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

The largest cog has 21 teeth and the others have 17, 12 and 10 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?

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 21 that is an integer after division by 17, 12 and 10.

Thus:

21     21     21
-- and -- and -- all need to be integers (and not fractions).
17     12     10

An easy way to do this would be to multiply by 17 x 12 x 10 = 2,040 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 - 21 = 3 x 7

Cog 2 - 17 = 17
Cog 3 - 12 = 2 x 2 x 3
Cog 4 - 10 = 2 x 5

Using these prime factors we can write the fractions as:

3 x 7 and   3 x 7   and 3 x 7
-----     ---------     ----- and these need to be integers
17      2 x 2 x 3     2 x 5

We can now simplify one of the fractions to give:

3 x 7 and   7   and 3 x 7
-----     -----     ----- and these need to be integers
17       2 x 2     2 x 5

To remove the 17 on the first fraction we can multiply throughout by 17 to give:

3 x 7 and 7 x 17  and 3 x 7 x 17
------      ---------- and these need to be integers
2 x 2         2 x 5

We can now ignore the first fraction as it is an integer.

To remove the 2 x 2 on the second fraction we can multiply throughout by 2 x 2 to give:

7 x 17 and 2 x 2 x 3 x 7 x 17
------------------ and these need to be integers
2 x 5

We can ignore the second fraction now as it is an integer, and then simplify the third fraction to give:

2 x 3 x 7 x 17 and this needs to be an integer
--------------
5

We now finally multiply by 5 to make the last fraction an integer.

We have therefore multiplied by 17, 4, and 5.

17 x 4 x 5 = 340 revolutions. As required.

The easy way from above of 2,040 is exactly 6 times this answer.

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