\) to the right side:\\n   \\\\[\\n   \\\\sqrt{x^2 + 165} = 7 + \\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n3. Square both sides to eliminate the square root on the left:\\n   \\\\[\\n   (\\\\sqrt{x^2 + 165})^2 = (7 + \\\\sqrt{x^2 - 52})^2\\n   \\\\]\\n   Simplifying both sides, we get:\\n   \\\\[\\n   x^2 + 165 = 49 + 14\\\\sqrt{x^2 - 52} + (x^2 - 52)\\n   \\\\]\\n\\n4. Combine like terms on the right side:\\n   \\\\[\\n   x^2 + 165 = x^2 - 52 + 49 + 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n   Simplifying further:\\n   \\\\[\\n   x^2 + 165 = x^2 - 3 + 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n5. Subtract \\\\(x^2\\\\) from both sides:\\n   \\\\[\\n   165 = -3 + 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n6. Add 3 to both sides to isolate the term with the square root:\\n   \\\\[\\n   168 = 14\\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n7. Divide both sides by 14:\\n   \\\\[\\n   12 = \\\\sqrt{x^2 - 52}\\n   \\\\]\\n\\n8. Square both sides again to eliminate the square root:\\n   \\\\[\\n   12^2 = x^2 - 52\\n   \\\\]\\n   Simplifying:\\n   \\\\[\\n   144 = x^2 - 52\\n   \\\\]\\n\\n9. Add 52 to both sides to solve for \\\\(x^2\\\\):\\n   \\\\[\\n   196 = x^2\\n   \\\\]\\n\\n10. Take the positive square root (since \\\\(x\\\\) is positive):\\n    \\\\[\\n    x = \\\\sqrt{196} = 14\\n    \\\\]\\n\\n11. Verify the solution by substituting \\\\(x = 14\\\\) back into the original equation:\\n    \\\\[\\n    \\\\sqrt{14^2 + 165} - \\\\sqrt{14^2 - 52} = \\\\sqrt{196 + 165} - \\\\sqrt{196 - 52} = \\\\sqrt{361} - \\\\sqrt{144} = 19 - 12 = 7\\n    \\\\]\\n    The solution checks out.\\n\\nThus, the only positive solution is:\\n\\\\[\\n\\\\boxed{14}\\n\\\\]<end_working_out><SOLUTION>14</SOLUTION><|endoftext|>\""
            ]
          },
          "execution_count": 9,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "tokenizer.apply_chat_template(dataset[\"Messages\"][0], tokenize = False)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "iHV9BXYiIYaq"
      },
      "source": [
        "Let's truncate the pre fine-tuning dataset to `max_seq_length/2` since we don't want too long reasoning traces.\n",
        "\n",
        "Note this might take 2 minutes!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "MBHFlRbae9_s",
        "outputId": "f0d89109-b397-4e1d-9a52-098167508609"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "(59, 5)"
            ]
          },
          "execution_count": 10,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "dataset[\"N\"] = dataset[\"Messages\"].apply(lambda x: len(tokenizer.apply_chat_template(x)))\n",
        "\n",
        "dataset = dataset.loc[dataset[\"N\"] <= max_seq_length/2].copy()\n",
        "dataset.shape"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "E6NkUCAGIj8N"
      },
      "source": [
        "We then tokenize the messages and convert it to a Hugging Face compatible dataset format:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "3rgdtiV_f5hx",
        "outputId": "764c505a-52da-4512-8e25-f9ac7261786e"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "Dataset({\n",
              "    features: ['expected_answer', 'problem', 'generated_solution', 'Messages', 'N', 'text', '__index_level_0__'],\n",
              "    num_rows: 59\n",
              "})"
            ]
          },
          "execution_count": 11,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "from datasets import Dataset\n",
        "\n",
        "dataset[\"text\"] = tokenizer.apply_chat_template(dataset[\"Messages\"].values.tolist(), tokenize = False)\n",
        "dataset = Dataset.from_pandas(dataset)\n",
        "dataset"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "bAQJjQrYKzOk"
      },
      "source": [
        "Let's now pre fine-tune the model so it follows our custom GRPO formatting!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 77,
          "referenced_widgets": [
            "3e62f48b4aad425fafc115e6c57de6a4",
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            "08b378b542234293b57ae8fbd63e4613",
            "902823b5472e4f54bd5573328f06021e",
            "b2b0a3e1e994479abaf8ac4bfd263004",
            "b30e4c9f78134ed0b52bb4d34bc62af1",
            "4d3e6fd0123b4262946145c39f2a7601",
            "146309965bf04eeabfb6c02d821bf321",
            "45841c30e99c430799346e2546106ea7",
            "2bd4b8a7151c4d108b422fff4d99fb3e",
            "c36c2fa5594e4fabb4651cb018a91804"
          ]
        },
        "id": "woYi0SSygpqp",
        "outputId": "faef19e7-228a-40b1-aa03-77122b4a2dc4"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "3e62f48b4aad425fafc115e6c57de6a4",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Unsloth: Tokenizing [\"text\"] (num_proc=2):   0%|          | 0/59 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "from trl import SFTTrainer, SFTConfig\n",
        "trainer = SFTTrainer(\n",
        "    model = model,\n",
        "    tokenizer = tokenizer,\n",
        "    train_dataset = dataset,\n",
        "    args = SFTConfig(\n",
        "        dataset_text_field = \"text\",\n",
        "        per_device_train_batch_size = 1,\n",
        "        gradient_accumulation_steps = 1, # Use GA to mimic batch size!\n",
        "        warmup_steps = 5,\n",
        "        num_train_epochs = 2, # Set this for 1 full training run.\n",
        "        learning_rate = 2e-4, # Reduce to 2e-5 for long training runs\n",
        "        logging_steps = 5,\n",
        "        optim = \"adamw_8bit\",\n",
        "        weight_decay = 0.01,\n",
        "        lr_scheduler_type = \"linear\",\n",
        "        seed = 3407,\n",
        "        report_to = \"none\", # Use this for WandB etc\n",
        "    ),\n",
        ")"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 956
        },
        "id": "l4-2v_bLhZuE",
        "outputId": "82af442c-124f-4bae-e179-0b7c9d67f39b"
      },
      "outputs": [
        {
          "name": "stderr",
          "output_type": "stream",
          "text": [
            "==((====))==  Unsloth - 2x faster free finetuning | Num GPUs used = 1\n",
            "   \\\\   /|    Num examples = 59 | Num Epochs = 2 | Total steps = 118\n",
            "O^O/ \\_/ \\    Batch size per device = 1 | Gradient accumulation steps = 1\n",
            "\\        /    Data Parallel GPUs = 1 | Total batch size (1 x 1 x 1) = 1\n",
            " \"-____-\"     Trainable parameters = 66,060,288/4,088,528,384 (1.62% trained)\n"
          ]
        },
        {
          "data": {
            "text/html": [
              "\n",
              "    <div>\n",
              "      \n",
              "      <progress value='118' max='118' style='width:300px; height:20px; vertical-align: middle;'></progress>\n",
              "      [118/118 02:45, Epoch 2/2]\n",
              "    </div>\n",
              "    <table border=\"1\" class=\"dataframe\">\n",
              "  <thead>\n",
              " <tr style=\"text-align: left;\">\n",
              "      <th>Step</th>\n",
              "      <th>Training Loss</th>\n",
              "    </tr>\n",
              "  </thead>\n",
              "  <tbody>\n",
              "    <tr>\n",
              "      <td>5</td>\n",
              "      <td>0.644900</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>10</td>\n",
              "      <td>0.639600</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>15</td>\n",
              "      <td>0.419300</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>20</td>\n",
              "      <td>0.389600</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>25</td>\n",
              "      <td>0.422200</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>30</td>\n",
              "      <td>0.448400</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>35</td>\n",
              "      <td>0.475100</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>40</td>\n",
              "      <td>0.419100</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>45</td>\n",
              "      <td>0.445300</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>50</td>\n",
              "      <td>0.328000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>55</td>\n",
              "      <td>0.381800</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>60</td>\n",
              "      <td>0.453100</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>65</td>\n",
              "      <td>0.248700</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>70</td>\n",
              "      <td>0.244500</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>75</td>\n",
              "      <td>0.300300</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>80</td>\n",
              "      <td>0.226000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>85</td>\n",
              "      <td>0.211500</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>90</td>\n",
              "      <td>0.258900</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>95</td>\n",
              "      <td>0.191900</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>100</td>\n",
              "      <td>0.261000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>105</td>\n",
              "      <td>0.255500</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>110</td>\n",
              "      <td>0.217400</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>115</td>\n",
              "      <td>0.180700</td>\n",
              "    </tr>\n",
              "  </tbody>\n",
              "</table><p>"
            ],
            "text/plain": [
              "<IPython.core.display.HTML object>"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "Unsloth: Will smartly offload gradients to save VRAM!\n"
          ]
        },
        {
          "data": {
            "text/plain": [
              "TrainOutput(global_step=118, training_loss=0.3486394861997184, metrics={'train_runtime': 170.8892, 'train_samples_per_second': 0.691, 'train_steps_per_second': 0.691, 'total_flos': 2374193075607552.0, 'train_loss': 0.3486394861997184})"
            ]
          },
          "execution_count": 13,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "trainer.train()"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "DRMBNUBgLC8T"
      },
      "source": [
        "Let's check if the model has learnt to follow the custom format:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "9HJxrS76h3Ds",
        "outputId": "312e41cd-cb12-4786-eb32-cfebf8d1ca59"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "You are given a problem.\n",
            "Think about the problem and provide your working out.\n",
            "Place it between <start_working_out> and <end_working_out>.\n",
            "Then, provide your solution between <SOLUTION></SOLUTION><|endoftext|>Jenifer has 82 cents in pennies and nickels. Her younger brother mistook all her nickels for dimes and counted the total as $1.47. How many pennies does Jenifer have?<start_working_out>Okay, let's see. Jenifer has 82 cents in pennies and nickels. Her brother thought all the nickels were dimes and counted the total as $1.47. I need to find out how many pennies she has. Hmm, let's break this down.\n",
            "\n",
            "First, I need to set up some equations. Let's say the number of pennies is P and the number of nickels is N. Since pennies are worth 1 cent each and nickels are 5 cents each, the total value is P + 5N = 82 cents. That's the first equation.\n",
            "\n",
            "Now, her brother thought all the nickels were dimes. Dimes are 10 cents each. So, he counted the total as $1.47, which is 147 cents. So, the equation for his count would be P + 10N = 147. Wait, but that's the second equation.\n",
            "\n",
            "So now I have two equations:\n",
            "1. P + 5N = 82\n",
            "2. P + 10N = 147\n",
            "\n",
            "I need to solve these two equations to find P. Let me subtract the first equation from the second to eliminate P. So, (P + 10N) - (P + 5N) = 147 - 82. That simplifies to 5N = 65. Then, dividing both sides by 5, N = 13. So there are 13 nickels.\n",
            "\n",
            "Now, plug N back into the first equation to find P. So, P + 5*13 = 82. That's P + 65 = 82. Subtract 65 from both sides, so P = 17. Therefore, Jenifer has 17 pennies.\n",
            "\n",
            "Wait, let me check that. If she has 17 pennies and 13 nickels, that's 17 + 65 = 82 cents, which matches. And if her brother counts the nickels as dimes, that's 17 + 130 = 147 cents, which is $1.47. That checks out. So yeah, the answer should be 17 pennies.\n",
            "To solve the problem, we start by defining the variables:\n",
            "- Let \\( P \\) be the number of pennies.\n",
            "- Let \\( N \\) be the number of nickels.\n",
            "\n",
            "We know two things:\n",
            "1. The total value of the pennies and nickels is 82 cents.\n",
            "2. If all nickels were mistaken for dimes, the total value would be $1.47 (or 147 cents).\n",
            "\n",
            "We can set up the following system of equations based on this information:\n",
            "1. \\( P + 5N = 82 \\)\n",
            "2. \\( P + 10N = 147 \\)\n",
            "\n",
            "To find the values of \\( P \\) and \\( N \\), we can subtract the first equation from the second to eliminate \\( P \\):\n",
            "\\[\n",
            "(P + 10N) - (P + 5N) = 147 - 82\n",
            "\\]\n",
            "This simplifies to:\n",
            "\\[\n",
            "5N = 65\n",
            "\\]\n",
            "Solving for \\( N \\), we get:\n",
            "\\[\n",
            "N = \\frac{65}{5} = 13\n",
            "\\]\n",
            "\n",
            "Now that we know \\( N = 13 \\), we substitute this value back into the first equation to find \\( P \\):\n",
            "\\[\n",
            "P + 5(13) = 82\n",
            "\\]\n",
            "This simplifies to:\n",
            "\\[\n",
            "P + 65 = 82\n",
            "\\]\n",
            "Solving for \\( P \\), we get:\n",
            "\\[\n",
            "P = 82 - 65 = 17\n",
            "\\]\n",
            "\n",
            "Thus, Jenifer has \\(\\boxed{17}\\) pennies.<end_working_out><SOLUTION>17</SOLUTION><|endoftext|>\n"
          ]
        }
      ],
      "source": [
        "text = tokenizer.apply_chat_template(\n",
        "    dataset[0][\"Messages\"][:2],\n",
        "    tokenize = False,\n",
        "    add_generation_prompt = True, # Must add for generation\n",
        ")\n",
        "\n",
        "from transformers import TextStreamer\n",
        "_ = model.generate(\n",
        "    **tokenizer(text, return_tensors = \"pt\").to(\"cuda\"),\n",
        "    temperature = 0,\n",
        "    max_new_tokens = 1024,\n",
        "    streamer = TextStreamer(tokenizer, skip_prompt = False),\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "AtZ3qGOALF95"
      },
      "source": [
        "Yes it did follow the formatting! Great! Let's remove some items before the GRPO step"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "YWSZ0DET7bob",
        "outputId": "b02a0eff-3910-49ca-f900-a09a520770a2"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "0"
            ]
          },
          "execution_count": 15,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "del dataset\n",
        "torch.cuda.empty_cache()\n",
        "import gc\n",
        "gc.collect()"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "7KGgPgk_5S8r"
      },
      "source": [
        "### Data Prep\n",
        "<a name=\"Data\"></a>\n",
        "\n",
        "We're using Hugging Face's [Open R1 Math dataset](https://huggingface.co/datasets/open-r1/DAPO-Math-17k-Processed). You can also utilize OpenAI's famous [GSM8K dataset](https://huggingface.co/datasets/openai/gsm8k)"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 201,
          "referenced_widgets": [
            "c811b38ac4b84edca4f8992795364254",
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            "51761db5c13b4d418614165d033ac6f5",
            "e489590b36ff4d2386ada63aade4a79c",
            "0775473857a643c093764a2ca588d5c7",
            "db544ad9ccd3495882b1a3d0f7644ded",
            "a1eadf3bb73c4a9fb7ded669df942e29",
            "f40d34dd33144818a896ce31cc95e222",
            "1e301887f50942d28ea1c9b7b2478331",
            "19978cc961894e63bc4d1a914a277234",
            "9443539f42e54bf79847f679c7d1d2f3",
            "28e48534b89a403da0c07339682fa74e",
            "05ae7521c459456a80369a22e1ee88a8",
            "c1c1c813346947f298261d558576cd30",
            "c0fb4a9d32214edebbc9443287ef08f0",
            "21547cb7f692473c8166e88154915dff",
            "26983a12f6384fa6bbe345651db6e79e",
            "02deca4036cc480f9ed6c2ef3d29fb73",
            "068a889330bd4e039714ea86d8484bf6",
            "70f8efe0424b4ba7bae3aeb7591df22a",
            "68229c85da5b43698abdeb474bfe80c8",
            "041067884fb540fca3c0f29f5bb50914",
            "9e85547dc0f8443cb16bfbda4a5cf463",
            "2720be9bf09146179f5d0e48a90632a5",
            "99f0d155bdc741678854e082378a866b",
            "62e6423319c3493a8a6788918522b890",
            "6cc43eb82b2e4ebb9bcc13a7591da5b2",
            "c04bb6548e494830b3bbe145b1c61c37",
            "53d45528cf5c43ee9f5781e5cdecde97",
            "d9bb7dd467804742ba2cb93a14b7bde0",
            "1f1665a6bb1b46c091e6ffc08c21dbfc",
            "739707b035094350859d1143f5ae1b0f",
            "7241eafe1fb34ca89437566845dd5d22"
          ]
        },
        "id": "o7-eUrQn-OzE",
        "outputId": "5a096f76-ed3e-46b4-a599-360c857fa46a"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "c811b38ac4b84edca4f8992795364254",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "README.md:   0%|          | 0.00/3.44k [00:00<?, ?B/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "28e48534b89a403da0c07339682fa74e",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "en/train-00000-of-00001.parquet:   0%|          | 0.00/5.23M [00:00<?, ?B/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "9e85547dc0f8443cb16bfbda4a5cf463",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Generating train split:   0%|          | 0/14116 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
            "text/plain": [
              "Dataset({\n",
              "    features: ['prompt', 'solution', 'data_source', 'source_prompt', 'ability', 'reward_model', 'extra_info'],\n",
              "    num_rows: 14116\n",
              "})"
            ]
          },
          "execution_count": 16,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "from datasets import load_dataset\n",
        "dataset = load_dataset(\"open-r1/DAPO-Math-17k-Processed\", \"en\", split = \"train\")\n",
        "dataset"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "3b00gUsS-ROW"
      },
      "source": [
        "Let's look at the first row:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 87
        },
        "id": "siopxjG8-ReF",
        "outputId": "35fc68d9-f4fd-45e8-a265-5de46bfeeac3"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "'In triangle $ABC$, $\\\\sin \\\\angle A = \\\\frac{4}{5}$ and $\\\\angle A < 90^\\\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\\\angle BAD = \\\\angle DAC$ and $\\\\angle BDC = 90^\\\\circ$. Suppose that $AD = 1$ and that $\\\\frac{BD}{CD} = \\\\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\\\frac{a\\\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.'"
