If x and y are positive integers, what is the remainder when 5^x is divided by y?

In order to answer this question, we need some information about x and y. For some divisors of y, the value of x wouldn't matter: if y = 5, the remainder is always 0, because any 5^x will be divisible by 5. If y = 2, the remainder will be 1, because 5^x will always be odd. For other values of y, though, we'd need to know the value of x (or some pattern to the values of x) to answer the question.

(1) x is an even integer.
This tells us nothing about y. In the examples listed above, y could be 5, and the remainder could be 0. Or y could be 2, and the remainder = 1. Insufficient.

(2) y = 3.
Will the remainder always be the same when any power of 5 is divided by 3? Test cases:
5^1 = 5. 5/3 = 1 R. 2
5^2 = 25. 25/3 = 8 R. 1
5^3 = 25. 125/3 = 41 R. 2
5^4 = 625. 625/3 = 208 R. 1
We can see a pattern: the remainder alternates between 1 (when x is odd) and 2 (when x is even). Knowing the value of y alone is insufficient to know what the remainder will be.

(1) & (2) together
Based on the pattern we established in statement 2, the remainder when 5^x is divided by 3 will always be 2 if x is even. Sufficient.

The answer is C.