025-1000-digit-fibonacci-number.ipynb

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    "# 1000-digit Fibonacci number\n",
    "## Problem 25 \n",
    "\n",
    "<div class=\"problem_content\" role=\"problem\">\n",
    "<p>The Fibonacci sequence is defined by the recurrence relation:</p>\n",
    "<blockquote>F<sub><i>n</i></sub> = F<sub><i>n</i>&#8722;1</sub> + F<sub><i>n</i>&#8722;2</sub>, where F<sub>1</sub> = 1 and F<sub>2</sub> = 1.</blockquote>\n",
    "<p>Hence the first 12 terms will be:</p>\n",
    "<blockquote>F<sub>1</sub> = 1<br/>\n",
    "F<sub>2</sub> = 1<br/>\n",
    "F<sub>3</sub> = 2<br/>\n",
    "F<sub>4</sub> = 3<br/>\n",
    "F<sub>5</sub> = 5<br/>\n",
    "F<sub>6</sub> = 8<br/>\n",
    "F<sub>7</sub> = 13<br/>\n",
    "F<sub>8</sub> = 21<br/>\n",
    "F<sub>9</sub> = 34<br/>\n",
    "F<sub>10</sub> = 55<br/>\n",
    "F<sub>11</sub> = 89<br/>\n",
    "F<sub>12</sub> = 144</blockquote>\n",
    "<p>The 12th term, F<sub>12</sub>, is the first term to contain three digits.</p>\n",
    "<p>What is the index of the first term in the Fibonacci sequence to contain 1000 digits?</p>\n",
    "</div>"
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    "def fibonacci():\n",
    "    a, b = 1, 1\n",
    "    yield a\n",
    "    yield b\n",
    "    while True:\n",
    "        a, b = b, a + b\n",
    "        yield b"
   ]
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   "source": [
    "from math import log10, ceil\n",
    "\n",
    "def digits(n):\n",
    "    return int(ceil(log10(n)))"
   ]
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   "source": [
    "def solve(count):\n",
    "    for (i, n) in enumerate(fibonacci(), start=1):\n",
    "        if digits(n) == count:\n",
    "            return i"
   ]
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   "source": [
    "assert solve(3) == 12"
   ]
  },
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   "source": [
    "## Solution"
   ]
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       "4782"
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     "execution_count": 5,
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   ],
   "source": [
    "solve(1000)"
   ]
  }
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