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The BrainBashers mechanical computer is a very sensitive device, in consists of four sequential cog wheels which are in constant mesh.

The largest cog has 63 teeth and the others have 42, 35 and 27 respectively.

By accident, Daniel started to rotate the largest cog.

How many revolutions must he rotate the largest cog make before the computer is back in its starting position with all of the cogs where they started?

[Ref: ZTLC] © Kevin Stone

If we break each wheel into its prime factors, we get:

63 = 3 x 3 x 7
42 = 2 x 3 x 7
35 = 5 x 7
27 = 3 x 3 x 3

We now think of rotating the large wheel just once, and this is 3 x 3 x 7 teeth moved (3 x 21), and we can see that 42 tooth wheel also has a 3 x 7 (21 teeth) in it, with an extra 2. If we therefore rotate the 63 toothed wheel twice, the 42 will have rotated three times.

The answer involves cancelling any common factors from the large wheel. We can cancel 3, 3, 7 from any of the smaller ones to leave 2 (from the 42), 5 (from the 35) and 3 (from the 27). 2 x 5 x 3 = 30. QED.

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