Difference between revisions of "2005 AMC 8 Problems/Problem 20"

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==Solution==
 
==Solution==
Alice moves <math>5k</math> steps and Bob moves <math>9k</math> steps, where <math>k</math> is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, <math>14k</math>, is a multiple of <math>12</math>. Since <math>14</math> has a factor <math>2</math>, <math>k</math> must have a factor of <math>6</math>. The smallest number of turns that is a multiple of <math>6</math> is <math>\boxed{\textbf{(A)}\ 6}</math>.
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Alice moves <math>5k</math> steps and Bob moves <math>9k</math> steps, where <math>k</math> is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, <math>14k</math>, is a multiple of <math>12</math>. Since this number must be a multiple of <math>12</math>, as stated in the previous sentence, <math>14</math> has a factor <math>2</math>, <math>k</math> must have a factor of <math>6</math>. The smallest number of turns that is a multiple of <math>6</math> is <math>\boxed{\textbf{(A)}\ 6}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=19|num-a=21}}
 
{{AMC8 box|year=2005|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:04, 3 August 2020

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 this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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