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

(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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{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #15]] and [[2008 AMC 10A Problems/Problem 24|2008 AMC 10A #24]]}}
 
==Problem==
 
==Problem==
 
Let <math>k={2008}^{2}+{2}^{2008}</math>. What is the units digit of <math>k^2+2^k</math>?
 
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>
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<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8</math>
  
 
==Solution==
 
==Solution==
 
<math>k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}</math>.  
 
<math>k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}</math>.  
  
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>.  
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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 2^4 \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>.
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Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow \boxed{D}</math>.
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==See Also==
 +
{{AMC12 box|year=2008|ab=A|num-b=14|num-a=16}}
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{{AMC10 box|year=2008|ab=A|num-b=23|num-a=25}}
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{{MAA Notice}}

Revision as of 16:40, 13 January 2018

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 $6 \Rightarrow \boxed{D}$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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