ananthpatri wrote:

If x is a positive integer, is (x+y) even?

(1) K^x*K^y =1

(2) K is not equal to 1

Statement 1: K^x*K^y =1

K^(x+y) = 1, therefore

K = 1, then x+y can odd/even

K > 1, then x+y = 0, hence even

K = -1, then x+y is even

K < -1, then x+y = 0 hence even

Statement 1 is not sufficient.

Statement 2: K is not equal to 1.

No information about x or y. Statement 2 is not sufficient.

Combining, we get K = -1, > 1 or < -1, hence x+y is even

Combining is sufficient.

Answer C.

Thanks,

GyM