If x and y are positive integers, and 4x^2=3y,

Hi M7MBA,

We're told that X and Y are POSITIVE INTEGERS and 4X^2 = 3Y. We're asked which of the following MUST be a multiple of 9. While this question is 'quirky', it can be solved with a little logic and Prime Factorization.

To start, we should prime factor 4X^2 = 3Y...

(2)(2)(X)(X) = (3)(Y)

Since these two calculations are EQUAL to one another, each 'side' has to include the SAME prime factors. The two 2s on the 'left side' must ALSO be on the 'right side' and the 3 on the 'right side' must ALSO be on the 'left side. However, since there are two Xs, there's no way for a factor to appear just once - it would have to appear TWICE. While there are lots of different possible pairs of values for X and Y, since both variables have to be positive integers, the minimum possible values would be...

X = 3, Y = 12
this would give us....
(2)(2)(3)(3) = (3)(3)(2)(2)

The other possible solutions are all MULTIPLES of those minimum values.

I. X^2... X has to be 3 or a multiple of 3, so X^2 will always be 9 or a multiple of 9.
II. Y^2.... Y has to be 12 or a multiple of 12, so Y^2 will always be 144 or a multiple of 144; thus, it will always be a multiple of 9
III. (X)(Y)... (X)(Y) will always be a multiple of 36, so it will always be a multiple of 9

Final Answer: E

GMAT assassins aren't born, they're made,
Rich