Prove directly: If m,n is an integer and 6m-7n-11 is odd then m or n is even

6m - 7n - 11 = 2k + 1, \text{ where } k \in \mathbb{Z}

6m - 7n = 2k + 12

6m = 7n + 2k + 12 = 7n + 2(k + 6)

Since the left-hand side is even (divisible by 2) but the right-hand side is odd, it implies that n must be even in order for 7n to be even and 6m to be even as well.

Hence, if 6m - 7n - 11 is odd, then n must be even.

\boxed{n \text{ is even}}