(A) Suppose 3n is even, this implies that the product of 3n is a multiple of two, let's say 2(3k) where k is a real number. Rearranging that, we have 3(2k), let 2k = n. Therefore, n is even.
(B) Suppose xyz is odd, then let x = 2k, y = 2p + 1 and z = 2q + 1. xyz = 2k(2p + 1)(2q + 1) and this is obviously even since it is a multiple of 2.
