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

## 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 2

We can also solve this problem by using patterns. Notice that when the powers of $5$ are listed out, you get $25$, $125$ , $625$. This proves that the last $2$ digits of $2015^{2016}$ have to be $25$. Now, looking at $2017$, we see that only the $17$ matters. Subtracting, we get our answer $08$.

Therefore, our answer is $\textbf{(A)}\ 0$.

## See Also

 2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 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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