            ]
          },
          "execution_count": 17,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "dataset[0][\"prompt\"]"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 35
        },
        "id": "KGupRQqD-Wcf",
        "outputId": "cd5d25dc-3e63-4692-ee5e-cea88c4ae3a9"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "'34'"
            ]
          },
          "execution_count": 18,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "dataset[0][\"solution\"]"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "CmnXj6hn-Ydi"
      },
      "source": [
        "In GSM8K, ee notice all answers like about have a ####, so we extract it. But for the Open R1 dataset, we can skip the below."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 35
        },
        "id": "8JJGXKdJ-Zl_",
        "outputId": "74eb22f0-72c1-42a5-bbb1-6d1863c13b72"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "'34'"
            ]
          },
          "execution_count": 19,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "def extract_hash_answer(text):\n",
        "    # if \"####\" not in text: return None\n",
        "    # return text.split(\"####\")[1].strip()\n",
        "    return text\n",
        "extract_hash_answer(dataset[0][\"solution\"])"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "K30CygaU-dir"
      },
      "source": [
        "Let's map the dataset! and see the first row:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 452,
          "referenced_widgets": [
            "0a87ca8327df48099b3691bfa446ef3a",
            "9abf2f4547c9483a99f983d7ba26b1cf",
            "ed21acc8e597481fb1b5dfda4adacaa7",
            "cfdb8549eed24409b2aa09271863c06e",
            "db8c351189b44767a9aca3534e5cc93e",
            "4886339fcdee4cb3990e458061c33a74",
            "20f7043b723746519cef74899dfa4153",
            "97119d7743e34110af6c48fbd14104ac",
            "81eaf78c3c294383b43c2efdfa5e2225",
            "b66ab6459f9b4d15b364a9e94d161534",
            "a7c1f428da3448688b6336aa14c3859d"
          ]
        },
        "id": "qyEVI972-d3n",
        "outputId": "8efe505b-fd40-4940-a55e-519fa4260d65"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "0a87ca8327df48099b3691bfa446ef3a",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Map:   0%|          | 0/14116 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
            "text/plain": [
              "{'prompt': [{'content': 'You are given a problem.\\nThink about the problem and provide your working out.\\nPlace it between <start_working_out> and <end_working_out>.\\nThen, provide your solution between <SOLUTION></SOLUTION>',\n",
              "   'role': 'system'},\n",
              "  {'content': 'In triangle $ABC$, $\\\\sin \\\\angle A = \\\\frac{4}{5}$ and $\\\\angle A < 90^\\\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\\\angle BAD = \\\\angle DAC$ and $\\\\angle BDC = 90^\\\\circ$. Suppose that $AD = 1$ and that $\\\\frac{BD}{CD} = \\\\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\\\frac{a\\\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.',\n",
              "   'role': 'user'}],\n",
              " 'solution': '34',\n",
              " 'data_source': 'math_dapo',\n",
              " 'source_prompt': [{'content': 'Solve the following math problem step by step. The last line of your response should be of the form Answer: $Answer (without quotes) where $Answer is the answer to the problem.\\n\\nIn triangle $ABC$, $\\\\sin \\\\angle A = \\\\frac{4}{5}$ and $\\\\angle A < 90^\\\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\\\angle BAD = \\\\angle DAC$ and $\\\\angle BDC = 90^\\\\circ$. Suppose that $AD = 1$ and that $\\\\frac{BD}{CD} = \\\\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\\\frac{a\\\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.\\n\\nRemember to put your answer on its own line after \"Answer:\".',\n",
              "   'role': 'user'}],\n",
              " 'ability': 'MATH',\n",
              " 'reward_model': {'ground_truth': '34', 'style': 'rule-lighteval/MATH_v2'},\n",
              " 'extra_info': {'index': '9a9b6eb4-a1cb-49d1-8c1e-62eaf2f74079'},\n",
              " 'answer': '34'}"
            ]
          },
          "execution_count": 20,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "dataset = dataset.map(lambda x: {\n",
        "    \"prompt\" : [\n",
        "        {\"role\": \"system\", \"content\": system_prompt},\n",
        "        {\"role\": \"user\",   \"content\": x[\"prompt\"]},\n",
        "    ],\n",
        "    \"answer\": extract_hash_answer(x[\"solution\"]),\n",
        "})\n",
        "dataset[0]"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "-9m8eR9T-gMh"
      },
      "source": [
        "We create a regex format to match the reasoning sections and answers:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "iQwjTjNz-gY_",
        "outputId": "7364b9dc-668d-42ef-fc3a-df768285591f"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "re.compile(r'<end_working_out>.*?<SOLUTION>(.+?)</SOLUTION>[\\s]{0,}(?:<\\|endoftext\\|>)?[\\s]{0,}$',\n",
              "re.MULTILINE|re.DOTALL|re.UNICODE)"
            ]
          },
          "execution_count": 21,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "import re\n",
        "\n",
        "# Add optional EOS token matching\n",
        "solution_end_regex = r\"</SOLUTION>[\\s]{0,}\" + \\\n",
        "    \"(?:\" + re.escape(tokenizer.eos_token) + \")?\"\n",
        "\n",
        "match_format = re.compile(\n",
        "    rf\"{reasoning_end}.*?\"\\\n",
        "    rf\"{solution_start}(.+?){solution_end_regex}\"\\\n",
        "    rf\"[\\s]{{0,}}$\",\n",
        "    flags = re.MULTILINE | re.DOTALL\n",
        ")\n",
        "match_format"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "OycMneOq-iNC"
      },
      "source": [
        "We verify it works:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "ndzHnQ_6-jHt",
        "outputId": "6b211d52-6e49-4a44-82b6-3a34c643da4d"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "['\\n2\\n']"
            ]
          },
          "execution_count": 22,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "match_format.findall(\n",
        "    \"Let me think!<end_working_out>\"\\\n",
        "    f\"<SOLUTION>\\n2\\n</SOLUTION>\",\n",
        ")"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "eRMDAzDk2x6t",
        "outputId": "c4f3c11c-7c8d-4ea6-8df3-ce06180dfac8"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "['  2  ']"
            ]
          },
          "execution_count": 23,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "match_format.findall(\n",
        "    \"<start_working_out>Let me think!<end_working_out>\"\\\n",
        "    f\"<SOLUTION>  2  </SOLUTION>\\n\\n\",\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "weOjmO5l-kl3"
      },
      "source": [
        "We now want to create a reward function to match the format exactly - we reward it with 3 points if it succeeds:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "qgFNXORy-lpO"
      },
      "outputs": [],
      "source": [
        "def match_format_exactly(completions, **kwargs):\n",
        "    scores = []\n",
        "    for completion in completions:\n",
        "        score = 0\n",
        "        response = completion[0][\"content\"]\n",
        "        # Match if format is seen exactly!\n",
        "        if match_format.search(response) is not None: score += 3.0\n",
        "        scores.append(score)\n",
        "    return scores"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "Gf69i2WT-m4K"
      },
      "source": [
        "If it fails, we want to reward the model if it at least follows the format partially, by counting each symbol:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "cUfHzCVx-nGK"
      },
      "outputs": [],
      "source": [
        "def match_format_approximately(completions, **kwargs):\n",
        "    scores = []\n",
        "    for completion in completions:\n",
        "        score = 0\n",
        "        response = completion[0][\"content\"]\n",
        "        # Count how many keywords are seen - we penalize if too many!\n",
        "        # If we see 1, then plus some points!\n",
        "\n",
        "        # No need to reward <start_working_out> since we always prepend it!\n",
        "        # score += 0.5 if response.count(reasoning_start) == 1 else -1.0\n",
        "        score += 0.5 if response.count(reasoning_end)   == 1 else -1.0\n",
        "        score += 0.5 if response.count(solution_start)  == 1 else -1.0\n",
        "        score += 0.5 if response.count(solution_end)    == 1 else -1.0\n",
        "        scores.append(score)\n",
        "    return scores"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "9wAUWwtE-s6n"
      },
      "source": [
        "Finally, we want to extract the generated answer, and reward or penalize it! We also reward it based on how close the answer is to the true one via ratios:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "hmtI_8gg-uIE"
      },
      "outputs": [],
      "source": [
        "def check_answer(prompts, completions, answer, **kwargs):\n",
        "    question = prompts[0][-1][\"content\"]\n",
        "    responses = [completion[0][\"content\"] for completion in completions]\n",
        "\n",
        "    extracted_responses = [\n",
        "        guess.group(1)\n",
        "        if (guess := match_format.search(r)) is not None else None \\\n",
        "        for r in responses\n",
        "    ]\n",
        "\n",
        "    scores = []\n",
        "    for guess, true_answer in zip(extracted_responses, answer):\n",
        "        score = 0\n",
        "        if guess is None:\n",
        "            scores.append(-2.0)\n",
        "            continue\n",
        "        # Correct answer gets 5 points!\n",
        "        if guess == true_answer:\n",
        "            score += 5.0\n",
        "        # Match if spaces are seen, but less reward\n",
        "        elif guess.strip() == true_answer.strip():\n",
        "            score += 3.5\n",
        "        else:\n",
        "            # We also reward it if the answer is close via ratios!\n",
        "            # Ie if the answer is within some range, reward it!\n",
        "            try:\n",
        "                ratio = float(guess) / float(true_answer)\n",
        "                if   ratio >= 0.9 and ratio <= 1.1: score += 2.0\n",
        "                elif ratio >= 0.8 and ratio <= 1.2: score += 1.5\n",
        "                else: score -= 2.5 # Penalize wrong answers\n",
        "            except:\n",
        "                score -= 4.5 # Penalize\n",
        "        scores.append(score)\n",
        "    return scores"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "atMyfhXh-v3R"
      },
      "source": [
        "Also sometimes it might not be 1 number as the answer, but like a sentence for example \"The solution is $20\" -> we extract 20.\n",
        "\n",
        "We also remove possible commas for example as in 123,456"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "AVW0kL8q-wL5",
        "outputId": "a972c5f9-4cf4-46f9-bd0b-c12d9e01de65"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "['0.34']\n",
            "['123,456']\n",
            "['-0.234']\n",
            "['17']\n"
          ]
        }
      ],
      "source": [
        "match_numbers = re.compile(\n",
        "    solution_start + r\".*?[\\s]{0,}([-]?[\\d\\.\\,]{1,})\",\n",
        "    flags = re.MULTILINE | re.DOTALL\n",
        ")\n",
        "print(match_numbers.findall(\"<SOLUTION>  0.34  </SOLUTION>\"))\n",
        "print(match_numbers.findall(\"<SOLUTION>  123,456  </SOLUTION>\"))\n",
        "print(match_numbers.findall(\"<SOLUTION>  -0.234  </SOLUTION>\"))\n",
        "print(match_numbers.findall(\"<SOLUTION>17</SOLUTION>\"))"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "RbfaaAywNHHh"
      },
      "source": [
        "We now prepare our main function which will print out the generated responses and the true answer, along with another reward function which converts text to float via `float` and sees if it's the same."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "GjBFrttr-y1_"
      },
      "outputs": [],
      "source": [
        "global PRINTED_TIMES\n",
        "PRINTED_TIMES = 0\n",
        "global PRINT_EVERY_STEPS\n",
        "PRINT_EVERY_STEPS = 5\n",
        "\n",
        "def check_numbers(prompts, completions, answer, **kwargs):\n",
        "    question = prompts[0][-1][\"content\"]\n",
        "    responses = [completion[0][\"content\"] for completion in completions]\n",
        "\n",
        "    extracted_responses = [\n",
        "        guess.group(1)\n",
        "        if (guess := match_numbers.search(r)) is not None else None \\\n",
        "        for r in responses\n",
        "    ]\n",
        "\n",
        "    scores = []\n",
        "    # Print only every few steps\n",
        "    global PRINTED_TIMES\n",
        "    global PRINT_EVERY_STEPS\n",
        "    if PRINTED_TIMES % PRINT_EVERY_STEPS == 0:\n",
        "        print(\n",
        "            '*'*20 + f\"Question:\\n{question}\", f\"\\nAnswer:\\n{answer[0]}\", f\"\\nResponse:\\n{responses[0]}\", f\"\\nExtracted:\\n{extracted_responses[0]}\"\n",
        "        )\n",
        "    PRINTED_TIMES += 1\n",
        "\n",
        "    for guess, true_answer in zip(extracted_responses, answer):\n",
        "        if guess is None:\n",
        "            scores.append(-2.5)\n",
        "            continue\n",
        "        # Convert to numbers\n",
        "        try:\n",
        "            true_answer = float(true_answer.strip())\n",
        "            # Remove commas like in 123,456\n",
        "            guess       = float(guess.strip().replace(\",\", \"\"))\n",
        "            scores.append(3.5 if guess == true_answer else -1.5)\n",
        "        except:\n",
        "            scores.append(0)\n",
        "            continue\n",
        "    return scores"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "fgOR3wJ_AyLr"
      },
      "source": [
        "Get the top 90% prompt length so we don't accidentally truncate them!\n",
        "\n",
        "Ie we'll remove the top 10% long prompts."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 189,
          "referenced_widgets": [
            "45529456ca08414fb6941d5d9c0620ba",
            "73ef0bf5f1fe4617a05c0dcea97d2694",
            "d16e3bbdf6f4487c8ce52a0045c388e7",
            "f950c74d41464996b06a84d15a3ad726",
            "bb70890c0f0f495a9b92a3ee6b25981d",
            "f10f9ba304fb4f3fb16078c17afc4732",
            "f62e720173534370a3c136e5741b764c",
            "84250230c06e436cbc7b7ad27714d467",
            "3bff794ce935432cacc12ff1e59895ab",
            "192818ea28e9415ebb28a77c763af419",
            "f8e295340e3d4babaac21e38c7adf8c7",
            "5a6e164f1bbd40bf874c94142f8d0912",
            "866fca106ba140eb9d9909c6d76dcc4a",
            "ca8137bdf68a45689c06942e8d72fde8",
            "affdb144cf7d4250abba63fb73749e79",
            "a2c1209c193c408e80a6a49c24f09edc",
            "0daefb9c3dd44f149b5e5265c721f0e0",
            "d0747440cf3243aab108a974878ab6e4",
            "762b33136da34e3cb201cff3a142da09",
            "642154039dd940bd85225057ecbad572",
            "c5cee8d441bf48c9add4b6537f6440e5",
            "8cb3184ee7f6498b81ab3a3fafbffbeb"
          ]
        },
        "id": "6EgAi4Q5fGE-",
        "outputId": "5f660416-82a2-4cf9-933a-b4c8e9335411"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "45529456ca08414fb6941d5d9c0620ba",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Map:   0%|          | 0/14116 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "You are given a problem.\n",
            "Think about the problem and provide your working out.\n",
            "Place it between <start_working_out> and <end_working_out>.\n",
            "Then, provide your solution between <SOLUTION></SOLUTION><|endoftext|>In triangle $ABC$, $\\sin \\angle A = \\frac{4}{5}$ and $\\angle A < 90^\\circ$. Let $D$ be a point outside triangle $ABC$ such that $\\angle BAD = \\angle DAC$ and $\\angle BDC = 90^\\circ$. Suppose that $AD = 1$ and that $\\frac{BD}{CD} = \\frac{3}{2}$. If $AB + AC$ can be expressed in the form $\\frac{a\\sqrt{b}}{c}$ where $a, b, c$ are pairwise relatively prime integers, find $a + b + c$.<start_working_out>\n"
          ]
        },
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "5a6e164f1bbd40bf874c94142f8d0912",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Map:   0%|          | 0/14116 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "Max Length =  201\n"
          ]
        }
      ],
      "source": [
        "tokenized = dataset.map(\n",
        "    lambda x: {\"tokens\" : tokenizer.apply_chat_template(x[\"prompt\"], add_generation_prompt = True, tokenize = True)},\n",
        "    batched = True,\n",
        ")\n",
        "print(tokenizer.decode(tokenized[0][\"tokens\"]))\n",
        "tokenized = tokenized.map(lambda x: {\"L\" : len(x[\"tokens\"])})\n",
        "\n",
        "import numpy as np\n",
        "maximum_length = int(np.quantile(tokenized[\"L\"], 0.9))\n",
        "print(\"Max Length = \", maximum_length)\n",
        "\n",
        "# Filter only samples smaller than 90% max length\n",
        "dataset = dataset.select(np.where(np.array(tokenized[\"L\"]) <= maximum_length)[0])\n",
        "del tokenized"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "9-IOMhVg-2AM"
      },
      "source": [
        "<a name=\"Train\"></a>\n",
        "### Train the model\n",
        "\n",
        "Now set up GRPO Trainer and all configurations!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "ptqkXK2D4d6p",
        "outputId": "a79c06e7-afb3-45b5-bde3-77dbf20a7686"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "Unsloth: We now expect `per_device_train_batch_size` to be a multiple of `num_generations`.\n",
            "We will change the batch size of 1 to the `num_generations` of 4\n"
          ]
        }
      ],
      "source": [
        "max_prompt_length = maximum_length + 1 # + 1 just in case!\n",
        "max_completion_length = max_seq_length - max_prompt_length\n",
        "\n",
        "from vllm import SamplingParams\n",
        "vllm_sampling_params = SamplingParams(\n",
        "    min_p = 0.1,\n",
        "    top_p = 1.0,\n",
        "    top_k = -1,\n",
        "    seed = 3407,\n",
        "    stop = [tokenizer.eos_token],\n",
        "    include_stop_str_in_output = True,\n",
        ")\n",
        "\n",
        "from trl import GRPOConfig, GRPOTrainer\n",
        "training_args = GRPOConfig(\n",
        "    vllm_sampling_params = vllm_sampling_params,\n",
        "    temperature = 1.0,\n",
        "    learning_rate = 5e-6,\n",
        "    weight_decay = 0.01,\n",
        "    warmup_ratio = 0.1,\n",
        "    lr_scheduler_type = \"linear\",\n",
        "    optim = \"adamw_8bit\",\n",
        "    logging_steps = 1,\n",
        "    per_device_train_batch_size = 1,\n",
        "    gradient_accumulation_steps = 1, # Increase to 4 for smoother training\n",
        "    num_generations = 4, # Decrease if out of memory\n",
        "    max_prompt_length = max_prompt_length,\n",
        "    max_completion_length = max_completion_length,\n",
        "    # num_train_epochs = 1, # Set to 1 for a full training run\n",
        "    max_steps = 100,\n",
        "    save_steps = 100,\n",
        "    report_to = \"none\", # Can use Weights & Biases\n",
        "    output_dir = \"outputs\",\n",
        "\n",
        "    # For optional training + evaluation\n",
        "    # fp16_full_eval = True,\n",
        "    # per_device_eval_batch_size = 4,\n",
        "    # eval_accumulation_steps = 1,\n",
        "    # eval_strategy = \"steps\",\n",
        "    # eval_steps = 1,\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "r9Mv8UZO5hz-"
      },
      "source": [
        "And let's run the trainer! If you scroll up, you'll see a table of rewards. The goal is to see the `reward` column increase!\n",
        "\n",
        "You might have to wait 150 to 200 steps for any action. You'll probably get 0 reward for the first 100 steps. Please be patient!\n",
        "\n",
        "| Step | Training Loss | reward    | reward_std | completion_length | kl       |\n",
        "|------|---------------|-----------|------------|-------------------|----------|\n",
        "| 1    | 0.000000      | 0.125000  | 0.000000   | 200.000000        | 0.000000 |\n",
        "| 2    | 0.000000      | 0.072375  | 0.248112   | 200.000000        | 0.000000 |\n",
        "| 3    | 0.000000      | -0.079000 | 0.163776   | 182.500000        | 0.000005 |\n"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 1000
        },
        "id": "vzOuSVCL_GA9",
        "outputId": "58bf64fe-2e53-4d28-a517-a1bf68ecc517"
      },
      "outputs": [
        {
          "name": "stderr",
          "output_type": "stream",
          "text": [
            "==((====))==  Unsloth - 2x faster free finetuning | Num GPUs used = 1\n",
            "   \\\\   /|    Num examples = 12,709 | Num Epochs = 1 | Total steps = 100\n",
            "O^O/ \\_/ \\    Batch size per device = 4 | Gradient accumulation steps = 1\n",
            "\\        /    Data Parallel GPUs = 1 | Total batch size (4 x 1 x 1) = 4\n",
            " \"-____-\"     Trainable parameters = 66,060,288/4,088,528,384 (1.62% trained)\n"
          ]
        },
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "********************Question:\n",
            "Compute the number of positive integers that divide at least two of the integers in the set $\\{1^1,2^2,3^3,4^4,5^5,6^6,7^7,8^8,9^9,10^{10}\\}$. \n",
            "Answer:\n",
            "22 \n",
            "Response:\n",
            "Okay, so I need to find the number of positive integers that divide at least two of the numbers in the set {1^1, 2^2, 3^3, ..., 10^10}. Hmm, let's start by understanding what the problem is asking. We have a set where each element is a number raised to its own power, from 1 to 10. The goal is to find how many positive integers are divisors of at least two of these numbers. \n",
            "\n",
            "First, I think I should list out the elements of the set to see what they are. Let me compute each one:\n",
            "\n",
            "1. 1^1 = 1\n",
            "2. 2^2 = 4\n",
            "3. 3^3 = 27\n",
            "4. 4^4 = 256\n",
            "5. 5^5 = 3125\n",
            "6. 6^6 = 46656\n",
            "7. 7^7 = 823543\n",
            "8. 8^8 = 16777216\n",
            "9. 9^9 = 387420489\n",
            "10. 10^10 = 10000000000\n",
            "\n",
            "Wait, those are all the elements. Now, I need to find all positive integers that divide at least two of these numbers. The key here is that a divisor of two numbers must be a common divisor. So perhaps the problem reduces to finding the common divisors of these numbers taken two at a time?\n",
            "\n",
            "But wait, the problem says \"at least two,\" which could mean any pair, triplet, etc., but the wording is \"divide at least two,\" which I interpret as divisors of two or more numbers. So maybe it's all integers that appear in the compound of at least two numbers in the set? \n",
            "\n",
            "Alternatively, maybe it's the union of all common divisors for each pair. Let me think. Since the problem is asking for numbers that divide at least two, that would include all common divisors of any two numbers, three numbers, etc., up to all ten numbers. But the question is how many such integers there are.\n",
            "\n",
            "Hmm, maybe I should approach this by finding all divisors of pairs, triples, etc., and then counting the unique ones. But that seems complicated because as the number of elements increases, the number of subsets increases exponentially. Wait, but since the set is relatively small (only 10 elements), maybe we can compute them step by step.\n",
            "\n",
            "Alternatively, perhaps there's a smarter way. Let me think about the prime factorizations. Maybe the divisors that are common to at least two numbers are those that are divisors of the greatest common divisor (GCD) of the pair? Wait, no, because even if two numbers have a GCD, but their GCD might not divide other numbers in the set. For example, take 4 and 8. Their GCD is 4, which is a divisor of both, but is there a number that divides at least two and is not the GCD of any pair? Let me think of another example. Take 6 and 10. GCD is 2. Are there other divisors? Yes, 1. So 1 is a divisor of both 6 and 10. So divisors of the GCD include all lower divisors of the GCD, but some divisors might not be GCDs of any pair but still divide other numbers.\n",
            "\n",
            "Wait, perhaps the problem is equivalent to finding the number of integers that are in the intersection of the divisors of at least two numbers. So for example, if we can find the common divisors of each pair and then count the unique ones across all pairs, that should give the answer. Given the size of the set (10 elements), maybe there are not too many pairs. Let's see: 10 choose 2 is 45 pairs. But enumerating all 45 might be time-consuming, but possibly manageable.\n",
            "\n",
            "Alternatively, maybe there's a pattern or a shortcut. Let me think about the numbers. Each element is n^n. So their prime factorizations involve each prime raised to up to n. For example, 6^6 = (2*3)^6 = 2^6 * 3^6. So the prime factors are all the primes up to n in the range 1 to 10. So primes up to 10 are 2,3,5,7. Each n^n will have these primes with exponents up to n. Therefore, the divisors of any n^n are all combinations of these primes with exponents from 0 up to n.\n",
            "\n",
            "So the problem reduces to finding all combinations of these primes (and their exponents) that divide at least two n^n. That is, for each prime, the maximum exponent in the two numbers. For example, take 2. In 4=2^2 and 8=2^3, the maximum exponent for 2 is 3. So in any pair, for each prime, take the maximum exponent in both numbers. Then the tuple of maximum exponents for all primes will be the exponents of a number that divides both.\n",
            "\n",
            "Therefore, the problem is equivalent to finding all tuples (e2, e3, e5, e7) where each e_p is the maximum exponent of p in the factorization of two numbers, and then counting the number of such tuples possible across all pairs.\n",
            "\n",
            "But wait, but the divisors must satisfy the condition that there exists at least one pair where the exponents match in such a tuple. So if we fix the tuple, we need to find pairs where for each prime, the maximum exponent is at least the exponent in the other. For example, if the tuple is (2,1,0,0), then there must be at least some pair where the exponents are 2 and 1 (or both 1) for each prime. \n",
            "\n",
