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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