Count the Zer00000000000000000000000s

The number of trailing zeros of $n!$ is given by the largest $k$ such that $10^k|n$ . But since $10=2\cdot 5$ , $k=\operatorname{min}(v_2(n),v_5(n))$ , where for any prime number p, $v_p(n)=\operatorname{max}(k: p^k|n)$ . Furthermore it is clear that $v_2(n!)\geq v_5(n!)$ for any $n$ , since many more even numbers contribute than multiples of 5. So the answer we are looking for is $v_5(2020!)$ . Using Legendre's formula, we have $v_p(n!)=\frac{n-s_p(n)}{p-1}$ , where $s_p(n)$ is the sum of the digits of $n$ when expressed in base $p$ . Now write $2020=3\cdot 5^4+5^3+4\cdot5=31040_5$ , so that $s_5(2020)=8$ . Finally, our answer is $\frac{2020-8}{5-1}=503$ .

The number of trailing zeros of $n!$ is given by $z(n) = \displaystyle \sum_{k=1}^\infty \left \lfloor \frac n{5^k} \right \rfloor$ , where $\lfloor \cdot \rfloor$ denotes the floor function . For $n=2020$ , we have:

$\begin{aligned} z(2020) & = \left \lfloor \frac {2020}5 \right \rfloor + \left \lfloor \frac {2020}{25} \right \rfloor + \left \lfloor \frac {2020}{125} \right \rfloor + \left \lfloor \frac {2020}{625} \right \rfloor \\ & = 404 + 80 + 16 + 3 \\ & = \boxed {503} \end{aligned}$

Bonus: It is obvious that $2020!$ is divisible by $9$ , hence the sum of all the digits of $2020!$ is also divisible by $9$ .