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 1

Notice that, for $n\ge 2$ , $2015^n\equiv 15^n$ 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

In a very similar fashion, we find that $2015^{2016} \equiv 15^{2016} \pmod{100}$ , which equals $225^{1008}$ . Next, since every power (greater than $0$ ) of every number ending in $25$ will end in $25$ (which can easily be verified), we get $225^{1008} \equiv 25 \pmod{100}$ . (In this way, we don't have to worry about the exponent very much.) Finally, $2017 \equiv 17 \pmod{100}$ , and thus $2015^{2016}-2017 \equiv 25-17 \equiv 08 \pmod{100}$ , as above.

Video Solution

https://youtu.be/GVPF6CcCugU

~savannahsolver

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 1

Solution 2

Video Solution

See Also