2012 AMC 10A Problems/Problem 16

Problem

Anna, Stephanie, and James all start running around a track at 8:00. Anna completes a lap every 4 minutes, Stephanie finishes a lap every 7 minutes, and James finishes a lap every 6 minutes. What is the earliest time when all three meet back at the beginning?

Solution 1

First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.

$4.8x-4.4x=500 \Rightarrow x=1250$

Because $4.4(1250)=5500$ 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 $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\boxed{\textbf{(C)}\ 2,500}$ seconds.

Solution 2

Working backwards from the answers starting with the smallest answer, if they had run $1000$ seconds, they would have run $4400, 4800, 5000$ meters, respectively. The first two runners have a difference of $400$ meters, which is not a multiple of $500$ (one lap), so they are not in the same place.

If they had run $1250$ seconds, the runners would have run $5500, 6000, 6250$ meters, respectively. The last two runners have a difference of $250$ meters, which is not a multiple of $500$ .

If they had run $2500$ seconds, the runners would have run $11000, 12000, 12500$ meters, respectively. The distance separating each pair of runners is a multiple of $500$ , so the answer is $\boxed{\textbf{(C)}\ 2,500}$ seconds.

Solution 3

Let $t$ be the time run in seconds, then the difference in meters run between the three runners is $0.2t, 0.4t, 0.6t$ . For them to be at the same location all of them need to be multiples of 500. It is now easy to see that $0.2t=500, 0.4t=1000, 0.6t=1500$ , so $t=\boxed{\textbf{(C)}\ 2,500}$ .

Solution 4

After $t$ seconds, respectively the runners would've ran $4.4t, 4.8t,$ and $5t$ meters. Their current positions on the track are these values $\pmod{500}$ . We're trying to find the value of $t$ such that $4.4t \equiv 4.8t \equiv 5t \pmod{500}$ Subtracting $4.4t$ on all sides, we get $0 \equiv 0.4t \equiv 0.6t \pmod{500}$ Now, we must find a value for $t$ such that both $0.6t$ and $0.4t$ are simultaneously multiples of $500$ .

Plugging in $500$ for $0.4t$ we get $t=1250$ , but this does not work for $0.6t$ ( $750$ isn't a multiple of $500$ ). Plugging in $0.4t=1000$ , we get $t=2500$ , and this does work for $0.6t$ .

Therefore, $t=2500$ and the answer is $\textbf{(C) } 2500$ .

— @adihaya (talk) 15:21, 19 February 2016 (EST)

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 $t$ seconds, respectively the runners would've ran $4.4t, 4.8t,$ and $5t$ meters. These three values are congruent $\pmod{500}$ , so $4.4t \equiv 4.8t \equiv 5t \pmod{500}$ . Subtract 4.4t from all three sides to get 0, 0.4t, and 0.6t are congruent. Now all we need to find is a value of t for which 0.4t and 0.6t are congruent mod 500. Subtract 0.4t from both sides to get 0.2t and 0 are congruent mod 500, or that 0.2t=t/5 is a multiple of 500. Let t=500k, so we want 100k to be a multiple of 500, or k to be a multiple of 5. Therefore, the smallest value of t is when k=5, and when t=500k=500(5)= $2500 \space (\text{C})$

- Solution by mathchampion1

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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