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 $60+1=61$ , and 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, $n-1=lcm(4,5,6)=69$ , so $n=60+1=61$ which is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

Video Solution

https://youtu.be/SwgXyYFIuxM

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

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2017 AMC 8 Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also