$32 \cdot 34$ a six-digit number $\bar{a} 9991 b$ can be divided by 12, how many such six-digit numbers are there?
(A) 4.
(B) 6.
(C) 8.
(D) 12.
(Zhejiang Province Junior High School Mathematics Competition, 1992)
[Solution] Let $k=\overline{a 1991 b}$, given $12 \mid k$, then $3 \mid k$, and $4 \mid k$. For $4 \mid k$, the last two digits $\overline{1 b}$ of $k$ must be divisible by $4$, so $b=2$ or $6$.
For $3 \mid k$, then $k \mid(a+1+9+9+1+b)$, which means $k \mid (a+b+20)$. If $b=2$, then $a$ can be $2, 5, 8$; if $b=6$, then $a$ can be $1, 4, 7$. In total, there are 6 such numbers. Therefore, the answer is $(B)$.