As a running example, consider the case when both "u" and "v" are even. The 6 possible pairs are: (6, 8), (6, 10), (6, 12), (8, 10), (8, 12), and (10, 12). Since the sums "u" + "v" and "v" + "u*" are even, the sums "a" + "b" + "c" and "d" + "e" + "f" should be even as well. The only way that the sum of three integers will result in an even number is when 1) two of them are odd and one is even, or 2) when all three are even. The fact that the two corner cells are even means that we have only 2 even numbers at our disposal. Thus, the second statement is not compatible with this fact. Hence, it must be the case that the first statement is true: two of the three numbers should be odd, while one be even.