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

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==Solution 1==
 
==Solution 1==
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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>.
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By [url = https://artofproblemsolving.com/wiki/index.php?title=Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url], <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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 14:02, 30 August 2017

Problem

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

$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$

Solution 1

By [url = https://artofproblemsolving.com/wiki/index.php?title=Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url], $N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}$ when N is relatively prime to 5. However, this happens with probability $\boxed{\textbf{(D) } \frac 45}$.

Solution 2

Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ . The pattern for $0$ is $0$, no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$. Doing the same for the rest of the digits, we find that the units digits of $1^{16}$, $2^{16}$ ,$3^{16}$, $4^{16}$ ,$6^{16}$, $7^{16}$ ,$8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$, so $\boxed{\textbf{(D) } \frac 45}$.


2017 AMC 10B (Problems • Answer Key • Resources)
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Problem 15
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