Brent@GMATPrepNow wrote:If x and y are positive integers, what is the value of x?

1) 3x + 2y = 12
2) |9x² + 30xy + 25y²| = 21²

Answer: D

Source: www.gmatprepnow.com
Difficulty level: 650

Nice question.

Statement 1: Certainly doesn't look like it's sufficient at first glance. But anytime something feels too good to be true on the GMAT, you want to slow down and examine your assumptions. We know that 2y must be EVEN. Because 12 is also EVEN, we know that 3x (and thus x) has be EVEN. Because x is a positive integer, the smallest value that will work here is x = 2, in which case y = 3. If x = 4, y would have to be 0, but y can't be 0, as it must be a positive integer. So statement 1 tells us definitively that x = 2. Sufficient.

Statement 2: The left side of the equation can be factored to produce: |(3x + 5y)²| = 21²
Taking the square root of both sides, we get 3x + 5y = 21 (Notice that we don't have to concern ourselves with -21, or worry about the absolute value, as x and y are both positive integers, and thus 3x + 5y will clearly produce a positive number. Again, the only set of integers that will work is x = 2 and y = 3. Statement 2 also tells us definitively that x = 2, so this is also sufficient. The answer is D

I'm thinking:

S1:

3x + 2y = 12

(3/2)x + y = 6

Since y and 6 are both integers, (3/2)x must be an integer too. That means x is a multiple of 2, so it must be 2 or 4.

If x = 2, then y = 3, and we're fine.

If x = 4, then y = 0, but that's forbidden, since y is positive.

If x > 4, then y < 0, which is also forbidden, so our only solution is x = 2, y = 4.

S2:

|9x² + 30xy + 25y²| = 21²

or

9x² + 30xy + 25y² = ±21²

(3x + 5y)² = ±21²

Since the left side is positive, the right side must be positive:

(3x + 5y)² = 21²

So (3x + 5y) = 21 or (3x + 5y) = -21. But the second case is impossible, since x and y are both positive.

That leaves 3x + 5y = 21. If y = 1, 2, or 4, x is not an integer. If y > 4, then x is negative. We're left with y = 3, forcing x = 2, the only solution.