Data Sufficiency

sukhman wrote:Is the positive integer p even?
(1) p² + p is even
(2) 4p + 2 is even

Target question: Is positive integer p even?

Statement 1: p² + p is even
This doesn't tell us anything, because p² + p is even for ANY integer p.
We know this because p² + p = p(p+1). Notice that p and p+1 are two consecutive integers, which means one must be odd and the other must be even.
Since the product of an odd and even number is ALWAYS even, statement 1 provides no additional information.
So, it is NOT SUFFICIENT.

Aside: We can also plug in some numbers to show that statement 1 is NOT SUFFICIENT.
case a: If p = 1, then p² + p = 2, which is even. Here p is ODD
case b: If p = 2, then p² + p = 6, which is even. Here p is EVEN

Statement 2: 4p + 2 is even.
This doesn't tell us anything, because 4p + 2 is even for ANY integer p.
We know this because 4p + 2 = 2(2p+1), and if we multiply 2 by any integer, the product is ALWAYS even.
Since statement 2 provides no additional information, it is NOT SUFFICIENT.

Statements 1 and 2 combined
We don't really need to do anything here because both statements failed to provide any new (useful) information.
So, the correct answer must be E

If we want proof, we can see that the following cases both satisfy statements 1 and 2:
case a: p = 1 (since p² + p will be even, and since 4p + 2 will be even Here p is ODD
case b: p = 2, (since p² + p will be even, and since 4p + 2 will be even Here p is ODD

Cheers,
Brent