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Consider a sprocket has $s$ teeth, and a chain has $h$ holes. Think about a bicycle chain and the sprocket wheel in the back of the chain.

Sprockets are numbered $0 \ldots s-1$ and holes are numbers $0 \ldots h-1$. We assume $h > s$ and and clearly $s > 1$.

Lets look at discrete time intervals - one sprocket movement is one time unit. At time $t = 0$ we have sprocket $0$ in hole $0$.

What I want to know is at what time $t > 0$ will the $0$ sprocket again find itself to be in the $0$ hole.

I am wondering how would one go about finding an answer for this interesting problem. What kind of mathematics would one use?

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    $\begingroup$ Look at the least common multiple. $\endgroup$ – copper.hat Jan 10 '14 at 19:56
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The time that they will line up again will be $LCM(s,h)$, the least common multiple of $s$ and $h$.

The value of $t$ needs to be such that $t/s$ and $t/h$ are both integers. (At this point, both sprockets and holes have made exactly an integer number of revolutions.)

Another way of expressing this is that $t$ is an integer multiple of $s$, and that $t$ is an integer multiple of $h$. The first (smallest) nonzero value of $t$ that this happens at is the LCM of $s$ and $h$.

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