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Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

Thank you for your solutions. I received the most complete one from Clement Goh (Age 13) of the River Valley High School, Singapore.

This is a neat little problem and I have seen other similar ones that involve two numbers that have no common factor (relatively prime). If you cannot see the connection - draw two cogs with a relatively prime number of teeth. Colour a tooth on the smaller red and its starting point on the larger cog. Imagine you could rotate the smaller cog and mark the point on the larger cog that the red tooth would reach after each revolution. Now do the same with two cogs with a number of teeth that are not relatively prime.

Can you see it?

Here is Clement's solution

For this question, I made some cogwheels with different numbers of teeth.

I found out that the number of teeth in both cogwheels must not have the same factor.

For example: with 3 and 4 teeth respectively, the cogwheel with 3 teeth will be able to contact every gap on the cogwheel with 4 teeth.

But cog wheels with 3 and 6 teeth respectively will not be able to do so.

Therefore the values of p and q are:

3 : 4
3 : 5
3 : 7
3 : 8
3 : 10
3 : 11
4 : 5
4 : 7
4 : 9
4 : 11
5 : 6
5 : 7
5 : 8
5 : 9
5 : 11
5 : 12
6 : 7
6 : 11
7 : 8
7 : 9
7 : 10
7 : 11
7 : 12
8 : 9
8 : 11
9 : 10
9 : 11
9 : 12