Two standard dice are rolled. The product of the two numbers rolled is calculated. What is the probability that the ones digit of this product is 0 ?
(A) $\frac{11}{36}$
(B) $\frac{2}{9}$
(C) $\frac{1}{36}$
(D) $\frac{1}{6}$
(E) $\frac{5}{36}$
Solution 1

We begin by completing a table in which the ones digit of each possible product is listed.

For example, when the number on the first die is 3 and the number on the second die is 6 , the entry in the table is 8 since $3 \times 6=18$ and the ones digit of 18 is 8 .

Number on the Second Die

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-005.jpg?height=673&width=1130&top_left_y=1409&top_left_x=281)

Of the 36 possible outcomes in the table above, 6 outcomes have a ones digit that is equal to 0 . Thus, the probability that the ones digit of the product is 0 is $\frac{6}{36}=\frac{1}{6}$.

## Solution 2

Since the ones digit of the product is 0 , then the product is divisible by 5 and is even.

Since the possible numbers in the product are $1,2,3,4,5,6$, then one of the numbers rolled must be 5 .

Since the product is even (and 5 is not), then the other number rolled must be one of the three even numbers, namely $2,4,6$.

Thus, the possible pairs of numbers that can be rolled to give a product whose ones digit is 0 ,
are $(5,2),(5,4),(5,6)$ or $(2,5),(4,5),(6,5)$. (We note that the first number in the ordered pair represents the first number rolled, while the second number in the pair is the second number rolled.)

Since there are 6 possible rolls for each of the two dice, then there are $6 \times 6=36$ possible ordered pairs representing all possible outcomes.

Since 6 of these ordered pairs represent a product whose ones digit is 0 , then the required probability is $\frac{6}{36}=\frac{1}{6}$.

ANSWER: (D)