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

(Undo revision 76753 by Alberty44 (talk))
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\textbf{(E)}\ 8</math>
 
\textbf{(E)}\ 8</math>
  
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==Solution==
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==Solution 1==
  
 
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, 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>.
 +
 +
 +
==Solution 2==
 +
 +
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.
  
 
==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 23:02, 23 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 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 $\textbf{(A)}\ 0$.


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.

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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