$1.1996^{1996}$ The tens digit is ( ).
$\begin{array}{llll}\text { (A) } 1 & \text { (B) } 3 & \text { (C) } 5 & \text { (D) } 9\end{array}$
-1. (D).
Obviously, we only need to find the tens digit of $96^{1096}$. In the following discussion, $m \equiv n(\bmod k)$ means that $m$ and $n$ have the same remainder when divided by $k$ (referred to as “$m$ is congruent to $n$ modulo $k$”). It is easy to see that,
$$
\begin{array}{l}
96^{2} \equiv 16(\bmod 100) . \\
96^{4} \equiv 56(\bmod 100), \\
96^{6} \equiv 96(\bmod 100), \\
\therefore 96^{6+r} \equiv\left(96^{6}\right)^{k} \cdot 96^{r}(\bmod 100) \\
\quad \equiv 96^{4+}(\bmod 100) \\
\therefore 96^{1096}=96^{6 \times 33 z+4} \equiv 96^{336}(\bmod 100) \\
\equiv 96^{6 \times 56}(\bmod 100) \equiv 96^{56}(\bmod 100) \\
\equiv 96^{6 \times 9+2}(\bmod 100) \equiv 96^{11}(\bmod 100) \\
\equiv 96^{6 \times 1+5}(\bmod 100) \equiv 96^{6}(\bmod 100) \\
\equiv 96(\bmod 100) .
\end{array}
$$