2008 AMC 10A Problem 24

Only last digits matter, so track those.

1. Last digits of powers of $2$. Write out a few:

The last digits go $2, 4, 8, 6$ and then repeat. So when the exponent is a multiple of $4$, the power of $2$ ends in $6$.
2. Last digit of $k$. $2008^2$ ends like $8^2 = 64$, so in $4$. The exponent $2008$ is a multiple of $4$, so $2^{2008}$ ends in $6$. Then $k$ ends like $4 + 6 = 10$, so $k$ ends in $0$.
3. Last digit of $k^2$. A number ending in $0$ squared still ends in $0$.
4. Last digit of $2^k$. Here the last digit of $k$ does not help; we need to know where $k$ falls in the repeating block of $4$. Is $k$ a multiple of $4$?
- $2008$ is a multiple of $4$, so $2008^2$ is too.
- $2^{2008}$ is $2 \times 2 \times \cdots$, certainly a multiple of $4$.

So $k$ is a multiple of $4$, and by step 1, $2^k$ ends in $6$.
5. Add. $k^2 + 2^k$ ends like $0 + 6$, so in $6$.

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

The last digit of a sum or product depends only on the last digits of its parts. For a power, though, the exponent works differently: last digits of powers of $2$ repeat in blocks of $4$, so you need the exponent's remainder when divided by $4$, not its last digit. Keep those two ideas separate and giant numbers become easy.

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 $\boxed{\textbf{(D)}\ 6}$.