Difference between revisions of "2003 AMC 8 Problems/Problem 12"
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| − | All the possibilities where 6 is on any of the five sides is always divisible by six, and 1*2*3*4*5 is divisible by 6 since 2*3 is equal to six. So, the answer is (E) | + | ==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 <math>6</math>? | ||
| + | |||
| + | <math> \textbf{(A)}\ 1/3\qquad\textbf{(B)}\ 1/2\qquad\textbf{(C)}\ 2/3\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 1 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | All the possibilities where 6 is on any of the five sides is always divisible by six, and 1*2*3*4*5 is divisible by 6 since 2*3 is equal to six. So, the answer is <math>\textbf{(E)}1</math> because the outcome is always divisible by 6. | ||
{{AMC8 box|year=2003|num-b=11|num-a=13}} | {{AMC8 box|year=2003|num-b=11|num-a=13}} | ||
Revision as of 10:03, 25 November 2011
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 
?
Solution
All the possibilities where 6 is on any of the five sides is always divisible by six, and 1*2*3*4*5 is divisible by 6 since 2*3 is equal to six. So, the answer is 
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 | ||
