2017 AMC 8 Problems/Problem 12

Problem

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 1

Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The $\operatorname{LCM}(4,5,6)$ is $60$ . Since $60+1=61$ , that is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

Solution 2

Call the number we want to find $n$ . We can say that $n \equiv 1 \mod 4$ $n \equiv 1 \mod 5$ $n \equiv 1 \mod 6.$ We can also say that $n-1$ is divisible by $4,5$ , and $6.$ Therefore, $n-1=lcm(4,5,6)=60$ , so $n=60+1=61$ which is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

~PEKKA

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/UIrHmHT_jl4

~Education, the Study of Everything

Video Solution

https://youtu.be/SwgXyYFIuxM

https://youtu.be/kNIx_iDhVD4

~savannahsolver

https://www.youtube.com/watch?v=nOSqQjEv0U0 ~David

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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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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