131 lines
3.3 KiB
Plaintext
131 lines
3.3 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "eba5f5fb",
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"metadata": {},
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"source": [
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"# [Digit Factorials](https://projecteuler.net/problem=34)\n",
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"\n",
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"Numbers like 145 are called [factorions](https://en.wikipedia.org/wiki/Factorion).\n",
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"\n",
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"We can use similar logic as [problem 30](https://projecteuler.net/problem=30) to produce an upper bound for how many numbers to brute force. Let $f(n)$ be the sum of the factorials of the digits of $n$, and consider the largest seven digit number, 9999999. $f(9999999) = 7 \\times 9! = 2540160$. If we replaced any digits in 9999999, they would have to be smaller than 9, so the sum of the factorials of the digits would be smaller than 2540160. Therefore, for any $n \\leq 9999999$, $f(n) \\leq 2540160$, so for $n$ to *equal* $f(n)$, $n$ must be less than or equal to 2540160, as well.\n",
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"\n",
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"We can write a simple recursive function for calculating the sum of the factorials of the digits:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "06bc38a2",
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"metadata": {},
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"outputs": [],
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"source": [
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"from functools import cache\n",
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"from math import factorial\n",
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"\n",
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"@cache\n",
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"def sfd(n):\n",
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" q = n // 10\n",
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" if q == 0:\n",
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" return factorial(n)\n",
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" \n",
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" return factorial(n % 10) + sfd(q)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "c6c47e17",
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"metadata": {},
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"source": [
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"Now we just iterate over all the numbers less than our upper bound."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"id": "b4642720",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"{145, 40585}"
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]
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},
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"execution_count": 2,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"factorions = {n for n in range(3, 2540161) if sfd(n) == n}\n",
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"factorions"
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]
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},
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{
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"cell_type": "markdown",
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"id": "1257c102",
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"metadata": {},
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"source": [
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"Interestingly, there's only two factorions!"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"id": "caccb3bd",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"40730"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"sum(factorions)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "b25666f7",
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"metadata": {},
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"source": [
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"## Relevant sequences\n",
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"* Factorions: [A014080](https://oeis.org/A014080)\n",
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"\n",
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"#### Copyright (C) 2025 filifa\n",
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"\n",
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"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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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "SageMath 9.5",
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"language": "sage",
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"name": "sagemath"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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