2017 AMC 10B Problems/Problem 14

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 $\boxed{\textbf{(D) } \frac 45}$ .

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. 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 $5$ $\boxed{\textbf{(D) } \frac 45}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2017 AMC 10B Problems/Problem 14

Contents

Problem

Solution

Solution 2

See Also