Y62 ** Find the smallest natural number $n$, such that $n!$ ends with exactly 1987 zeros.
Obviously, if $n!$ ends with exactly 1987 zeros, then the number of times 5 appears in the prime factorization of $n!$ is 1987.

It is well known that the number of times 5 appears in $n!$ is $h(n)=\left[\frac{n}{5}\right]+\left[\frac{n}{5^{2}}\right]+\cdots+\left[\frac{n}{5^{k}}\right]$, where $k$ is the largest integer such that $5^{k} \leqslant n$.

Let $n=\sum_{i=0}^{k} a_{i} 5^{i}, a_{i} \in\{0,1,2,3,4\}$, then $h(n)=\sum_{i=1}^{k} a_{i} h\left(5^{i}\right)$, and $h\left(5^{m}\right)=1+$ $5+5^{2}+\cdots+5^{m-1}=\frac{1}{4}\left(5^{m}-1\right)$, so $h(5)=1, h\left(5^{2}\right)=6, h\left(5^{3}\right)=31$, $h\left(5^{4}\right)=156, h\left(5^{5}\right)=781$. Since $h(n)=1987$, and
$$
1987-2 \times 781+2 \times 156+3 \times 31+3 \times 6+2 \times 1 \text {, }
$$

Therefore, $n=2 \times 5^{5}+2 \times 5^{4}+3 \times 5^{3}+3 \times 5^{2}+2 \times 5=7960$. Hence, 7960 is the smallest $n$ sought.