5. For what positive integer $k$ does $\frac{20^{k}+18^{k}}{k!}$ achieve its maximum value?
(Hong Kong, China)
5. 19 .

Let $A_{k}=\frac{20^{k}+18^{k}}{k!}$. Then
$A_{k+1}=\frac{20^{k+1}+18^{k+1}}{(k+1)!}$,
$$
\frac{A_{k+1}}{A_{k}}=\frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)} \text {. }
$$

Notice that,
$$
\begin{array}{l}
\frac{A_{k+1}}{A_{k}}>1 \Leftrightarrow \frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}>1 \\
\Leftrightarrow 20^{k+1}+18^{k+1}>(k+1)\left(20^{k}+18^{k}\right) \\
\Leftrightarrow 20^{k}(19-k)+18^{k}(17-k)>0 .
\end{array}
$$

Upon inspection, when $k=1,2, \cdots, 18$, equation (1) holds; when $k=19,20, \cdots$, equation (1) does not hold.
Therefore, when $k=19$, $A_{k}$ reaches its maximum value.