My Math Forum The number of a number in the prime factorization of a factorial

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 August 16th, 2018, 04:14 AM #1 Newbie   Joined: Aug 2018 From: TÃ¼rkiye Posts: 1 Thanks: 0 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? Thanks a lot. Last edited by skipjack; August 16th, 2018 at 01:26 PM.
 August 16th, 2018, 05:12 AM #2 Member   Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 Because 5 must appear at least once in every fifth number from 1 to n. Sent from my HUAWEI SCL-L04 using Tapatalk Thanks from CSandMathsGuy
 August 16th, 2018, 07:49 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,474 Thanks: 2039 So it would be at least floor(n/5), not necessarily exactly floor(n/5).
 August 16th, 2018, 12:14 PM #4 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 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. Thanks from Sebastian Garth

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