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

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

by M7MBA » Fri Mar 09, 2018 9:46 am
If x and y are positive integers, and 4x^2=3y, then which of the following must be a multiple of 9?

I. x^2
II. y^2
III. xy

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

The OA is the option E.

I am really confused here. How can I conclude that the 3 given expressions are multiple of 9? I don't know how to solve this. Experts, I ask for your help. <i class="em em-disappointed"></i>

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by [email protected] » Fri Mar 09, 2018 12:47 pm
Hi M7MBA,

We're told that X and Y are positive INTEGERS, and 4(X^2) = 3Y. We're asked which of the following MUST be a multiple of 9. This question is built around Prime Factorization - and can actually be solved with some 'concept knowledge' and a bit of logic.

To start, we know that the two sides are EQUAL. They can be re-written as....

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

For those two sides to be EQUAL, they must have the same 'pieces.' Notice that there are two 2s on the 'left side'.... that means there must be two 2s on the 'right side' SOMEWHERE. Those two 2s MUST be "in" the Y (so whatever Y ends up being, it must include those two 2s when you prime factor it).

In that same way, the one 3 on the 'right side' must also appear on the 'left side.' However, if we include one 3 inside of an X, two 3s will actually be created (one in each of the Xs) so we would need to include a second 3 on the 'right side'. That would give us:

(2)(2)(3 and possibly other primes)(3 and possibly other primes) = (3)(2 and 2 and 3 and possibly other primes)

Thus, we now know that:
1) X is a multiple of 3
2) Y is some multiple of 12

With that knowledge, we can now assess the Roman Numerals:
I. X^2 --> since X is a multiple of 3, X^2 would be a multiple of 9. Roman Numeral 1 IS true.

II. Y^2 --> since Y is a multiple of 12, it's clearly a multiple of 3. Y^2 would thus be a multiple of 9. Roman Numeral 2 IS true.

III. XY --> since we're multiplying a multiple of 3 and a multiple of 12, we will always end up with a multiple of 9. Roman Numeral 3 IS true.

Final Answer: E

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