$32 \cdot 27$ Among all 4-digit numbers, the number of those that can be simultaneously divided by $2, 3, 5, 7$ and 11 is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] Since $2,3,5,7,11$ are pairwise coprime, any number that can be simultaneously divided by them must also be divisible by $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11=2310$.
Among the four-digit numbers, those divisible by 2310 are $2310, 4620, 6930, 9240$. Therefore, the answer is $(D)$.