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August 16th, 2018, 04:14 AM   #1
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The number of a number in the prime factorization of a factorial

How do we find, for example, how many '5's there are in the prime factorization of n! ? I've read that it's floor(n/5), but why is that?
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Last edited by skipjack; August 16th, 2018 at 01:26 PM.
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August 16th, 2018, 05:12 AM   #2
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Because 5 must appear at least once in every fifth number from 1 to n.

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August 16th, 2018, 07:49 AM   #3
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So it would be at least floor(n/5), not necessarily exactly floor(n/5).
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August 16th, 2018, 12:14 PM   #4
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The answer is $\sum_{k=1}^{\infty}\lfloor \frac{n}{5^k}\rfloor$. Note that the sum is finite since terms where $5^k \gt n$ are all 0.
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