Difference between revisions of "2012 AMC 8 Problems/Problem 15"

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<math> \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}\text{61 and 65}\qquad\textbf{(E)}\hspace{.05in}\text{66 and 99} </math>
 
<math> \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}\text{61 and 65}\qquad\textbf{(E)}\hspace{.05in}\text{66 and 99} </math>
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==See Also==
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{{AMC8 box|year=2012|num-b=14|num-a=16}}

Revision as of 10:28, 24 November 2012

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

$\textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}\text{61 and 65}\qquad\textbf{(E)}\hspace{.05in}\text{66 and 99}$

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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