eulerbooks/notebooks/problem0069.ipynb

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    "# [Totient Maximum](https://projecteuler.net/problem=69)\n",
    "\n",
    "This problem can be solved by hand.\n",
    "\n",
    "The [totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function) can be computed as\n",
    "$$\\phi(n) = n\\prod_{p|n} \\left(1 - \\frac{1}{p}\\right)$$\n",
    "Therefore\n",
    "$$\\frac{n}{\\phi(n)} = \\prod_{p|n}\\left(\\frac{p}{p-1}\\right)$$\n",
    "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",
    "\n",
    "Furthermore, if $p < q$ for primes $p$ and $q$, then\n",
    "$$\\frac{p}{p-1} > \\frac{q}{q-1}$$\n",
    "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",
    "$$2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17 = 510510$$\n",
    "\n",
    "Alternatively, you can just ask SageMath - its implementation of $\\phi(n)$ is fast enough to brute force this problem. However, you might run into difficulties writing your own implementation that's this fast."
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       "510510"
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   "source": [
    "max(range(1, 1000001), key=lambda n: n / euler_phi(n))"
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    "## Relevant sequences\n",
    "* Totient: [A000010](https://oeis.org/A000010)\n",
    "* Primorial numbers: [A002110](https://oeis.org/A002110)\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)."
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