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

(Putting the box in so that people could navigate from this page.)
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So the answer is <math>\textbf{(A)}\ 0</math>.
 
So the answer is <math>\textbf{(A)}\ 0</math>.
  
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solution by Wwang
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==Solution 2==
 +
 
 +
We can also solve this problem by using patterns.
 +
Notice that when the powers of <math>5</math> are listed out, you get <math>25</math>, <math>125</math> , <math>625</math>.
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This proves that the last <math>2</math> digits of <math>2015^{2016}</math> have to be <math>25</math>.
 +
Now, looking at  <math>2017</math>, we see that only the <math>17</math> matters. Subtracting,
 +
we get our answer <math>08</math>.
 +
 
 +
Therefore, our answer is
 +
<math>\textbf{(A)}\ 0</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:46, 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 $2015^n$ 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 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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