Perfect Square Type Question

[EMAN]

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

[sunnyjohn]

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)

[scoobydooby]

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.

[viju9162]

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

[mp2437]

Answer is E. 48^2 = 2304, which is divisible by 72. n = 48, so largest divisor is also 48.

Not sure why you are limiting yourself to 12^2 = 144.

[viju9162]

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?

[EMAN]

Thanks for the posts guys. The correct answer is actually B. I will post the detailed explanation later today.

[sunnyjohn]

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..??

[Stuart@KaplanGMAT]

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 answer is certainly B, 12.

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.

[Stuart@KaplanGMAT]

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

As an aside, here's the algebraic solution:

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.