2003 AMC 8 Problems/Problem 12

Problem

When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces than can be seen is divisible by $6$ ?

$\textbf{(A)}\ 1/3\qquad\textbf{(B)}\ 1/2\qquad\textbf{(C)}\ 2/3\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 1$

Solution

All the possibilities where $6$ is on any of the five sides is always divisible by six, and $1 \times 2 \times 3 \times 4 \times 5$ is divisible by $6$ since $2 \times 3 = 6$ . So, the answer is $\boxed{\textbf{(E)}\ 1}$ because the outcome is always divisible by $6$ .

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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

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

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2003 AMC 8 Problems/Problem 12

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