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A small gear with 17 teeth is meshed into a large gear with 60 teeth. The large gear starts rotating at one revolution per minute. How long will it take until the small gear is back to its original position and the large gear is one quarter of a revolution past its original position?

I can just set this up as a linear combination, right? Is that the easiest way to solve this? If so, what is the general method for solving such a problem?

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  • $\begingroup$ i think any problem like this can be set up as a linear combination problem though i am not sure whether it is the easiest here take x to mean position change by x gear then x is 0mod17 and x is 15mod60 and 60 and 17 are coprime. $\endgroup$ – happymath Mar 20 '14 at 15:34
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Let the required number of minutes be $x$.

Now, $x = a + \frac{1}{4}$ for integer $a$. This is because the $60$-tooth gear must end up $+\frac{1}{4}$ revolutions away from its original position.

The speed of revolution for the $17$ tooth gear is $\frac{60}{17}$ RPM. Hence the number of revolutions will be $\frac{60}{17}x$ after $x$ minutes. But we want an integer number of revolutions (so that it ends back where it started off). So $x$ must be a multiple of $17$, i.e. $x = 17b$ for integer $b$.

Hence, we have

$$a + \frac{1}{4} = 17b$$ $$4a + 1 = 68 b$$

Now, considering both sides modulo $2$, we see that there are no integer solutions in $a,b$. Hence we deduce that, surprisingly, it is not possible for the described situation to happen.


In fact, we can generalize to say that it will not be possible for the 60-tooth gear to end up, for integers $n > 1$, $\frac{1}{n}$ revolutions away from its original position with the 17-tooth gear to end up exactly back to where it was.

We can further generalize to say that for the similar setup with a $x$ and $y$ tooth gear, for coprime integers $x,y$, it is not possible for either one to end up, for integers $n > 1$, $\frac{1}{n}$ revolutions away from its original position, with the other one ending up back where it was.

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  • $\begingroup$ I was looking at this from a different perspective. I was thinking that because they are relatively prime to each other, then that would allow us to obtain any integer we wanted by exhausting different linear combinations of them. $\endgroup$ – GaMbiTaaaa Mar 20 '14 at 19:37
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I think you're looking at this wrong. Let $r$ be the number of revolutions of the small gear, then the big gear does $\frac{17r}{60}$ of a revolution. Then, with respect to the big gear, what you want is to solve $17r ≡ 15$ (mod 60), since each revolution of the small gear will move the big gear 17 teeth, and you want to end up at $\frac{1}{4}$ of a revolution or 15 teeth past the starting point. Then you will have the number of small gear revolutions, which you can translate to big gear revolutions to get the time this will take.

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