messages to break hash modulus 2^n
If k is even


Choose a string m
Choose any block of m other than the last
Flip the most signficant bit of that block.


They collide because you effictively shift all bits of that block one place to the left. If $x \equiv y \mod 2^{n-1}$ then $2x \equiv 2y \mod 2^n$.

Additionally if $k$ is even all strings with their last $n$ blocks equal collide. This is because the previous blocks have a coefficient congruent to 0 mod $2^n$. (They're multiplied by a multiple of $2^n$.)