Brent@GMATPrepNow wrote:When 15x is divided by 2y, the quotient is x, and the remainder is x. If x and y are positive integers, which of the following must be true?

A) x is odd
B) x < y
C) 3y < 20x
D) x < 3
E) y < 8

Answer: E

Source: GMAT Prep Now
Difficulty level: 650+

I was just about to lament the absence of new Brent originals.

Let's trot our trusty formula: dividend = quotient*divisor + remainder

When 15x is divided by 2y, the quotient is x, and the remainder is x, so we know:

dividend = 15x
divisor = 2y
quotient = x
remainder = x

Plugging into our formula, we get 15x = x*2y + x
We know x is positive, so let's divide through by x to get: 15 = 2y + 1; 14 = 2y; y = 7.
Well, if y = 7, E will have to be true.

(Notice that the only restriction for x is that it must be less than 2y, as the remainder, by definition, is always less than the divisor. So we know that x < 2*7, or x < 14. Any integers falling between 0 and 14 will work as values for x.)

Let's turn to our standby, Dividend = Quotient*Divisor + Remainder.

With that, 15x = x*2y + x. Since x ≠ 0 (we're told it's positive), we can divide by x: 15 = 2y + 1. From there, y = 7, and we're done.

Ugh, I copied David's solution there, my bad. Let me add an alternative!

When in doubt, try to find answers. Since x is positive, let's start with the minimum x, x = 1. That gives us 15 / 2y = 1, with remainder 1, so y = 7.

That's useful, but unfortunately ALL of the answers are still in play, so let's keep going with x = 2. That gives us 30 / 2y = 2, with remainder 2 ... and again y = 7. Hmmm ....

Since it's looking suspicious, let's try to eliminate all the other answers in one shot by trying a huge x value, x = 100. That gives us 1500 / 2y = 100, with remainder 100. Again y = 7, and we're stone cold sure of it now, the answer is E.