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Number properties

by sana.noor » Sun May 19, 2013 11:23 pm
If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p^3 is
(A) 2^3
(B) 2^6
(C) 3^3
(D) 6^3
(E) 12^2

OA is D
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by Atekihcan » Mon May 20, 2013 12:13 am
p² must be of the form 12k = (2²)*3*k

As p is an integer, k must be of the form 3n², for some positive integer n.
So, p² = (2²)*3*(3n²) = (2*3*n)²
So, p = (6n)
So, p³ = (6n)³ = Multiple of 6³

Answer : D

Alternatively, we can proceed as follows also...
If p² is divisible by 12, p must contain at least one 2 and one 3.
So, p must be a multiple of 6.
So, p³ must be a multiple of 6³
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by GMATGuruNY » Mon May 20, 2013 3:16 am
sana.noor wrote:If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p^3 is
(A) 2^3
(B) 2^6
(C) 3^3
(D) 6^3
(E) 12^2

OA is D
The correct answer choice MUST divide p³ -- even if p³ is its MINIMUM POSSIBLE VALUE.
Since p is an integer, p² is a perfect square:
1, 4, 9, 16, 25, 36...
Of these values, the least possible multiple of 12 is 36.
If p²=36, then p=6 and p³=6³.
Thus, the minimum possible value for p³ is 6³.
Clearly, 6³ is divisible by answer choice D -- the greatest of the five answer choices.

The correct answer is D.
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