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

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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  09 Feb 2017, 17:24
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amitvmane 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, then 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 the larges [#permalink] New post  11 Feb 2017, 09:12
ScottTargetTestPrep

Thank you for your reply!

But I still dont get why the Q says the largest and not the smallest. 48 divides 12, so why not 48?
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  14 Feb 2017, 16:57
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iliavko wrote:
ScottTargetTestPrep

Thank you for your reply!

But I still dont get why the Q says the largest and not the smallest. 48 divides 12, so why not 48?


I think you may have misinterpreted the phrase "integer that must divide n." You interpreted it as “the integer must be divisible by n.” By your interpretation, if 12 is divisible by n, 48 is also divisible by n; this would be correct, had the wording of the question been as you interpreted it.

The phrase "integer that must divide n" really means “n must be divisible by that integer.” So if n is divisible by 12 (which means n/12 = integer), it doesn't mean n is divisible by 48 (i.e., it doesn't mean n/48 will be an integer). For example, if n = 12, n is divisible by 12, but n is not divisible by 48.

And thus, since we determined in the original question that n is a multiple of 12, n could be as small as 12, and the largest integer that must divide into 12 is 12. Does that answer your question?
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  14 Dec 2017, 13:56
ScottTargetTestPrep wrote:
amitvmane 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, then n/12 = integer, and thus the largest positive integer that must divide n is 12.

Answer: B


Hey Scott,

I think we may have covered it buy why can't we make n=72 therefore n^2 = (72)(72) ... why are we trying to use the SMALLEST perfect square .. using n=72 follows the rules set out in the question... ???
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  14 Dec 2017, 20:42
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YYZ wrote:
ScottTargetTestPrep wrote:
amitvmane 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, then n/12 = integer, and thus the largest positive integer that must divide n is 12.

Answer: B


Hey Scott,

I think we may have covered it buy why can't we make n=72 therefore n^2 = (72)(72) ... why are we trying to use the SMALLEST perfect square .. using n=72 follows the rules set out in the question... ???


The question asks to find the largest positive integer that MUST divide n. So, which ALWAYS divides n, if n^2 is divisible by 72. Now, while n COULD be divisible by any integer, for example, by by 48, 72, 1,000,000, ... it MUST be divisible only by factors of 12. Why? Because the least value of n for which n^2 is divisible by 144 is 12.

Hope it's clear.
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  23 May 2019, 12:33
ScottTargetTestPrep Bunuel i have understood why and how 12 is the answer, but why are we assuming that in order to find largest integer that must divide n, we first need to find the least/minimum possible value of n and not the maximum possible value of n ???

if i take maximum possible value of n ie 48, then 48 will be the largest positive integer that will divide n as per the same logic..

PLEASE HELP me understand this !

Look forward to hearing from you


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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  27 May 2019, 18:56
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aditliverpoolfc wrote:
ScottTargetTestPrep Bunuel i have understood why and how 12 is the answer, but why are we assuming that in order to find largest integer that must divide n, we first need to find the least/minimum possible value of n and not the maximum possible value of n ???

if i take maximum possible value of n ie 48, then 48 will be the largest positive integer that will divide n as per the same logic..

PLEASE HELP me understand this !

Look forward to hearing from you


ScottTargetTestPrep


First of all, there is no maximum possible value of n since n can be as large as possible. For example, if a value of n is 24, you can multiply it by any positive integer greater than 1 to make it even larger. Secondly, we are finding the largest number that must divide n,; let’s say that number is d. That is, no matter what the value of n is, d must divide into n. Therefore, we need to find the smallest value of n such that d will divide into it.
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  14 Oct 2020, 23:59
The lowest number which is divisible by '72' and is also a perfect square is 144.

=> \(n^2\) = 144 and hence n = 12.Largest positive integer that divides '12' is '12'.

