Difference between revisions of "2003 AMC 8 Problems/Problem 12"

(Solution)
 
(12 intermediate revisions by 10 users not shown)
Line 1: Line 1:
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 the outcome is always divisible by 6.
+
==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}}
 
{{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 $6$?

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

Solution

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 (ProblemsAnswer KeyResources)
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. AMC logo.png