2008 AMC 10A Problem 24

Units digit of $k$. $2008^2$ ends in the units digit of $8^2 = 64$, namely $4$. Powers of $2$ end in $2, 4, 8, 6$ repeating with period $4$; since $2008$ is a multiple of $4$, $2^{2008}$ ends in $6$. So $k$ ends in $4 + 6 = 10$, i.e. in $0$.

Units digit of $k^2$. A number ending in $0$ has a square ending in $0$.

Units digit of $2^k$. This depends on $k \bmod 4$, not on its units digit. Both $2008^2$ and $2^{2008}$ are multiples of $4$, so $k \equiv 0 \pmod 4$, and $2^k$ ends in $6$ (the last entry of the cycle $2, 4, 8, 6$).

Combine. $k^2 + 2^k$ ends in $0 + 6 = 6$.

The answer is $\boxed{\textbf{(D)}\ 6}$.

Units digits are arithmetic mod $10$, and for powers the base's cycle length is what matters: $2^n \bmod 10$ has period $4$, so the exponent must be read mod $4$. Keep the two reductions separate: reduce the base information mod $10$ and the exponent information mod $4$. Here $k$ is both a multiple of $10$ and a multiple of $4$, which pins everything down.

• Using the units digit of k (0) to decide the units digit of 2^k; the exponent must be reduced mod 4, not mod 10.
• Taking $2^{2008}$ to end in $8$ or $2$ by miscounting the cycle; $2008 \equiv 0 \pmod 4$ lands on the fourth entry, $6$.