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

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If n is a positive integer and n^2 is divisible by 72, then [#permalink] New post 18 Feb 2010, 02:34
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Feb 2012, 04:22, edited 1 time in total.
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Re: Properties of Numbers- Any short cuts? [#permalink] New post 18 Feb 2010, 07:27
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If n is a positive integer and n-squared is divisible by 72, then the largest positive integer that must divide n is
A. 6
B. 12
C. 24
D. 36
E. 48

The largest positive integer that must divide n, means for the least value of n which satisfies the given statement in the question. The lowest square of an integer, which is multiple of 72 is 144 --> n^2=144=12^2=72*2 --> n_{min}=12. Largest factor of 12 is 12.

OR:

Given: 72k=n^2, where k is an integer \geq1 (as n is positive).

72k=n^2 --> n=6\sqrt{2k}, as n is an integer \sqrt{2k}, also must be an integer. The lowest value of k, for which \sqrt{2k} is an integer is when k=2 --> \sqrt{2k}=\sqrt{4}=2 --> n=6\sqrt{2k}=6*2=12

Answer: B.

Similar problem:
division-factor-88388.html#p666722

Hope it's helps.
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Problem Solving Question [#permalink] New post 10 May 2010, 20:33
I was hoping to get some clarification on Problem 169 from Quantitative Review 2nd Ed:

Q: 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

n^2 is divisible by 72, but it must also be greater than 72. If n is an integer, then n^2 must be a perfect square. The factorization of 72 is (8)(9), so if it is multiplied by 2, it will be (2)(8)(9) = (16)(9) = 144, a perfect square. So n^2 must be at least 144 or a multiple of 144, which means that n must be 12 or a multiple of 12.

I know that Quantitative Review also has 12 as the answer, but I had a question: Since n must be 12 or a multiple of 12, why is it that 48 isn't a solution since its a multiple of 12 and 48 divides 48 and is also the greatest number amongst the solutions, especially because the question does not state 'largest integer other than n that divides n'? What is the concept that I am not getting?

Please help.
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Re: Problem Solving Question [#permalink] New post 10 May 2010, 21:17
B.

Prime factorization of 72 --> 2^3 * 3^2

n^2 is divisible by 2^3 * 3^2
the largest positive integer that must divide n is:2^2 * 3 --> 12
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Re: Problem Solving Question [#permalink] New post 18 May 2010, 08:47
ok,see
in order to find he largest positive integer that must divide , means for lowest value of n^2 which is 144 , or N comes out to be 12, and now if you devide this by 48 then it would not come out to be an integer. Hence he largest integer must be 12.
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Re: Problem Solving Question [#permalink] New post 22 May 2010, 02:35
abc123def wrote:
I was hoping to get some clarification on Problem 169 from Quantitative Review 2nd Ed:

Q: 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

n^2 is divisible by 72, but it must also be greater than 72. If n is an integer, then n^2 must be a perfect square. The factorization of 72 is (8)(9), so if it is multiplied by 2, it will be (2)(8)(9) = (16)(9) = 144, a perfect square. So n^2 must be at least 144 or a multiple of 144, which means that n must be 12 or a multiple of 12.

I know that Quantitative Review also has 12 as the answer, but I had a question: Since n must be 12 or a multiple of 12, why is it that 48 isn't a solution since its a multiple of 12 and 48 divides 48 and is also the greatest number amongst the solutions, especially because the question does not state 'largest integer other than n that divides n'? What is the concept that I am not getting?

Please help.



This question testing MUST or COULD

In this case n can be 12 or 36 or 48

But if take n =12 which is one of the condition (least possible value of n), and divide it by any integer greater than 12. the resulting number can never be a integer (question is asking the largest possible value that MUST divide n, in all cases). Hence it can only be 12

But if the question asks about the largest possible value which COULD divide n, in that case answer can be 48 (largest value in all answer). though it can be bigger than 48 also.

Hope this helps
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Re: Properties of Numbers- Any short cuts? [#permalink] New post 08 Jun 2010, 07:22
nsp007, can you please explain how you got your last step??
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Re: Integers [#permalink] New post 03 Nov 2010, 01:30
student26 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
In order to find the largest integer that must divide n, since there is no upper bound on n, we should choose the smallest possible value of n.
Given the prime factorisation of 72, it is easy to see, the smallest n^2 divisible by 72 would be n^2=2^4*3^2, hence the smallest choice of n would be n=2^2*3=12
Hence, for all possible n, the smallest value is 12
Hence, for all possible n, 12 always divides n, and is the largest such value to work for all n

Answer : (b)
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number properties question from OG-quant [#permalink] New post 05 Jan 2011, 07:25
I don't really understand the explanation that is given in the OG quant book, would someone please explain?

If n is a positive integer, and n^2 is divisible by 72, then the largest possible integer that must divide n is?

A) 6
B) 12
C) 24
D) 36
E) 48

Thanks!
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Arithmetic Properties-QR-169 [#permalink] New post 18 Feb 2011, 14:38
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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Official guide - quant review - question [#permalink] New post 01 Mar 2011, 13:38
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

6
12
24
36
48
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Re: Official guide - quant review - question [#permalink] New post 01 Mar 2011, 13:42
Re: Official guide - quant review - question   [#permalink] 01 Mar 2011, 13:42
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