2012 AMC 8 Problems/Problem 15

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?

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

Video Solution

https://youtu.be/rQUwNC0gqdg?t=172

https://www.youtube.com/watch?v=Vfsb4nwvopU ~David

https://youtu.be/hOnw5UtBSqI ~savannahsolver

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. To calculate the least common multiple, we need to find the prime factorization for each of them. $3 = 3^{1}$ , $4 = 2^{2}$ , $5 = 5^{1}$ , and $6 = 2^{1}*{3^{1}}$ . So the least common multiple of the four numbers is $2^{2}*{3^{1}*{5^{1}}} = 60$ , 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 $\boxed{\textbf{(D)}\ 61\text{ and }65}$ .

~NXC

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2012 AMC 8 Problems/Problem 15

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Problem

Video Solution

Solution

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