{ "cells": [ { "cell_type": "markdown", "id": "829a3994", "metadata": {}, "source": [ "# [Powerful Digit Counts](https://projecteuler.net/problem=63)\n", "\n", "This problem can be solved with a calculator.\n", "\n", "For $x^n$ to have $n$ digits, the following inequality must hold:\n", "$$10^{n-1} \\leq x^n < 10^n$$\n", "If we solve for $x$, we get\n", "$$10^\\frac{n-1}{n} \\leq x < 10$$\n", "In other words, if $10^\\frac{n-1}{n} \\leq x < 10$ for some $x$, then $x^n$ will have $n$ digits. Therefore, the number of $n$ digit numbers that are also $n$th powers is simply the number of integers between $10^\\frac{n-1}{n}$ and 10. This can be computed as\n", "$$\\lfloor 10 - 10^\\frac{n-1}{n} \\rfloor$$\n", "At $n=22$, the lower bound is greater than 9, so we only need to consider values between $n=1$ and $n=21$.\n", "\n", "Evaluating this expression for $n=1,2,3,\\ldots,21$ and summing, we get\n", "$$9 + 6 + 5 + 4 + 3 + 3 + 2 + 2 + 2 + 2 + 1 \\times 11 = 49$$\n", "\n", "## Relevant sequences\n", "* Numbers with $k$ digits that are also $k$th powers: [A132722](https://oeis.org/A132722)\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)." ] } ], "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 }