2001 AMC 8 Problems/Problem 18

Problem

Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?

$\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}$

Solution

This is equivalent to asking for the probability that at least one of the numbers is a multiple of $5$ , since if one of the numbers is a multiple of $5$ , then the product with it and another integer is also a multiple of $5$ , and if a number is a multiple of $5$ , then since $5$ is prime, one of the factors must also have a factor of $5$ , and $5$ is the only multiple of $5$ on a die, so one of the numbers rolled must be a $5$ . To find the probability of rolling at least one $5$ , we can find the probability of not rolling a $5$ and subtract that from $1$ , since you either roll a $5$ or not roll a $5$ . The probability of not rolling a $5$ on either dice is $(\frac{5}{6})(\frac{5}{6})=\frac{25}{36}$ . Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $5$ , is $1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}}$

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2001 AMC 8 Problems/Problem 18

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