2017 AMC 8 Problems/Problem 12

Revision as of 21:05, 14 December 2017 by Nukelauncher (talk | contribs) (Solution)

Problem 12

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

$\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124$

Solution

The $\lcm(4,5,6)$ (Error compiling LaTeX. ! Undefined control sequence.) is 60. Since $60+1=61$, and that is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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