$30 \cdot 4$ has 99 natural numbers greater than 1, their sum is 300. If 9 of these numbers are each decreased by 2, and the remaining 90 numbers are each increased by 1. Then the product of the resulting 99 numbers must be
(A) odd.
(B) even.
(C) prime.
(D) a perfect square.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] According to the problem, if we subtract 2 from each of the 9 numbers, then a total of 18 is subtracted.
Also, if we add 1 to each of the 90 numbers, then a total of 90 is added.
Thus, the new sum of the 99 numbers is $300 + 90 - 18 = 372$.
Since the sum of these 99 (odd number) new numbers is even, there must be at least one even number among them.
$\therefore \quad$ The product of these 99 new numbers must be even.
Therefore, the answer is $(B)$.