            "Therefore, the number of such tuples is the number of ways the maximum exponents can align across at least one pair. Hmm, perhaps the total number of possible tuples is the product of (upper bound on exponent +1) for each prime. The upper bound for each prime p is the maximum exponent in the set, which is the maximum n for that prime p. For example, 2 can have exponents up to 10, 3 up to 10, etc. Since the numbers are n^n, so for prime p, the maximum exponent is 10. Wait, no, for each number n, if the number is p^k where p is the prime, then k can be up to n. But in our set, the maximum n is 10, so for each prime, the maximum exponent in any number is 10. Therefore, for each tuple, each e_p can be from 0 to 10. So the total number of tuples is 11^4, since there are 4 primes (2,3,5,7) and each can be 11 values (0-10). \n",
            "\n",
            "But wait, the problem is asking for tuples where there exists at least one pair with that tuple. So some tuples might not be achievable because they require exponents that are too high in some numbers. For example, a tuple where one of the exponents is 10. But in the factorization of each number, the maximum exponent for each prime is 10. So every tuple where each e_p is between 0 and 10 is possible. Therefore, the total number of such tuples is 11^4. \n",
            "\n",
            "But wait, is that correct? Let me verify. Suppose we fix a tuple like (2,1,0,0). Then does there exist a pair where in one number the exponents of the primes are at least 2,1,0,0? Yes, for example, take 10^10=2^10 3^0 5^0 7^0 and 4^4=2^4 3^0 5^0 7^0. Wait, no, because in 4, the exponent for 2 is 4, which is less than 2. So maybe the maximum is the upper bound. Therefore, for each prime, if the tuple requires an exponent of 2, then there must be some number where that prime is raised to at least 2. Since the maximum possible exponent in the set is 10, each e_p can range from 0 to 10. Therefore, each prime in the tuple can independently be any integer from 0 to 10. Thus the total number of tuples is 11^4. \n",
            "\n",
            "But wait, but in reality, some ex \n",
            "Extracted:\n",
            "None\n"
          ]
        },
        {
          "data": {
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              "\n",
              "    <div>\n",
              "      \n",
              "      <progress value='100' max='100' style='width:300px; height:20px; vertical-align: middle;'></progress>\n",
              "      [100/100 2:54:46, Epoch 0/1]\n",
              "    </div>\n",
              "    <table border=\"1\" class=\"dataframe\">\n",
              "  <thead>\n",
              " <tr style=\"text-align: left;\">\n",
              "      <th>Step</th>\n",
              "      <th>Training Loss</th>\n",
              "      <th>reward</th>\n",
              "      <th>reward_std</th>\n",
              "      <th>completion_length</th>\n",
              "      <th>kl</th>\n",
              "      <th>rewards / match_format_exactly</th>\n",
              "      <th>rewards / match_format_approximately</th>\n",
              "      <th>rewards / check_answer</th>\n",
              "      <th>rewards / check_numbers</th>\n",
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              "  </thead>\n",
              "  <tbody>\n",
              "    <tr>\n",
              "      <td>1</td>\n",
              "      <td>0.006200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
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              "      <td>0.000000</td>\n",
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              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
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              "    <tr>\n",
              "      <td>2</td>\n",
              "      <td>0.005200</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1754.000000</td>\n",
              "      <td>0.130613</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>3</td>\n",
              "      <td>0.006300</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1826.000000</td>\n",
              "      <td>0.156329</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>4</td>\n",
              "      <td>0.007100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.176596</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>5</td>\n",
              "      <td>0.007500</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1297.500000</td>\n",
              "      <td>0.188479</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>6</td>\n",
              "      <td>0.004800</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.119617</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>7</td>\n",
              "      <td>0.006200</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1679.000000</td>\n",
              "      <td>0.154963</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>8</td>\n",
              "      <td>0.004200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.105323</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>9</td>\n",
              "      <td>0.006100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.152696</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>10</td>\n",
              "      <td>0.004900</td>\n",
              "      <td>-0.875000</td>\n",
              "      <td>9.672771</td>\n",
              "      <td>1784.750000</td>\n",
              "      <td>0.123577</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>-0.875000</td>\n",
              "      <td>-0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>11</td>\n",
              "      <td>0.006300</td>\n",
              "      <td>11.000000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1222.000000</td>\n",
              "      <td>0.157895</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>4.250000</td>\n",
              "      <td>2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>12</td>\n",
              "      <td>0.006200</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1797.500000</td>\n",
              "      <td>0.153804</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>13</td>\n",
              "      <td>0.008500</td>\n",
              "      <td>6.750000</td>\n",
              "      <td>7.216878</td>\n",
              "      <td>1503.250000</td>\n",
              "      <td>0.212452</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>1.250000</td>\n",
              "      <td>1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>14</td>\n",
              "      <td>0.130100</td>\n",
              "      <td>3.625000</td>\n",
              "      <td>6.250000</td>\n",
              "      <td>1309.500000</td>\n",
              "      <td>3.253437</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.625000</td>\n",
              "      <td>-0.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>15</td>\n",
              "      <td>0.004900</td>\n",
              "      <td>7.875000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1581.250000</td>\n",
              "      <td>0.121786</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>3.250000</td>\n",
              "      <td>2.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>16</td>\n",
              "      <td>0.004200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.104844</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>17</td>\n",
              "      <td>0.004400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.109389</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>18</td>\n",
              "      <td>0.005700</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.141450</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>19</td>\n",
              "      <td>0.005900</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.146412</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>20</td>\n",
              "      <td>0.006500</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.162441</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>21</td>\n",
              "      <td>0.005500</td>\n",
              "      <td>4.750000</td>\n",
              "      <td>10.070584</td>\n",
              "      <td>1403.500000</td>\n",
              "      <td>0.138529</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>1.375000</td>\n",
              "      <td>0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>22</td>\n",
              "      <td>0.006400</td>\n",
              "      <td>11.000000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1244.250000</td>\n",
              "      <td>0.159613</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>4.250000</td>\n",
              "      <td>2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>23</td>\n",
              "      <td>0.006500</td>\n",
              "      <td>-3.500000</td>\n",
              "      <td>4.618802</td>\n",
              "      <td>1539.500000</td>\n",
              "      <td>0.162264</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>-2.250000</td>\n",
              "      <td>-2.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>24</td>\n",
              "      <td>0.005600</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1672.000000</td>\n",
              "      <td>0.140189</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>25</td>\n",
              "      <td>0.005600</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1187.000000</td>\n",
              "      <td>0.140135</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>26</td>\n",
              "      <td>0.007500</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>10.070584</td>\n",
              "      <td>1610.000000</td>\n",
              "      <td>0.188116</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>27</td>\n",
              "      <td>0.005900</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1727.750000</td>\n",
              "      <td>0.147394</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>28</td>\n",
              "      <td>0.005900</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.146671</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>29</td>\n",
              "      <td>0.004600</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.115949</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>30</td>\n",
              "      <td>0.003700</td>\n",
              "      <td>-6.000000</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.093672</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>31</td>\n",
              "      <td>0.005800</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1076.750000</td>\n",
              "      <td>0.145769</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>32</td>\n",
              "      <td>0.007100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.177763</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>33</td>\n",
              "      <td>0.010600</td>\n",
              "      <td>8.750000</td>\n",
              "      <td>4.907477</td>\n",
              "      <td>956.250000</td>\n",
              "      <td>0.266028</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>3.250000</td>\n",
              "      <td>1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>34</td>\n",
              "      <td>0.003300</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1833.500000</td>\n",
              "      <td>0.083449</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>35</td>\n",
              "      <td>0.006900</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1325.500000</td>\n",
              "      <td>0.172706</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>36</td>\n",
              "      <td>0.003600</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.089926</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>37</td>\n",
              "      <td>0.006600</td>\n",
              "      <td>3.875000</td>\n",
              "      <td>8.469307</td>\n",
              "      <td>1525.750000</td>\n",
              "      <td>0.165132</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>1.750000</td>\n",
              "      <td>-0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>38</td>\n",
              "      <td>0.006700</td>\n",
              "      <td>7.875000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1720.250000</td>\n",
              "      <td>0.166728</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>3.250000</td>\n",
              "      <td>2.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>39</td>\n",
              "      <td>0.005500</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.138440</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>40</td>\n",
              "      <td>0.003400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.085036</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>41</td>\n",
              "      <td>0.005600</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1749.000000</td>\n",
              "      <td>0.138822</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>42</td>\n",
              "      <td>0.022800</td>\n",
              "      <td>-0.875000</td>\n",
              "      <td>9.672771</td>\n",
              "      <td>1105.000000</td>\n",
              "      <td>0.570743</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>-0.875000</td>\n",
              "      <td>-0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>43</td>\n",
              "      <td>0.004600</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1777.000000</td>\n",
              "      <td>0.115114</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>44</td>\n",
              "      <td>0.003400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.085207</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>45</td>\n",
              "      <td>0.006200</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1760.250000</td>\n",
              "      <td>0.155049</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>46</td>\n",
              "      <td>0.005100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.128713</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>47</td>\n",
              "      <td>0.005600</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1204.750000</td>\n",
              "      <td>0.140702</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>48</td>\n",
              "      <td>0.005000</td>\n",
              "      <td>7.875000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1461.250000</td>\n",
              "      <td>0.125023</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>3.250000</td>\n",
              "      <td>2.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>49</td>\n",
              "      <td>0.005100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.127187</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>50</td>\n",
              "      <td>0.004300</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.107273</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>51</td>\n",
              "      <td>0.005600</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1628.000000</td>\n",
              "      <td>0.140713</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>52</td>\n",
              "      <td>0.006200</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1696.500000</td>\n",
              "      <td>0.155484</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>53</td>\n",
              "      <td>0.005200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.130052</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>54</td>\n",
              "      <td>0.005700</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.143681</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>55</td>\n",
              "      <td>0.005900</td>\n",
              "      <td>-1.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1718.250000</td>\n",
              "      <td>0.147326</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>-1.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>56</td>\n",
              "      <td>0.004400</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1381.250000</td>\n",
              "      <td>0.110959</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>57</td>\n",
              "      <td>0.004600</td>\n",
              "      <td>6.750000</td>\n",
              "      <td>7.216878</td>\n",
              "      <td>1100.000000</td>\n",
              "      <td>0.116166</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>1.250000</td>\n",
              "      <td>1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>58</td>\n",
              "      <td>0.006100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.151746</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>59</td>\n",
              "      <td>0.005000</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.124102</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>60</td>\n",
              "      <td>0.005400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.134021</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>61</td>\n",
              "      <td>0.005400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.134724</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>62</td>\n",
              "      <td>0.004600</td>\n",
              "      <td>-3.500000</td>\n",
              "      <td>4.618802</td>\n",
              "      <td>1739.250000</td>\n",
              "      <td>0.114914</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>-2.250000</td>\n",
              "      <td>-2.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>63</td>\n",
              "      <td>0.004600</td>\n",
              "      <td>-6.500000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>1840.500000</td>\n",
              "      <td>0.115412</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-2.250000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>64</td>\n",
              "      <td>0.003900</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1700.500000</td>\n",
              "      <td>0.096654</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>65</td>\n",
              "      <td>0.005800</td>\n",
              "      <td>13.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1177.000000</td>\n",
              "      <td>0.144043</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>5.000000</td>\n",
              "      <td>3.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>66</td>\n",
              "      <td>0.005700</td>\n",
              "      <td>10.875000</td>\n",
              "      <td>4.250000</td>\n",
              "      <td>1196.500000</td>\n",
              "      <td>0.141380</td>\n",
              "      <td>3.000000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>4.125000</td>\n",
              "      <td>2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>67</td>\n",
              "      <td>0.004700</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1684.000000</td>\n",
              "      <td>0.117414</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>68</td>\n",
              "      <td>0.004600</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.115974</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>69</td>\n",
              "      <td>0.005800</td>\n",
              "      <td>5.750000</td>\n",
              "      <td>9.699656</td>\n",
              "      <td>1510.500000</td>\n",
              "      <td>0.145899</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>2.375000</td>\n",
              "      <td>0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>70</td>\n",
              "      <td>0.004900</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.122652</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>71</td>\n",
              "      <td>0.003700</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.092026</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>72</td>\n",
              "      <td>0.003600</td>\n",
              "      <td>0.250000</td>\n",
              "      <td>9.836158</td>\n",
              "      <td>1810.500000</td>\n",
              "      <td>0.089310</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>73</td>\n",
              "      <td>0.003100</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.078275</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>74</td>\n",
              "      <td>0.004700</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1662.750000</td>\n",
              "      <td>0.117625</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>75</td>\n",
              "      <td>0.005400</td>\n",
              "      <td>1.875000</td>\n",
              "      <td>6.250000</td>\n",
              "      <td>1588.250000</td>\n",
              "      <td>0.134210</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>-1.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>76</td>\n",
              "      <td>0.003400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.085360</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>77</td>\n",
              "      <td>0.006400</td>\n",
              "      <td>5.750000</td>\n",
              "      <td>9.699656</td>\n",
              "      <td>1511.500000</td>\n",
              "      <td>0.159980</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>2.375000</td>\n",
              "      <td>0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>78</td>\n",
              "      <td>0.004400</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1780.250000</td>\n",
              "      <td>0.109536</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>79</td>\n",
              "      <td>0.006600</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1650.250000</td>\n",
              "      <td>0.164220</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>80</td>\n",
              "      <td>0.004400</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.109638</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>81</td>\n",
              "      <td>0.005800</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1718.500000</td>\n",
              "      <td>0.146145</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>82</td>\n",
              "      <td>0.006000</td>\n",
              "      <td>-5.500000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>1838.500000</td>\n",
              "      <td>0.149491</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.250000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>83</td>\n",
              "      <td>0.004900</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1801.000000</td>\n",
              "      <td>0.122010</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>84</td>\n",
              "      <td>0.003300</td>\n",
              "      <td>-4.875000</td>\n",
              "      <td>3.772157</td>\n",
              "      <td>1534.500000</td>\n",
              "      <td>0.081980</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.500000</td>\n",
              "      <td>-2.125000</td>\n",
              "      <td>-2.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>85</td>\n",
              "      <td>0.004500</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1736.000000</td>\n",
              "      <td>0.112571</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>86</td>\n",
              "      <td>0.004200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.105930</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>87</td>\n",
              "      <td>0.004700</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.117618</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>88</td>\n",
              "      <td>0.003800</td>\n",
              "      <td>-2.375000</td>\n",
              "      <td>10.250000</td>\n",
              "      <td>1771.750000</td>\n",
              "      <td>0.094815</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-1.875000</td>\n",
              "      <td>-0.250000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>89</td>\n",
              "      <td>0.004200</td>\n",
              "      <td>-1.500000</td>\n",
              "      <td>2.273030</td>\n",
              "      <td>1682.500000</td>\n",
              "      <td>0.103934</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.750000</td>\n",
              "      <td>-2.750000</td>\n",
              "      <td>-1.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>90</td>\n",
              "      <td>0.003200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.078818</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>91</td>\n",
              "      <td>0.003700</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.092524</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>92</td>\n",
              "      <td>0.005700</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1567.250000</td>\n",
              "      <td>0.142832</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>93</td>\n",
              "      <td>0.005300</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>11.835680</td>\n",
              "      <td>1706.500000</td>\n",
              "      <td>0.131643</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.750000</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>94</td>\n",
              "      <td>0.005100</td>\n",
              "      <td>-3.875000</td>\n",
              "      <td>2.926175</td>\n",
              "      <td>1557.500000</td>\n",
              "      <td>0.128169</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>-0.375000</td>\n",
              "      <td>-3.250000</td>\n",
              "      <td>-1.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>95</td>\n",
              "      <td>0.004000</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.100573</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>96</td>\n",
              "      <td>0.006600</td>\n",
              "      <td>2.750000</td>\n",
              "      <td>8.568353</td>\n",
              "      <td>1248.500000</td>\n",
              "      <td>0.164374</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>0.625000</td>\n",
              "      <td>-0.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>97</td>\n",
              "      <td>0.007200</td>\n",
              "      <td>4.750000</td>\n",
              "      <td>10.070584</td>\n",
              "      <td>1428.250000</td>\n",
              "      <td>0.179731</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>1.375000</td>\n",
              "      <td>0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>98</td>\n",
              "      <td>0.005000</td>\n",
              "      <td>5.875000</td>\n",
              "      <td>9.681382</td>\n",
              "      <td>1680.250000</td>\n",
              "      <td>0.125890</td>\n",
              "      <td>2.250000</td>\n",
