Difference between revisions of "2016 AMC 10B Problems/Problem 8"

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==Solution==
 
==Solution==
  
Notice that <math>2015^{n}</math> is <math>25 (mod 100)</math> when n is even and <math>75 (mod 100)</math> when n is odd. (Check for yourself).  Since 2016 is even, <math>2015^{2016} \equiv 25 (mod 100)</math> and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv 08 (mod  100)</math>.
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Notice that <math>2015^n</math> is congruent to <math>25\pmod{100}</math> when <math>n</math> is even and <math>75\pmod{100}</math> when <math>n</math> is odd. (Check for yourself).  Since <math>2016</math> is even, <math>2015^{2016} \equiv 25\pmod{100}</math> and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}</math>.
  
So the answer is <math>\textbf{(A)}\ 0 \qquad</math>
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So the answer is <math>\textbf{(A)}\ 0</math>.
  
 
solution by Wwang
 
solution by Wwang

Revision as of 10:37, 21 February 2016

Problem

What is the tens digit of $2015^{2016}-2017?$

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$

Solution

Notice that $2015^n$ is congruent to $25\pmod{100}$ when $n$ is even and $75\pmod{100}$ when $n$ is odd. (Check for yourself). Since $2016$ is even, $2015^{2016} \equiv 25\pmod{100}$ and $2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}$.

So the answer is $\textbf{(A)}\ 0$.

solution by Wwang

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