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

Revision as of 13:22, 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 $2015^n\equiv 15^n$ 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 $\textbf{(A)}\ 0$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 ==Solution==
-Notice that, for <math>n\ge 2</math>, <math>2015^n\equiv 15^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>.
+Notice that <math>2015^n\equiv 15^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</math>.

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

Revision as of 13:22, 21 February 2016

Problem

Solution

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