Trailing Zeros of Factorials

Mar 11, 2016 · Gian Sass

Since I’m bored to hell in programming class, my teacher, who is also my physics teacher, gave me the task to find out the trailing zeros of $100!$ and then for $n!$ . I then began searching for patterns in factorials starting from $1!$ to 10!. Interestingly, $5! = 120$ has one trailing zero and $10! =3628800$ has two trailing zeros. By induction you could determine the trailing zeros of factorials to be expressed by $\lfloor \frac{n}{5} \rfloor$ , where $\lfloor x \rfloor$ is the floor function.

But if you go down the pattern a bit further you find that $25! =15511210043330985984000000$ which has six zeros! The heck? This contradicts our idea. Let’s investigate a bit further.

In a factorial, the amount of fives getting multiplied adds a zero. For example $10! = 1\times2\times3\times4\times5\times6\times7\times8\times9\times10$ . If we take a look we see that there are two 5s hidden here, $5$ and $10=5\times2$ . But $25!$ contains the factor $25$ which is $5^2$ ! So in that case, we find that there are six fives, because $25=5\times5$ .

With this additional information we can determine the number of trailing zeros of $25!$ to be $\lfloor \frac{25}{5}\rfloor +\lfloor \frac{25}{5^2} \rfloor = 6$

But we are not finished yet, there is a pattern here and we should be able to define it for any $n!$ .

Given a number $n$ , the trailing zeros of $n!$ is the sum of $n$ divided by all of its prime factors of 5.
$\big f(n) = \sum\limits_{i=1}^{k} \lfloor \frac{n}{5^i} \rfloor$

where $k$ has to be chosen such that $5^{k+1} > n$ .

For example let’s calculate the trailing zeros of $100!$ . We find $5^3 = 125 > 100$ leaving us with $f(100) = \lfloor \frac{100}{5}\rfloor +\lfloor \frac{100}{25}\rfloor = 24$ .