              "      <td>0.375000</td>\n",
              "      <td>2.500000</td>\n",
              "      <td>0.750000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>99</td>\n",
              "      <td>0.004800</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.120776</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>100</td>\n",
              "      <td>0.005200</td>\n",
              "      <td>-7.500000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1846.000000</td>\n",
              "      <td>0.129945</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>-3.000000</td>\n",
              "      <td>-2.000000</td>\n",
              "      <td>-2.500000</td>\n",
              "    </tr>\n",
              "  </tbody>\n",
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            "********************Question:\n",
            "The sum of $\\lfloor x \\rfloor$ for all real numbers $x$ satisfying the equation $16 + 15x + 15x^2 = \\lfloor x \\rfloor^3$ is: \n",
            "Answer:\n",
            "33 \n",
            "Response:\n",
            "Okay, so I need to find the sum of the floor of x for all real numbers x that satisfy the equation 16 + 15x + 15x² = ⌊x⌋³. Hmm, let me think.\n",
            "\n",
            "First, let me understand what the equation is saying. The equation relates a quadratic in x, 16 + 15x + 15x², to the cube of the floor of x. The floor function ⌊x⌋ gives the greatest integer less than or equal to x. So, for each solution x, I need to compute ⌊x⌋ and then sum those values.\n",
            "\n",
            "Alright, let me try to find the possible values of ⌊x⌋. Since ⌊x⌋ is an integer, maybe I can denote it as n, where n is an integer. Then the equation becomes 16 + 15x + 15x² = n³.\n",
            "\n",
            "But wait, x is a real number, but since we're taking its floor, n ≤ x < n+1. So perhaps I can substitute x with n + y, where y is the fractional part, 0 ≤ y < 1. Then the equation becomes:\n",
            "\n",
            "16 + 15(n + y) + 15(n + y)² = n³\n",
            "\n",
            "Let me expand that. First, 15(n + y) is 15n + 15y. Then (n + y)² is n² + 2ny + y². So multiplying by 15 gives 15n² + 30ny + 15y². Adding the 16, the equation is:\n",
            "\n",
            "16 + 15n + 15y + 15n² + 30ny + 15y² = n³\n",
            "\n",
            "But since y is between 0 and 1, all the terms with y are positive terms. Maybe I can rearrange the equation to solve for n. Let's collect like terms:\n",
            "\n",
            "15n² + 15n + 16 + 15y + 30ny + 15y² = n³\n",
            "\n",
            "Hmm, let me subtract 15n² + 15n + 16 from both sides to get:\n",
            "\n",
            "30ny + 15y² + 15y = n³ - 15n² - 15n - 16\n",
            "\n",
            "Factor out y on the left side:\n",
            "\n",
            "y(30n + 15y + 15) = n³ - 15n² - 15n - 16\n",
            "\n",
            "Now, since 0 ≤ y < 1, the left side is between 0 and 30n + 15 + 15 = 30n + 30. The right side is n³ - 15n² - 15n - 16. So for there to be a solution, the right side must be between 0 and 30n + 30. That might help narrow down possible integer values of n.\n",
            "\n",
            "Alternatively, perhaps it's easier to test integer values of n. Since the equation involves n³ on the right, which can be large, maybe n is small. Let me try plugging in small integers for n and see if the equation holds.\n",
            "\n",
            "Let's start with n=0. Then the equation becomes 16 + 15x + 15x² = 0. Let's see if there are real solutions. The equation is 15x² + 15x + 16 = 0. The discriminant is 225 - 4*15*16 = 225 - 960 = -735, which is negative. So no real solutions here. Thus, n=0 is not possible.\n",
            "\n",
            "Next, n=1. Then the equation is 16 + 15x + 15x² = 1. Rearranged: 15x² + 15x +15 = 0. Divided by 15: x² + x +1 =0. Discriminant is 1 - 4 = -3, again no real solutions. So n=1 doesn't work either.\n",
            "\n",
            "n=2: equation becomes 16 + 15x + 15x² = 8. So 15x² + 15x -4 = 0. Let's solve: discriminant is 225 + 240 = 465. So x = [-15 ± sqrt(465)]/30. Since sqrt(465) is approximately 21.564, we get x ≈ ( -15 + 21.564)/30 ≈ 0.219 and x ≈ ( -15 - 21.564)/30 ≈ -2.156. But we need 1 ≤ x <2. So x ≈ 0.219 is too small, and -2.156 is less than 0. So no solution here for n=2.\n",
            "\n",
            "n=3: equation becomes 16 + 15x + 15x² = 27. So 15x² + 15x -11 = 0. Discriminant is 225 -660 = -435. No real solutions. So n=3 doesn't work.\n",
            "\n",
            "n=4: equation is 16 + 15x + 15x² = 64. So 15x² + 15x -48 = 0. Divided by 15: x² + x -3.2 = 0. Discriminant: 1 + 12.8 = 13.8. So x ≈ (-1 ± sqrt(13.8))/2. So x ≈ (-1 + 3.715)/2 ≈ 1.357 and x ≈ (-1 - 3.715)/2 ≈ -2.357. Since 3 ≤ x <4, x ≈ 1.357 is too low, and -2.357 is less than 0. So no solution here for n=4.\n",
            "\n",
            "Hmm, maybe I should try n=-1. Let's see. For n=-1, the equation is 16 + 15x + 15x² = -1. So 15x² + 15x +17 = 0. Discriminant: 225 -1020 = -795. No real solutions here. So n=-1 doesn't work.\n",
            "\n",
            "n=-2. Equation: 16 + 15x + 15x² = -8. So 15x² + 15x +24 = 0. Divided by 15: x² + x +1.6 = 0. Discriminant: 1 -6.4 = -5.4. No real solutions. So n=-2 is out.\n",
            "\n",
            "Wait, maybe I should try n=1 again, but check the calculations. Wait, no. So far, negative integers don't work. Let me check n=1 again. The equation was 15x² + 15x +15=0, which simplifies to x² + x +1=0. The solutions would be complex, so no real solutions. So n=1 is not possible.\n",
            "\n",
            "Wait, perhaps n=0 was not possible, and increasing n, the left side 15x² + 15x +16 grows faster than n³. Let me check n=3 again. Equation: 15x² +15x +16=27. So 15x²+15x-11=0. The solutions are approximately x ≈0.393 and x≈-1.093. But n=3 requires 3 ≤ x<4. So x≈0.393 is too low, and -1.093 is less than 0. So no solution here. Similarly, n=4 gave x≈1.357, which is too low. So perhaps there's no solution for integer n? But the problem states that there are solutions. Maybe I made a mistake?\n",
            "\n",
            "Wait, let me check n=1 again. The equation is 16 +15x + 15x² =1 \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "Letters $A, B, C,$ and $D$ represent four different digits from 0, 1, 2, 3, ..., 9. If $\\frac{A+B}{C+D}$ is an integer that is as large as possible, what is the value of $A+B$? \n",
            "Answer:\n",
            "17 \n",
            "Response:\n",
            "Okay, so I need to solve this problem where four different digits A, B, C, and D represent four different digits from 0 to 9. The goal is to maximize the expression (A + B)/(C + D) and find the value of A + B. Hmm, let's break this down step by step.\n",
            "\n",
            "First, I need to remember that A, B, C, and D are all different digits, so no repeats. And they range from 0 to 9. The fraction (A + B)/(C + D) needs to be as large as possible, and since we're dealing with integers, that means (A + B) is a multiple of (C + D). The goal is to maximize the numerator (A + B), so that's the top part. Therefore, to maximize (A + B)/(C + D), A and B should be the largest possible digits since their sum is the numerator. The maximum digits available are 9 and 8. So if A is 9 and B is 8, then A + B would be 17. But wait, does that work? Let me check.\n",
            "\n",
            "If A=9, B=8, then (9 + 8)/(C + D) = 17/(C + D). For this to be an integer, C + D must be a divisor of 17. Since 17 is a prime number, the possible sum of C and D can only be 1 or 17. But since C and D are digits and all four digits are different, if C + D were 17, the maximum possible sum would be 9+8=17, but then A and B would be the other two digits, which are 1 and 0. Wait, but 17 is too large because the maximum possible sum for C+D would be 9+8=17, but then A and B are 1 and 0, which would make A + B = 1. That's too small. So if C+D is 17, then A and B have to be the smallest possible digits, which would make A+B very small. That's not maximizing the fraction.\n",
            "\n",
            "Therefore, the only other option is C+D=1. If C+D=1, then the digits would have to be 0 and 1. So D is 0 and C is 1. Then A and B are the remaining digits, which can be 8 and 9. So A+B would be 17. For example, A=9, B=8, C=1, D=0. Then (9+8)/(1+0) = 17/1 = 17, which is an integer and the maximum possible since increasing C+D would make the denominator larger, reducing the value of the fraction. So that seems right.\n",
            "\n",
            "Wait, let me double-check if there's any other combination where A and B are larger. Like maybe using some larger digits? If A and B are 9 and 7, then A+B is 16. Then C+D would have to divide 16. The possible values could be 1, 2, 4, 8, 16. But since the maximum sum of C and D is 9+8=17, 16 is possible. If C+D=16, then the digits could be 9 and 7, but then A and B would have to be 0 and 1. But again, in that case, A+B is 0+1=1, which is smaller than 17. So the higher we make C+D, the smaller A+B becomes. Therefore, the maximum occurs when C+D is as small as possible, which is 1, making A+B=17. Yep, that makes sense. So the answer should be 17.\n",
            "To maximize the fraction \\(\\frac{A+B}{C+D}\\), where \\(A, B, C,\\) and \\(D\\) are distinct digits from 0 to 9, we need to maximize the numerator \\(A+B\\) while ensuring that the denominator \\(C+D\\) is a divisor of \\(A+B\\).\n",
            "\n",
            "First, we note that the maximum possible value for \\(A+B\\) is achieved when \\(A\\) and \\(B\\) are the largest distinct digits available, which are 9 and 8. Thus, \\(A = 9\\) and \\(B = 8\\) gives \\(A + B = 17\\).\n",
            "\n",
            "Next, we need to check if \\(C + D\\) can be a divisor of 17. Since 17 is a prime number, its only divisors are 1 and 17. \n",
            "\n",
            "1. If \\(C + D = 1\\), the only possible distinct digits are 0 and 1. Therefore, \\(C = 0\\) and \\(D = 1\\) or vice versa. This gives \\(C + D = 1\\), which is a divisor of 17. In this case, \\(A + B = 17\\) and the fraction \\(\\frac{17}{1} = 17\\) is an integer.\n",
            "2. If \\(C + D = 17\\), the only possible distinct digits that sum to 17 are 9 and 8. However, this would mean \\(A, B, C,\\) and \\(D\\) are all the same digits (9 and 8), which is not allowed since they must be distinct.\n",
            "\n",
            "Therefore, the only feasible option is when \\(C + D = 1\\), which maximizes \\(A + B = 17\\).\n",
            "\n",
            "Thus, the value of \\(A + B\\) that maximizes the fraction \\(\\frac{A+B}{C+D}\\) is \\(\\boxed{17}\\).<end_working_out><SOLUTION>17</SOLUTION> \n",
            "Extracted:\n",
            "17\n",
            "********************Question:\n",
            "Let $f(x)=\\int_0^{x} e^{t} (\\cos t+\\sin t)\\, dt$, $g(x)=\\int_0^{x} e^{t} (\\cos t-\\sin t)\\, dt$. \n",
            "\n",
            "For a real number $a$, find:\n",
            "\\[ \\sum_{n=1}^{\\infty} \\frac{e^{2a}}{\\{f^{(n)}(a)\\}^2+\\{g^{(n)}(a)\\}^2}. \\] \n",
            "Answer:\n",
            "1 \n",
            "Response:\n",
            "Okay, let's tackle this problem step by step. I need to find the sum from n=1 to infinity of e^(2a) divided by the sum of the squares of the nth derivatives of f and g evaluated at a. Hmm, let's start by understanding what f and g are.\n",
            "\n",
            "Given that f(x) is the integral from 0 to x of e^t(cos t + sin t) dt, and g(x) is the integral from 0 to x of e^t(cos t - sin t) dt. So, both are integrals of exponential functions multiplied by combinations of cosine and sine. Let me see if I can express these integrals in a closed form to make finding their derivatives easier.\n",
            "\n",
            "Wait, when you integrate e^t times a linear combination of sin and cos, maybe there's a pattern like integrating by parts multiple times. Let me recall that integrating e^t(sin t) would involve applying integration by parts twice, and similar for cos t. Let's compute the integral of e^t(cos t + sin t).\n",
            "\n",
            "Let's set u = cos t + sin t, then du = (-sin t + cos t)dt. Wait, no, that's the derivative of the numerator. Hmm, perhaps a better approach is to integrate e^t(d/dt of (cos t + sin t)). Since d/dt (cos t + sin t) = sin t + cos t, which is the same as the integrand. So integrating e^t d/dt(cos t + sin t) would be e^t(cos t + sin t) evaluated from 0 to x. Let me check:\n",
            "\n",
            "Integrate [e^t d/dt (cos t + sin t) dt] from 0 to x. Using the Fundamental Theorem of Calculus, this is e^x(cos x + sin x) - e^0(cos 0 + sin 0) = e^x(cos x + sin x) - 1. So f(x) = e^x(cos x + sin x) - 1. Similarly, for g(x), since the integrand is e^t(cos t - sin t), integrating e^t d/dt(cos t - sin t) gives e^x(cos x - sin x) - 1. So g(x) = e^x(cos x - sin x) - 1.\n",
            "\n",
            "Great, so now I can easily compute the derivatives of f and g. Let's compute f^(n)(x). Since f(x) is e^x(cos x + sin x) - 1, taking derivatives term by term. Let me compute the derivatives of e^x(cos x + sin x). Using the product rule, the derivative is e^x(cos x + sin x) + e^x(-sin x + cos x) = e^x(cos x + sin x - sin x + cos x) = e^x(2cos x). So f'(x) = 2e^x cos x. Then f''(x) = 2e^x cos x - 2e^x sin x = 2e^x(cos x - sin x). Similarly, each time we differentiate, the coefficient doubles and the signs alternate. So f^(n)(x) = 2^n e^x cos x for odd n, and (even n terms), and similarly for g^(n)(x).\n",
            "\n",
            "Wait, but actually, since cos and sin derivatives alternate in sign and coefficient, perhaps there's a pattern. Let's compute the first few derivatives to see.\n",
            "\n",
            "f(x) = e^x(cos x + sin x) -1\n",
            "\n",
            "f'(x) = e^x(cos x + sin x) + e^x(-sin x + cos x) = 2e^x cos x\n",
            "\n",
            "f''(x) = 2e^x cos x - 2e^x sin x = 2e^x(cos x - sin x)\n",
            "\n",
            "f'''(x) = 2e^x(cos x - sin x) - 2e^x(-sin x - cos x) = 2e^x(cos x - sin x + sin x + cos x) = 4e^x cos x\n",
            "\n",
            "f^(4)(x) = 4e^x cos x - 4e^x sin x = 4e^x(cos x - sin x)\n",
            "\n",
            "So it seems like f^(n)(x) = 2^n e^x cos x when n is odd, and 2^n e^x(cos x - sin x) when n is even. Similarly, g(x) would be the same but with the sign on sin term. So g^(n)(x) = 2^n e^x cos x when n is even, and 2^n e^x(cos x - sin x) when n is odd.\n",
            "\n",
            "Therefore, for each n, we have two terms: f^(n)(a) and g^(n)(a). Let's compute their squares and their sum.\n",
            "\n",
            "If n is odd, then both f^(n)(a) and g^(n)(a) are 2^n e^a cos a. So the sum of squares is 4*2^{2n} e^{2a} cos^2 a. Then the term in the sum is e^{2a} / (4*2^{2n} e^{2a} cos^2 a) = 1 / (4*2^{2n} cos^2 a). \n",
            "\n",
            "If n is even, then f^(n)(a) is 2^n e^a (cos a - sin a), and g^(n)(a) is 2^n e^a cos a. So the sum of squares is 2^{2n} e^{2a} (cos a - sin a)^2 + 2^{2n} e^{2a} cos^2 a. Simplifying the expression inside the parentheses: (cos a - sin a)^2 + cos^2 a = cos^2 a - 2sin a cos a + sin^2 a + cos^2 a = (1 - 2sin a cos a + 1) + cos^2 a =2 - 2sin a cos a + cos^2 a = 2 - sin 2a + cos^2 a. But cos^2 a + sin^2 a =1, so that's 2 - sin 2a + (1 - sin^2 a) = 3 - sin 2a - sin^2 a. Hmm, that seems complicated. Maybe I took a wrong approach.\n",
            "\n",
            "Wait, perhaps instead of trying to compute each derivative explicitly, there's a pattern or a formula. Let me recall that when dealing with \\( f(x) = \\int_0^x e^{t} \\cos t \\, dt \\) and \\( g(x) = \\int_0^x e^{t} \\sin t \\, dt \\), their derivatives alternate between functions like \\( e^x \\sin x \\) and \\( e^x \\cos x \\) with coefficients increasing by powers of 2. So for example, \\( f^{(n)}(x) = 2^n e^x \\sin x \\) and \\( g^{(n)}(x) = 2^n e^x \\cos x \\). But wait, no, because integrating \\( e^t \\sin t \\) gives \\( \\frac{1}{2} e^t (\\sin t - \\cos t) \\), so the derivatives alternate. Let me check with n=1. Since f'(x) = 2 cos x, which would be \\( -g(a) \\) if g=... But hold on, maybe the derivatives of f and g are related to \\( \\sin x \\) and \\( \\cos x \\) with coefficients doubling. Yeah, let's think differently. Maybe consider the original integrals as being related to the solutions of a differential equation.\n",
            "\n",
            "Alternatively, perhaps using complex exponentials would simplify things. Let's express cos and sin as exponentials. Let f(x) = Integral [e^t (e^{it} + e^{-it})] from 0 to x, then f(x) = Integral [e^{(1+i)t} + e^{(1-i)t}] from 0 to x. So f(x) = (1/(1+i) + 1/(1-i)) (e^{(1+i)x} + e^{(1-i)x}) - (1/(1+i) + 1/(1-i)) (1 + 1) = [(1-i + 1+i)/(1 - i^2)] (e^{(1+i)x} + e^{(1-i)x}) - 2 = [0/(1+1)] (e^{(1+i)x} + \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "How many distinct ordered pairs of positive integers $(m,n)$ are there so that the sum of the reciprocals of $m$ and $n$ is $\\frac14$? \n",
            "Answer:\n",
            "5 \n",
            "Response:\n",
            "Okay, so I need to find all distinct ordered pairs of positive integers (m, n) such that 1/m + 1/n equals 1/4. Hmm, let me think. Let's start by writing down the equation: 1/m + 1/n = 1/4. To make this easier, maybe I can find a common denominator. So, adding the fractions on the left side gives (n + m)/(mn) = 1/4. So, (m + n)/(mn) = 1/4. Then, cross-multiplying gives 4(m + n) = mn. Let me rearrange that equation to make it more manageable. Subtracting 4m and 4n from both sides gives mn - 4m - 4n = 0. Now, I need to find positive integer solutions (m, n) to this equation. Maybe I can factor this. Let me add 16 to both sides to complete the rectangle. Wait, how does that work? Adding 16 gives mn - 4m - 4n + 16 = 16. Then, the left side factors as (m - 4)(n - 4) = 16. Alright, so now I need to find all pairs of positive integers (m - 4, n - 4) that multiply to 16. Let's list the factor pairs of 16. The positive integer factor pairs are (1,16), (2,8), (4,4), (8,2), and (16,1). Each of these pairs will give a solution for (m, n). Let's convert them back to (m, n) by adding 4 to each component. \n",
            "\n",
            "For (1,16): m = 5, n = 20. So (5,20) is a solution. \n",
            "\n",
            "For (2,8): m = 6, n = 12. So (6,12) is another solution. \n",
            "\n",
            "For (4,4): m = 8, n = 8. So (8,8) is a solution. \n",
            "\n",
            "For (8,2): m = 12, n = 6. So (12,6) is a solution. \n",
            "\n",
            "For (16,1): m = 20, n = 5. So (20,5) is a solution. \n",
            "\n",
            "Wait, so those are all the pairs where both m and n are greater than 4. But the problem asks for distinct ordered pairs. Since (6,12) and (12,6) are different, as are (5,20) and (20,5), and so on, all these pairs are distinct. Let me count them. Starting from the first one, there are 5 pairs here. But wait, let me double-check that I didn't miss any factor pairs. The factors of 16 are 1, 2, 4, 8, 16. So the pairs are (1,16), (2,8), (4,4), (8,2), (16,1). That's 5 pairs. Therefore, there are 5 distinct ordered pairs. So the answer should be 5. But let me make sure I didn't make a mistake with the algebra. Starting from 4(m + n) = mn, then subtracting 4m and 4n gives mn - 4m - 4n = 0. Adding 16: mn - 4m - 4n +16 = 16. Then factoring as (m - 4)(n - 4). Yep, that's correct. And the factor pairs of 16 give the solutions. So confident now. The answer is 5.\n",
            "To solve the problem of finding the number of distinct ordered pairs of positive integers \\((m, n)\\) such that \\(\\frac{1}{m} + \\frac{1}{n} = \\frac{1}{4}\\), we start with the equation:\n",
            "\n",
            "\\[\n",
            "\\frac{1}{m} + \\frac{1}{n} = \\frac{1}{4}\n",
            "\\]\n",
            "\n",
            "First, we find a common denominator for the left-hand side:\n",
            "\n",
            "\\[\n",
            "\\frac{m + n}{mn} = \\frac{1}{4}\n",
            "\\]\n",
            "\n",
            "Next, we cross-multiply to clear the fractions:\n",
            "\n",
            "\\[\n",
            "4(m + n) = mn\n",
            "\\]\n",
            "\n",
            "Rearranging the equation, we get:\n",
            "\n",
            "\\[\n",
            "mn - 4m - 4n = 0\n",
            "\\]\n",
            "\n",
            "To factor this equation, we add 16 to both sides to complete the rectangle:\n",
            "\n",
            "\\[\n",
            "mn - 4m - 4n + 16 = 16\n",
            "\\]\n",
            "\n",
            "This can be factored as:\n",
            "\n",
            "\\[\n",
            "(m - 4)(n - 4) = 16\n",
            "\\]\n",
            "\n",
            "Now, we need to find all pairs of positive integers \\((m-4, n-4)\\) that multiply to 16. The factor pairs of 16 are:\n",
            "\n",
            "\\[\n",
            "(1, 16), (2, 8), (4, 4), (8, 2), (16, 1)\n",
            "\\]\n",
            "\n",
            "For each pair \\((a, b)\\), we have \\(m = a + 4\\) and \\(n = b + 4\\). Converting each pair to \\((m, n)\\):\n",
            "\n",
            "1. For \\((1, 16)\\): \\(m = 5\\), \\(n = 20\\)\n",
            "2. For \\((2, 8)\\): \\(m = 6\\), \\(n = 12\\)\n",
            "3. For \\((4, 4)\\): \\(m = 8\\), \\(n = 8\\)\n",
            "4. For \\((8, 2)\\): \\(m = 12\\), \\(n = 6\\)\n",
            "5. For \\((16, 1)\\): \\(m = 20\\), \\(n = 5\\)\n",
            "\n",
            "Thus, the distinct ordered pairs \\((m, n)\\) are:\n",
            "\n",
            "\\[\n",
            "(5, 20), (6, 12), (8, 8), (12, 6), (20, 5)\n",
            "\\]\n",
            "\n",
            "There are 5 distinct ordered pairs. Therefore, the number of distinct ordered pairs \\((m, n)\\) is:\n",
            "\n",
            "\\[\n",
            "\\boxed{5}\n",
            "\\]<end_working_out><SOLUTION>5</SOLUTION> \n",
            "Extracted:\n",
            "5\n",
            "********************Question:\n",
            "Suppose that $n$ persons meet in a meeting, and that each of the persons is acquainted with exactly $8$ others. Any two acquainted persons have exactly $4$ common acquaintances, and any two non-acquainted persons have exactly $2$ common acquaintances. Find all possible values of $n$. \n",
            "Answer:\n",
            "21 \n",
            "Response:\n",
            "Okay, so I need to solve a problem about people at a meeting with specific acquaintance relationships. Let me try to break it down.\n",
            "\n",
            "The problem states that there are n people. Each person knows exactly 8 others. If two people are acquainted, they have exactly 4 common acquaintances. If two people are not acquainted, they have exactly 2 common acquaintances. I need to find all possible values of n.\n",
            "\n",
            "Hmm, this seems like a problem that might involve graph theory. Maybe I can model this as a graph where each person is a vertex, and each acquaintance is an edge. So, each vertex has degree 8. The conditions about common acquaintances might correspond to some properties of the graph.\n",
            "\n",
            "Let me recall some graph theory concepts. In graph theory, the number of common neighbors between two vertices is related to the edges not in the path between them. Maybe there's a formula or a known graph that fits these conditions.\n",
            "\n",
            "Wait, the conditions here seem specific. Let me think about known graphs. The problem mentions that two acquainted people have 4 common acquaintances, and two not-acquainted have 2 common acquaintances. That's a symmetric property. Maybe this is a strongly regular graph?\n",
            "\n",
            "A strongly regular graph has parameters (n, k, λ, μ), where:\n",
            "- n is the number of vertices,\n",
            "- k is the number of edges per vertex,\n",
            "- λ is the number of common neighbors for any two adjacent vertices,\n",
            "- μ is the number of common neighbors for any two non-adjacent vertices.\n",
            "\n",
            "Comparing to the problem:\n",
            "- k = 8 (each person knows 8 others),\n",
            "- λ = 4 (acquainted people have 4 common acquaintances),\n",
            "- μ = 2 (non-acquainted people have 2 common acquaintances).\n",
            "\n",
            "So, if this is a strongly regular graph, then the parameters are (n, 8, 4, 2). The problem is asking for all possible n. So maybe I need to find all values of n for which a strongly regular graph with these parameters exists.\n",
            "\n",
            "I remember there's a formula for strongly regular graphs. The equation is:\n",
            "\n",
            "k(k - λ - 1) = (n - k - 1)μ\n",
            "\n",
            "Let me plug in the values:\n",
            "\n",
            "k = 8, λ = 4, μ = 2.\n",
            "\n",
            "So, 8(8 - 4 - 1) = (n - 8 - 1)2\n",
            "\n",
            "Simplify the left side: 8*(3) = 24\n",
            "\n",
            "Right side: (n - 9)*2 = 2n - 18\n",
            "\n",
            "Set equal: 24 = 2n - 18\n",
            "\n",
            "Solve for n: 2n = 42, so n = 21\n",
            "\n",
            "Wait, so that gives n = 21. But the problem asks for all possible values of n. Does that mean 21 is the only solution? Maybe there are other strongly regular graphs with different parameters, but in this case, this one gives a unique n.\n",
            "\n",
            "But wait, I should check if this graph exists. Maybe there's another way to verify.\n",
            "\n",
            "Another approach: Maybe use the Moore bound for strongly regular graphs. The Moore bound gives the maximum number of vertices for a given degree and girth. But I'm not sure how to apply it here directly.\n",
            "\n",
            "Alternatively, perhaps I can use the fact that strongly regular graphs have a certain spectrum or eigenvalues. The equation I used is derived from the Hoffman-Singleton theorem, which gives a relationship between the parameters. Since the equation leads to n=21, that's likely the solution.\n",
            "\n",
            "But to be thorough, maybe I should check if there are other possible values. Since the equation gave a unique solution, maybe 21 is the only value. But perhaps there are other strongly regular graphs with different parameters. But according to the equation derived from the friendship theorem (a corollary of the Hoffman-Singleton theorem), this specific combination is unique. So the answer should be n=21.\n",
            "\n",
            "Wait, but maybe the problem allows for different graphs? Let me think. The conditions are very specific, like each vertex has degree 8, and the number of common neighbors is fixed. That would restrict n uniquely. Therefore, n must be 21.\n",
            "\n",
            "Therefore, the only possible value is 21.<end_working_out><SOLUTION>21</SOLUTION> \n",
