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

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==Solution==
 
==Solution==
To find the answer to this problem, we need to find the least common multiple of <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math> and add <math>2</math> to the result. The least common multiple of the four numbers is <math>60</math>, and by adding <math>2</math>, we find that that such number is <math>62</math>. Now we need to find the only given range that contains <math>62</math>. The only such range is answer <math>{\textbf{(D)}</math>, and so our final answer is <math> \boxed{\textbf{(D)}\ 61\text{ and }65} </math>.
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To find the answer to this problem, we need to find the least common multiple of <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math> and add <math>2</math> to the result. The least common multiple of the four numbers is <math>60</math>, and by adding <math>2</math>, we find that that such number is <math>62</math>. Now we need to find the only given range that contains <math>62</math>. The only such range is answer <math>\textbf{(D)}</math>, and so our final answer is <math> \boxed{\textbf{(D)}\ 61\text{ and }65} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=14|num-a=16}}
 
{{AMC8 box|year=2012|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:07, 18 March 2015

Problem

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

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

Solution

To find the answer to this problem, we need to find the least common multiple of $3$, $4$, $5$, $6$ and add $2$ to the result. The least common multiple of the four numbers is $60$, and by adding $2$, we find that that such number is $62$. Now we need to find the only given range that contains $62$. The only such range is answer $\textbf{(D)}$, and so our final answer is $\boxed{\textbf{(D)}\ 61\text{ and }65}$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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