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

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So the answer is <math>\textbf{(A)}\ 0</math>.
 
So the answer is <math>\textbf{(A)}\ 0</math>.
  
solution by Wwang
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==Solution 2==
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We can also solve this problem by using patterns.
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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>.
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Now, looking at  <math>2017</math>, we see that only the <math>17</math> matters. Subtracting,
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we get our answer <math>08</math>.
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Therefore, our answer is
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<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 12:46, 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 (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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