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2017 AMC 10B Problems/Problem 14

Revision as of 01:19, 26 December 2017 by Wannabecharmander (talk | contribs)

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 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. 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 case of $n(\text{mod }5)$


If $n \equiv 0(\text{mod }5)$, then $n^{16} \equiv 0^{16} \equiv 0 (\text{mod }5)$


If $n \equiv 1(\text{mod }5)$, then $n^{16} \equiv 1^{16} \equiv 1 (\text{mod }5)$


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


If $n \equiv 3(\text{mod }5)$, then $n^{16} \equiv (n^4)^4 \equiv 81^4 \equiv (1)^4 \equiv 1 (\text{mod }5)$


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


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


2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions

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