# 2009 AMC 8 Problems/Problem 13

## Problem

A three-digit integer contains one of each of the digits $1$, $3$, and $5$. What is the probability that the integer is divisible by $5$? $\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$

## Solution

The three digit numbers are $135,153,351,315,513,531$. The numbers that end in $5$ are divisible are $5$, and the probability of choosing those numbers is $\boxed{\textbf{(B)}\ \frac13}$.

## Alternate Solution

The number is odd if and only if the number ends in $5$ (also $0$, but that case can be ignored, as none of the digits are $0$) If we randomly arrange the three digits, the probability of the last digit being $5$ is $\boxed{\textbf{(B)}\ \frac13}$.

Note: The last sentence is true because there are $3$ randomly-arrangeable numbers)

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