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>. | + | 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 10:04, 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?
Solution
Alice moves 
steps and Bob moves 
steps, where 
is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, 
, is a multiple of 
. Since this number must be a multiple of , as stated in the previous sentence, 
has a factor 
, must have a factor of 
. The smallest number of turns that is a multiple of is 
.
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
