2008 AMC 12A Problem 15

The units digit of $2008^2$ is $4.$ Since $2008$ is a multiple of $4,$ the units digit of $2^{2008}$ is $6.$ Thus $k$ has units digit $0,$ and so does $k^2.$ Both $2008^2$ and $2^{2008}$ are multiples of $4,$ so $k$ is a multiple of $4.$ Therefore the units digit of $2^k$ is $6.$ The units digit of $k^2 + 2^k$ is then $0 + 6 = 6.$ Thus, D is the correct answer.