22. For each positive integer $n$, define $A_{n}=\frac{20^{n}+11^{n}}{n!}$, where $n!=1 \times 2 \times \cdots \times n$. Find the value of $n$ that maximizes $A_{n}$.
22. Answer: 19 .
$$
\frac{A_{n+1}}{A_{n}}=\frac{1}{n+1} \frac{20^{n+1}+11^{n+1}}{20^{n}+11^{n}}-\frac{20+11 \cdot\left(\frac{11}{20}\right)^{n}}{(n+1)\left(1+\left(\frac{11}{20}\right)^{n}\right)}
$$

Then $A_{n+1}10+\frac{9}{1+\left(\frac{11}{20}\right)^{n}}$; and $A_{n+1}>A_{n}$ if $nA_{n+1}$.
If $10 \leq n \leq 18$, then $n \leq 10+8<10+\frac{9}{1+\left(\frac{11}{20}\right)^{n}}$; if $n<10$, then $n<10+\frac{9}{1+\left(\frac{11}{20}\right)^{n}}$. Hence, $n \leq 18$ implies $A_{n}<A_{n+1}$.