Tessellate S.T.E.M.S. (2019) - Computer Science - School - Set 1 - Objective Problem 2

For any natural number $N,$ let $f(N)$ denote the number of trailing zeroes of $N!\text{.}$ Then we can use the known formula $f(N) = \sum_{k=1}^\infty \left\lfloor \frac{N}{5^k} \right\rfloor$ which directly implies $f(N) < \sum_{k=1}^\infty \frac{N}{5^k} = \frac{N}{4}$ and can be used to show $f(N) \geq \frac{N}{4} - \log_5N - 1.$ Then setting $f(N) = 2019$ and solving $N$ yields $8076 < N \leq 8102$

Additionally, for any positive integer $k,$ $f(5k-1) < f(5k) = f(5k+1) = f(5k+2) = f(5k+3) = f(5k+4) < f(5k+5)$ which means we only need to test integers $N$ which are multiples of five, so we only need to test $N \in \{8080, 8085, 8090, 8095\}$

In particular, we can see that $f(8090) = f(8091) = f(8092) = f(8093) = f(8094) = 2019$ and $N < 8090 \implies f(N) < 2019 \\ N > 8094 \implies f(N) > 2019$ so that exactly $\boxed{5}$ natural numbers have factorials with $2019$ trailing zeroes.