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2008 AMC 12A Problems/Problem 15

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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 D.

See Also

2008 AMC 12A (Problems • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
2008 AMC 10A (Problems • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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