From c92aff27da5f75a7ef21d17b73bfd6a007cabe45 Mon Sep 17 00:00:00 2001
From: filifa <filifa.breath652@8alias.com>
Date: Mon, 14 Apr 2025 22:30:45 -0400
Subject: [PATCH] add problem 30

---
 problem0030.ipynb | 75 +++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 75 insertions(+)
 create mode 100644 problem0030.ipynb

diff --git a/problem0030.ipynb b/problem0030.ipynb
new file mode 100644
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--- /dev/null
+++ b/problem0030.ipynb
@@ -0,0 +1,75 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "cbbd0d3f",
+   "metadata": {},
+   "source": [
+    "# [Digit Fifth Powers](https://projecteuler.net/problem=30)\n",
+    "\n",
+    "These sorts of numbers are called [perfect digital invariants](https://en.wikipedia.org/wiki/Perfect_digital_invariant) (that page talks about the concept for arbitrary bases, but we obviously only need to worry about base 10).\n",
+    "\n",
+    "It's not difficult to implement the PDI function from the page linked above, and to check numbers one by one to see if they are a PDI, but how do we know when to stop searching? Well, let $f(n)$ be the fifth power PDI function, and consider the largest possible six digit number, 999999. The sum of the fifth powers of the digits is $f(999999) = 6 \\times 9^5 = 354294$. If we replaced any digits in 999999, they would have to be smaller than 9, so the power digit sum would be smaller than 354294. Therefore, for any $n \\leq 999999$, $f(n) \\leq 354294$, so for $n$ to *equal* $f(n)$, $n$ must be less than or equal to 354294, as well.\n",
+    "\n",
+    "(There's a formal proof on the page linked above that shows the upper bound can be reduced even further, but I think the logic is harder to follow than what's presented here, and using this bound solves the problem quickly enough.)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "e8ab0179",
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "443839\n"
+     ]
+    }
+   ],
+   "source": [
+    "def F(n, p):\n",
+    "    total = 0\n",
+    "    while n != 0:\n",
+    "        total += (n % 10) ** p\n",
+    "        n //= 10\n",
+    "    \n",
+    "    return total\n",
+    "\n",
+    "\n",
+    "print(sum(n for n in range(2, 354295) if n == F(n, 5)))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f0754e9f",
+   "metadata": {},
+   "source": [
+    "## Relevant sequences\n",
+    "* Perfect digital invariants: [A252648](https://oeis.org/A252648)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "SageMath 9.5",
+   "language": "sage",
+   "name": "sagemath"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.11.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}