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by eyelikecheese » Sun Jan 23, 2011 5:41 pm
Just by using the numbers in the answer choice.

Say X is 12

12^2=144, which is divisible by 12.

12^3 is also divisible by 12 and 12^2

Although I initially thought it was 6^3

Anyone else?
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by Night reader » Sun Jan 23, 2011 5:54 pm
eyelikecheese wrote:Just by using the numbers in the answer choice.

Say X is 12

12^2=144, which is divisible by 12.

12^3 is also divisible by 12 and 12^2

Although I initially thought it was 6^3

Anyone else?
so you have answer 12^2?
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by eyelikecheese » Sun Jan 23, 2011 6:41 pm
Yes. I assume the other logical answer would be 6^3. This is correct in that it fits with the conditions of the stem. BUT it is not the largest integer that can divide p^3
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by anshumishra » Sun Jan 23, 2011 6:45 pm
Night reader 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
p^2 = 12 * K = 2^2 * 3^1 * K

Since p is a +ve integer, hence p^2 must have at least one more power of 3, otherwise the square root of p^2 wouldn't be a +ve integer.

So, p^2 = 2^2 * 3^2 * K' = 6^2 * K' => p = 6* k"

Hence, p^3 = 6^3 * k'''
D
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by Night reader » Sun Jan 23, 2011 6:48 pm
eyelikecheese wrote:Yes. I assume the other logical answer would be 6^3. This is correct in that it fits with the conditions of the stem. BUT it is not the largest integer that can divide p^3
you mentioned p^3 :) and the number must be indeed raised to the cubic power
so with 144 you can't break down this to p^3
whereas 6^2=36 is divisible by 12, p^2=6^2, it could be 24 too only if 24 can be brought down to p^2
so p^2=36 --> p=6 and p^3=6^3
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by eyelikecheese » Sun Jan 23, 2011 9:30 pm
I initially had 6^3, but I guess I overthought the question because I noticed the 12^2 right under there and figured there was a way to make it happen.

Is the reason 12^2 cannot be the answer is that "12^3 has larger numbers that can be divided into it, whereas in 6^3, the largest number that will divide is 6^3?
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by Night reader » Sun Jan 23, 2011 9:34 pm
eyelikecheese wrote:I initially had 6^3, but I guess I overthought the question because I noticed the 12^2 right under there and figured there was a way to make it happen.

Is the reason 12^2 cannot be the answer is that "12^3 has larger numbers that can be divided into it, whereas in 6^3, the largest number that will divide is 6^3?
yes
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by Geva@EconomistGMAT » Mon Jan 24, 2011 12:16 am
I think the confusing (and slightly un-GMAT) bit with this question is that 6^3 = 196 is actually greater than 12^2=144. GMAT questions with numerical answer will usually present them in ascending order, but here this is not the case. So even if we find that p^3 must be divisible by 12^2 (which is untrue), the fact that p^3 must be divisible by 6^3 makes D the right answer in any case.
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by sanju09 » Mon Jan 24, 2011 12:48 am
Night reader 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

Great question! If p^2 is divisible by 12, then p must have 2 and 3 as its prime factors. Hence, the largest positive integer that must divide p^3 will be p^3 itself; and the smallest p that can be assumed under the limited conditions is 6, and hence the required integer is [spoiler]6^3.


D[/spoiler]
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