# Difference between revisions of "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 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

Notice that we can rewrite $N^{16}$ as $(N^{4})^4$. By Fermat's Little Theorem, we know that $N^{(5-1)} \equiv 1 \pmod {5}$ if $N \not \equiv 0 \pmod {5}$. Therefore for all $N \not \equiv 0 \pmod {5}$ we have $N^{16} \equiv (N^{4})^4 \equiv 1^4 \equiv 1 \pmod 5$. Since $1\leq N \leq 2020$, and $2020$ is divisible by $5$, $\frac{1}{5}$ of the possible $N$ are divisible by $5$. Therefore, $N^{16} \equiv 1 \pmod {5}$ with probability $1-\frac{1}{5},$ or $\boxed{\textbf{(D) } \frac 45}$.

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

## Solution 3 (Casework)

We can use modular arithmetic for each residue of $n \pmod 5$

If $n \equiv 0 \pmod 5$, then $n^{16} \equiv 0^{16} \equiv 0 \pmod 5$

If $n \equiv 1 \pmod 5$, then $n^{16} \equiv 1^{16} \equiv 1 \pmod 5$

If $n \equiv 2 \pmod 5$, then $n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5$

If $n \equiv 3 \pmod 5$, then $n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5$

If $n \equiv 4 \pmod 5$, then $n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 5$

In $4$ out of the $5$ cases, the result was $1 \pmod 5$, and since each case occurs equally as $2020 \equiv 0 \pmod 5$, the answer is $\boxed{\textbf{(D) }\frac{4}{5}}$

~ pi_is_3.14

## Video Solution 2

~savannahsolver

 2017 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 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