Difference between revisions of "2003 AMC 8 Problems/Problem 12"
m (→Problem) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==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 | + | 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 that 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> | <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> | ||
Revision as of 19:20, 5 July 2020
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 that can be seen is divisible by 
?
Solution
All the possibilities where is on any of the five sides is always divisible by six, and 
is divisible by since 
. So, the answer is 
because the outcome is always divisible by .
See Also
| 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. 
