$33 \cdot 3$ less the units digit of $1984^{1984}-1$ is
(A) 3 .
(B) 4 .
(C) 5 .
(D) 6 .
(China Jilin Province Seven Cities and Prefectures Junior High School Mathematics Competition, 1984)
[Solution] Note that if $\langle a\rangle$ represents the unit digit of the natural number $a$, then
$$
\left\langle 1984^{1984}\right\rangle=\left\langle 4^{1984}\right\rangle \text {. }
$$

And $\left\langle 4^{1}\right\rangle=4,\left\langle 4^{2}\right\rangle=6,\left\langle 4^{3}\right\rangle=4,\left\langle 4^{4}\right\rangle=6, \cdots$
That is, when $n$ is odd, $\left\langle 4^{n}\right\rangle=4$; when $n$ is even, $\left\langle 4^{n}\right\rangle=6$.
Thus, $\left\langle 4^{1984}\right\rangle=6$.
Therefore, $\left\langle 1984^{1984}\right\rangle=\left\langle 4^{1984}\right\rangle=6$. That is, the unit digit of $1984^{1984}-1$ is 5. Hence, the answer is $(C)$.