026-reciprocal-cycles.ipynb

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    "# Reciprocal cycles\n",
    "## Problem 26 \n",
    "\n",
    "<div class=\"problem_content\" role=\"problem\">\n",
    "<p>A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:</p>\n",
    "<blockquote>\n",
    "<table><tr><td><sup>1</sup>/<sub>2</sub></td><td>=&#160;</td><td>0.5</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>3</sub></td><td>=&#160;</td><td>0.(3)</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>4</sub></td><td>=&#160;</td><td>0.25</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>5</sub></td><td>=&#160;</td><td>0.2</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>6</sub></td><td>=&#160;</td><td>0.1(6)</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>7</sub></td><td>=&#160;</td><td>0.(142857)</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>8</sub></td><td>=&#160;</td><td>0.125</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>9</sub></td><td>=&#160;</td><td>0.(1)</td>\n",
    "</tr><tr><td><sup>1</sup>/<sub>10</sub></td><td>=&#160;</td><td>0.1</td>\n",
    "</tr></table></blockquote>\n",
    "<p>Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that <sup>1</sup>/<sub>7</sub> has a 6-digit recurring cycle.</p>\n",
    "<p>Find the value of <i>d</i> &lt; 1000 for which <sup>1</sup>/<sub><i>d</i></sub> contains the longest recurring cycle in its decimal fraction part.</p>\n",
    "</div>"
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   "source": [
    "def solve():\n",
    "    # Your code here\n",
    "    \n"
   ]
  },
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   "source": [
    "## Solution"
   ]
  },
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   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "solve()"
   ]
  }
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