Difference between revisions of "2017 AMC 10B Problems/Problem 14"

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==Problem==
 
==Problem==
An integer <math>N</math> is selected at random in the range <math>1\leq N \leq 2020</math> . What is the probablilty that the remainder when <math>N^16</math> is divided by <math>5</math> is <math>1</math>?
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An integer <math>N</math> is selected at random in the range <math>1\leq N \leq 2020</math> . What is the probablilty that the remainder when <math>N^{16}</math> is divided by <math>5</math> is <math>1</math>?
 
==Solution==
 
==Solution==
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By Fermat's Little Theorem, <math>N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}</math> when N is relatively prime to 5. However, this happens with probability <math>\boxed{\textbf{(D) } \frac 45}</math>.
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:29, 16 February 2017

Problem

An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probablilty that the remainder when $N^{16}$ is divided by $5$ is $1$?

Solution

By Fermat's Little Theorem, $N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}$ when N is relatively prime to 5. However, this happens with probability $\boxed{\textbf{(D) } \frac 45}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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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