            "Extracted:\n",
            "21\n",
            "********************Question:\n",
            "On rectangular coordinates, point $A = (1,2)$, $B = (3,4)$. $P = (a, 0)$ is on $x$-axis. Given that $P$ is chosen such that $AP + PB$ is minimized, compute $60a$. \n",
            "Answer:\n",
            "100 \n",
            "Response:\n",
            "Okay, so I need to find the point P on the x-axis that minimizes the sum of the distances AP and PB, where A is (1,2) and B is (3,4). Then, I have to compute 60 times the x-coordinate of P. Hmm, let me think.\n",
            "\n",
            "First, since P is on the x-axis, its coordinates are (a, 0). So, the distances AP and PB are sqrt((a-1)^2 + (0-2)^2) and sqrt((a-3)^2 + (0-4)^2), respectively. But minimizing their sum seems a bit tricky. Maybe there's a geometric trick here?\n",
            "\n",
            "Wait, I remember something about reflecting points over axes to find the shortest path. Let me recall. If we reflect point B over the x-axis, its reflection would be B' with coordinates (3, -4). Then, the minimal path from A to B via the x-axis would be a straight line from A to B', crossing the x-axis at P. The distance AP + PB would then be equal to the straight-line distance AB'. So perhaps P is the intersection of line AB' with the x-axis.\n",
            "\n",
            "Let me check that. If I find the equation of the line passing through A (1,2) and B' (3,-4), then set y=0 to find the x-coordinate of P. Let's compute the slope. The slope m is ( -4 - 2 ) / (3 - 1) = (-6)/2 = -3. So the equation is y - 2 = -3(x - 1), which simplifies to y = -3x + 5. Setting y=0, we get 0 = -3x + 5, so x = 5/3. Therefore, P is at (5/3, 0).\n",
            "\n",
            "So 60a would be 60 * (5/3) = 100. Let me verify that this gives the minimal distance. Since the straight-line distance from A to B' is shorter than any other path via points on the x-axis, this must be the minimal case. Therefore, the answer should be 100.\n",
            "\n",
            "Wait, another thought: is there another way? Like using calculus? Maybe taking the derivative of the distance function with respect to a? Let me try that. The distance AP + PB is sqrt((a-1)^2 +4) + sqrt((a-3)^2 +16). Let's denote this as D(a). Then, D(a) = sqrt((a-1)^2 +4) + sqrt((a-3)^2 +16). To minimize D, take the derivative D'(a) and set it to zero.\n",
            "\n",
            "Differentiating term by term: the derivative of sqrt((a-1)^2 +4) is [(a-1)/sqrt((a-1)^2 +4)], and similarly, the derivative of the second term is [ (a-3)/sqrt((a-3)^2 +16) ]. Setting D'(a) = 0 gives [(a-1)/sqrt((a-1)^2 +4)] = -[ (a-3)/sqrt((a-3)^2 +16) ]. Hmm, this seems complicated. Let's see if we can cross-multiply. But notice that the left is positive when a>1 and negative when a<1, whereas the right is positive when a>3 and negative when a<3. So at the minimum, a must be between 1 and 3. From our earlier geometric approach, we found a =5/3, which is between 1 and 3. Given that this is the only critical point in that interval, that should be the minimizing value. So the answer must be 100.\n",
            "To minimize the sum of the distances \\( AP + PB \\) where \\( P = (a, 0) \\) is on the x-axis, we can use the method of reflection. Reflect point \\( B = (3, 4) \\) over the x-axis to get \\( B' = (3, -4) \\). The shortest path from \\( A = (1, 2) \\) to \\( B \\) via the x-axis will be a straight line from \\( A \\) to \\( B' \\), intersecting the x-axis at \\( P \\).\n",
            "\n",
            "First, find the equation of the line passing through \\( A = (1, 2) \\) and \\( B' = (3, -4) \\). The slope \\( m \\) is given by:\n",
            "\\[\n",
            "m = \\frac{-4 - 2}{3 - 1} = \\frac{-6}{2} = -3\n",
            "\\]\n",
            "\n",
            "Using the point-slope form of the line equation \\( y - y_1 = m(x - x_1) \\) with point \\( A = (1, 2) \\):\n",
            "\\[\n",
            "y - 2 = -3(x - 1)\n",
            "\\]\n",
            "\\[\n",
            "y - 2 = -3x + 3\n",
            "\\]\n",
            "\\[\n",
            "y = -3x + 5\n",
            "\\]\n",
            "\n",
            "To find the x-coordinate of \\( P \\), set \\( y = 0 \\):\n",
            "\\[\n",
            "0 = -3x + 5\n",
            "\\]\n",
            "\\[\n",
            "3x = 5\n",
            "\\]\n",
            "\\[\n",
            "x = \\frac{5}{3}\n",
            "\\]\n",
            "\n",
            "Thus, \\( P = \\left( \\frac{5}{3}, 0 \\right) \\). The value of \\( 60a \\) is:\n",
            "\\[\n",
            "60 \\times \\frac{5}{3} = 100\n",
            "\\]\n",
            "\n",
            "Therefore, the answer is:\n",
            "\\[\n",
            "\\boxed{100}\n",
            "\\]<end_working_out><SOLUTION>100</SOLUTION> \n",
            "Extracted:\n",
            "100\n",
            "********************Question:\n",
            "A circle is inscribed in a regular octagon with area $2024$. A second regular octagon is inscribed in the circle, and its area can be expressed as $a + b\\sqrt{c}$, where $a$, $b$, $c$ are integers and $c$ is square-free. Compute $a + b + c$. \n",
            "Answer:\n",
            "1520 \n",
            "Response:\n",
            "Okay, so I need to solve this problem about a regular octagon and a circle inscribed in it. Then, there's another octagon inscribed in that circle. The goal is to find the area of the second octagon in the form \\( a + b\\sqrt{c} \\) and compute \\( a + b + c \\). The first octagon has an area of 2024.\n",
            "\n",
            "First, let me recall some properties of regular polygons and circles. When a circle is inscribed in a regular polygon, each side of the polygon touches the circle. Similarly, when another regular polygon is inscribed in the circle, each vertex of the polygon lies on the circle. For both octagons to be regular, all their sides and angles are equal.\n",
            "\n",
            "I think the key here is to relate the side lengths of the first octagon to the radius of the inscribed circle, and then use that relationship to find the side length of the second octagon. Once I have the side length, I can compute its area using the formula for the area of a regular octagon.\n",
            "\n",
            "The formula for the area of a regular octagon with side length \\( s \\) is \\( 2(1 + \\sqrt{2})s^2 \\). But since we have the area of the first octagon, which is 2024, maybe I can solve for \\( s \\) first, then use that to find the radius, and then the side length of the second octagon.\n",
            "\n",
            "Wait, but how does the side length relate to the radius in a regular octagon? For a regular polygon inscribed in a circle, the side length \\( s \\) relates to the radius \\( R \\) by the formula:\n",
            "\\[\n",
            "s = 2R \\sin\\left(\\frac{\\pi}{n}\\right)\n",
            "\\]\n",
            "where \\( n \\) is the number of sides. For an octagon, \\( n = 8 \\), so:\n",
            "\\[\n",
            "s = 2R \\sin\\left(\\frac{\\pi}{8}\\right)\n",
            "\\]\n",
            "\n",
            "But the first octagon has an area of 2024. The area formula for a regular octagon is \\( 2(1 + \\sqrt{2})s^2 \\). So, setting that equal to 2024:\n",
            "\\[\n",
            "2(1 + \\sqrt{2})s^2 = 2024\n",
            "\\]\n",
            "Solving for \\( s^2 \\):\n",
            "\\[\n",
            "s^2 = \\frac{2024}{2(1 + \\sqrt{2})} = \\frac{1012}{1 + \\sqrt{2}}\n",
            "\\]\n",
            "To rationalize the denominator:\n",
            "\\[\n",
            "s^2 = \\frac{1012(1 - \\sqrt{2})}{(1 + \\sqrt{2})(1 - \\sqrt{2})} = \\frac{1012(1 - \\sqrt{2})}{1 - 2} = \\frac{1012(1 - \\sqrt{2})}{-1} = 1012(\\sqrt{2} - 1)\n",
            "\\]\n",
            "So,\n",
            "\\[\n",
            "s = \\sqrt{1012(\\sqrt{2} - 1)}\n",
            "\\]\n",
            "But perhaps there's a simpler way. Maybe instead of finding \\( s \\) directly, we can find the radius \\( R \\) first. Since the circle is inscribed in the first octagon, the radius \\( R \\) is actually the apothem of the octagon. The apothem \\( a \\) of a regular polygon is related to the side length \\( s \\) by:\n",
            "\\[\n",
            "a = \\frac{s}{2 \\tan(\\pi/n)}\n",
            "\\]\n",
            "For the octagon, \\( n = 8 \\), so:\n",
            "\\[\n",
            "a = \\frac{s}{2 \\tan(\\pi/8)}\n",
            "\\]\n",
            "But since the area is \\( 2(1 + \\sqrt{2})s^2 \\), and \\( a = \\frac{s}{2 \\tan(\\pi/8)} \\), maybe we can express the area in terms of \\( a \\). Wait, but perhaps it's easier to compute the radius first using the area formula.\n",
            "\n",
            "The area formula \\( A = \\frac{1}{2} \\times \\text{perimeter} \\times \\text{apothem} \\) applies to all regular polygons. For the octagon with area 2024:\n",
            "\\[\n",
            "2024 = \\frac{1}{2} \\times 8s \\times a = 4 s a\n",
            "\\]\n",
            "So,\n",
            "\\[\n",
            "s a = 506\n",
            "\\]\n",
            "We also have \\( a = \\frac{s}{2 \\tan(\\pi/8)} \\). Let me compute \\( \\tan(\\pi/8) \\). Since \\( \\pi/8 = 22.5^\\circ \\), and \\( \\tan(22.5^\\circ) = \\sqrt{2} - 1 \\). So:\n",
            "\\[\n",
            "a = \\frac{s}{2(\\sqrt{2} - 1)} = \\frac{s (\\sqrt{2} + 1)}{2 (\\sqrt{2} - 1)(\\sqrt{2} + 1)} = \\frac{s (\\sqrt{2} + 1)}{2 (2 - 1)} = \\frac{s (\\sqrt{2} + 1)}{2}\n",
            "\\]\n",
            "Then, from \\( s a = 506 \\):\n",
            "\\[\n",
            "s \\cdot \\frac{s (\\sqrt{2} + 1)}{2} = 506 \\implies \\frac{s^2 (\\sqrt{2} + 1)}{2} = 506 \\implies s^2 (\\sqrt{2} + 1) = 1012 \\implies s^2 = \\frac{1012}{\\sqrt{2} + 1}\n",
            "\\]\n",
            "Rationalizing the denominator:\n",
            "\\[\n",
            "s^2 = \\frac{1012 (\\sqrt{2} - 1)}{(\\sqrt{2} + 1)(\\sqrt{2} - 1)} = \\frac{1012 (\\sqrt{2} - 1)}{2 - 1} = 1012 (\\sqrt{2} - 1)\n",
            "\\]\n",
            "So,\n",
            "\\[\n",
            "s = \\sqrt{1012 (\\sqrt{2} - 1)}\n",
            "\\]\n",
            "Now, the radius \\( R \\) of the circumscribed circle is equal to the distance from the center to a vertex, which is \\( R = \\frac{s}{2 \\sin(\\pi/8)} \\). But since \\( \\sin(\\pi/8) = \\frac{\\sqrt{2 - \\sqrt{2}}}{2} \\), this might complicate things. Alternatively, since the second octagon is inscribed in the same circle, its side length can be found directly in terms of \\( R \\).\n",
            "\n",
            "Alternatively, there's a formula for the side length of a regular octagon inscribed in a circle. The side length \\( s_{\\text{in}} \\) of a regular octagon inscribed in a circle of radius \\( R \\) is:\n",
            "\\[\n",
            "s_{\\text{in}} = 2R \\sin\\left(\\frac{\\pi}{8}\\right) = R \\sqrt{2 - \\sqrt{2}}\n",
            "\\]\n",
            "So, to find \\( a + b\\sqrt{c} \\) for the area of the second octagon, we need to compute \\( R^2 \\) first.\n",
            "\n",
            "From the first octagon, the relationship between the side length \\( s \\) and the apothem \\( a \\) is used to find \\( s \\). Then, the radius \\( R \\) is related to \\( s \\) by \\( R = \\frac{s}{2 \\sin(\\pi/8)} \\).\n",
            "\n",
            "Wait, perhaps a better approach is to express everything in terms of the radius. Let's denote the radius of the circle as \\( R \\). For the first octagon, the apothem \\( a \\) is equal to \\( R \\). The area formula for a regular octagon is \\( A = 2(1 + \\sqrt{2})s^2 \\), where \\( s \\) is the side length. But since the apothem \\( a = R \\), we can write:\n",
            "\\[\n",
            "s = 2R \\sin(\\pi/8)\n",
            "\\]\n",
            "Then,\n",
            "\\[\n",
            "a = \\frac{s}{2 \\tan(\\pi/8)} = \\frac{2R \\sin(\\pi/8)}{2 \\tan(\\pi/8)} = R \\sin(\\pi/8) \\cdot \\frac{1}{\\tan(\\pi/8)} = R \\cos(\\pi/ \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "A circle centered at $O$ has radius $1$ and contains the point $A$. Segment $AB$ is tangent to the circle at $A$ and $\\angle AOB = \\theta$. If point $C$ lies on $\\overline{OA}$ and $\\overline{BC}$ bisects $\\angle ABO$, find the value of $OC$. The original answer is in the form \\frac{k}{1+m\\sin(\\theta)}, please give the value of k + m. \n",
            "Answer:\n",
            "2 \n",
            "Response:\n",
            "Okay, let me see. So the problem is about a circle with center O and radius 1. It contains point A, and segment AB is tangent to the circle at A. Then, angle AOB is θ. Point C is on OA, and BC bisects angle ABO. We need to find the length OC in terms of θ, and then determine k + m where the answer is in the form k/(1 + m sin(θ)). Hmm, okay, let's start breaking this down.\n",
            "\n",
            "First, since AB is tangent to the circle at A, the radius OA is perpendicular to AB. So triangle OAB is a right triangle with right angle at A. The angle at O is θ, so the other angle, angle OBA, is 90° - θ. But wait, BC bisects angle ABO, which is actually angle OBA. So the bisected angle is (90° - θ)/2. That means angle OBC is (90° - θ)/2.\n",
            "\n",
            "Now, in triangle OAB, since we know the angles and the radius is 1, we can find the lengths of the sides. The distance AB is the tangent segment, which can be found using trigonometry. In right triangle OAB, tan(θ) = AB / OA. Since OA is the radius, OA = 1. So AB = tan(θ). Wait, no. Let me double-check. In triangle OAB, angle at A is 90°, angle at O is θ, so angle at B is 90° - θ. Therefore, AB is opposite angle O, which is θ, and OA is adjacent to angle O. So AB = OA * tan(θ) = 1 * tan(θ) = tan(θ). Yes, that makes sense.\n",
            "\n",
            "Now, point C is on OA, so let's say OC = x. Then CA = OA - OC = 1 - x. We need to find x.\n",
            "\n",
            "Since BC bisects angle ABO, which is (90° - θ)/2, we can use the Angle Bisector Theorem. The Angle Bisector Theorem states that the bisector of an angle in a triangle divides the opposite side in the ratio of the adjacent sides. So in triangle OAB, the bisector BC divides OA in the ratio OB : AB. Wait, let's write that down. The theorem says that the ratio OC : CA = AB : OB.\n",
            "\n",
            "But in triangle OAB, AB is the tangent, which is tan(θ), and OB is the hypotenuse. Wait, in right triangle OAB, the hypotenuse is the line from O to B. Since OA is 1 and angle at O is θ, then OB = OA / cos(θ) = 1/cos(θ). Right?\n",
            "\n",
            "So according to the Angle Bisector Theorem: OC / CA = AB / OB. Plugging in the values we have, OC / (1 - OC) = tan(θ) / (1/cos(θ)).\n",
            "\n",
            "Simplify the right side: tan(θ)/(1/cos(θ)) = tan(θ)*cos(θ) = (sin(θ)/cos(θ))*cos(θ) = sin(θ). So the equation becomes OC / (1 - OC) = sin(θ).\n",
            "\n",
            "Now, solve for OC. Cross-multiplying gives OC = sin(θ)(1 - OC). Expanding: OC = sin(θ) - OC sin(θ). Add OC sin(θ) to both sides: OC + OC sin(θ) = sin(θ). Factor out OC: OC(1 + sin(θ)) = sin(θ). Then, divide both sides by (1 + sin(θ)) to get OC = sin(θ)/(1 + sin(θ)).\n",
            "\n",
            "So, the problem asks for the value of OC in the form k/(1 + m sin(θ)). Comparing, we have OC = (sin(θ))/(1 + sin(θ)), so k = 1 and m = 1. Therefore, k + m = 2.\n",
            "\n",
            "Wait, let me verify the steps again. I think that's correct. The Angle Bisector Theorem was applied correctly here, and the calculations seem right. So the answer should be 2.\n",
            "To find the length \\( OC \\) in the given geometric configuration, we start by noting the following:\n",
            "\n",
            "1. The circle is centered at \\( O \\) with radius 1, and \\( A \\) is a point on the circle.\n",
            "2. Segment \\( AB \\) is tangent to the circle at \\( A \\), so \\( \\angle OAB = 90^\\circ \\).\n",
            "3. \\( \\angle AOB = \\theta \\), making \\( \\angle OBA = 90^\\circ - \\theta \\).\n",
            "4. \\( \\overline{BC} \\) bisects \\( \\angle ABO \\), so \\( \\angle ABC = \\angle CBO = \\frac{1}{2}(90^\\circ - \\theta) \\).\n",
            "\n",
            "Using the Angle Bisector Theorem in \\( \\triangle OAB \\), we have:\n",
            "\\[\n",
            "\\frac{OC}{CA} = \\frac{AB}{OB}\n",
            "\\]\n",
            "\n",
            "Since \\( AB \\) is the tangent segment:\n",
            "\\[\n",
            "AB = \\tan(\\theta)\n",
            "\\]\n",
            "\n",
            "And \\( OB \\) is the hypotenuse of the right triangle \\( \\triangle OAB \\):\n",
            "\\[\n",
            "OB = \\frac{1}{\\cos(\\theta)}\n",
            "\\]\n",
            "\n",
            "Applying the Angle Bisector Theorem:\n",
            "\\[\n",
            "\\frac{OC}{1 - OC} = \\frac{\\tan(\\theta)}{\\frac{1}{\\cos(\\theta)}} = \\tan(\\theta) \\cdot \\cos(\\theta) = \\sin(\\theta)\n",
            "\\]\n",
            "\n",
            "Let \\( OC = x \\). Then:\n",
            "\\[\n",
            "\\frac{x}{1 - x} = \\sin(\\theta)\n",
            "\\]\n",
            "\n",
            "Cross-multiplying gives:\n",
            "\\[\n",
            "x = \\sin(\\theta)(1 - x)\n",
            "\\]\n",
            "\n",
            "Expanding and solving for \\( x \\):\n",
            "\\[\n",
            "x = \\sin(\\theta) - x \\sin(\\theta)\n",
            "\\]\n",
            "\\[\n",
            "x + x \\sin(\\theta) = \\sin(\\theta)\n",
            "\\]\n",
            "\\[\n",
            "x(1 + \\sin(\\theta)) = \\sin(\\theta)\n",
            "\\]\n",
            "\\[\n",
            "x = \\frac{\\sin(\\theta)}{1 + \\sin(\\theta)}\n",
            "\\]\n",
            "\n",
            "Thus, \\( OC = \\frac{\\sin(\\theta)}{1 + \\sin(\\theta)} \\). Comparing this with the form \\( \\frac{k}{1 + m \\sin(\\theta)} \\), we identify \\( k = 1 \\) and \\( m = 1 \\). Therefore, \\( k + m = 2 \\).\n",
            "\n",
            "The value of \\( k + m \\) is:\n",
            "\\[\n",
            "\\boxed{2}\n",
            "\\]<end_working_out><SOLUTION>2</SOLUTION> \n",
            "Extracted:\n",
            "2\n",
            "********************Question:\n",
            "In the addition shown below $A$, $B$, $C$, and $D$ are distinct digits. How many different values are possible for $D$?  \n",
            "$$  \n",
            "\\begin{array}{cccccc}&A&B&B&C&B\\\\+&B&C&A&D&A\\\\\\hline&D&B&D&D&D\\end{array}\n",
            "$$  \n",
            "Provide the number of different possible values for $D$. \n",
            "Answer:\n",
            "7 \n",
            "Response:\n",
            "Okay, so let's tackle this problem step by step. I need to find how many different possible values the digit D can take in this addition problem. The problem involves four distinct digits labeled A, B, C, D, each representing a different number from 0 to 9. The addition is set up horizontally with two numbers being added together to form a larger number with the same last digit as D in the units column.\n",
            "\n",
            "Let me write down the addition again to make sure I understand:\n",
            "\n",
            "```\n",
            "  A B B C B\n",
            "+ B C A D A\n",
            "-----------\n",
            "  D B D D D\n",
            "```\n",
            "\n",
            "So the first number is ABCB (read as A followed by B, B, C, B) and the second number is BCADA. The sum is a five-digit number DBDDE, where E is the units digit. But interestingly, the units digit of the sum is D. So, looking at the units columns, we have the bottom row: B + A = D (possibly with a carryover).\n",
            "\n",
            "But wait, the problem says all digits are distinct, so A, B, C, D are all different. Let me think about the units place first. We have B + A in the units column. The result must end with D. So, depending on whether there's a carryover, we can write:\n",
            "\n",
            "If there's a carryover, then B + A = D + 10\n",
            "If there's no carryover, then B + A = D\n",
            "\n",
            "But the sum of two single-digit numbers can range from 0 to 18. Since digits are at least 0, the maximum sum is 9 + 8 = 17 (since all digits are distinct, so they can't wrap around all nine digits). So the maximum D can be is 9, and if there's a carryover, you add 10 to it.\n",
            "\n",
            "But let's check the possible values for D. Since digits are distinct, the possible digits cycle from 0 to 9. But since the digits are distinct, some may be excluded.\n",
            "\n",
            "Another approach might be to consider the entire addition column by column, starting from the rightmost digit. Let's look at the first column (units place):\n",
            "\n",
            "B + A = D (mod 10) with possible carryover. Then the second column from the right would have B + C + possible carryover from the first column equal to the next D in the sum.\n",
            "\n",
            "Wait, but the sum in the third column is D. So let me denote the columns left to right as the units, tens, hundreds, thousands, ten-thousands.\n",
            "\n",
            "Units column: B + A = D (mod 10) with possible carryover 1.\n",
            "\n",
            "Now, second column (tens place): C + A + carryover from first column should equal the second D in the sum.\n",
            "\n",
            "But the problem states that the sum is D B D D D. So the sum has three D's after the first two columns. This seems tricky. Maybe I can set up equations.\n",
            "\n",
            "Let me label the columns as Column 1 (units), Column 2 (tens), Column 3 (hundreds), Column 4 (thousands), Column 5 (ten-thousands).\n",
            "\n",
            "From Column 1: B + A = D + 10k, where k is either 0 or 1 (carryover). Since B and A are single digits (0-9), the maximum sum is 9+8=17, so possible D values are 0 to 9 with carryover 1 if the sum is 10 or more.\n",
            "\n",
            "From Column 2 (tens place): C + A + k = D + 10m, where m is the carryover to the next column. Since D is a single digit, m can be 0 or 1.\n",
            "\n",
            "Wait, but Column 3 is the hundreds place: B + C + m = D + 10n. And so on. But the problem states the sum is D B D D D. Let's align the columns:\n",
            "\n",
            "For the third column (hundreds place): B + C + m = D + 10n\n",
            "\n",
            "But since the sum in the hundreds place is D (the third digit from the right), n must be 0. So B + C + m = D\n",
            "\n",
            "Similarly, the fourth column: C + B + n = D + 10p\n",
            "\n",
            "Wait, but from the hundreds place, we have n=0. So this column would have C + B = D + 10p. But since C and B are single digits, the maximum sum is 18, so p can be 0 or 1.\n",
            "\n",
            "But wait, the problem states the entire sum is D B D D D. So let me see if I can set up a system of equations.\n",
            "\n",
            "From the units column: B + A = D + 10k, where k is either 0 or 1.\n",
            "\n",
            "From the tens column: C + A + k = 10m + D, where m is the carryover to the hundreds place (m is either 0 or 1).\n",
            "\n",
            "From the hundreds column: B + C + m = 10n + D, but since the sum in the hundreds place is D, n must be 0. So B + C + m = D.\n",
            "\n",
            "From the thousands column: C + B + n = 10p + D, again n=0 (since the sum is D at this column), so C + B = 10p + D. But since C and B are single digits, p must be 0. So C + B = D.\n",
            "\n",
            "Wait a second, now we have two equations: B + C = D and B + C + m = D. This implies that m=0. So the carryover from the tens column is 0. Then, going back to the tens column equation: C + A + 0 = D, so C + A = D.\n",
            "\n",
            "From the units column: B + A = D + 10k. Since C + A = D, we can substitute A = D - C into the units column equation: B + (D - C) = D + 10k, simplifying to B - C = 10k. Since B and C are digits, the only possible solution is k=0 and B = C. But the problem states that all digits A, B, C are distinct, so B cannot equal C. This seems like a contradiction.\n",
            "\n",
            "Therefore, there must be a carryover from the units column, so k=1. Then, the units column equation becomes B + A = D + 10. Since digits are from 0 to 9, the maximum B+A can be is 18. Therefore, B + A = D + 10. Then, subtracting, we get A - 10 = D. Since A and D are digits, the only possible value here is A=9 and D=9. But wait, but the problem states that all digits are distinct. So if A=9 and D=9, then we have a non-distinct digit, which violates the problem's condition.\n",
            "\n",
            "Wait, that's a problem. So there's no solution? But let me check again. Maybe I made a mistake.\n",
            "\n",
            "The equations were:\n",
            "\n",
            "Units column: B + A = D + 10k (k=0 or 1)\n",
            "\n",
            "Tens column: C + A + k = 10m + D (m=0 or 1)\n",
            "\n",
            "Hundreds column: B + C + m = D\n",
            "\n",
            "Thousands column: C + B = D (since sum digit is D)\n",
            "\n",
            "Wait, so from the thousands column, C + B = D, so D = B + C. Then substitute into the hundreds column equation: B + C + m = B + C, which implies m=0. So there is no carryover from the hundreds to the ten-thousands column.\n",
            "\n",
            "Then, going back to the tens column: C + A + 0 = 10m + D, so C + A = D. But also, D = B + C, so substituting into C + A = D gives C + A = B + C, which simplifies to A = B. But the problem states that A and B are distinct. So this leads to a contradiction.\n",
            "\n",
            "Therefore, the only possibility is that k=1 in the units column. But then, B + A = D + 10. Subtracting D from both sides: B + A - D = 10. Since A, B, D are distinct digits, the minimum sum of B + A is 1+2=3, which is less than 10. But since D is a digit, D can be 0 to 9. \n",
            "\n",
            "Let me substitute D + 10k \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "Let $a_1 = 2021$ and for $n \\geq 1$, let $a_{n+1} = \\sqrt{4 + a_n}$. Then $a_5$ can be written as \\[ \\sqrt{\\frac{m+\\sqrt{n}}{2}} + \\sqrt{\\frac{m-\\sqrt{n}}{2}} \\] where $m$ and $n$ are positive integers. Find $10m + n$. \n",
            "Answer:\n",
            "45 \n",
            "Response:\n",
            "Okay, so I have this problem here. It starts with \\( a_1 = 2021 \\) and then for each \\( n \\geq 1 \\), the next term is defined as \\( a_{n+1} = \\sqrt{4 + a_n} \\). The question is asking for \\( a_5 \\) to be expressed in a specific form involving square roots and then find \\( 10m + n \\) where \\( m \\) and \\( n \\) are positive integers.\n",
            "\n",
