The product of first twelve positive integers is divisible

Hi All,

Bunuel's solution for this question is spot-on, so I won't rehash any of that math here. There is a variation on this question though that is worth noting: sometimes the issue is that a specific prime does not show up "enough" in the product of the larger number.

Consider the exact same question with one answer changed:

The product of first twelve positive integers is divisible by all of the the following EXCEPT?

(A) 210
(B) 125
(C) 88
(D) 75
(E) 60

In this situation, all of the answer choices include prime factors that are in 12!, but since 125=(5)(5)(5), it's NOT divisible into 12!

The reason: 12! only includes two 5's (in the 5 and the 10).

GMAT assassins aren't born, they're made,
Rich

Can't we use another shot cut here...

as in we know 12! has 10 in it and will certainly having a zero in the end. Rule out (A) and (D). Similarly, C as well since 5 is there. I know we cannot say certainly it will not be divisible by looking into it but POE can be done. Also 11 is there so if we open up the factorial - 12x11x10x9x8 and so on. We see easily 8 being divisible after 11 being divided by 8x11 from denominator.
Same fashion 34, we don't see it getting divisible by any number apart from 17, which is not there.

(A)210
(B)88
(C)75
(D)60
(E)34

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