Is the positive integer n a multiple of 40?

1) 20 is a factor of n^2

2) (n^3)/128 is an integer.

I like to think of these “is X a multiple of Y?” questions as: “does X contain *ALL* of Y’s *factors*?”

40 = 2 x 2 x 2 x 5, or three 2’s and one 5.

This is a Yes or No question.

If Yes, then “n” will have three 2’s and one 5 as factors.

If No, then “n” will be account for all factors.

We don’t care whether the answer is actually “yes” or “no” (no horse in this race! ), but we just need it to be 100% certain.

1) 20 is a factor of n^2

Okay, so n^2 = n x n, and they are saying that n x n will have all the factors of 20. 20 = 2 x 2 x 5

So basically 2 x 2 x 5 would evenly divide into n x n. But hang on a minute, we have two n’s in that hypothetical numerator, and the factors don’t evenly split up!

This tells us that “n” must have AT LEAST one 2 and one 5 as factors. So, it’s possible the answer is YES, if n also has two more 2’s, but what if n = 10? Then we’d get a NO answer. This is insufficient. Cross off answer choices A and D.

Let’s apply the same logic to Statement 2:

2) n^3/128 is an integer.

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2, since we’re splitting these eight 2’s amongst three n’s, it must be that we’re missing a 2, and that each “n” has AT LEAST three 2’s in it. What we don’t know anything about is whether or not it has a 5. If it does, the answer is YES. If it doesn’t, the answer is NO. This is insufficient; cross off answer choice B.

Combined:

“n” must have one 5 as per Statement 1, and “n” must have three 2’s as per Statement 2. Therefore the answer will always be YES, that n is a multiple of 40. (Also, remember that every number is factor and a multiple of itself. So let’s say n = 40. 40 is a multiple of 40, so that would give us a YES answer.)

The answer is (C).

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