Can anyone give me a generalized way to find the number of zeroes trailing at the end of $n!$ ?
2026-03-31 19:16:12.1774984572
Find the number of trailing zeros in $n!$.
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The number of zeroes at the end of $n!$ is
$$f(n) = \large \sum_{i=1}^{\log_5n} \lfloor\frac{n}{5^i}\rfloor$$
In simpler terms, it's basically the largest power of $5$ that divide $n!$.
Explanation: Okay. Well, a number is divisible by $10$ if it is divisible by both $5$ and $2$. The largest power of $10$ that divides it is simply the minimum of the largest power of $5$ that divides it or $2$ the largest power of $2$ that divides it. It is obvious that the power of $2$ dividing $n!$ is larger than the power of $5$ dividing $n!$. So, we find the largest power of $5$ that divides $n!$. Now what you might do is simply divide $n$ by $5$ which is incorrect. You see, there exist numbers like $25$ which are divisible by $5^2$. So, you have to count them again.
Then find $f(n!)$. You are done.