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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]

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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 31 Jan 2014, 07:21, edited 2 times in total.
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Re: Properties of Numbers- Any short cuts? [#permalink]

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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

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Problem Solving Question [#permalink]

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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]

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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]

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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]

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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: Integers [#permalink]

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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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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 15 Apr 2014, 01:42
72 x 1 = 72

72 x 2 = 144

\(144 = 12^2\)

Largest possible which can divide 12 is 12

Answer = B
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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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Given: 72k=n^2, where k is an integer >=1(as n is positive).

K cannot be = 1 since n is an integer and 72 is not perfect square.
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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 31 Dec 2015, 01:03
I think this is a high-quality question and I agree with explanation.

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 16 Mar 2016, 01:34
asyahamed 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


Excellent Question
Here N must have 2 and 3 as its primes so N=6*p for some p
now n^2/72 =integer
hence n has the least value of 12
so B is the answer.
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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 16 Mar 2016, 01:50
asyahamed 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

The first square number, which is divisible by 72 is 144. It implies n is 12

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 20 Mar 2017, 00:39
asyahamed 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 = 3^2 x 2^3
since n^2 has a factor 2^3 we can say that 2^4 will also be a factor of this and 3^2 will be the factor of this

therefore n^2 has a definite factor, which is 144

n will be having a definite factor of 12.

Option B

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 28 Mar 2017, 05:07
Bunuel wrote:
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:
http://gmatclub.com/forum/division-fact ... ml#p666722

Hope it's helps.



I am little (a lot) confused with the term 'largest integer'. Does it mean that x may have more prime factors other than 2 and 3 but the largest value using these prime factors is 12?
Even 24 can be written using these two prime factors. In that case k=2^3
Can you please explain this term and what happens if it would have been the ' smallest integer'?
Thanks

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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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New post 28 Mar 2017, 05:31
Since n^2 is divisible by 72 ,
72 has prime factors ( 2*2*2*3*3)
Since 72 divides n^2 so n must have the prime factors of √72
That is n should contain( 2*2*3)
Which is 12
So the largest positive integer that divides n will be B 12


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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

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asyahamed 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


We are given that n^2/72 = integer, or n^2/[(2^3)(3^2)] = integer.

However, since n^2 is a perfect square, we need to make 72, or (2^3)(3^2), a perfect square. Since all perfect squares consist of unique primes, each raised to an even exponent, the smallest perfect square that divides into n^2 is (2^4)(3^2) = 144.

Since n^2/144 = integer, n/12 = integer, and thus the largest positive integer that must divide n is 12.

Answer: B
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Re: If n is a positive integer and n^2 is divisible by 72, then   [#permalink] 30 Mar 2017, 16:37
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