OG- Quants review (Green).169

The answer should be 48.
I can prove it too!n square is divisible by 72.
so the largest integer that divides n is 48.
check: 48 square = 2304 that is divisible by 72 and 48 is divisible by 48.

Voila, I just proved Gmat OG wrong!!

Naah, I think the GMAC is playing with the words here.
The addition of the word must changes the whole meaning of the sentence.(Must means necessarily should be divisible).
What I mean over here is that the minimum value that can be placed instead of n square is 144 and that makes n = 12. As the largest possible integer that can divide any possibility of n is 12.
Hence, the answer has to be 12.
Hope, you got the point, revert back if you don't get whats in my head!

An explanation as to how can we get the answer with larger/ bigger digits would be appreciated!

friend, i think u make absolute sense. good catch. thanks again.

General soln:
n*n=72*y
=36*2*y
=6*6*2*2*x ( as y has to be twice of some square, say x)

If you take square root of b/s you'll get,
n=12*z (where z is the square root of x)
So we know for sure that n must be divisible by 12 whatever be the value of z.
Hence the answer has to be 12.

You are simply looking for a multiple of 72 that is a perfect square. 1*72 is not a perfect square so that is out.
try 2* 72 = 144. 144 is a perfect square and your answer is 12.

There are lots more multiples of 72 that are perfect squares, but the one that MUST divide evenly is 12, because anything larger would not be a square root of 144.

b. 12

The general approach: 72 = 2*2*2*3*3 = 2*(2^2)*(3^2). If n^2 square is divisible by 72, n must be divisible by SQRT(2)*2*3. Leaving out SQRT(2) since we're only concerned w/ integers, n must be divisible by 2*3=6. Since the rest of the options are multiples of 6, find the largest one which works with the smallest multiple of 72 that actually is a square of a positive integer. 72*2=144 (lucky us it's only the second one...!) and the largest number in the list that divides it neatly is 12.