21. Suppose that a function $M(n)$, where $n$ is a positive integer, is defined by
$$
M(n)=\left\{\begin{array}{ll}
n-10 & \text { if } n>100 \\
M(M(n+11)) & \text { if } n \leq 100
\end{array}\right.
$$

How many solutions does the equation $M(n)=91$ have?
21. Answer: 101.
If $n \geq 102$, then $M(n)=n-10 \geq 92$.
$$
M(91)=M(M(102))=M(92)=M(M(103))=M(93)=\cdots=M(101)=91 \text {. }
$$

For each $k=1, \ldots, 10, M(80+k)=M(M(91+k))=M(91)=91$, and thus
$$
\begin{array}{l}
M(70+k)=M(M(81+k))=M(91)=91, \\
M(k)=M(M(11+k))=M(91)=91 . \\
\end{array}
$$

Hence, all integers from 1 to 101 are solutions to $M(n)=91$.