Difference between revisions of "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)

See Also

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

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