2017 AMC 8 Problems/Problem 12: Difference between revisions

Revision as of 13:02, 18 January 2021

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

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 $\boxed{\textbf{(D)}\ \text{60 and 79}}.$

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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	==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?		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?

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2017 AMC 8 Problems/Problem 12: Difference between revisions

Revision as of 13:02, 18 January 2021

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Problem

Solution

Video Solution

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