If p is a positive integer, is p^2 divisible by 96?

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BTGmoderatorDC wrote:
Tue Jun 02, 2020 7:35 pm
If p is a positive integer, is p^2 divisible by 96?

(1) p is a multiple of 8.
(2) p^2 is a multiple of 12.

OA C

Source: Manhattan Prep
Let's take each statement one by one.

(1) p is a multiple of 8.

Say p = 8m, where m = a positive integer

Thus, p^2 = 64m^2.

Case 1: Say m = 1, then p^2 = 64. We see that p^2 = 64 is not divisible by 96.
Case 2: Say m = 3, then p^2 = 64*9 = 96*6. We see that p^2 = 96*6 is divisible by 96.

No unique answer. Insufficient.

(2) p^2 is a multiple of 12.

Say p^2 = 12n, where n = a positive integer

Case 1: Say n = 1, then p^2 = 12. We see that p^2 = 12 is not divisible by 96.
Case 2: Say n = 8, then p^2 = 96. We see that p^2 = 96 is divisible by 96.

No unique answer. Insufficient.

(1) and (2) together

From (1), we know that p^2 is divisible by 64 and from (2), we know that p^2 is divisible by 12; thus, we conclude that p^2 is divisible by the LCM of 64 & 12 or p^2 is divisible by 192 (LCM of 64 & 12). Since 192 is divisible by 96, the answer is yes. Sufficient

The correct answer: C

Hope this helps!

-Jay
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