{
  "aliases": [
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  "body": "\n## Statement\n\nLet $p$ be a prime number and let $n$ be a natural number. Let $\\nu_p$ be the [[p-adic valuation|04VCPS30ZE]].\nThen\n$$\\nu_p(n!) = \\sum_{i=1}^{\\infty} \\left\\lfloor \\frac{n}{p^i} \\right\\rfloor.$$\nThe sum is finite, because $\\lfloor n / p^i \\rfloor = 0$ when $p^i \u003e n$.\n\n## Remarks\n\nAn equivalent form: $\\nu_p(n!) = \\dfrac{n - s_p(n)}{p - 1}$, where $s_p(n)$ is the sum of the digits of $n$ in base $p$.\nFor example, $\\nu_2(10!) = 5 + 2 + 1 = 8$, and $\\nu_5(100!) = 20 + 4 = 24$. So $100!$ ends with $24$ zeros.\nThe proof counts, for each $i$, the multiples of $p^i$ among $1, \\dots, n$.\n\n## Explanations\n\n### L1 human\n\nTo count the factors $p$ in $n!$, count the multiples of $p$ up to $n$. Then add the multiples of $p^2$, which give a second factor, and so on.\nFor $10!$ and $p = 3$: there are $3$ multiples of $3$ and $1$ multiple of $9$, so $3^4$ is the largest power of $3$ that divides $10!$.\n",
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    "name": "Legendre's formula",
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  "name": "Legendre's formula",
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