$11 \cdot 34$ Determine the smallest natural number $n$, such that $n!$ ends with exactly 1987 zeros.
(28th International Mathematical Olympiad Candidate Question 1987)
【Solution】Let $S(n)$ be the sum of the digits of $n$ in base $p$. We first prove that the exponent of the prime $p$ in $n!$ is equal to
$$
\frac{n-S(n)}{p-1}
$$

For this, let the base $p$ representation of $n$ be
$$
\begin{aligned}
n & =\left(a_{k} a_{k-1} \cdots a_{1} a_{0}\right)_{p} \\
& =a_{k} p^{k}+a_{k-1} p^{k-1}+\cdots+a_{1} p+a_{0} . \\
S(n) & =a_{k}+a_{k-1}+\cdots+a_{1}+a_{0} .
\end{aligned}
$$

Then the exponent of the prime $p$ in $n!$ is
$$
\begin{aligned}
& {\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\cdots+\left[\frac{n}{p^{k}}\right] } \\
= & \left(a_{k} p^{k-1}+a_{k-1} p^{k-2}+\cdots+a_{1}\right)+\left(a_{k} p^{k-2}+a_{k-1} p^{k-3}+\cdots+a_{2}\right) \\
& +\cdots+\left(a_{k} p+a_{k-1}\right)+a_{k} \\
= & a_{k} \frac{1-p^{k}}{1-p}+a_{k-1} \frac{1-p^{k-1}}{1-p}+\cdots+a_{1} \\
= & \frac{\left(a_{k} p^{k}+a_{k-1} p^{k-1}+\cdots+a_{1} p+a_{0}\right)-\left(a_{k}+a_{k-1}+\cdots+a_{1}+a_{0}\right)}{p-1} \\
= & \frac{n-S(n)}{p-1} .
\end{aligned}
$$

The key to solving this problem is to calculate the exponent of the prime factor 5 in $n!$, i.e., the value of $n$ for which the exponent of 5 in $n!$ is exactly 1987.
Since
$$
\begin{array}{l}
1987 \cdot 4=7948, \\
7948=(223243)_{5}, \\
S(7948)=16, \\
\frac{7948-16}{5-1}=1983 .
\end{array}
$$

Since 7950 is divisible by $5^{2}$, and 7955, 7960 are only divisible by 5.
Thus, the exponent of 5 in 7960! is exactly 1987.
That is,
$$
n=7960 .
$$