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# If n is a positive integer and n^2 is divisible by 72, then

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Senior Manager
Joined: 10 Mar 2008
Posts: 342
If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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12 Sep 2008, 15:07
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If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ?

a. 6
b. 12
c. 24
d. 36
e. 48

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Intern
Joined: 05 Sep 2008
Posts: 13
Re: PS: Number properties [#permalink]

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12 Sep 2008, 15:14
IMO b

finding the largest pos integer is basically finding smallest multiple of 72.
in this case is 72x2 which is 144
=12
Senior Manager
Joined: 31 Jul 2008
Posts: 280
Re: PS: Number properties [#permalink]

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12 Sep 2008, 15:16
12 :

the minimum value for n^2 is 72*2 = 144

so whatever might be the case the largest value that MUST divide N has to be 12
Manager
Joined: 22 Jul 2008
Posts: 144
Re: PS: Number properties [#permalink]

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19 Sep 2008, 09:28
The largest value of n could actually be 72 as well but that is not given as an option. Therefore, I would say, 36. However, if I have to choose the integer that MUST divide n, then it has to be 2*2*3 =12
Intern
Joined: 04 Aug 2008
Posts: 11
Re: PS: Number properties [#permalink]

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19 Sep 2008, 10:06
if n^2 is divisible by 72 and if question is to ask largest integer that must divide N would be N itself, if you see in the choice all these digits square are divisible by 72 then largest should be 48 , why 12? Sorry i haven't got it yet. none of the logic.
Retired Moderator
Joined: 05 Jul 2006
Posts: 1741
Re: PS: Number properties [#permalink]

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19 Sep 2008, 16:17
vksunder wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ?

a. 6
b. 12
c. 24
d. 36
e. 48

72 = 2^3 * 3^2 , THUS N^2 = 2^3 * 3^2 * X WHERE X = 2B AND B IS A SQUARE OF AN INTIGER

THUS n MUST BE IN THE FORM = 2^2 * 3 * B
DIVIDE ALL VALUES BY 12 ANS SEE WHAT YEILDS A PERFECT SQUARE ....E
VP
Joined: 21 Jul 2006
Posts: 1490
Re: PS: Number properties [#permalink]

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19 Sep 2008, 18:22
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vksunder wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ?

a. 6
b. 12
c. 24
d. 36
e. 48

$$72 = (2^3)(3^2)$$

We are told that $$n^2$$ is divisible by $$72$$. Because $$n^2$$ is a perfect square, all its prime factors must have even exponents. Because 72 has a prime factor 2 with an odd exponent, which is 3, 72 lacks another 2 for the whole product to be at least a square of an integer. So if you multiply 72*2, you will get the perfect square 144. Therefore $$sqrt(144)$$ = 12, which is n. The largest positive integer that will divide n is itself, so it should be 12.

Answer B
Senior Manager
Joined: 10 Mar 2008
Posts: 342
Re: PS: Number properties [#permalink]

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21 Sep 2008, 10:32
Great explanation Tarek. Thanks! +1

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: PS: Number properties   [#permalink] 21 Sep 2008, 10:32
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# If n is a positive integer and n^2 is divisible by 72, then

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