goodyear2013 wrote:

If x and y are integers, is y divisible by 2?

(1) \(x^y=1\)

(2) x is negative.

the question really asks if y is even.

i spent some extra time thinking = is 0 divisible by 2? yes, 0 is divisible by 2, as there is no remainder.

1. x^y = 1

1^2 = 1 - yes

1^3 = 1 - no

1 alone is insufficient.

2. x < 0 - not sufficient.

1+2 -> if the base is negative, then the only way it can be positive if the power is even.

thus, it can be x^0=1, x^2 =1, x^4 = 1.

so yes, y is even. and since 0 is divisible by 2, then C is the correct answer.