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    "# [Pandigital Prime](https://projecteuler.net/problem=41)\n",
    "\n",
    "There's only $1! + 2! + 3! + \\cdots + 9! = 409113$ $n$-digit pandigital numbers. This is a small enough number to brute force, but we can easily optimize even further by applying a [divisibility rule](https://en.wikipedia.org/wiki/Divisibility_rule).\n",
    "\n",
    "If the digits of a number sum to a multiple of 3, that number is divisible by 3. Since:\n",
    "* the digits of every 5-digit pandigital number will sum to 15\n",
    "* the digits of every 6-digit pandigital number will sum to 21\n",
    "* the digits of every 8-digit pandigital number will sum to 36\n",
    "* the digits of every 9-digit pandigital number will sum to 45\n",
    "\n",
    "all of these numbers will be divisible by 3, and therefore not be prime. Consequently, to find the largest $n$-digit pandigital prime, we only need to check 4-digit and 7-digit pandigital numbers."
   ]
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    "from itertools import permutations\n",
    "\n",
    "pandigitals = set()\n",
    "for n in (4, 7):\n",
    "    for permutation in permutations(range(1, n + 1)):\n",
    "        k = sum(10^i * d for (i, d) in enumerate(reversed(permutation)))\n",
    "        pandigitals.add(k)"
   ]
  },
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   "source": [
    "Now just sort largest-to-smallest and find the first prime."
   ]
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   "source": [
    "for p in reversed(sorted(pandigitals)):\n",
    "    if is_prime(p):\n",
    "        break\n",
    "\n",
    "p"
   ]
  },
  {
   "cell_type": "markdown",
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   "source": [
    "## Relevant sequences\n",
    "* Pandigital numbers: [A352991](https://oeis.org/A352991)\n",
    "* Pandigital primes: [A216444](https://oeis.org/A216444)\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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