When three consecutive positive integers are multiplied together, the answer is always
(A) odd
(B) a multiple of 6
(D) a multiple of 4
(E) a multiple of 5
(C) a multiple of 12
According to the problem, we want to find the choice that works no matter what three consecutive positive integers we choose.

Thus, we try $1(2)(3)$ to see if we can eliminate any of the given choices.

Now, $1(2)(3)=6$. 6 is not a multiple of 4,5 or 12 , and is not odd.

Therefore, only "a multiple of 6 " is possible.

(In fact, the product of three consecutive positive integers will always be a multiple of 6 , because at least one of the three integers is even and one of the three must be a multiple of 3.)

ANSWER: (B)

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