Difference between revisions of "2003 AMC 8 Problems/Problem 12"
Glowinglol (talk | contribs) (Created page with "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)- 1 because ...") |
Superstriker (talk | contribs) (→Solution) |
||
| (13 intermediate revisions by 10 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | ==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 <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== | ||
| + | We have six cases: each different case, every one where a different number cannot be seen. The rolls that omit numbers one through five are all something times six: an example would be where the number you cannot see is one, so the product should be 2 x 3 x 4 x 5 x 6, and so product should be divisible by six. The roll that omits six on the other hand is 1 x 2 x 3 x 4 x 5, which has 2 x 3, also equal to six. We can see that all of them have a factor of 6 and therefore are divisible by six, so the solution should be E,1. | ||
| + | {{AMC8 box|year=2003|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:41, 19 February 2024
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
We have six cases: each different case, every one where a different number cannot be seen. The rolls that omit numbers one through five are all something times six: an example would be where the number you cannot see is one, so the product should be 2 x 3 x 4 x 5 x 6, and so product should be divisible by six. The roll that omits six on the other hand is 1 x 2 x 3 x 4 x 5, which has 2 x 3, also equal to six. We can see that all of them have a factor of 6 and therefore are divisible by six, so the solution should be E,1.
| 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. 
