# 2008 AMC 12A Problems/Problem 15

The following problem is from both the 2008 AMC 12A #15 and 2008 AMC 10A #24, so both problems redirect to this page.

## Problem

Let $k={2008}^{2}+{2}^{2008}$. What is the units digit of $k^2+2^k$? $\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

## Solution $k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$.

So, $k^2 \equiv 0 \pmod{10}$. Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$.

Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$. So the units digit is $6 \Rightarrow \boxed{D}$.

### Note

Another way to get $k \equiv 0 \pmod{10}$ is to find the cycles of the digits.

For $2008^2$, we only have to care about the last digit $8$ since $8^2$ itself is a two digit and we want the last digit. The last digit of $2008^2$ is obviously $4.$

For $2^{2008}$, note that the last digit cycles thru the pattern ${2, 4, 8, 6}$. (You can easily do this by simply calculating the first powers of $2$.)

Since $2008$ is a multiple of $4$, the last digit of $2^{2008}$ is evidently $6.$

Continue as follows.

~mathboy282

## Solution 2 (Video solution)

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 