Difference between revisions of "2008 AMC 12A Problems/Problem 15"

Revision as of 23:01, 19 February 2008

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

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

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

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 $6 \Rightarrow D$ .

2008 AMC 12A (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

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 So, <math>k^2 \equiv 0 \pmod{10}</math>. Since <math>k \equiv 2008^2+2^{2008} \equiv 0 \pmod{4}</math>, <math>2^k \equiv 2^4 \equiv 6 \pmod{10}</math>.
 Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow D</math>.
+==See Also==
+{{AMC12 box|year=2008|num-b=14|num-a=16|ab=A}}