54 teeth 18 teeth how many rotations to get back to original position

Question

The front year of a bicycle has 54 teeth the back gear has AT&T how many complete rotation must the smaller gear make for both years to be aligned in the original starting position

Answer 1

What you're looking for here first is the least common multiple of both numbers - this is the smallest number that would result in both gears making complete rotations, and so ending up where they began.

 

54 has factors 2*3*3*3

18 has factors 2*3*3

 

The least common multiple which includes all factors from both is in fact 2*3*3*3, just 54 again.  This is the number of teeth which must be moved past for both gears to return to their original position.

 

The number of complete rotations this is for the smaller gear can be determined by dividing by the number of teeth the smaller gear has.  54/18 = 3, so it must make three complete rotations for both gears to be returned to their original positions.