            "Right, so I need to compute \\( a_5 \\) by applying the recursive formula four times (since \\( a_1 \\) is the starting point). But calculating each term step by step might be tedious. Maybe there's a pattern or a simplification here? Let me see.\n",
            "\n",
            "First, let's compute the first few terms to see if a pattern emerges.\n",
            "\n",
            "Given \\( a_1 = 2021 \\), then:\n",
            "\n",
            "\\( a_2 = \\sqrt{4 + 2021} = \\sqrt{2025} = 45 \\)\n",
            "\n",
            "Next, \\( a_3 = \\sqrt{4 + 45} = \\sqrt{49} = 7 \\)\n",
            "\n",
            "Then \\( a_4 = \\sqrt{4 + 7} = \\sqrt{11} \\)\n",
            "\n",
            "And finally, \\( a_5 = \\sqrt{4 + \\sqrt{11}} \\)\n",
            "\n",
            "So \\( a_5 = \\sqrt{4 + \\sqrt{11}} \\).\n",
            "\n",
            "The problem states that \\( a_5 \\) can be written in the form \\( \\sqrt{\\frac{m+\\sqrt{n}}{2}} + \\sqrt{\\frac{m-\\sqrt{n}}{2}} \\), and we need to find \\( 10m + n \\).\n",
            "\n",
            "Hmm, so I need to express \\( \\sqrt{4 + \\sqrt{11}} \\) in that specific form. Let me think about how to manipulate this expression.\n",
            "\n",
            "Maybe I can set each square root of the binomial inside the other terms. Let me denote the expression as:\n",
            "\n",
            "\\( \\sqrt{4 + \\sqrt{11}} = \\sqrt{\\frac{m + \\sqrt{n}}{2}} + \\sqrt{\\frac{m - \\sqrt{n}}{2}} \\)\n",
            "\n",
            "If I square both sides, perhaps that will help. Let me try that.\n",
            "\n",
            "Let \\( x = \\sqrt{\\frac{m + \\sqrt{n}}{2}} + \\sqrt{\\frac{m - \\sqrt{n}}{2}} \\)\n",
            "\n",
            "Then \\( x^2 = \\left( \\sqrt{\\frac{m + \\sqrt{n}}{2}} \\right)^2 + \\left( \\sqrt{\\frac{m - \\sqrt{n}}{2}} \\right)^2 + 2 \\times \\sqrt{\\frac{m + \\sqrt{n}}{2}} \\times \\sqrt{\\frac{m - \\sqrt{n}}{2}} \\)\n",
            "\n",
            "Wait, that's the expansion of \\( (a + b)^2 = a^2 + b^2 + 2ab \\). So,\n",
            "\n",
            "\\( x^2 = \\frac{m + \\sqrt{n}}{2} + \\frac{m - \\sqrt{n}}{2} + 2 \\times \\sqrt{ \\frac{(m + \\sqrt{n})(m - \\sqrt{n})}{4} } \\)\n",
            "\n",
            "Simplifying the first two terms:\n",
            "\n",
            "\\( \\frac{m + \\sqrt{n}}{2} + \\frac{m - \\sqrt{n}}{2} = \\frac{2m}{2} = m \\)\n",
            "\n",
            "Then the third term simplifies inside the square root:\n",
            "\n",
            "\\( (m + \\sqrt{n})(m - \\sqrt{n}) = m^2 - n \\)\n",
            "\n",
            "So the third term becomes \\( 2 \\times \\sqrt{ \\frac{m^2 - n}{4} } = 2 \\times \\frac{ \\sqrt{m^2 - n} }{2} = \\sqrt{m^2 - n} \\)\n",
            "\n",
            "Therefore,\n",
            "\n",
            "\\( x^2 = m + \\sqrt{m^2 - n} \\)\n",
            "\n",
            "But according to the problem, \\( x = \\sqrt{4 + \\sqrt{11}} \\), so squaring both sides gives:\n",
            "\n",
            "\\( \\sqrt{4 + \\sqrt{11}}^2 = 4 + \\sqrt{11} \\)\n",
            "\n",
            "Therefore, we have the equation:\n",
            "\n",
            "\\( m + \\sqrt{m^2 - n} = 4 + \\sqrt{11} \\)\n",
            "\n",
            "For these two expressions to be equal, the rational and irrational parts must match separately. Therefore:\n",
            "\n",
            "\\( m = 4 \\) and \\( \\sqrt{m^2 - n} = \\sqrt{11} \\)\n",
            "\n",
            "Substituting \\( m = 4 \\) into the second equation:\n",
            "\n",
            "\\( \\sqrt{16 - n} = \\sqrt{11} \\)\n",
            "\n",
            "Squaring both sides gives \\( 16 - n = 11 \\), so \\( n = 5 \\)\n",
            "\n",
            "Therefore, \\( m = 4 \\) and \\( n = 5 \\)\n",
            "\n",
            "The problem asks for \\( 10m + n \\). Plugging in the values:\n",
            "\n",
            "\\( 10m + n = 10 \\times 4 + 5 = 40 + 5 = 45 \\)\n",
            "\n",
            "So the answer should be 45. Let me check if this makes sense.\n",
            "\n",
            "Wait, but when I substituted back, everything seemed to match: \\( m = 4 \\), then \\( n = 5 \\), and plugging into the original expression gives \\( \\sqrt{4 + \\sqrt{11}} \\), which matches. So I think that's correct.\n",
            "\n",
            "Therefore, the answer is \\(\\boxed{45}\\).<end_working_out><SOLUTION>45</SOLUTION> \n",
            "Extracted:\n",
            "45\n",
            "********************Question:\n",
            "Phoenix hiked the Rocky Path Trail last week. It took four days to complete the trip. The first two days she hiked a total of 26 miles. The second and third days she averaged 12 miles per day. The last two days she hiked a total of 28 miles. The total hike for the first and third days was 22 miles. How many miles long was the trail? \n",
            "Answer:\n",
            "54 \n",
            "Response:\n",
            "Okay, let's try to figure out how long the Rocky Path Trail is that Phoenix hiked over four days. I need to find the total miles she covered each day and then sum them up. The problem gives me a few pieces of information:\n",
            "\n",
            "1. First two days: 26 miles total.\n",
            "2. Second and third days: averaged 12 miles each day. Wait, averaged means the total for those two days is 24 miles (12 miles/day multiplied by 2 days).\n",
            "3. Last two days: 28 miles total.\n",
            "4. First and third days: 22 miles total.\n",
            "\n",
            "I need to find the miles for each day. Let me label them as Day 1, Day 2, Day 3, Day 4. The variables will be x1, x2, x3, x4. Then, I can set up equations based on the given information.\n",
            "\n",
            "From the first piece: Day 1 + Day 2 = 26 miles.\n",
            "So, x1 + x2 = 26. Equation (1)\n",
            "\n",
            "Next, the second and third days average 12 miles per day. So, over those two days, total is 24 miles.\n",
            "Thus, x2 + x3 = 24. Equation (2)\n",
            "\n",
            "Then, the last two days total 28 miles. That's Day 3 + Day 4 = 28. Equation (3)\n",
            "\n",
            "Also, the first and third days total 22 miles. So, x1 + x3 = 22. Equation (4)\n",
            "\n",
            "Now, I have four equations:\n",
            "\n",
            "1. x1 + x2 = 26\n",
            "2. x2 + x3 = 24\n",
            "3. x3 + x4 = 28\n",
            "4. x1 + x3 = 22\n",
            "\n",
            "I need to solve for each variable. Let me see. Starting with equations (1) and (4).\n",
            "\n",
            "Equation (1): x1 + x2 = 26\n",
            "\n",
            "Equation (4): x1 + x3 = 22\n",
            "\n",
            "If I subtract Equation (4) from Equation (1), I get:\n",
            "(x1 + x2) - (x1 + x3) = 26 - 22\n",
            "Which simplifies to:\n",
            "x2 - x3 = 4. Equation (5)\n",
            "\n",
            "Now, I have Equation (2): x2 + x3 = 24\n",
            "And Equation (5): x2 - x3 = 4\n",
            "\n",
            "If I add these two equations together:\n",
            "(x2 + x3) + (x2 - x3) = 24 + 4\n",
            "2x2 = 28\n",
            "So, x2 = 14\n",
            "\n",
            "Now that I have x2, I can plug it back into Equation (5) to find x3:\n",
            "14 - x3 = 4\n",
            "x3 = 14 - 4 = 10\n",
            "\n",
            "Now, from Equation (4): x1 + x3 = 22\n",
            "So, x1 + 10 = 22\n",
            "x1 = 22 - 10 = 12\n",
            "\n",
            "Now, using Equation (2): x2 + x3 = 24\n",
            "We already know x2=14 and x3=10, which checks out.\n",
            "\n",
            "Then, from Equation (3): x3 + x4 = 28\n",
            "We know x3=10, so:\n",
            "10 + x4 = 28\n",
            "x4 = 18\n",
            "\n",
            "Now, let's check all the values:\n",
            "Day 1: 12 miles, Day 2: 14 miles, Day 3: 10 miles, Day 4: 18 miles.\n",
            "\n",
            "Check the original conditions:\n",
            "1. First two days: 12+14=26 ✔️\n",
            "2. Second and third days: 14+10=24 ✔️\n",
            "3. Last two days: 10+18=28 ✔️\n",
            "4. First and third days: 12+10=22 ✔️\n",
            "\n",
            "Everything adds up correctly. So the total miles of the trail is the sum of all four days:\n",
            "12 + 14 + 10 + 18 = 54 miles. That seems right. I don't see any mistakes here. To verify again, I'll perform the addition step-by-step: 12 + 14 is 26, then 26 + 10 is 36, and finally 36 + 18 is 54. Yeah, that's correct. So the trail was 54 miles long.\n",
            "To determine the total length of the Rocky Path Trail that Phoenix hiked over four days, we can set up and solve a system of equations based on the given information:\n",
            "\n",
            "1. \\( x_1 + x_2 = 26 \\) (First two days)\n",
            "2. \\( x_2 + x_3 = 24 \\) (Second and third days average 12 miles per day)\n",
            "3. \\( x_3 + x_4 = 28 \\) (Last two days)\n",
            "4. \\( x_1 + x_3 = 22 \\) (First and third days)\n",
            "\n",
            "We start by solving the system of equations. From equation (1) and equation (4), we subtract equation (4) from equation (1):\n",
            "\n",
            "\\[\n",
            "(x_1 + x_2) - (x_1 + x_3) = 26 - 22\n",
            "\\]\n",
            "\n",
            "This simplifies to:\n",
            "\n",
            "\\[\n",
            "x_2 - x_3 = 4\n",
            "\\]\n",
            "\n",
            "We now have equation (5):\n",
            "\n",
            "\\[\n",
            "x_2 - x_3 = 4\n",
            "\\]\n",
            "\n",
            "We also have equation (2):\n",
            "\n",
            "\\[\n",
            "x_2 + x_3 = 24\n",
            "\\]\n",
            "\n",
            "Adding equations (2) and (5) together:\n",
            "\n",
            "\\[\n",
            "(x_2 + x_3) + (x_2 - x_3) = 24 + 4\n",
            "\\]\n",
            "\n",
            "This simplifies to:\n",
            "\n",
            "\\[\n",
            "2x_2 = 28 \\implies x_2 = 14\n",
            "\\]\n",
            "\n",
            "Next, we substitute \\( x_2 = 14 \\) back into equation (5):\n",
            "\n",
            "\\[\n",
            "14 - x_3 = 4 \\implies x_3 = 10\n",
            "\\]\n",
            "\n",
            "Now, substituting \\( x_3 = 10 \\) into equation (4):\n",
            "\n",
            "\\[\n",
            "x_1 + 10 = 22 \\implies x_1 = 12\n",
            "\\]\n",
            "\n",
            "Finally, substituting \\( x_3 = 10 \\) into equation (3):\n",
            "\n",
            "\\[\n",
            "10 + x_4 = 28 \\implies x_4 = 18\n",
            "\\]\n",
            "\n",
            "We have the following values for the days:\n",
            "- Day 1: \\( x_1 = 12 \\) miles\n",
            "- Day 2: \\( x_2 = 14 \\) miles\n",
            "- Day 3: \\( x_3 = 10 \\) miles\n",
            "- Day 4: \\( x_4 = 18 \\) miles\n",
            "\n",
            "Summing these values gives the total hike length:\n",
            "\n",
            "\\[\n",
            "12 + 14 + 10 + 18 = 54\n",
            "\\]\n",
            "\n",
            "Thus, the total length of the trail is \\(\\boxed{54}\\) miles.<end_working_out><SOLUTION>54</SOLUTION> \n",
            "Extracted:\n",
            "54\n",
            "********************Question:\n",
            "Right triangle $ABC$ has side lengths $BC=6$, $AC=8$, and $AB=10$. A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. The original answer is in \\(\\frac{k}{m}\\) format, please give the value of k + m. \n",
            "Answer:\n",
            "47 \n",
            "Response:\n",
            "Okay, let's try to solve this geometry problem step by step. I need to find the values of k and m where k/m is the answer, but first, I must figure out the actual coordinates or positions of the points and the circles involved. The problem is about a right triangle ABC with sides 6, 8, 10, and two circles with specific tangencies and points passing through vertices. \n",
            "\n",
            "First, let me sketch the triangle. Since it's a right triangle with sides 6, 8, 10, that should be a 6-8-10 right triangle, with the right angle at C. So, if I place point C at the origin (0,0), then B would be at (6,0) and A at (0,8). Wait, let me check: the hypotenuse would be AB, which is 10. So yes, that fits. So, C(0,0), B(6,0), A(0,8).\n",
            "\n",
            "Now, there's a circle centered at O tangent to BC at B and passing through A. Since BC is along the x-axis from (0,0) to (6,0), and the circle is tangent there at B, the center of the circle must be directly above B. So if the circle passes through A(0,8), then the distance from O to A should be the radius. But since it's tangent at B, the radius is equal to the y-coordinate of O. So if O is at (6,y), then the distance from (6,y) to (0,8) is the radius. But since the circle is tangent at B, the radius is also the distance from O to B, which is y. So the distance from O to A is sqrt((6-0)^2 + (y-8)^2) = y. So sqrt(36 + (y-8)^2) = y. Squaring both sides: 36 + (y-8)^2 = y^2. Expanding: 36 + y^2 -16y +64 = y^2. Simplifying: 100 -16y = 0. So 16y = 100, y = 25/4. So the circle centered at O is at (6, 25/4), and its radius is 25/4.\n",
            "\n",
            "Next, there's another circle centered at P tangent to AC at A and passing through B. AC is along the y-axis from (0,0) to (0,8). So the circle is tangent at A, meaning P is directly horizontally from A. So P is at (x,8). Since it passes through B(6,0), the distance from P to B is the radius, which is also the distance from P to A. So sqrt((x-6)^2 + (8-0)^2) = x. Squaring both sides: (x-6)^2 +64 = x^2. Expanding: x^2 -12x +36 +64 = x^2. Simplifying: -12x +100 = 0. So 12x = 100, x = 25/3. So P is at (25/3, 8). The radius is then 25/3.\n",
            "\n",
            "Now, I need to find the distance between O and P. O is at (6, 25/4) and P is at (25/3, 8). The distance formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. So let me convert these to have the same denominator. 6 is 24/4, 25/4 is (25*3)/(4*3) = 75/12. 25/3 is (25*4)/(3*4) = 100/12. Wait, that doesn't align with the original fractions. Maybe keeping them as fractions would work better. So O is at (24/4, 75/12), P at (100/12, 96/12). Then, the horizontal distance is |100/12 - 24/4| = |100/12 - 72/12| = 28/12 = 7/3. The vertical distance is |96/12 - 75/12| = 21/12 = 7/4. Distance between O and P is sqrt[(7/3)^2 + (7/4)^2] = sqrt[(49/9) + (49/16)] = sqrt[49*(1/9 +1/16)] = sqrt[49*(16+9)/144] = sqrt[49*25/144] = 7*5/12 = 35/12. So the distance is 35/12. Therefore, k=35 and m=12, so k+m=47. The answer is 47.\n",
            "To solve the given problem, we first need to establish the coordinates of the points in the right triangle \\(ABC\\). Since \\(ABC\\) is a right triangle with sides \\(BC = 6\\), \\(AC = 8\\), and \\(AB = 10\\), we can place the points as follows:\n",
            "- \\(C\\) at \\((0,0)\\),\n",
            "- \\(B\\) at \\((6,0)\\),\n",
            "- \\(A\\) at \\((0,8)\\).\n",
            "\n",
            "Next, we consider the circle centered at \\(O\\) that is tangent to \\(BC\\) at \\(B\\) and passes through \\(A\\). Since \\(BC\\) is along the x-axis and the circle is tangent to \\(BC\\) at \\(B\\), the center \\(O\\) must be directly above \\(B\\). Therefore, the coordinates of \\(O\\) are \\((6, y)\\), where \\(y\\) is the radius of the circle. Since the circle passes through \\(A\\), the distance from \\(O\\) to \\(A\\) is equal to the radius \\(y\\). Thus, we have:\n",
            "\\[\n",
            "\\sqrt{(6-0)^2 + (y-8)^2} = y\n",
            "\\]\n",
            "Squaring both sides, we get:\n",
            "\\[\n",
            "36 + (y-8)^2 = y^2\n",
            "\\]\n",
            "Expanding and simplifying:\n",
            "\\[\n",
            "36 + y^2 - 16y + 64 = y^2 \\implies 100 - 16y = 0 \\implies y = \\frac{100}{16} = \\frac{25}{4}\n",
            "\\]\n",
            "So, the coordinates of \\(O\\) are \\((6, \\frac{25}{4})\\).\n",
            "\n",
            "Next, we consider the circle centered at \\(P\\) that is tangent to \\(AC\\) at \\(A\\) and passes through \\(B\\). Since \\(AC\\) is along the y-axis and the circle is tangent to \\(AC\\) at \\(A\\), the center \\(P\\) must be directly to the right of \\(A\\). Therefore, the coordinates of \\(P\\) are \\((x, 8)\\), where \\(x\\) is the radius of the circle. Since the circle passes through \\(B\\), the distance from \\(P\\) to \\(B\\) is equal to the radius \\(x\\). Thus, we have:\n",
            "\\[\n",
            "\\sqrt{(x-6)^2 + (8-0)^2} = x\n",
            "\\]\n",
            "Squaring both sides, we get:\n",
            "\\[\n",
            "(x-6)^2 + 64 = x^2\n",
            "\\]\n",
            "Expanding and simplifying:\n",
            "\\[\n",
            "x^2 - 12x + 36 + 64 = x^2 \\implies 100 - 12x = 0 \\implies x = \\frac{100}{12} = \\frac{25}{3}\n",
            "\\]\n",
            "So, the coordinates of \\(P\\) are \\((\\frac{25}{3}, 8)\\).\n",
            "\n",
            "Now, we need to find the distance between \\(O\\) and \\(P\\). The coordinates of \\(O\\) are \\((6, \\frac{25}{4})\\) and the coordinates of \\(P\\) are \\((\\frac{25}{3}, 8)\\). The distance formula gives:\n",
            "\\[\n",
            "\\sqrt{\\left(6 - \\frac{25}{3}\\right)^2 + \\left(\\frac{25}{4} \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "What is the area of the region in the $xy-$plane that satisfies \\[\\lfloor x \\rfloor \\lfloor y \\rfloor = 16\\, ?\\] \n",
            "Answer:\n",
            "10 \n",
            "Response:\n",
            "Okay, so I need to find the area of the region in the xy-plane where the product of the floor of x and the floor of y is 16. Hmm, let me break this down.\n",
            "\n",
            "First, I remember that the floor function, ⌊a⌋, gives the greatest integer less than or equal to a. So, for example, ⌊3.7⌋ is 3, and ⌊-2.3⌋ is -3. The equation here is ⌊x⌋ ⌊y⌋ = 16. Since both ⌊x⌋ and ⌊y⌋ are integers (because the floor function always returns an integer), their product is 16, which is also an integer. So I need to find all pairs of integers (m, n) such that m * n = 16, and then determine the corresponding regions in the plane where x is in [m, m+1) and y is in [n, n+1).\n",
            "\n",
            "Let me list the pairs (m, n) where m and n are integers and their product is 16. Since 16 is positive, both m and n must be either positive or both negative.\n",
            "\n",
            "Positive pairs:\n",
            "1. (1, 16)\n",
            "2. (2, 8)\n",
            "3. (4, 4)\n",
            "4. (8, 2)\n",
            "5. (16, 1)\n",
            "\n",
            "Negative pairs:\n",
            "1. (-1, -16)\n",
            "2. (-2, -8)\n",
            "3. (-4, -4)\n",
            "4. (-8, -2)\n",
            "5. (-16, -1)\n",
            "\n",
            "Wait, let me check if I missed any. The factors of 16 are 1, 2, 4, 8, 16, and their negatives. So yes, those are all the possible pairs where m and n are integers multiplying to 16.\n",
            "\n",
            "Now, for each pair (m, n), the area defined by the inequality ⌊x⌋ = m and ⌊y⌋ = n is a square with side length 1. Because x is in [m, m+1) and y is in [n, n+1). So each of these regions is a 1x1 square, and there are 10 pairs here (5 positive and 5 negative). Therefore, the total area would be 10 squares, each with area 1, so 10*1 = 10. But wait, let me make sure I'm not double-counting any regions or missing any cases.\n",
            "\n",
            "Looking at the problem again: the equation is ⌊x⌋ ⌊y⌋ = 16. So for each solution (m, n), there is a distinct square in the plane. Since these squares don't overlap (as each (m, n) pair has a unique rectangle where x is between m and m+1, and y between n and n+1), the total area is indeed the sum of the areas of these squares. Each square has area 1, and there are 10 such pairs, so the total area is 10. \n",
            "\n",
            "I think that's it. Let me verify with a smaller example. If the equation was ⌊x⌋⌊y⌋ = 4, then the pairs would be (1,4), (2,2), (4,1), (-1,-4), (-2,-2), (-4,-1). That's 6 pairs, and each gives a 1x1 square, so area 6. Which makes sense. So scaling it up, for 16, it's 10 pairs, so area 10. Seems correct.\n",
            "To solve the problem of finding the area of the region in the \\(xy\\)-plane that satisfies \\(\\lfloor x \\rfloor \\lfloor y \\rfloor = 16\\), we proceed as follows:\n",
            "\n",
            "1. **Identify the integer pairs \\((m, n)\\) such that \\(m \\cdot n = 16\\):**\n",
            "   - Positive pairs: \\((1, 16)\\), \\((2, 8)\\), \\((4, 4)\\), \\((8, 2)\\), \\((16, 1)\\)\n",
            "   - Negative pairs: \\((-1, -16)\\), \\((-2, -8)\\), \\((-4, -4)\\), \\((-8, -2)\\), \\((-16, -1)\\)\n",
            "\n",
            "2. **Determine the corresponding regions in the \\(xy\\)-plane:**\n",
            "   - For each pair \\((m, n)\\), the region is defined by \\(m \\leq \\lfloor x \\rfloor < m+1\\) and \\(n \\leq \\lfloor y \\rfloor < n+1\\). This corresponds to a square with vertices at \\((m, n)\\), \\((m, n+1)\\), \\((m+1, n)\\), and \\((m+1, n+1)\\).\n",
            "   - Each of these squares has an area of \\(1 \\times 1 = 1\\).\n",
            "\n",
            "3. **Count the number of such squares:**\n",
            "   - There are 5 positive pairs and 5 negative pairs, giving a total of \\(5 + 5 = 10\\) pairs.\n",
            "\n",
            "4. **Calculate the total area:**\n",
            "   - Since each square has an area of 1, the total area is \\(10 \\times 1 = 10\\).\n",
            "\n",
            "Thus, the area of the region in the \\(xy\\)-plane that satisfies \\(\\lfloor x \\rfloor \\lfloor y \\rfloor = 16\\) is \\(\\boxed{10}\\).<end_working_out><SOLUTION>10</SOLUTION> \n",
            "Extracted:\n",
            "10\n",
            "********************Question:\n",
            "An event occurs periodically over $x$ consecutive years, followed by a break of $y$ consecutive years. The event occurred in the years 1964, 1986, 1996, and 2008, and did not occur in 1976, 1993, 2006, or 2013. Determine the first year in which the event will occur again. \n",
            "Answer:\n",
            "2018 \n",
            "Response:\n",
            "Okay, so I need to figure out when this event will happen again. The problem says it occurs every 'x' years, then there's a break of 'y' years, and some years in between. Hmm, let's try to find the pattern here.\n",
            "\n",
            "First, let's list the years where the event happened: 1964, 1986, 1996, and 2008. The years it didn't happen are 1976, 1993, 2006, 2013. So, the event occurs in certain years and skips others. The goal is to find the length of the cycle, which is (x + y) years.\n",
            "\n",
            "Let's see the differences between the given years where the event occurred. Let me calculate the differences between consecutive years.\n",
            "\n",
            "From 1964 to 1986: 1986 - 1964 = 22 years. The first cycle is 22 years long.\n",
            "\n",
            "Then from 1986 to 1996: 1996 - 1986 = 10 years. So the next cycle is 10 years.\n",
            "\n",
            "Wait, but the problem says the event follows a pattern of 'x' years, then 'y' years. So maybe the cycles themselves have lengths x and y. Let me check another interval. From 1996 to 2008: 2008 - 1996 = 12 years. Hmm, that's different from the previous cycles.\n",
            "\n",
            "So maybe the cycles vary? Or maybe there's a different pattern? Let me check the breaks. The problem mentions the break is 'y' years. The first break after 1964 (the last occurrence before 1976) is from 1964 to 1976. Let's compute that: 1976 - 1964 = 12 years. So the first period is 22 years, then a break of 12 years. Then the next event is 1986, which is 1964 + 22. Then from 1986 to 1996 is 10 years, so the next period is x=10 and y=10 because there's a break of 10 years. Wait, but then from 1996 to 2008 is 12 years, which doesn't fit that 10-year pattern.\n",
            "\n",
            "This seems inconsistent. Maybe there's a typo in the problem, or perhaps the pattern isn't strictly periodic with a fixed x and y. Alternatively, maybe the cycle resets after the first occurrence. But looking at the given years, the first cycle is 22 years, then a break of 12 years, then a 10-year cycle with a 10-year break, then 12 years again. This doesn't form a clear periodic pattern.\n",
            "\n",
            "Wait, perhaps the problem expects us to find the Least Common Multiple (LCM) of the cycle lengths. From the years 1964 to 1986 is 22 years, 1986 to 1996 is 10 years, and the breaks are 12 or 10. If we consider the cycles as 22 and 10, then the LCM of 22 and 10 would be 110.\n",
            "\n",
            "Let me test that. If the cycle is 22 years, the event would next occur in 1964 + 110 = 2074. Let's check if that makes sense with the breaks.\n",
            "\n",
            "But looking back at the given years: after 1996, the next event is 2008. Let's compute 1996 + 22 = 2018, which is after the given years. Then after 2018, another 22 years would be 2040. Since 2040 is after the given years, maybe the cycle repeats. Let me confirm.\n",
            "\n",
            "Wait, but 2018 - 2008 is 10 years, which doesn't match the 22-year cycle. So perhaps the cycle isn't purely 22 or 10. Maybe it's a different pattern. Alternatively, perhaps the problem expects us to find the smallest 'x' and 'y' such that both patterns fit.\n",
            "\n",
            "Wait a second, maybe the problem has a typo. Maybe one of the years where the event didn't occur is supposed to be a year where it did. For example, maybe 1993 should be 1996? But that's not given. Alternatively, maybe the breaks are not consistent.\n",
            "\n",
            "Alternatively, perhaps the cycle is 10 years, then a break of 12 years, repeat. Let's check:\n",
            "\n",
            "If the cycle is 10 years, then the first event after 2008 would be 2008 + 10 = 2018. Then another 10 years after that is 2028, and so on. Let's see if that fits with the given data.\n",
            "\n",
            "But looking back at the years where the event didn't occur: 1976, 1993, 2006, 2013. If the cycle is 10 years, then the event occurs every 10 years. Let's check:\n",
            "\n",
            "1964 + 10 = 1974 (but in 1976, it didn't occur, which doesn't match the problem statement).\n",
            "\n",
            "Wait, so if the cycle is 10 years, then 1964 + 10 = 1974, so the event should have happened in 1974. But according to the problem, it didn't occur in 1976. So maybe the cycle is not exactly 10 years. But maybe there's a difference in the first or last occurrence.\n",
            "\n",
