2004 AMC 8 Problems/Problem 19

Problem

A whole number larger than $2$ leaves a remainder of $2$ when divided by each of the numbers $3, 4, 5,$ and $6$ . The smallest such number lies between which two numbers?

$\textbf{(A)}\ 40\ \text{and}\ 49 \qquad \textbf{(B)}\ 60 \text{ and } 79 \qquad \textbf{(C)}\ 100\ \text{and}\ 129 \qquad \textbf{(D)}\ 210\ \text{and}\ 249\qquad \textbf{(E)}\ 320\ \text{and}\ 369$

Solution

The smallest number divisible by $3,4,5,$ and $6$ , or their least common multiple, can be found to be $60$ . When $2$ is added to a multiple of number, its remainder when divided by that number is $2$ . The number we are looking for is therefore $62$ , and between $\boxed{\textbf{(B)}\ 60\ \text{and}\ 79}$ .

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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2004 AMC 8 Problems/Problem 19

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