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

Revision as of 00:35, 26 April 2008

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

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

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

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

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

2008 AMC 10A (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

@@ Line 1: / Line 1: @@
+{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #15]] and [[2008 AMC 10A Problems/Problem 24|2008 AMC 10A #24]]}}
 ==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{(D)}\ 6 \qquad \textbf{(E)}\ 8</math>
+<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8</math>
 ==Solution==
@@ Line 12: / Line 13: @@
 ==See Also==
-{{AMC12 box|year=2008|num-b=14|num-a=16|ab=A}}
+{{AMC12 box|year=2008|ab=A|num-b=14|num-a=16}}
+{{AMC10 box|year=2008|ab=A|num-b=23|num-a=25}}