2018 AMC 8 Problems/Problem 21

Problem 21

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution

Looking at the values, we notice that $11-7=4$ , $9-5=4$ and $6-2=4$ . This means we are looking for a value that is four less than a mulitple of 11, 9 and 6. The least common multiple of these number is $11*3^2*2=198$ , so the numbers that fulfill this can be written as $198k-4$ , where k is a positive integer. This value is only a three digit integer when $k$ is $1, 2, 3, 4$ or $5$ , which output $194, 392, 590, 788,$ and $986$ respectively. Thus we have 5 values, so our answer is $\boxed{E}$

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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2018 AMC 8 Problems/Problem 21

Problem 21

Solution