In how many zeros does $100!$ end?

To find the number of trialing zeros of $100!$ is equivalent to finding the power of 10 $p_{10}$ of the factor $10^{p_{10}}$ in $100!$ . Since the power of 2 is always greater than the power of 5 in $100!$ or $p_2 > p_5$ , $\implies p_{10} = p_{5}$ . A way to find $p_k$ in $n!$ is given by:

$\begin{aligned} p_k & = \sum_{j=1}^\infty \left \lfloor \frac n{k^j} \right \rfloor \\ \implies p_5 & = \sum_{j=1}^\infty \left \lfloor \frac {100}{5^j} \right \rfloor = \left \lfloor \frac {100}{5} \right \rfloor + \left \lfloor \frac {100}{25} \right \rfloor = 20 + 4 = \boxed{24} \end{aligned}$

Notation: $\lfloor \cdot \rfloor$ denotes the floor function .

Pretty simple. I used the same math to build up a computer program.. See it here