Answer B
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  05 May 2021, 04:24
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Solution:

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

72k=n^2

=> n=6√2k

n is an integer

=> √2k, also must be an integer

The minimum value of k, for which √2 k is an integer is at k=2

At k=2,

√2k

=√4=2

n=6√(2k)

=>n=6∗2

=12 (option b)

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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  03 Aug 2021, 08:09
BrentGMATPrepNow wrote:
amitvmane 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

---------------ASIDE #1--------------------------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:

If N is a factor by k, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)

---------------ASIDE #2--------------------------------------
IMPORTANT CONCEPT: The prime factorization of a perfect square will have an even number of each prime

For example: 400 is a perfect square.
400 = 2x2x2x2x5x5. Here, we have four 2's and two 5's
This should make sense, because the even number of primes allows us to split the primes into two EQUAL groups to demonstrate that the number is a square.
For example: 400 = 2x2x2x2x5x5 = (2x2x5)(2x2x5) = (2x2x5)²

Likewise, 576 is a perfect square.
576 = 2x2x2x2x2x2x3x3 = (2x2x2x3)(2x2x2x3) = (2x2x2x3)²
--------NOW ONTO THE QUESTION!------------------

Given: n² is divisible by 72 (in other words, there's a 72 hiding in the prime factorization of n²)
So, n² = (2)(2)(2)(3)(3)(?)(?)(?)(?)(?)... [the ?'s represent other possible primes in the prime factorization of n²]
Since we have an ODD number of 2's in the prime factorization, we can be certain that there is at least one more 2 in the prime factorization.
So, we know that n² = (2)(2)(2)(3)(3)(2)(?)(?)(?)(?)
So, while there MIGHT be tons of other values in the above prime factorization, we do know that there MUST BE at least four 2's and two 3's.
Now do some grouping to get: n² = [(2)(2)(3)(?)(?)...][(2)(2)(3)(?)(?)...]
From this we can see that n = (2)(2)(3)(?)(?)...

Question: What is the largest positive integer that must divide n?
(2)(2)(3) = 12.
So, the largest positive integer that must divide n is 12

Cheers,
Brent


I'm confused - why can't you throw an additional "2" to each so that n = 24 or any of the other answer choices. The question asks for the LARGEST positive integer after all.
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If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  09 Sep 2021, 04:21
GK002 wrote:
I'm confused - why can't you throw an additional "2" to each so that n = 24 or any of the other answer choices. The question asks for the LARGEST positive integer after all.


That is a good question and something I feel should be addressed as part of the solution
Quote:
"Why are we taking the smallest number and calling it largest".


I have tried to explain in the video (forward to 1:36 & 3:27) however if there are still questions, please comment. In the video:

01:36 Understanding what "largest" & "must divide" mean
03:27 Why smallest in the series is the "largest that must divide"
04:50 Step-by-Step strategy to solve this problem


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If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  02 Feb 2022, 14:44
Bunuel wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

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 \(k=2\) --> \(\sqrt{2k}=\sqrt{4}=2\) --> \(n=6\sqrt{2k}=6*2=12\)



Why is the lowest value of k for which\(\sqrt{2k}\) is an integer relevant? If \(n=6\sqrt{2*o}\), for some odd number \(o>3\), then n wouldn't be divisible by 12.

This seems to only work because if 2k is a perfect square, it must be divisible by 4 (since all even squares are all divisible by 4)
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  11 Mar 2022, 02:15
amitvmane 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



1. the largest positive integer that must divide n is n (itself)
2. Factorize 72 = 2x6x6 >> notice that there are two six, since the numerator is n^2, I assume that there must be another 2 in it
n^2 = 2x2x6x6, thus n=2x6 = 12
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  29 May 2022, 07:05
Bunuel wrote:
https://gmatclub.com/forum/if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html

Bunuel

This redirects to the same as the current topic

Bunuel wrote:
https://gmatclub.com/forum/if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html

This is the current topic link

We may further add the link https://gmatclub.com/forum/x-is-divisib ... 06841.html
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  04 Jun 2022, 04:03
Bunuel wrote:
SOLUTION

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 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.

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chetan2u

in 2nd solution ,Bunuel has choosen lowest value of k ...why so?? k can be 8(k=8)

Please help me understand logic

thx
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Re: If n is a positive integer and n^2 is divisible by 72, then the larges [#permalink] New post  04 Jun 2022, 05:26
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CEO2021 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

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.


chetan2u

in 2nd solution ,Bunuel has choosen lowest value of k ...why so?? k can be 8(k=8)

Please help me understand logic

thx



Hi,

We are looking at the integer that MUST divide n. So you would take the least possible value of that number as n could be anything.
For example, say n is 36, so n^2 becomes 36*36 or 72*18, so divisible by 72.
But n being 36 is not divisible by \(6\sqrt{2*8}\) or 24.

Thus you take the least value of k, so that number surely divides n.
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