Greatest Common Factor and Least Common Multiple

GCF and LCM

Motivation Problem for LCM

How are multiples involved in the following scenario?

A machine contains two gears.   One gear has 12 teeth and the other has 20 teeth.   The gears are aligned by a mark drawn from the center of the first gear to the center of the second gear.   What is the smallest number of revolutions of the first gear necessary to realign the mark?

We begin by listing, for each gear, the number of teeth that would pass the starting mark after the first revolution, second revolution, third revolution, … .

 First gear:  12, 24, 36, 48, 60 , 72, 84, 96, 108, 120 , 132, 144, 156, 168, 180 , 192, …

Second gear: 20, 40, 60 , 80, 100, 120 , 140, 160, 180 , 200, 220, 240 , 260, 280, 300 ,…

Notice that we are listing the multiples of 12 and 20. The first gear will be aligned with the second gear after five revolutions. (Also, note the second gear will have made three revolutions.)

This question was answered using the least common multiple. Notice that the mark on the first gear returns to its original position after each full revolution. This will happen when it has gone through 12 teeth, 24 teeth, 36 teeth, etc. The mark on the second gear returns to its original position after it has made any number of full revolutions. This will happen when it has gone through 20 teeth, 40 teeth, 60 teeth, 80 teeth, etc.

Notice that 12, 24, 36, … are multiples of 12 and 20, 40, 60, … are multiples of 20.  

The mark will be realigned only when a value that is both a multiple 12 and of 20 is reached.   The mark will be realigned for the first time when the first such multiple is reached. Therefore, we are looking for the Least Common Multiple of 12 and 20.  

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