            "Alternatively, perhaps the cycle is 12 years, then a break of 10 years. Let's check that.\n",
            "\n",
            "If the cycle is 12 years, then 1964 + 12 = 1976, which is a year it didn't occur. So that's a problem. So that can't be right.\n",
            "\n",
            "Alternatively, maybe the cycle is 20 years. Let's see: 1964 + 20 = 1984, which is not given. So that's not right.\n",
            "\n",
            "Alternatively, maybe the cycle is 19 years. 1964 + 19 = 1983, which is before 1964. No, that's not possible.\n",
            "\n",
            "Alternatively, maybe the cycle is 22 years. Let's see: 1964 + 22 = 1986, which matches. Then 1986 + 22 = 2008. Then 2008 + 22 = 2030. But the problem says it occurred in 1996 and 2008. So the cycle isn't strictly 22 years.\n",
            "\n",
            "Wait, maybe the first cycle is 22 years, then a new cycle starts. For example, 1964 (event), 1986 (event), 1996 (event), so between 1986 to 1996 is 10 years. So maybe the pattern is a 22-year cycle, then a 10-year cycle. Let's see:\n",
            "\n",
            "So 1964 + 22 = 1986, then 1986 + 10 = 1996. Then next event would be 1996 + 22 = 2018. Then 2018 + 10 = 2028, and so on.\n",
            "\n",
            "But let's check against the non-events. The years where it didn't occur are 1976, 1993, 2006, 2013. According to this pattern, after 1996, the next event would be 2018. So 2018 - 2008 = 10 years, which matches the second cycle. Then 2018 + 22 = 2040. So the cycle is 22, then 10, then 22, then 10, \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "Compute the sum of the series \\( \\sum_{k=0}^{\\infty} \\frac{1}{(4 k+1)(4 k+2)(4 k+3)(4 k+4)}=\\frac{1}{1 \\cdot 2 \\cdot 3 \\cdot 4}+\\frac{1}{5 \\cdot 6 \\cdot 7 \\cdot 8}+\\cdots  \\). The original answer is in a special format：$\\frac{\\ln m}n-\\frac\\pi{p}.$ provide the sum of the denominators in the expression.  \n",
            "Answer:\n",
            "30 \n",
            "Response:\n",
            "Okay, let's see. The problem here is to compute the sum of the series \\(\\sum_{k=0}^{\\infty} \\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\). Let me start by simplifying the general term in the series.\n",
            "\n",
            "First, notice that the denominator is a product of four consecutive integers. Maybe partial fraction decomposition can help here. Let me assume that the fraction \\(\\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\) can be expressed as the sum of simpler fractions. For example:\n",
            "\n",
            "\\[\n",
            "\\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \\frac{A}{4k+1} + \\frac{B}{4k+2} + \\frac{C}{4k+3} + \\frac{D}{4k+4}\n",
            "\\]\n",
            "\n",
            "To find the coefficients \\(A, B, C, D\\), I need to solve the equation:\n",
            "\n",
            "\\[\n",
            "1 = A(4k+2)(4k+3)(4k+4) + B(4k+1)(4k+3)(4k+4) + C(4k+1)(4k+2)(4k+4) + D(4k+1)(4k+2)(4k+3)\n",
            "\\]\n",
            "\n",
            "This seems complicated. Maybe there's a better approach. Let me think differently. Notice that the denominator can be rewritten as \\((4k+1)(4k+4) \\times (4k+2)(4k+3) = (4k+4)(4k+1) \\times (4k+2)(4k+3)\\). Maybe partial fractions over two denominators?\n",
            "\n",
            "Alternatively, consider that \\((4k+n)(4k+n+1)(4k+n+2)(4k+n+3)\\) is related to factorials or double factorials. Wait, perhaps expressing it in terms of beta function or gamma function would work. But I'm not sure.\n",
            "\n",
            "Another thought: Maybe the numerator is the derivative of the denominator? But that's not directly applicable here since the derivatives don't align neatly.\n",
            "\n",
            "Let me try expanding the denominator. For example, take \\(k=0\\):\n",
            "\n",
            "\\[\n",
            "(4(0)+1)(4(0)+2)(4(0)+3)(4(0)+4) = 1 \\times 2 \\times 3 \\times 4 = 24\n",
            "\\]\n",
            "\n",
            "Next, \\(k=1\\):\n",
            "\n",
            "\\[\n",
            "5 \\times 6 \\times 7 \\times 8 = 1680\n",
            "\\]\n",
            "\n",
            "Wait, so each term is \\(\\frac{1}{24}, \\frac{1}{1680}, \\frac{1}{28560}, \\ldots\\). Hmm, these seem to grow rapidly. The denominators are \\((4k+1)(4k+2)(4k+3)(4k+4)\\), which are increasing factorials-like terms.\n",
            "\n",
            "Maybe the series is telescoping. But looking at the terms, they don't seem to cancel out immediately. Alternatively, perhaps there's a generating function approach.\n",
            "\n",
            "Wait, the problem states the result is in the form \\(\\frac{\\ln m}{n} - \\frac{\\pi}{p}\\). So maybe the sum involves both logarithmic and trigonometric functions. This suggests a possible connection to integrals.\n",
            "\n",
            "Let me consider the integral representation. For example, \\(\\frac{1}{n} = \\int_0^1 x^{n-1} dx\\). Maybe applying this to each term in the series.\n",
            "\n",
            "So, for a general term \\(\\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\), we can write it as:\n",
            "\n",
            "\\[\n",
            "\\frac{1}{4k+1} - \\frac{1}{4k+2} - \\frac{1}{4k+3} + \\frac{1}{4k+4}\n",
            "\\]\n",
            "\n",
            "Wait, but that's not correct. Let me try again. Let's compute the partial fractions directly.\n",
            "\n",
            "Let’s denote \\(n = 4k\\), so \\(4k+1 = n+1\\), \\(4k+2 = n+2\\), \\(4k+3 = n+3\\), \\(4k+4 = n+4\\). Then the term is \\(\\frac{1}{(n+1)(n+2)(n+3)(n+4)}\\).\n",
            "\n",
            "Using partial fractions:\n",
            "\n",
            "\\[\n",
            "\\frac{1}{(n+1)(n+2)(n+3)(n+4)} = \\frac{A}{n+1} + \\frac{B}{n+2} + \\frac{C}{n+3} + \\frac{D}{n+4}\n",
            "\\]\n",
            "\n",
            "Multiplying both sides by the denominator, we get:\n",
            "\n",
            "\\[\n",
            "1 = A(n+2)(n+3)(n+4) + B(n+1)(n+3)(n+4) + C(n+1)(n+2)(n+4) + D(n+1)(n+2)(n+3)\n",
            "\\]\n",
            "\n",
            "To solve for \\(A, B, C, D\\), we can substitute specific values of \\(n\\) to find the coefficients.\n",
            "\n",
            "For example, let \\(n = -3\\):\n",
            "\n",
            "Left side is 1, right side is \\(A(1)(0)(1) = 0\\). Not useful. Hmm, maybe another approach.\n",
            "\n",
            "Alternatively, we can equate coefficients of like terms. Let's expand the right side:\n",
            "\n",
            "\\[\n",
            "A(n^3 + 9n^2 + 26n + 24) + B(n^3 + 8n^2 + 21n + 12) + C(n^3 + 7n^2 + 18n + 8) + D(n^3 + 6n^2 + 15n + 6)\n",
            "\\]\n",
            "\n",
            "Combining like terms:\n",
            "\n",
            "\\[\n",
            "(A+B+C+D)n^3 + (9A+8B+7C+6D)n^2 + (26A+21B+18C+15D)n + (24A+12B+8C+6D) = 1n^0\n",
            "\\]\n",
            "\n",
            "Setting coefficients equal to zero:\n",
            "\n",
            "1. \\(A + B + C + D = 0\\)\n",
            "2. \\(9A + 8B + 7C + 6D = 0\\)\n",
            "3. \\(26A + 21B + 18C + 15D = 0\\)\n",
            "4. \\(24A + 12B + 8C + 6D = 1\\)\n",
            "\n",
            "This system can be solved step by step. From equation 1, \\(D = -A - B - C\\). Substitute into the other equations.\n",
            "\n",
            "Plugging into equation 2:\n",
            "\n",
            "\\[\n",
            "9A + 8B + 7C + 6(-A - B - C) = 0 \\Rightarrow 3A + 2B + C = 0 \\quad \\text{(2a)}\n",
            "\\]\n",
            "\n",
            "Equation 3:\n",
            "\n",
            "\\[\n",
            "26A + 21B + 18C + 15(-A - B - C) = 0 \\Rightarrow 11A + 6B + 3C = 0 \\quad \\text{(3a)}\n",
            "\\]\n",
            "\n",
            "Equation 4:\n",
            "\n",
            "\\[\n",
            "24A + 12B + 8C + 6(-A - B - C) = 1 \\Rightarrow 18A + 6B + 2C = 1 \\quad \\text{(4a)}\n",
            "\\]\n",
            "\n",
            "Now, solve equations 2a and 3a:\n",
            "\n",
            "From 2a: \\(3A = -2B - C \\Rightarrow A = -\\frac{2B + C}{3}\\)\n",
            "\n",
            "Substitute into 3a:\n",
            "\n",
            "\\[\n",
            "11(-\\frac{2B + C}{3}) + 6B + 3C = 0 \\Rightarrow -\\frac{22B + 11C}{3} + 6B + 3C = 0\n",
            "\\]\n",
            "\n",
            "Multiply everything by 3:\n",
            "\n",
            "\\[\n",
            "-22B - 11C + 18B + 9C = 0 \\Rightarrow -4B -2C = 0 \\Rightarrow B = -\\frac{C}{2}\n",
            "\\]\n",
            "\n",
            "Now, substitute back into 2a:\n",
            "\n",
            " \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out.  What is the probability that all beads in the bag are red after three such replacements? Express your answer as a common fraction.The answer is in the form \frac{m}{n}, where gcd(m, n) = 1. Please provide the value of m + n. \n",
            "Answer:\n",
            "41 \n",
            "Response:\n",
            "Okay, let's see. There's a problem here involving a bag with two red and two green beads. Every time I pick a bead, I replace it with a red bead. I need to figure out the probability that all beads in the bag are red after three such replacements. The answer should be in a simplified fraction m/n, and then I need to find m + n.\n",
            "\n",
            "Alright, let me start by understanding the process. Initially, the bag has two red (R) and two green (G) beads. Each replacement involves picking a bead, changing its color to red, and putting it back. After three replacements, the question is asking for the probability that all beads are red.\n",
            "\n",
            "Hmm, okay. Let me work this out step by step.\n",
            "\n",
            "First, what are the possible outcomes for each replacement? Since the bead can be either red or green when picked, each replacement has two outcomes: red or green. But wait, if it's green, after replacement it becomes red. So the number of green beads either stays the same or decreases, while red beads either stay the same or increase.\n",
            "\n",
            "Wait, but depending on the color picked, the number of green beads changes. Let me denote the state of the bag as (r, g), where r is the number of red beads and g is the number of green beads. The initial state is (2, 2).\n",
            "\n",
            "Each replacement involves adding one red bead to the total number of beads, regardless of which color is pulled out. So after each replacement, the total number of beads increases by one. Since we're doing three replacements, the total number of beads will be 2 initial + 3 replacements = 5 beads. The goal is to end up with 5 red beads (since we want all beads to be red after replacements).\n",
            "\n",
            "But the tricky part is that each replacement affects the composition. Let me think about the possible paths that lead to (5, 0). \n",
            "\n",
            "Actually, since each replacement adds one red bead, the total number of red beads increases by one each time. So, starting from 2 red beads, after three replacements, there will be 2 + 3 = 5 red beads. So the desired outcome is that all three replacements were green beads being pulled, which makes the current green bead become red. Therefore, the probability would be the probability that all three picks are green.\n",
            "\n",
            "Wait, but let me check if this is correct. If at any point a red bead is picked, it's replaced with a red bead, so the number of red beads doesn't change in that case. But if a green bead is picked, it's replaced with a red bead, increasing the red count by one. So to have all beads become red after three replacements, each replacement must have been a green bead being picked, right? Because otherwise, the number of green beads would not decrease, and thus we wouldn't end up with all red beads.\n",
            "\n",
            "Therefore, the problem reduces to the probability that three consecutive picks are all green. Since the color of beads changes after each pick (each pick is independent and has a certain probability of being green), but the problem states that after replacement, all beads are red regardless of the color picked. Wait, that's important. So the replacement step ensures that any green bead picked becomes red. Therefore, the key is that each replacement has a possibility to turn a green bead into a red one. But the total number of red beads increases by one each time, regardless of the original color picked. So the condition for all beads to be red after three replacements is that each replacement effectively turns a green bead into red. Therefore, the probability would be the probability that all three picks were green.\n",
            "\n",
            "But wait, but in reality, the composition of the bag changes with each pick. Initially, there are 2 red and 2 green, so the probability of picking green first is 2/4 = 1/2. After replacing the green with red, the number of green beads decreases by one, and the total beads increase by one. So the next pick has 3 red and 3 green (since one green was replaced with red, making the total beads 6). So the probability of picking green again is 3/6 = 1/2. Then, replacing that green with red, the number of green becomes 2, and total beads 7. The next pick has 4 red and 2 green, so the probability of picking green is 2/6 = 1/3. Then replacing that green with red, the number of green becomes 1, and total beads 8. The last pick has 5 red and 1 green, so the probability of picking green is 1/6. Then replacing that green with red, the number of green becomes 0, and total beads 9. Therefore, after three replacements, if all three picks were green, then all beads become red. So the probability is (1/2)*(1/2)*(1/3)*(1/6) = (1/2)^3 = 1/8?\n",
            "\n",
            "But let me verify each step carefully.\n",
            "\n",
            "Initial state: 2 red, 2 green\n",
            "\n",
            "1. Pick a green bead: 2/4 = 1/2 chance. Replace with red. New state: 3 red, 1 green. Total beads: 6.\n",
            "2. Pick a green bead: 1/6 chance. Replace with red. New state: 4 red, 0 green. Total beads: 7.\n",
            "3. Pick a green bead: 0/6 chance. But wait, there are no green beads left. This can't happen. So this path is impossible. So maybe my initial assumption was wrong.\n",
            "\n",
            "Ah, okay, so perhaps the numbering of the replacements matters. Let me think differently.\n",
            "\n",
            "Each replacement involves adding one red bead, regardless of the color picked. So after k replacements, there are k replaced beads (from green to red). Therefore, the number of red beads in the bag is 2 (initial) + k (replacements) = 2 + k. The number of green beads is the original number minus those that were replaced, which is 2 - (number of green beads picked). But each replacement only happens if a green bead is picked. Therefore, the number of red beads after k replacements is 2 + k, and the number of green beads is 2 - g, where g is the number of green beads picked in the three replacements. For all beads to be red, the number of green beads picked must be zero. So the probability that all three picks are green is (1/2)*(1/2)*(1/2) = 1/8.\n",
            "\n",
            "Wait, but earlier, the second replacement had a 1/6 chance, which is not 1/2. Oh right, because after the first replacement, the state changes. So the probability is not 1/2 for each pick. Therefore, the correct probability is (1/2)*(1/3)*(1/7) = 1/42.\n",
            "\n",
            "Wait, no. After each replacement, the total number of beads increases by one, and the number of green beads either stays the same or decreases by one. So the probability for each replacement depends on the current state. Let's try another way.\n",
            "\n",
            "Let's denote the state after each replacement. Let G be the number of green beads. Initially, G=2.\n",
            "\n",
            "After the first replacement:\n",
            "- Probability of picking green: 2/4 = 1/2. Then G becomes 1, total beads 5.\n",
            "- Probability of picking red: 2/4 = 1/2. Then G remains 2, total beads 5.\n",
            "So the state after first replacement is either (1,4) or (2,6) depending on the pick.\n",
            "\n",
            "Then the second replacement:\n",
            "If the previous state was (1,4), then probability of picking green: 4/5. Then G becomes 3, total 6.\n",
            "If the previous state was (2,6), then probability of picking green: 6/6 = 1. Then G becomes 1, total 7.\n",
            "So the possible states after second replacement depend on the sequence of picks.\n",
            "\n",
            "Third replacement:\n",
            "Again, it depends on the previous state. If G is now 3, then picking green turns it into 2. If G is 1, then picking green turns it into 0. We want G=0 after all three replacements.\n",
            "\n",
            "So let's compute the probability step by step.\n",
            "\n",
            "Initial state: (2,2)\n",
            "\n",
            "**First replacement**\n",
            "\n",
            "Probability of picking green (G becomes 1, total 5): 2/4 = 1/2\n",
            "Probability of picking red (G remains 2, total 5): 2/4 = 1/2\n",
            "\n",
            "**Second replacement**\n",
            "\n",
            "If after first replacement, G=1:\n",
            "Probability of picking green (G becomes 2, total 6): 1/ \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "A triangle has an area of 114 square units and sides of integer length. What is the perimeter of the triangle? \n",
            "Answer:\n",
            "76 \n",
            "Response:\n",
            "Okay, let's see. I need to find the perimeter of a triangle with an area of 114 square units that has integer side lengths. Hmm, perimeter would be the sum of the three sides, right? So the challenge here is that the triangle has whole number sides, so the lengths must be integers, and the area is 114.\n",
            "\n",
            "First, I remember that the area of a triangle can be calculated using Heron's formula. Heron's formula is especially useful when you know all three sides of a triangle. Heron's formula states that the area is the square root of (s times (s - a) times (s - b) times (s - c)), where s is the semi-perimeter, which is half of the perimeter. \n",
            "\n",
            "So, if the sides are integers a, b, and c, then the semi-perimeter s = (a + b + c)/2. The area formula would be sqrt[s(s - a)(s - b)(s - c)] = 114. Hmm, that's a bit complicated, but maybe I can work with it.\n",
            "\n",
            "Since the sides are integers, I need to find three integers that satisfy this equation. Let me think. One approach could be to try different integer values for the sides and check if they work. But how do I start? Maybe start with small integers and go up?\n",
            "\n",
            "But trial and error might take a while. Maybe there's a smarter way. Alternatively, since the area is 114, which is 19 * 6, maybe that can help. But I'm not sure how to connect that to the sides. \n",
            "\n",
            "Another thought: perhaps assume the triangle is right-angled. If it's a right-angled triangle, the area is (a * b)/2 = 114, so a * b = 228. Then the sides a, b, and the hypotenuse c must satisfy a^2 + b^2 = c^2. Let me check if 228 can be factored into two integers whose squares add up to another perfect square.\n",
            "\n",
            "Let's list factor pairs of 228: (1, 228), (2, 114), (3, 76), (4, 57), (6, 38), (12, 19). Now check if each pair (a, b) satisfies a^2 + b^2 being a perfect square.\n",
            "\n",
            "Start with (12, 19): 12^2 + 19^2 = 144 + 361 = 505, which is not a perfect square.\n",
            "Next (6, 38): 6^2 + 38^2 = 36 + 1444 = 1480, not a perfect square.\n",
            "Then (4, 57): 4^2 + 57^2 = 16 + 3249 = 3265, not a perfect square.\n",
            "Next (3, 76): 3^2 + 76^2 = 9 + 5776 = 5785, not a perfect square.\n",
            "Then (2, 114): 2^2 + 114^2 = 4 + 12996 = 13000, not a perfect square.\n",
            "Finally (1, 228): 1^2 + 228^2 = 1 + 51984 = 51985, not a perfect square.\n",
            "\n",
            "So none of these pairs work. Therefore, the triangle can't be a right-angled triangle with integer sides. Back to Heron's formula then.\n",
            "\n",
            "Let's denote the sides as a, b, c, all integers. Then s = (a + b + c)/2, and s(s - a)(s - b)(s - c) = 114^2 = 12996. So (a + b + c)(b + c - a)(a + c - b)(a + b - c) = 12996. This seems messy, but maybe I can factor 12996 into four integers that correspond to these expressions.\n",
            "\n",
            "Let's factor 12996. 12996 divided by 4 is 3249, which is 57^2. So 12996 = 4 * 57^2. Hmm, not sure how that helps directly. Maybe try to find a, b, c such that (a + b + c) is even since s is an integer. Let's try different combinations.\n",
            "\n",
            "Suppose the sides are 17, 15, and 19. Let's check if they work.\n",
            "\n",
            "Semi-perimeter s = (17 + 15 + 19)/2 = 24. Then the area is sqrt[24(24 - 17)(24 - 15)(24 -19)] = sqrt[24*7*9*5] = sqrt[7560]. Wait, 7560 divided by 114 is 66, which is close to 66.66, so 66 isn't the exact area. So this triangle doesn't work.\n",
            "\n",
            "Next, try sides 20, 21, 23. Semi-perimeter is 32. Area = sqrt[32*12*11*9] = sqrt[38016]. 38016 divided by 114 is 333.47..., so not 114. Doesn't fit.\n",
            "\n",
            "Let me try 21, 20, 22. Semi-perimeter is 31.5. But sides must be integers, so this can't be a triangle. So maybe 21, 20, 23? Semi-perimeter 32. Area sqrt[32*11*3*5] = sqrt[5280]. 5280 divided by 114 is approximately 46.32, so different. Not working.\n",
            "\n",
            "Wait, maybe I need a different approach. Let me think about the formula differently. The area can also be expressed as (1/2)ab sin C = 114. But without knowing the angles, that might not help. Alternatively, perhaps using the formula area = (a * b * c)/(4R), where R is the circumradius. But without R, this doesn't help.\n",
            "\n",
            "Alternatively, perhaps use the fact that the triangle must satisfy the triangle inequality: sum of any two sides must be greater than the third. So a + b > c, a + c > b, b + c > a.\n",
            "\n",
            "Given that the area is 114, which is 19 * 6, maybe the sides are related to 19 and 6. Let me think. If the triangle were a triangle with sides proportional to 19 and 6, but since they're integers, perhaps multiples.\n",
            "\n",
            "Wait, maybe 19, 6, and the third side. Let's see. If one side is 19 and another is 6, then the third side must satisfy 19 + 6 > c, 19 + c > 6, 6 + c > 19, so c < 25, c >13. So c is an integer between 14 and 24.\n",
            "\n",
            "Then the semi-perimeter s = (19 + 6 + c)/2 = (25 + c)/2. The area is sqrt[(25 + c)/2 * (c - 3)/2 * (c + 3)/2 * (25 - c)/2] = 114. This seems complicated, but maybe c can be a multiple of 19 or 6.\n",
            "\n",
            "Let me try c = 19. Then sides 19, 6, 19. Semi-perimeter is 22.5. Since sides must be integers, this can't work. c=19 would make the triangle isosceles with sides 19, 6, 19, but 19 + 6 = 25 >19, 19 + 19 >6, 6 + 19 >19. So yes, valid triangle. Area: sqrt[22.5*(16.5)*(17.5)*(-3.5)]. Hmm, one of the terms is negative, which isn \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "Find the $2019^{th}$ strictly positive integer $n$ such that $\\binom{2n}{n}$ is not divisible by $5$. \n",
            "Answer:\n",
            "37805 \n",
            "Response:\n",
            "Okay, so I need to find the 2019th strictly positive integer \\( n \\) such that the binomial coefficient \\( \\binom{2n}{n} \\) is not divisible by 5. Hmm, let's start by recalling some basics about binomial coefficients and divisibility by primes like 5.\n",
            "\n",
            "First, I remember that when we say a binomial coefficient \\( \\binom{2n}{n} \\) is not divisible by 5, it means that when we express \\( \\binom{2n}{n} \\) in terms of factorials, none of the factors in the denominator (after canceling with the numerator) introduce a factor of 5. So, let me think about how often the prime number 5 appears in the factorials involved.\n",
            "\n",
            "From my knowledge, the p-adic valuation \\( v_p(k!) \\) which is the exponent of the prime \\( p \\) in the prime factorization of \\( k! \\) is given by:\n",
            "\n",
            "\\[ v_p(k!) = \\left\\lfloor \\frac{k}{p} \\right\\rfloor + \\left\\lfloor \\frac{k}{p^2} \\right\\rfloor + \\left\\lfloor \\frac{k}{p^3} \\right\\rfloor + \\ldots \\]\n",
            "\n",
            "Since we're dealing with 5, the formula would be:\n",
            "\n",
            "\\[ v_5((2n)!)=\\left\\lfloor \\frac{2n}{5} \\right\\rfloor + \\left\\lfloor \\frac{2n}{25} \\right\\rfloor + \\left\\lfloor \\frac{2n}{125} \\right\\rfloor + \\ldots \\]\n",
            "\n",
            "\\[ v_5(n!)=\\left\\lfloor \\frac{n}{5} \\right\\rfloor + \\left\\lfloor \\frac{n}{25} \\right\\rfloor + \\left\\lfloor \\frac{n}{125} \\right\\rfloor + \\ldots \\]\n",
            "\n",
            "Then, \\( v_5(\\binom{2n}{n}) = v_5((2n)!)-2v_5(n!) \\). For \\( \\binom{2n}{n} \\) to **not** be divisible by 5, this value has to be zero.\n",
            "\n",
            "So, the condition is:\n",
            "\n",
            "\\[ \\left\\lfloor \\frac{2n}{5} \\right\\rfloor + \\left\\lfloor \\frac{2n}{25} \\right\\rfloor + \\left\\lfloor \\frac{2n}{125} \\right\\rfloor + \\ldots - 2\\left( \\left\\lfloor \\frac{n}{5} \\right\\rfloor + \\left\\lfloor \\frac{n}{25} \\right\\rfloor + \\left\\lfloor \\frac{n}{125} \\right\\rfloor + \\ldots \\right) = 0 \\]\n",
            "\n",
            "Simplifying terms, let's take each term:\n",
            "\n",
            "When you compare \\( \\left\\lfloor \\frac{2n}{5} \\right\\rfloor \\) and \\( 2\\left\\lfloor \\frac{n}{5} \\right\\rfloor \\), they are equal when \\( n \\) is not a multiple of 5 because, for example, if \\( n = 5k \\), then \\( \\left\\lfloor \\frac{2*5k}{5} \\right\\rfloor = 2k \\) and \\( 2 \\left\\lfloor \\frac{5k}{5} \\right\\rfloor = 2k \\). But when \\( n \\) is not a multiple of 5, say \\( n=5k+1 \\), then \\( \\left\\lfloor \\frac{2(5k+1)}{5} \\right\\rfloor = 2k \\), and \\( 2* \\left\\lfloor \\frac{5k+1}{5} \\right\\rfloor = 2k \\) also. Wait, that's not right. Let me check with examples. If \\( n=1 \\), which is not a multiple of 5, \\( \\left\\lfloor \\frac{2}{5} \\right\\rfloor = 0 \\), and \\( 2* \\left\\lfloor \\frac{1}{5} \\right\\rfloor = 0 \\). So it works here. If \\( n=6 \\), \\( \\left\\lfloor \\frac{12}{5} \\right\\rfloor = 2 \\), and \\( 2* \\left\\lfloor \\frac{6}{5} \\right\\rfloor = 2 \\). Hmm, so for these terms, it seems to match. Let's check \\( n=5 \\): \\( \\left\\lfloor \\frac{10}{5} \\right\\rfloor = 2 \\), and \\( 2* \\left\\lfloor \\frac{5}{5} \\right\\rfloor = 2 \\). So in fact, the first term will match because either n is a multiple of 5 or not. \n",
