elementary number theory - Is there a way to figure out the $n$ such that $n!$ ends with exactly $k$ zeroes?

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?