2012 AMC 10A Problems/Problem 16
Problem
Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
Solution 1
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let be the time these runners run in seconds.
Because is a multiple of 500, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of and results in the 5.0 m/s runner to end up on the start line. Every seconds, that fastest runner goes meters. In seconds, he goes meters. Therefore the runners run seconds.
Solution 2
Working backwards from the answers starting with the smallest answer, if they had run seconds, they would have run meters, respectively. The first two runners have a difference of meters, which is not a multiple of (one lap), so they are not in the same place.
If they had run seconds, the runners would have run meters, respectively. The last two runners have a difference of meters, which is not a multiple of .
If they had run seconds, the runners would have run meters, respectively. The distance separating each pair of runners is a multiple of , so the answer is seconds.
Solution 3
Let be the time run in seconds, then the difference in meters run between the three runners is . For them to be at the same location all of them need to be multiples of 500. It is now easy to see that , so .
Solution 4
After seconds, respectively the runners would've ran and meters. Their current positions on the track are these values . We're trying to find the value of such that Subtracting on all sides, we get Now, we must find a value for such that both and are simultaneously multiples of .
Plugging in for we get , but this does not work for ( isn't a multiple of ). Plugging in , we get , and this does work for .
Therefore, and the answer is .
- Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer
Solution 5
Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition... After seconds, respectively the runners would've ran and meters. These three values are congruent , so . Subtract from all three sides to get and are congruent. Now all we need to find is a value of for which and are congruent . Subtract from both sides to get and are congruent mod , or that is a multiple of . Let , so we want to be a multiple of , or to be a multiple of . Therefore, the smallest value of is when , and when
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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