            "\n",
            "But wait, the next term is \\( \\left\\lfloor \\frac{2n}{25} \\right\\rfloor \\) vs \\( 2\\left\\lfloor \\frac{n}{25} \\right\\rfloor \\). Now, if \\( n \\) is a multiple of 25, say \\( n=25k \\), both terms are \\( 2k \\). But if \\( n=25k + m \\), where \\( 1 \\leq m \\leq 24 \\), then:\n",
            "\n",
            "\\[ \\left\\lfloor \\frac{2(25k + m)}{25} \\right\\rfloor = \\left\\lfloor 2k + \\frac{2m}{25} \\right\\rfloor = 2k \\] if \\( m < 12.5 \\) or \\( 2k +1 \\) if \\( m \\geq 12.5 \\). And \\( 2\\left\\lfloor \\frac{25k + m}{25} \\right\\rfloor = 2( k + \\left\\lfloor \\frac{m}{25} \\right\\rfloor ) = 2k \\) if \\( m <25 \\). \n",
            "\n",
            "So the difference is 1 if \\( m \\geq 12.5 \\), but since \\( m \\) is an integer, if \\( m \\geq 13 \\), it adds 1. Therefore, when \\( n \\) is in the range \\( [12, 24] \\) modulo 25, it increases the exponent by 1. So only when \\( m <13 \\) does it match. Thus, the first few numbers don't work, but higher numbers do, with a pattern every 25 numbers.\n",
            "\n",
            "So the pattern should be that every 25 consecutive numbers, there are 13 numbers where \\( \\binom{2n}{n} \\) is divisible by 5. Because in your range, only when \\( n \\) is a multiple of 25 the difference is zero, otherwise, some will increment. Therefore, the ones where it's not divisible by 5 are the remaining numbers, which is 25 - 13 = 12 per 25 numbers. \n",
            "\n",
            "To get the 2019th number, I assume this pattern continues indefinitely. So, for every block of 25 numbers, 12 satisfy the condition. Therefore, the total numbers would be multiples of 12. 2019 divided by 12 is 168.25. So it's 168 blocks multiplied by 12, which is 2016, so the next one would be 2019 = 168*12 +9. So within the next block of 25, the 9th number.\n",
            "\n",
            "Thus, starting from the beginning, every 25 numbers, there are 12. So the one we find is \\( 24*25 + 9 = 609 \\). Wait, let's check the indices. If we start at n=1, after every 25, we add 25. So 168*25 is the end of the 168th block, which is 4200. Then we need the 9th number after that, so 4200 +9 = 4209. But wait, because in each block \n",
            "Extracted:\n",
            "None\n",
            "********************Question:\n",
            "Dragoons take up $1 \\times 1$ squares in the plane with sides parallel to the coordinate axes such that the interiors of the squares do not intersect. A dragoon can fire at another dragoon if the difference in the $x$-coordinates of their centers and the difference in the $y$-coordinates of their centers are both at most $6$, regardless of any dragoons in between. For example, a dragoon centered at $(4, 5)$ can fire at a dragoon centered at the origin, but a dragoon centered at $(7, 0)$ cannot. A dragoon cannot fire at itself. What is the maximum number of dragoons that can fire at a single dragoon simultaneously? \n",
            "Answer:\n",
            "168 \n",
            "Response:\n",
            "Okay, so I need to figure out how many dragoons can fire at a single dragoon at the same time. Let's break this down.\n",
            "\n",
            "First, understand the problem. Each dragoon takes up a 1x1 square with sides parallel to the axes. They can't overlap. A dragoon can fire at another one if the difference in their x-coordinates and y-coordinates is at most 6. Also, they can't fire at themselves.\n",
            "\n",
            "So, for example, a dragoon at (4,5) can shoot at (0,0), but not at (7,0). Hmm, right, because the diff in x would be 4, which is allowed, but the diff in y? Wait, 5-0=5, which is also at most 6, so yes, that's fine. But (7,0) the diff in x is 3, which is allowed, but the diff in y would be 5, which is 5<6, so that's fine. But let me check the other way. If the dragoon is at (0,0) can it shoot at (4,5)? Yeah, because diff in x is 4≤6 and diff in y is 5≤6. So both directions work.\n",
            "\n",
            "So, the condition is that for every other dragoon, both the absolute difference in x and absolute difference in y must be ≤6 from the targeted dragoon. And they can't be located at the same spot (since diff is 0, which is within the limit).\n",
            "\n",
            "Alright, so if we have a target at (x₀, y₀), then any other dragoon at (x,y) where |x - x₀| ≤6 and |y - y₀| ≤6 can potentially fire at it. But we need to ensure that they don't overlap. Wait, the problem states that the interiors don't intersect, which means that if two dragoons are within 1 unit of each other, they can't both fire at the target. So, the condition is that the ranges in x and y are both from x₀-6 to x₀+6 and y₀-6 to y₀+6, but with boundaries of 1 unit apart because each dragoon is 1x1. So the effective range for x and y is from x₀-6.5 to x₀+6.5 and y₀-6.5 to y₀+6.5, because the centers of adjacent dragoons are 1 unit apart.\n",
            "\n",
            "Wait, let me think through that. If the target is at (5,5), then the possible positions for other dragoons are (1 to 11, 1 to 11). But since each dragoon's center must be at least 1 unit away from the target's edges in each direction, it's actually (2 to 10, 2 to 10). Because the target's square is 5x5, so the nearest points in each direction are 1 unit away, hence the range is 2 to 10. So that's 9 points along each axis. Therefore in a 9x9 grid, the number of dragoons is 9*9 = 81. But check if that's the maximum.\n",
            "\n",
            "Wait, the problem is to find the maximum number of dragoons that can fire at a single one. So we need to find the largest possible number of dragoons within the 6x6 grid around the target, each not overlapping with the target. So yes, 9 positions in each direction gives 81. But let me confirm if all 81 can fire. Since each dragoon is 1x1, and the target is at the center, the grid lines around the target allow for 9 on each side. So yes, that's 81. And since there's no overlap between these dragoons, each one can fire at the center without blocking others. Therefore, the maximum should be 81.\n",
            "\n",
            "But wait, the problem also says that a dragoon can't fire at itself, which is already taken care of since the centers are at (5,5), which is not included in the 2 to 10 range. So that's not a factor here. So 81 seems right. But let me make sure I'm not missing something. The key is that within a 13x13 area around the target (6 units left and right, 6 up and down), there are 9 possible positions in each direction, hence 9*9=81. So I think that's it.\n",
            "To determine the maximum number of dragoons that can fire at a single dragoon simultaneously, we need to consider the conditions under which a dragoon can fire at another. Specifically, a dragoon at \\((x_0, y_0)\\) can fire at another dragoon at \\((x, y)\\) if \\(|x - x_0| \\leq 6\\) and \\(|y - y_0| \\leq 6\\). Additionally, the dragoons must not intersect, meaning their centers must be at least 1 unit apart.\n",
            "\n",
            "Given a dragoon at \\((x_0, y_0)\\), the possible positions for other dragoons that can fire at it form a 13x13 grid centered at \\((x_0, y_0)\\) with the boundaries extended by 6 units in both the x and y directions. However, since each dragoon is 1x1 and must be at least 1 unit away from the target's center, the valid positions form a 9x9 grid of points around the target dragoon.\n",
            "\n",
            "To visualize this, if the target dragoon is at \\((x_0, y_0)\\), the possible positions for other dragoons are within the range \\(2 \\leq x \\leq 10\\) and \\(2 \\leq y \\leq 10\\) (since the target's square is at the center and the boundaries extend 6 units in each direction, but the centers must be at least 1 unit away, so the range is from 2 to 10 in both directions).\n",
            "\n",
            "Thus, the number of valid positions for other dragoons is \\(9 \\times 9 = 81\\).\n",
            "\n",
            "Therefore, the maximum number of dragoons that can fire at a single dragoon simultaneously is \\(\\boxed{81}\\).<end_working_out><SOLUTION>81</SOLUTION> \n",
            "Extracted:\n",
            "81\n"
          ]
        },
        {
          "data": {
            "text/plain": [
              "TrainOutput(global_step=100, training_loss=0.0067273743636906145, metrics={'train_runtime': 10607.693, 'train_samples_per_second': 0.038, 'train_steps_per_second': 0.009, 'total_flos': 0.0, 'train_loss': 0.0067273743636906145})"
            ]
          },
          "execution_count": 31,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "# For optional training + evaluation\n",
        "# new_dataset = dataset.train_test_split(test_size = 0.01)\n",
        "\n",
        "trainer = GRPOTrainer(\n",
        "    model = model,\n",
        "    processing_class = tokenizer,\n",
        "    reward_funcs = [\n",
        "        match_format_exactly,\n",
        "        match_format_approximately,\n",
        "        check_answer,\n",
        "        check_numbers,\n",
        "    ],\n",
        "    args = training_args,\n",
        "    train_dataset = dataset,\n",
        "\n",
        "    # For optional training + evaluation\n",
        "    # train_dataset = new_dataset[\"train\"],\n",
        "    # eval_dataset = new_dataset[\"test\"],\n",
        ")\n",
        "trainer.train()"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "tlaUdxC_VHpz"
      },
      "source": [
        "<a name=\"Inference\"></a>\n",
        "### Inference\n",
        "Now let's try the model we just trained! First, let's first try the model without any GRPO trained:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 225,
          "referenced_widgets": [
            "a0959e1bca5a4c349cf0a960cac5a409",
            "a2a0258791614ce5b2fa6df769e13dfd",
            "417e93b26e894f9a95f042e143a656e8",
            "d1279e91e27b458c900d3643732485df",
            "57ce3d8d27a84a528b982b89140deeea",
            "59840055328441b38fbdad2592548b57",
            "ffcab2e96bd442c08336f871c27269ce",
            "ee8689f732384799be79eac6fb99ea91",
            "e7eb41f3bf4b4a068b6dc97cb067b43b",
            "d35d45abf5fa46beb8d940ea6d9dc1e1",
            "0146226f52b147119f45ea8da473cb22"
          ]
        },
        "id": "qtcz_lpbVC92",
        "outputId": "09698cd0-9005-428f-975b-97c8e1348fba"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "a0959e1bca5a4c349cf0a960cac5a409",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Processed prompts:   0%|          | 0/1 [00:00<?, ?it/s, est. speed input: 0.00 toks/s, output: 0.00 toks/s]"
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          "metadata": {},
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        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "' - Answers\\nEducation\\nMath and Arithmetic\\nWhat is the sqrt of 101?\\nWiki User\\n∙ 2009-06-05 01:02:40\\nBest Answer\\nCopy\\n10.05\\nWiki User\\n∙ 2009-06-05 01:02:40\\nThis answer is:\\n🙏\\n0\\n🤨\\n0\\n😮\\n0\\nWhat is the sqrt of 100?\\nThe square root of 100 is 10; 10x10=100\\nWhat is the Sqrt of 100?\\nSqrt (100) = +10 &amp; -10 since both (+10) X (+10) and (-10) X (-\\n10) equal 100\\nWhat is the sqrt of 10000?\\nIt is 100.\\nHow many meters in a kilometer?\\nThere are 1000 meters in a kilometer. 100 centimeter in a\\nmeter. 1,000 meters in a km. 100cm in a meter.\\nThe square root 100 is?\\n-10 +10\\nSqrt x equals 7 sqrt 2 find the value of x?\\nSqrt(x) = 7 Sqrt(2) x = 72 x 2 x = 49 x 2 x = 98\\nSqrt 100 simplified?\\nSqrt(100) = 10 is its simplest form.\\nWhat is the sqr root of 100?\\n10 or -10 because 10*10=100\\nand (-10)(-10)=100\\nSqr\\'s and cubs?\\nSqrt(100) = 10 and 102 = 100Sqrt(1000) = 31.6 and 31.62 = 10001Sqrt(24) = 4.9 and 4.92 = 24If you are unsure, just check them out by calculator.The \"cube\" of a number x means x multiplied by itself, twice: x x x or x cubed x3\\nHow do you find the square root of 100?\\n10\\nWhat is the square root of 100 in radical form?\\nThe square root of 100 is 10.That is already in radical form.\\nHow many m in 100 km?\\nThere are a thousand (1,000) metres in one (1) kilometre: 1 kilometre = 1,000 metres.There are a thousand (1,000) metres in one (1) kilometre: 1 kilometre = 1,000 metres.There are a thousand (1,000) metres in one (1) kilometre: 1 kilometre = 1,000 metres.There are a thousand (1,000) metres in one (1) kilometre: 1 kilometre = 1,000 metres.\\nWhat is 1 sqrt called?\\nThe square root of 1 is 1.\\nthe square root of 1 is 1, because 1&sup2; = 1.\\nWhat is 2 sqrt 10 x 2 sqrt 10?\\n40\\nCan you help me with math i am having problems with sqrt 102 divided by sqrt 45?\\nNo.\\nWhat is Sqrt 100?\\nThe square root of 100 is 10. :)\\nWhat does sqrt of 100 squared mean?\\n100\\nDoes sqrt 9 over 5 times 10 over sqrt 3 equal 3sqrt 30 over 5?\\nno.\\nWhat is the sqrt of 23?\\n2 sqrt 5 8 1/3 and 13 17/30\\nWhat is 5 sqrt 10 x 5 sqrt 10?\\n25 sqrt 100\\nWhat is the exact sqrt of 2 and what is the sqrt of 2?\\nsqrt2 and 1.414\\nWhat is the radicand if sqrt 101 is the radical and sqrt 101 is written in simple radical form?\\n101\\nWhat is the answer to sqrt 100?\\n10\\nWhat is a square root of 100?\\nThe square root of 100 is 10.\\n10 squared equals 100. The square root of 100 is 10.\\nSqrt 54?\\nThe square root of 54'"
            ]
          },
          "execution_count": 32,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "text = \"What is the sqrt of 101?\"\n",
        "\n",
        "from vllm import SamplingParams\n",
        "sampling_params = SamplingParams(\n",
        "    temperature = 1.0,\n",
        "    top_k = 50,\n",
        "    max_tokens = 1024,\n",
        ")\n",
        "output = model.fast_generate(\n",
        "    [text],\n",
        "    sampling_params = sampling_params,\n",
        "    lora_request = None,\n",
        ")[0].outputs[0].text\n",
        "\n",
        "output"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "Colxz9TAVMsi"
      },
      "source": [
        "And now with the LoRA we just trained with GRPO - we first save the LoRA first!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "AL-BcuB1VLIv"
      },
      "outputs": [],
      "source": [
        "model.save_lora(\"grpo_saved_lora\")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "a4LMOBl8boGX"
      },
      "source": [
        "Verify LoRA is actually trained!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "4SfdI-ERbpiw"
      },
      "outputs": [],
      "source": [
        "from safetensors import safe_open\n",
        "\n",
        "tensors = {}\n",
        "with safe_open(\"grpo_saved_lora/adapter_model.safetensors\", framework = \"pt\") as f:\n",
        "    # Verify both A and B are non zero\n",
        "    for key in f.keys():\n",
        "        tensor = f.get_tensor(key)\n",
        "        n_zeros = (tensor == 0).sum() / tensor.numel()\n",
        "        assert(n_zeros.item() != tensor.numel())"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "CwpbwnDBVRLg"
      },
      "source": [
        "Now we load the LoRA and test:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 225,
          "referenced_widgets": [
            "4da56ce2fac243808bfba15d02cdffe9",
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            "65f6ba1b6cfc4b2891984d32d7902618",
            "3271a4acd52e465aa6370045e40bff20",
            "e73145b3b4b04c19a7ad3061936153d7",
            "8b0c1ff7a53541a7b0a1a99bd2485b9c",
            "635e5142d133454cb85ffd26991ca8c8",
            "631095864293470e8205edef4e93da30",
            "d3b545cee66c4ea1accbd506811b18a2",
            "7d2acd645b8e4d0e841d9ca8eda1fbf4"
          ]
        },
        "id": "zf_OY5WMVOxF",
        "outputId": "b05f52c1-44e3-469a-b037-9ee544620495"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "4da56ce2fac243808bfba15d02cdffe9",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Processed prompts:   0%|          | 0/1 [00:00<?, ?it/s, est. speed input: 0.00 toks/s, output: 0.00 toks/s]"
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        {
          "data": {
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              "type": "string"
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            "text/plain": [
              "\"Okay, so I need to find the square root of 101. Hmm, let me think. I remember that square roots of numbers between perfect squares are irrational, meaning they can't be expressed as simple fractions. Let's see, the perfect squares around 101 are 100 and 121. Since 10² is 100 and 11² is 121, the square root of 101 must be between 10 and 11. \\n\\nWait, but how close is it to 10 or 11? Maybe I can estimate it a bit more. Let me try multiplying 10.05 by itself. Calculating (10.05)² would be 10.05 * 10.05. Let's break that down: 10 * 10 = 100, 10*0.05=0.5, 0.05*10=0.5, and 0.05*0.05=0.0025. Adding those up: 100 + 0.5 + 0.5 + 0.0025 = 101.0025. That's a bit more than 101. So maybe 10.05 is too high. \\n\\nLet me try 10.04... Let's compute (10.04)². That would be 10.04 * 10.04. Breaking it down: 100 + 40*0.04=1.6 + 0.04*0.04=0.0016. So 100 + 1.6 + 0.0016 = 101.6016. Still a bit high. So maybe 10.03. Let's try (10.03)²: 100 + 30*0.03=0.9 + 0.03*0.03=0.0009. Adding gives 100 + 0.9 + 0.0009 = 100.9009. That's less than 101. So the square root has to be between 10.03 and 10.04. \\n\\nLet me try 10.035. Then (10.035)²: 100 + 35*0.03=1.05 + 0.035*0.035. Wait, 0.035*0.035 is 0.001225. So total: 100 + 1.05 + 0.001225 = 101.051225. That's slightly more than 101. So now it's between 10.03 and 10.035. \\n\\nLet me try 10.032. Then (10.032)^2: 100 + 64*0.03=1.92 + 0.032*0.032=0.001024. So total: 100 + 1.92 + 0.001024 = 101.921024. Wait, that's higher. So 10.032 is too high. So the square root is between 10.03 and 10.032. And we need more precision. \\n\\nHmm, but calculators are available nowadays, right? But since I'm doing this manually, I can estimate. So the square root of 101 is approximately 10.049875. But since the problem likely expects a simplified radical form, which is √101, since it's not a perfect square. But if a decimal is preferred, then closer to 10.04987.\\nTo find the square root of 101, we note that it lies between the perfect squares 100 and 121, whose square roots are 10 and 11, respectively. Therefore, \\\\(\\\\sqrt{101}\\\\) must be between 10 and 11.\\n\\nTo refine the approximation, we can use the method of successive approximation. Let's start by testing values around the midpoint between 10 and 11.\\n\\n1. Calculate \\\\(10.06^2\\\\):\\n   \\\\[\\n   10.06 \\\\times 10.06 = 101.2036\\n   \\\\]\\n   This is slightly higher than 101.\\n\\n2. Calculate \\\\(10.05^2\\\\):\\n   \\\\[\\n   10.05 \\\\times 10.05 = 101.0025\\n   \\\\]\\n   This is slightly higher than 101 but closer.\\n\\n3. Calculate \\\\(10.04^2\\\\):\\n   \\\\[\\n   10.04 \\\\times 10.04 = 100.8016\\n   \\\\]\\n   This is slightly lower than 101.\\n\\nSince \\\\(10.05^2 = 101.0025\\\\) is slightly above 101 and \\\\(10.04^2 = 100.8016\\\\) is slightly below 101, the square root of 101 must be between 10.04 and 10.05. To get a more precise value, we can try a value between these two, such as 10.049.\\n\\n4. Calculate \\\\(10.049^2\\\\):\\n   \\\\[\\n   10.049 \\\\times 10.049 = 100.980401\\n   \\\\]\\n   This is very close but still slightly below 101.\\n\\n5. Calculate \\\\(10.0498^2\\\\):\\n   \\\\[\\n   10.0498 \\\\times 10.0498 = 100.99804004\\n   \\\\]\\n   This is still a bit below 101.\\n\\n6. Calculate \\\\(10.04987^2\\\\):\\n   \\\\[\\n   10.04987 \\\\times 10.04987 = 101.00000009\\n   \\\\]\\n   This is very close to 101.\\n\\nThus, the square root of 101 is approximately \\\\(10.0499875\\\\).\\n\\nHowever, since the problem does not specify the level of precision required, the most precise value we can provide is:\\n\\n\\\\[\\n\\\\boxed{10.049875}\\n\\\\]<end_working_out><SOLUTION>10.049875</SOLUTION>\""
            ]
          },
          "execution_count": 35,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "messages = [\n",
        "    {\"role\": \"system\", \"content\": system_prompt},\n",
        "    {\"role\": \"user\",   \"content\": \"What is the sqrt of 101?\"},\n",
        "]\n",
        "\n",
        "text = tokenizer.apply_chat_template(\n",
        "    messages,\n",
        "    add_generation_prompt = True, # Must add for generation\n",
        "    tokenize = False,\n",
        ")\n",
        "from vllm import SamplingParams\n",
        "sampling_params = SamplingParams(\n",
        "    temperature = 1.0,\n",
        "    top_k = 50,\n",
        "    max_tokens = 2048,\n",
        ")\n",
        "output = model.fast_generate(\n",
        "    text,\n",
        "    sampling_params = sampling_params,\n",
        "    lora_request = model.load_lora(\"grpo_saved_lora\"),\n",
        ")[0].outputs[0].text\n",
        "\n",
        "output"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "6aDgFfhFYIAS"
      },
      "source": [
        "Our reasoning model is much better - it's not always correct, since we only trained it for an hour or so - it'll be better if we extend the sequence length and train for longer!"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "-NUEmHFSYNTp"
      },
      "source": [
        "<a name=\"Save\"></a>\n",
        "### Saving to float16 for VLLM\n",
        "\n",
        "We also support saving to `float16` directly. Select `merged_16bit` for float16 or `merged_4bit` for int4. We also allow `lora` adapters as a fallback. Use `push_to_hub_merged` to upload to your Hugging Face account! You can go to https://huggingface.co/settings/tokens for your personal tokens."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "NjXGTkp7YNtB"
      },
      "outputs": [],
      "source": [
        "# Merge to 16bit\n",
        "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"merged_16bit\",)\n",
        "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"merged_16bit\", token = \"\")\n",
        "\n",
        "# Merge to 4bit\n",
        "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"merged_4bit\",)\n",
        "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"merged_4bit\", token = \"\")\n",
        "\n",
        "# Just LoRA adapters\n",
        "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"lora\",)\n",
        "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"lora\", token = \"\")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "52WMb3k_YPt8"
      },
      "source": [
        "### GGUF / llama.cpp Conversion\n",
        "To save to `GGUF` / `llama.cpp`, we support it natively now! We clone `llama.cpp` and we default save it to `q8_0`. We allow all methods like `q4_k_m`. Use `save_pretrained_gguf` for local saving and `push_to_hub_gguf` for uploading to HF.\n",
        "\n",
        "Some supported quant methods (full list on our [Wiki page](https://github.com/unslothai/unsloth/wiki#gguf-quantization-options)):\n",
        "* `q8_0` - Fast conversion. High resource use, but generally acceptable.\n",
        "* `q4_k_m` - Recommended. Uses Q6_K for half of the attention.wv and feed_forward.w2 tensors, else Q4_K.\n",
        "* `q5_k_m` - Recommended. Uses Q6_K for half of the attention.wv and feed_forward.w2 tensors, else Q5_K.\n",
        "\n",
        "[**NEW**] To finetune and auto export to Ollama, try our [Ollama notebook](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3_(8B)-Ollama.ipynb)"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {
        "id": "QyEjW-WuYQIm"
      },
      "outputs": [],
      "source": [
        "# Save to 8bit Q8_0\n",
        "if False: model.save_pretrained_gguf(\"model\", tokenizer,)\n",
        "# Remember to go to https://huggingface.co/settings/tokens for a token!\n",
        "# And change hf to your username!\n",
        "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, token = \"\")\n",
        "\n",
        "# Save to 16bit GGUF\n",
        "if False: model.save_pretrained_gguf(\"model\", tokenizer, quantization_method = \"f16\")\n",
        "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, quantization_method = \"f16\", token = \"\")\n",
        "\n",
        "# Save to q4_k_m GGUF\n",
        "if False: model.save_pretrained_gguf(\"model\", tokenizer, quantization_method = \"q4_k_m\")\n",
        "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, quantization_method = \"q4_k_m\", token = \"\")\n",
        "\n",
        "# Save to multiple GGUF options - much faster if you want multiple!\n",
        "if False:\n",
        "    model.push_to_hub_gguf(\n",
        "        \"hf/model\", # Change hf to your username!\n",
        "        tokenizer,\n",
        "        quantization_method = [\"q4_k_m\", \"q8_0\", \"q5_k_m\",],\n",
        "        token = \"\",\n",
        "    )"
      ]
    },
    {
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        "Now, use the `model-unsloth.gguf` file or `model-unsloth-Q4_K_M.gguf` file in llama.cpp or a UI based system like Jan or Open WebUI. You can install Jan [here](https://github.com/janhq/jan) and Open WebUI [here](https://github.com/open-webui/open-webui)\n",
        "\n",
        "And we're done! If you have any questions on Unsloth, we have a [Discord](https://discord.gg/unsloth) channel! If you find any bugs or want to keep updated with the latest LLM stuff, or need help, join projects etc, feel free to join our Discord!\n",
        "\n",
        "Some other links:\n",
        "1. Train your own reasoning model - Llama GRPO notebook [Free Colab](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3.1_(8B)-GRPO.ipynb)\n",
        "2. Saving finetunes to Ollama. [Free notebook](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3_(8B)-Ollama.ipynb)\n",
        "3. Llama 3.2 Vision finetuning - Radiography use case. [Free Colab](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3.2_(11B)-Vision.ipynb)\n",
        "6. See notebooks for DPO, ORPO, Continued pretraining, conversational finetuning and more on our [documentation](https://docs.unsloth.ai/get-started/unsloth-notebooks)!\n",
        "\n",
        "<div class=\"align-center\">\n",
        "  <a href=\"https://unsloth.ai\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/unsloth%20new%20logo.png\" width=\"115\"></a>\n",
        "  <a href=\"https://discord.gg/unsloth\"><img src=\"https://github.com/unslothai/unsloth/raw/main/images/Discord.png\" width=\"145\"></a>\n",
        "  <a href=\"https://docs.unsloth.ai/\"><img src=\"https://github.com/unslothai/unsloth/blob/main/images/documentation%20green%20button.png?raw=true\" width=\"125\"></a>\n",
        "\n",
        "  Join Discord if you need help + ⭐️ <i>Star us on <a href=\"https://github.com/unslothai/unsloth\">Github</a> </i> ⭐️\n",
        "</div>\n"
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