eulerbooks/notebooks/problem0043.ipynb

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   "source": [
    "# [Sub-string Divisibility](https://projecteuler.net/problem=43)\n",
    "\n",
    "If you *really* wanted to, you could work this out with pen and paper using [divisibility rules](https://en.wikipedia.org/wiki/Divisibility_rule). Alternatively, it's easy to brute force. First we write a function to test for the given property."
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    "def is_substring_divisible(digits):\n",
    "    for idx, divisor in enumerate((2, 3, 5, 7, 11, 13, 17), start=1):\n",
    "        x, y, z = digits[idx:idx+3]\n",
    "        n = 100*x + 10*y + z\n",
    "        if n % divisor != 0:\n",
    "            return False\n",
    "    \n",
    "    return True"
   ]
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    "Then we test every permutation of 0 through 9. There's $10! = 3628800$ such permutations, which may seem like a lot, and you could apply some logic to reduce the search space some (e.g. $d_4$ must be 0, 2, 4, 6, or 8), but the test runs relatively quickly for each permutation as-is."
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    {
     "data": {
      "text/plain": [
       "{1406357289, 1430952867, 1460357289, 4106357289, 4130952867, 4160357289}"
      ]
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   "source": [
    "from itertools import permutations\n",
    "\n",
    "numbers = set()\n",
    "for digits in permutations((0,1,2,3,4,5,6,7,8,9)):\n",
    "    if is_substring_divisible(digits):\n",
    "        n = sum(10^k * digit for (k, digit) in enumerate(reversed(digits)))\n",
    "        numbers.add(n)\n",
    "        \n",
    "numbers"
   ]
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    "Turns out there's only six of these numbers."
   ]
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       "16695334890"
      ]
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     "execution_count": 3,
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   "source": [
    "sum(numbers)"
   ]
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   "source": [
    "## Relevant sequences\n",
    "* Pandigital numbers: [A050278](https://oeis.org/A050278)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
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