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

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\textbf{(E)}\ 8</math>
 
\textbf{(E)}\ 8</math>
  
==Solution==
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==Solution 1==
  
We notice that <math>2015^{n}</math> is 25 (mod 100) when n is even and 75 (mod 100) when n is odd. (check for yourself).  Since 2016 is even, <math>2015^{2016}</math> is 25 (mod 100) and <math>2015^{2016}-2017 \equiv 25 - 17 \equiv 08 (mod 100)</math>
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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 \qquad</math>
 
  
solution by Wwang
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So the answer is <math>\textbf{(A)}\ \boxed{0}</math>.
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==Solution 2==
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In a very similar fashion, we find that <math>2015^{2016} \equiv 15^{2016} \pmod{100}</math>, which equals <math>225^{1008}</math>. Next, since every power (greater than <math>0</math>) of every number ending in <math>25</math> will end in <math>25</math> (which can easily be verified), we get <math>225^{1008} \equiv 25 \pmod{100}</math>. (In this way, we don't have to worry about the exponent very much.) Finally, <math>2017 \equiv 17 \pmod{100}</math>, and thus <math>2015^{2016}-2017 \equiv 25-17 \equiv 08 \pmod{100}</math>, as above.
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==Video Solution==
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https://youtu.be/GVPF6CcCugU
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~savannahsolver
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== Video Solution ==
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 https://youtu.be/zfChnbMGLVQ?t=2683
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~ pi_is_3.14
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==See Also==
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{{AMC10 box|year=2016|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 21:35, 25 January 2021

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 $\textbf{(A)}\ \boxed{0}$.

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

Video Solution

 https://youtu.be/zfChnbMGLVQ?t=2683

~ pi_is_3.14

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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