Given $n$, I can find the number of zeroes at the end of the decimal representation of $n!$ by $ \sum_{i=1}^\infty\left\lfloor\frac{n}{5^i}\right\rfloor. $
Is there a way to reverse this? That is, given $k$, is there a way to find out how many $n$ exist such that $n!$ has exactly $k$ zeroes at the end of its decimal representation besides making educated guesses and checking them?