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

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<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>
 
==Solution==
 
All the possibilities where <math>6</math> is on any of the five sides is always divisible by six, and <math>1 \times 2 \times 3 \times 4 \times 5</math> is divisible by <math>6</math> since <math>2 \times 3 = 6</math>. So, the answer is <math>\boxed{\textbf{(E)}\ 1}</math> because the outcome is always divisible by <math>6</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|num-b=11|num-a=13}}
 
{{AMC8 box|year=2003|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:48, 8 October 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 $6$?

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

See Also

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

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