3 Let $n$ be a natural number, $f(n)$ be the sum of the digits of $n^{2}+1$ (in decimal), $f_{1}(n)=f(n), f_{k+1}(n)$ $=f\left(f_{k}(n)\right)$, then the value of $f_{100}(1990)$ is $\qquad$.
311.
$$
\begin{array}{c}
f(1990)=20, f 2(1990)=f(20)=5, \\
f 3(1990)=f(5)=8, f 4(1990)=f(8)=11 \\
f_{5}(1990)=f(11)=5=f_{2}(1990)
\end{array}
$$

Therefore
$$
f_{100}(1990)=f_{4}(1990)=11 .
$$