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I have a machine which has four sequential cog wheels in constant mesh.

The largest cog has 81 teeth and the others have 52, 36 and 20 respectively.

What is the fewest number of revolutions the largest cog must make so that all of the cogs are back in their starting position?

?

 [Ref: ZAVS] © Kevin Stone

260 revolutions.

There are a number of ways of thinking about the solution, and we find this one the quickest way to find the answer.

The total number of teeth moved by Cog 1 will be wholly divisible by each cog in turn, therefore:

Revolutions x Cog 1 ÷ Cog2 is an integer
Revolutions x Cog 1 ÷ Cog3 is an integer
Revolutions x Cog 1 ÷ Cog4 is an integer

So we are after the first number of revolutions x 81 that is an integer after division by 52, 36 and 20.

Thus:

81     81     81
-- and -- and -- all need to be integers (and not fractions).
52     36     20

An easy way to do this would be to multiply by 52 x 36 x 20 = 37,440 revolutions, which would be a correct answer, but not necessarily the smallest answer.

A better way is to break each cog down into its prime factors, where Cog 1 has the largest number of teeth:

Cog 1 - 81 = 3 x 3 x 3 x 3

Cog 2 - 52 = 2 x 2 x 13
Cog 3 - 36 = 2 x 2 x 3 x 3
Cog 4 - 20 = 2 x 2 x 5

3 x 3 x 3 x 3 and 3 x 3 x 3 x 3 and 3 x 3 x 3 x 3 and these need to be integers
-------------     -------------      ------------
 2 x 2 x 13       2 x 2 x 3 x 3       2 x 2 x 5

Simplifying any fraction that can be simplified gives:

3 x 3 x 3 x 3 and 3 x 3 and 3 x 3 x 3 x 3 and these need to be integers
-------------     -----     -------------
 2 x 2 x 13       2 x 2       2 x 2 x 5

Multiplying throughout by 2 x 2 gives:

3 x 3 x 3 x 3 and 3 x 3 and 3 x 3 x 3 x 3 and these need to be integers
-------------               -------------
     13                            5

Multiplying throughout 13 gives:

3 x 3 x 3 x 3 and 3 x 3 x 13 and 3 x 3 x 3 x 3 x 13 and these need to be integers
                                 ------------------
                                         5

Multiplying throughout 5 gives:

3 x 3 x 3 x 3 x 5 and 3 x 3 x 13 x 5 and 3 x 3 x 3 x 3 x 13

They are all now integers, and we have therefore multiplied by 2 x 2 x 13 x 5.

2 x 2 x 13 x 5 = 260 revolutions. As required.

The easy way from above of 37,440 is exactly 144 times this answer.

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