rsarashi wrote:x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

A) x = w

B) x > w

C) x/y is an integer

D) w/z is an integer

E) x/z is an integer

w, x, y and z are all integers.
Test the SMALLEST POSSIBLE CASE.
Let z=1.

Since z=1, it must be possible that w/z and x/z are integers.
Eliminate D and E.

Since z=1, w is the sum of 1 consecutive integer, implying that w can be ANY INTEGER.
Thus, it must be possible that x=w or that x>w.
Eliminate A and B.

The correct answer is C.

It seems to me that the question really tests the ability to either quickly pick smart numbers that rule out four of the answers and/or the ability to quickly whip up an equation that will show the impossibility of one of the answers.

If I take the first approach, some easy sets are:

{1} vs {0,1} (eliminate A)

{1} vs {1,2} (eliminate B)

{1, 2, 3} (eliminate D)

{1, 2, 3, 4} vs {1, 2} (eliminate E)

If I take the second approach, I know that

x = n + (n + 1) + ... + (n + 2z - 1)

x = 2z * n + (1 + 2 + ... + 2z - 1)

x = 2zn + (2z - 1)*2z*(1/2) = 2zn + (2z - 1)*z

Since y = 2z, we know that x/y = (2nz + z * (2z - 1)) / 2z

Simplified, that becomes 2zn/2z + z*(2z - 1)/2z. The first term is an integer, but the second is not, so we'll never get an integer here for any integer z.

It's a great Q, with (at least) two clever and generally relevant paths to a solution - don't sell it short!

One other trick that just occurred to me:

If you're trying to backsolve from the answers, note that C implies E. (If x is divisible by 2z, then x must be divisible by z.)

With that in mind, you don't even need to bother with E!