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2008 AMC 12A Problems/Problem 15

Revision as of 22:01, 19 February 2008 by Xantos C. Guin (talk | contribs) (New page: ==Problem== Let <math>k={2008}^{2}+{2}^{2008}</math>. What is the units digit of <math>k^2+2^k</math>? <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(...)

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Problem

Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$ ?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

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

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

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