2012年AMC10竞赛A第16题
$$2012-AMC10A-16$$  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?  三名跑者在一条 500 米的环形跑道上从同一点同时开始跑步。 他们每个人都绕着赛道顺时针运行，保持每秒 4.4、$$4.8$$ 和 5.0 米的恒定速度。 一旦他们再次聚集在圆形路线上的某个地方，跑步者就会停下来。 跑者跑多少秒？
课内体系->知识点->等式与不等式->等式->方程组的解集 美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems
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.  Because $$4.8x-4.4x=500\Rightarrow x=1250$$ 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 $$5.0\rm m/s$$ 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 $$\boxed{\rm(C)\textasciitilde2500} $$seconds.  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 $$\boxed{\rm(C)\textasciitilde2500}$$ seconds.  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 $$t=\boxed{\rm(C)\textasciitilde2500}$$.  After $$t$$ seconds, respectively the runners would\textquotesingle ve ran $$4.4t$$, $$4.8t$$ and $$5t$$ meters. Their current positions on the track are these values $$(\rm mod\textasciitilde500)$$. We\textquotesingle re trying to find the value of $$t$$ such that $$4.4t\equiv 4.8t\equiv 5t\textasciitilde{} \rm (mod \textasciitilde500)$$ Subtracting $$4.4t$$ on all sides, we get $$0\equiv 0.4t\equiv 0.6t\textasciitilde{} \rm (mod \textasciitilde500)$$  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\textquotesingle 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 $$\rm(C)\textasciitilde2500$$.  $$\cdot$$ Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer.  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 $$\cdots$$~~After $$t$$ seconds, respectively the runners would\textquotesingle ve ran $$4.4t$$, $$4.8t$$, and $$5t$$ meters. These three values are congruent $$\rm(mod\textasciitilde500)$$, so $$4.4t\equiv 4.8t\equiv 5t\textasciitilde{} \rm (mod \textasciitilde500)$$. 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 $$0.2t=\frac t5$$ 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, thes mallest value of $$t$$ is when $$k=5$$, and when $$t=500k=500(5)=2500\rm(C)$$. 