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

(need powers of 15, not 5)
(Solution: n>=2)
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==Solution==
 
==Solution==
  
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>.
+
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>.
  
 
So the answer is <math>\textbf{(A)}\ 0</math>.
 
So the answer is <math>\textbf{(A)}\ 0</math>.

Revision as of 13:44, 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, 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 $\textbf{(A)}\ 0$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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

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