Art of Problem Solving

2016 AMC 10B Problems/Problem 8: Difference between revisions

Xturtlex (talk | contribs)
Xturtlex (talk | contribs)
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We notice that <math>2015^{n}</math> is 25 (mod 100) when n is even and 75 (mod 100) when n is odd. (check for yourself).  Since 2016 is even, <math>2015^{2016}</math> is 25 (mod 100) and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv 08 (mod 100)</math>
We notice that <math>2015^{n}</math> is 25 (mod 100) when n is even and 75 (mod 100) when n is odd. (check for yourself).  Since 2016 is even, <math>2015^{2016}</math> is 25 (mod 100) and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv 08 (mod 100)</math>
So the answer is <math>\textbf{(A)}\ 0 \qquad</math>
So the answer is <math>\textbf{(A)}\ 0 \qquad</math>
solution by Wwang
solution by Wwang

Revision as of 09:49, 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

We notice that $2015^{n}$ is 25 (mod 100) when n is even and 75 (mod 100) when n is odd. (check for yourself). Since 2016 is even, $2015^{2016}$ is 25 (mod 100) and $2015^{2016}-2017 \equiv 25 - 17 \equiv 08 (mod 100)$ So the answer is $\textbf{(A)}\ 0 \qquad$

solution by Wwang