add problem 63
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@ -0,0 +1,49 @@
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "829a3994",
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"metadata": {},
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"source": [
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"# [Powerful Digit Counts](https://projecteuler.net/problem=63)\n",
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"\n",
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"This problem can be solved with a calculator.\n",
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"\n",
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"For $x^n$ to have $n$ digits, the following inequality must hold:\n",
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"$$10^{n-1} \\leq x^n < 10^n$$\n",
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"If we solve for $x$, we get\n",
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"$$10^\\frac{n-1}{n} \\leq x < 10$$\n",
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"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",
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"$$\\lfloor 10 - 10^\\frac{n-1}{n} \\rfloor$$\n",
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"At $n=22$, the lower bound is greater than 9, so we only need to consider values between $n=1$ and $n=21$.\n",
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"\n",
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"Evaluating this expression for $n=1,2,3,\\ldots,21$ and summing, we get\n",
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"$$9 + 6 + 5 + 4 + 3 + 3 + 2 + 2 + 2 + 2 + 1 \\times 11 = 49$$\n",
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"\n",
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"## Relevant sequences\n",
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"* Numbers with $k$ digits that are also $k$th powers: [A132722](https://oeis.org/A132722)"
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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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