If $w, x, y$ and $z$ are consecutive positive integers, in the same order, which of the following expressions must be odd?
I. $w^x+z*(y+2)^2$
II. $w^x+z^y$
III. $2*w^3+3*x^2+5*y+6*z^2$

For II, think about when w is even vs odd. If w is even, w^x is even, and z^y parity depends on z. If w is odd, w^x is odd, so z^y must be even to sum to odd? Actually, try small numbers like 1,2,3,4.

Thanks, I'll try plugging in consecutive integers. I think I see it now.

Actually, check: if y is even, then y+2 is even, so (y+2)^2 is even. If y is odd, y+2 is odd, so (y+2)^2 is odd. And z is one more than y, so z's parity is opposite to y. So z*(y+2)^2 is always even. So I reduces to w^x parity.

Oh right, so then w^x is odd only when w is odd. So I is not always odd. Got it.

Yeah, I think so. For II, if w is even, w^x is even, z is odd, so z^y is odd, sum is odd. If w is odd, w^x is odd, z is even, so z^y is even, sum is odd. So II always odd. III is always odd because 2w^3 is even, 3x^2 parity matches x, 5y parity matches y, 6z^2 even. Since x and y are consecutive, one is even, one odd, so 3x^2+5y is odd+even or even+odd = odd. So III is odd. So E.