10-66 Let $n$ be a natural number, and $f(n)=n^{2}+1$ (in decimal) be the sum of the digits of $n^2 + 1$,
$$
\begin{array}{l}
f^{(1)}(n)=f(n), f^{(2)}(n)=f(f(n)), \cdots, f^{(k+1)}(n)=f\left(f^{(k)}(n)\right), \\
k \geqslant 1, \text { find } f^{(100)}(1990) .
\end{array}
$$
(Shanghai High School Mathematics Competition, 1990)
［Solution］Let $\mu(n)$ denote the sum of the digits of $n$. Then
$$
\begin{array}{l}
f(1990)=\mu\left(1990^{2}+1\right)=\mu(3960101)=20 . \\
f^{(2)}(1990)=f(20)=\mu\left(20^{2}+1\right)=\mu(401)=5, \\
f^{(3)}(1990)=f(5)=\mu\left(5^{2}+1\right)=\mu(26)=8, \\
f^{(4)}(1990)=f(8)=\mu\left(8^{2}+1\right)=\mu(65)=11, \\
f^{(5)}(1990)=f(11)=\mu\left(11^{2}+1\right)=\mu(122)=5 .
\end{array}
$$

Thus, for $k \geqslant 1$,
$$
\begin{array}{l}
f^{(3 k)}(1990)=8, \\
f^{(3 k+1)}(1990)=11, \\
f^{(3 k+2)}(1990)=5 .
\end{array}
$$

Since
$$
100 \equiv 1 \quad(\bmod 3)
$$
we have
$$
f^{(100)}(1990)=11
$$