n = SQRT(72.x)
n^2 == 72.x
n = 6* SQRT( 2*x)
n is postive integet so SQRT(2*x) has to be a number.
x =2 ==> 2x ==> 4
x =8 ==> 2x ==> 16
x =32==> 2x ==> 64
so n will be divislbe by 6*2 , 6*4, 6*8 etc...
Hence answer should be : 48 (E)
Perfect Square Type Question
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scoobydooby
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would go for B
given n^2/(2^3*3^2)=k (an integer)
since perfect squares have their prime factors raised to even powers, the smallest n^2 must be 2^4*3^2
=>smallest n= 2^2*3=12
so the largest positive integer that must divide n is 12.
given n^2/(2^3*3^2)=k (an integer)
since perfect squares have their prime factors raised to even powers, the smallest n^2 must be 2^4*3^2
=>smallest n= 2^2*3=12
so the largest positive integer that must divide n is 12.
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viju9162
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Answer is "B".. my reasoning is as follows:
If n^2 is divisible by 72.. then n^2 is a multiple of 72..
I will search for multiple of 72 which is a perfect square...
72*2 = 144
Therefore,n^2 =144, n = 12... the largest integer that divides n is 12
If n^2 is divisible by 72.. then n^2 is a multiple of 72..
I will search for multiple of 72 which is a perfect square...
72*2 = 144
Therefore,n^2 =144, n = 12... the largest integer that divides n is 12
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viju9162
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ya, i also thought about it..infact, we get lot of values for n ..
72 * 8 = 576 ...where n=24 ..
I got confused as what should be the "n" value?
72 * 8 = 576 ...where n=24 ..
I got confused as what should be the "n" value?
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sunnyjohn
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I think answer can not be B.
its simple and straight forward:
48*48/72 = 32
means n = 48
so highest +ve integer is 48.
why everything is thinking about 12^2 = 144..??
its simple and straight forward:
48*48/72 = 32
means n = 48
so highest +ve integer is 48.
why everything is thinking about 12^2 = 144..??
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Stuart@KaplanGMAT
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The answer is certainly B, 12.EMAN 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 question asks which is the largest positive integer that MUST divide n.
As was shown, n^2 = 144 is the smallest perfect square that's divisible by 72. Therefore, the smallest possible value of n is 12. Accordingly, the biggest positive integer that MUST be a factor of n is also 12.
Could n be 48? Sure. However, the question doesn't ask what's the largest possible integer that could be a factor of n; if that were the question, then there would be no answer, since n=infinity would certainly fit the question's parameters.
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Stuart@KaplanGMAT
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As an aside, here's the algebraic solution:EMAN 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
All perfect squares are composed of pairs of primes. So, let's start by breaking 72 down into primes:
72 = 2*2*2*3*3
If we break those primes into pairs, we get:
(2*2)*(3*3)*2
As we can see, there's a "dangling 2". In order to form the smallest possible perfect square that's a multiple of 72, we need to pair up that 2. So, the smallest possible perfect square that's a multiple of 72 is:
(2*2)*(3*3)*(2*2)
To find the root of that perfect square, we select one of each of our pairs of primes:
2*3*2 = 12
So, the smallest possible value of n is 12; accordingly, the largest positive integer that MUST be a factor of n is also 12.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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