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While watching YouTube videos, I came across a puzzle that had a pretty interesting solution and an underlying question behind it.

Question: Let $S$ be a set of gears that are connected to form a ring such that $n(S)\geq0$, $n(S)\in N$. For some values of $n(S)$, rotating one gear will cause all gears to rotate simultaneously, whereas for other values of $n(S)$, a conflicting rotation would make it impossible for all gears in the ring to rotate simultaneously. Thus, the natural question to ask would be what values of $n(S)$ would allow all gears to rotate simultaneously?

Solution: Simple trial and error as well as grunt work would imply that for all odd values of $n(S)$ such that $n(S)\geq3$, the gears would not be able to rotate simultaneously. However, for all even values of $n(S)$, the gears would be able to rotate simultaneously.

Even though the solution has been presented, the aspect that I am more curious about is how do I prove that the statement above is true?

Darrell Tan
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Choose any gear as $G_1$, the next in the ring as $G_2$ etc, and the original one will also be $G_{n+1}$. If $G_1$ turns clockwise, then $G_2$ turns anticlockwise, $G_3$ clockwise etc. Odd-numbered gears rotate clockwise, even-numbered gears rotate anticlockwise. $G_{n+1}$ must rotate clockwise because it is also $G_1$, so $n+1$ is odd. Hence $n$ is even.

The diameter (proportional to the number of teeth) does not matter. For the speed of the outside of each pair of gears must be equal as the gears do not allow slipping. Hence the rim speed of all gears is equal and $G_{n+1}$ turns at the same speed as $G_{1}$ as it must.

Peter
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