# Difference between revisions of "2017 AMC 8 Problems/Problem 12"

## 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

Finding the LCM of $4$, $5$, and $6$; we find it is $60$. Now, $60+1=61$, and that is in the range of $\boxed{\textbf{(D)}\ \text{60 and 79}}.$