To find the number of teeth Gear A must have so that when it rotates once, Gear B and Gear C have rotated some whole number of times, we need to determine the least common multiple (LCM) of the number of teeth on Gear B and Gear C.
Gear B has 12 teeth, and Gear C has 20 teeth.
First, let's find the prime factorizations of both numbers:
12 = 2^2 * 3
20 = 2^2 * 5
To find the LCM, we take the highest power of each prime factor that appears in either number:
LCM = 2^2 * 3 * 5 = 60
Therefore, for Gear B and Gear C to rotate some whole number of times when Gear A rotates once, Gear A must have 60 teeth.
To explain this reasoning, we use the concept of the LCM, which is the smallest multiple that both numbers share. In this case, we want to ensure that Gear B and Gear C will have rotated some whole number of times when Gear A completes one rotation. By finding the LCM of the number of teeth on Gear B and Gear C, we ensure that Gear A completes one rotation after the LCM number of rotations of Gear B and Gear C.
In this particular scenario, the LCM of 12 and 20 is 60. Therefore, Gear A must have 60 teeth for the desired condition to be met.