[Solution] From $a^{3}+3 a^{2}+2 a=a\left(a^{2}+3 a+2\right)=a(a+1)(a+2)$, we know that $2 \mid\left(a^{3}+3 a^{2}+2 a\right)$, and $31\left(a^{3}+3 a^{2}+2 a\right)$, which means $6!\left(a^{3}+3 a^{2}+2 a\right)$.
Also, 12 Y $1985 \cdot 1986 \cdot 1987$, which means
$$
12 y\left(1985^{3}+3 \cdot 1985^{2}+2 \cdot 1985\right) \text {. }
$$

Therefore, the largest integer that can divide each of the ten numbers is 6. Hence, the answer is $(C)$.
$29 \cdot 19$ has 10 numbers: $1983^{3}+3 \cdot 1983^{2}+2 \cdot 1983, 1984^{3}+3 \cdot 1984^{2}+2 \cdot 1984, \cdots, 1991^{3}+3 \cdot 1991^{2}+2 \cdot 1991, 1992^{3}+3 \cdot 1992^{2}+2 \cdot 1992$, the largest integer that can divide each of these 10 numbers is
(A) 2.
(B) 3.
(C) 6.
(D) 12.
(China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics league, 1992)