2005 AMC 8 Problems/Problem 20

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$ , is a multiple of $12$ . Since $14$ has a factor $2$ , $k$ must have a factor of $6$ . The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$ .

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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2005 AMC 8 Problems/Problem 20

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