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

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(Undo revision 28070 by Latenightworker13 (Talk))
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==Solution==
 
==Solution==
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<math>k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^8 \equiv 4+6 \equiv 0 \pmod{10}</math>.  
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<math>k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}</math>.  
  
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So, <math>k^2 \equiv 0 \pmod{10}</math>. Since <math>k = 2008^2+2^{2008}</math>  is a multiple of four and the units digit of powers of two repeat in cycles of four, <math>2^k \equiv 2^8 \equiv 6 \pmod{10}</math>.  
+
So, <math>k^2 \equiv 0 \pmod{10}</math>. Since <math>k = 2008^2+2^{2008}</math>  is a multiple of four and the units digit of powers of two repeat in cycles of four, <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>.
 
Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow D</math>.

Revision as of 23:02, 3 October 2008

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 • 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
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