# Counting trailing zeros for factorials [closed]

This question is not homework, just asked out of curiosity. I wondered how many zeroes could be found at the end of $1990!$ . I computed something that seemed to work and found out 439. So I computed it in Python, and it returned 494 zeroes, so I'm 55 short.

My reasoning went like this : I get 2 zeroes by series of 10 consecutive numbers. The one that can be divided by 10 gives one zero, and I get another one from the multiplication of the one ending with 2 and the one ending with 5. So 199*2 = 398. But I also have 2 zeroes by hundreds, one comming from the one divisible by 100, and one from XX20 * XX50. 19*2 = 38 more. Then same for thousands. I only get 3 zero more, from 1000, 200*500 and 1200*1500.

$398 + 38 + 3 = 439$

So here is my question:

How can I compute function $t0(n)$ where $t0(n)$ returns the number of trailing $0$ in $n! = factorial(n)$ ?

(Assume $factorial(n)$ is written in base 10, but even better for any base !)

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## closed as too localized by Pete L. Clark, Reid Barton, Scott Morrison♦Feb 24 '10 at 20:02

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This question is better for a site like: www.artofproblemsolving.com – Gerald Edgar Feb 24 '10 at 15:15

Take any integer $x$, and let $t,f$ represent the highest integers such that $2^t | x$ and $5^f | x$. Then the number of trailing zeros in the base 10 representation of $x$ is $z := \min\{t,f\}$. (One way to see this is to note it must be at least z since you have $(2*5)^z | x$, so you can write $x = 10^z * y$ where not both $2,5|y$, and so $y$ can't have any trailing zeros.)
Going back to factorials, $n!$ will always have $t > f$, so to count the zeros, you just have to count the fives. The expression you want is $$f = \sum_{i=1}^\infty \lfloor n / 5^i\rfloor.$$ But maybe there's a cleaner form.