Given number of trailing zeros in n!, find out the possible values of n.

Write $n$ in base $5$ as $n = a_0 + 5a_1 + 25a_2 + 125a_3 + \cdots$ where $0 \leq a_k \leq 4$. $$\left\lfloor \dfrac{n}5\right\rfloor = a_1 + 5a_2 + 25a_3 + \cdots$$ $$\left\lfloor \dfrac{n}{25}\right\rfloor = a_2 + 5a_3 + \cdots$$ $$\left\lfloor \dfrac{n}{125}\right\rfloor = a_3 + \cdots$$ Hence, $$\left\lfloor \dfrac{n}5\right\rfloor + \left\lfloor \dfrac{n}{25}\right\rfloor + \left\lfloor \dfrac{n}{125}\right\rfloor + \cdots = a_1 + 6 a_2 + 31 a_3 + \cdots$$ Note that the coefficients of the terms from $a_3$ are greater than $13$. Since the sum must give us $13$, we get that $a_k = 0$, for all $k \geq 3$. Hence, we need to find the number of integer solutions to $a_1 + 6a_2 =13$ with $0 \leq a_1,a_2 \leq 4$. The constraint $0 \leq a_1,a_2 \leq 4$ further implies that $a_2$ cannot be $0$ and $1$. Hence, $a_2 = 2$. This gives us $a_1 = 1$.

Hence, $n = a_0 + 5a_1 +25a_2 = a_0 + 5 \times 1 + 25 \times 2 = 55+a_0$ where $a_0 \in \{0,1,2,3,4\}$. Hence, the desired values of $n$ are $$\{55,56,57,58,59\}$$ The same idea in principle will work when the trailing number of zeros is any number not just $13$.

To get a very good estimate, note that the number of trailing $0$'s is $$\left\lfloor \frac{n}{5}\right\rfloor+ \left\lfloor \frac{n}{5^2}\right\rfloor+ \left\lfloor \frac{n}{5^3}\right\rfloor+\cdots.$$ This is less than the infinite sum $$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+\cdots.$$ The infinite geometric series has sum $\frac{n}{4}$. So if we want $z$ trailing zeros, then $n \gt 4z$.

A computation using $4z$ will tell us how much we need to go forward from $4z$. The difference between $\frac{n}{5^k}$ and $\left\lfloor \frac{n}{5^k}\right\rfloor$ is always less than $1$. So the amount we may need to go forward to find the appropriate $n$ (if there are any) is logarithmic in $z$.

Note that the first $n$ that works (if there is one) is a multiple of $5$. And then $n+1$, $n+2$, $n+3$, and $n+4$ are the others.

By that criterion, when $z=13$, our first candidate is $55$, the first multiple of $5$ after $(4)(13)$, and it works. Thereofore Of so do $56$, $57$, $58$, and $59$, and that's all.

Computations are equally straightforward for $z$ up to a few hundred. Calculate the number of trailing $0$'s for the first multiple of $5$ greater than $4z$, and make any minor adjustments necessary.