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 $\textbf{(A)}\ 0$ .

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 10 Problems and Solutions

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2016 AMC 10B Problems/Problem 8

Contents

Problem

Solution

Solution 2

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