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 1

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 $\boxed{\textbf{(B)}\ \frac13}$ .

Solution 2

The number is divisible by 5 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)

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AJHSME/AMC 8 Problems and Solutions

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2009 AMC 8 Problems/Problem 13

Contents

Problem

Solution 1

Solution 2

See Also