$33 \cdot 6$ If $n$ is a natural number, then the last digit of $n^{9999}-n^{5555}$
(A) is always 0.
(B) is sometimes 0, sometimes not 0.
(C) is the same as the last digit of $n$.
(D) cannot be determined.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] Let $\langle n\rangle$ denote the last digit of $n$, then
$$
\begin{array}{l}
\langle a-b\rangle=\left\{\begin{array}{ll}
\langle a\rangle-\langle b\rangle, & \text { if }\langle a\rangle>\langle b\rangle ; \\
10+\langle a\rangle-\langle b\rangle, & \text { if }\langle a\rangle<\langle b\rangle .
\end{array}\right. \\
\text { Also, }\left\langle a^{4 k+r}\right\rangle=\left\langle a^{r}\right\rangle, \\
\end{array}
$$

Thus, from $9999=4 \cdot 1111+5555$,
we have
$$
\left\langle n^{9999}\right\rangle=\left\langle n^{5555}\right\rangle,
$$

which means the last digit of the number is 0.
Therefore, the answer is $(A)$.