$32 \cdot 33$ A six-digit number $a 1988$ can be divided by 12, the number of such six-digit numbers is
(A) 9.
(B) 12.
(C) 15.
(D) 20.
(China Hubei Province Wuhan City Junior High School Mathematics Competition, 1988)
[Solution] Since $12=3 \cdot 4$, then $a 1988 b$ can be divisible by 3 and 4. Thus,
$$
3 \text { | }(a+1+9+8+8+b) \text {, }
$$

which means $3 \mid (26+a+b)$, or $3 \mid (27+a+b-1)$, hence
$$
3 \mid (a+b-1) \text {. }
$$

Also, $4 \mid \overline{8 b}$, so $b=0, 4$ or 8. This gives us the following table:
\begin{tabular}{c|c|c|c}
\hline$b$ & 0 & 4 & 8 \\
\hline$a$ & 1, or 4, or 7 & 3, or 6, or 9 & 2, or 5, or 8 \\
\hline
\end{tabular}

Thus, there are 9 such six-digit numbers.
Therefore, the answer is $(A)$.