$14 \cdot 49 k$ is a natural number, and $\frac{1001 \cdot 1002 \cdot \cdots \cdot 1985 \cdot 1986}{11^{k}}$ is an integer, what is the maximum value of $k$?
(China Beijing High School Grade 1 Mathematics Competition, 1986)
[Solution] The original expression can be transformed into
$$
\begin{aligned}
& \frac{1 \cdot 2 \cdot 3 \cdots 1000 \cdot 1001 \cdot 1002 \cdots 1985 \cdot 1986}{11^{k} \cdot 1000!} \\
= & \frac{1986!}{11^{k} \cdot 1000!} .
\end{aligned}
$$
The highest power of the factor 11 in 1986! is
$$
\begin{aligned}
& {\left[\frac{1986}{11}\right]+\left[\frac{1986}{11^{2}}\right]+\left[\frac{1986}{11^{3}}\right] } \\
= & 180+16+1 \\
= & 197 .
\end{aligned}
$$
The highest power of the factor 11 in 1000! is
$$
\begin{aligned}
& {\left[\frac{1000}{11}\right]+\left[\frac{1000}{11^{2}}\right] } \\
= & 90+8=98 .
\end{aligned}
$$

Therefore, the maximum value of $k$ is $197-98=99$.