2004 AMC 8 Problems/Problem 19

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

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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