add problem 69
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"cells": [
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{
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"cell_type": "markdown",
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"id": "d4c97c0c",
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"metadata": {},
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"source": [
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"# [Totient Maximum](https://projecteuler.net/problem=69)\n",
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"\n",
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"This problem can be solved by hand.\n",
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"\n",
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"The [totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function) can be computed as\n",
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"$$\\phi(n) = n\\prod_{p|n} \\left(1 - \\frac{1}{p}\\right)$$\n",
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"Therefore\n",
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"$$\\frac{n}{\\phi(n)} = \\prod_{p|n}\\left(\\frac{p}{p-1}\\right)$$\n",
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"This implies that $n/\\phi(n)$ increases as the number of distinct prime factors of $n$ increases, so to maximize this quotient, we want $n$ to have as many distinct prime factors as possible.\n",
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"\n",
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"Furthermore, if $p < q$ for primes $p$ and $q$, then\n",
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"$$\\frac{p}{p-1} > \\frac{q}{q-1}$$\n",
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"Put simply, this means that smaller prime factors will lead to a larger $n/\\phi(n)$. These two facts suggest we should try numbers that are the product of the first several primes, which are called [primorials](https://en.wikipedia.org/wiki/Primorial). Our answer then is the largest primorial less than 1000000:\n",
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"$$2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17 = 510510$$\n",
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"\n",
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"Alternatively, you can just ask SageMath - its implementation of $\\phi(n)$ is fast enough for this problem. However, you might run into difficulties writing your own implementation that's this fast."
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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": "da203e05",
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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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"510510"
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]
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},
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"execution_count": 1,
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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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"qs = {n: n/euler_phi(n) for n in range(1, 1000001)}\n",
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"max(qs, key=qs.get)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "5d99957e",
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"metadata": {},
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"source": [
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"## Relevant sequences\n",
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"* Totient: [A000010](https://oeis.org/A000010)\n",
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"* Primorial numbers: [A002110](https://oeis.org/A002110)"
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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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