Four gears are in constant mesh with each other. Gear A meshes only with Gear B, B meshes with A and C, C meshes with B and D, and D meshes only with C, like this:

A B C D

Gears A, B, C, and D have 35, 27, 84, and 34 teeth, respectively.

If a vertical stripe is painted on each gear, what is the minimum number of revolutions (>0) gear A would have to make before the four stripes are vertical again?

This is a simple Least Common Multiple problem, remember those? Luckily I recently took the GMAT, and spotted it immediately. So you break each down to it's primes, then pick the highest power of each prime from each number and multiply. That gives you the number of teeth, then simply divide by the number of teeth for A to get the solution.

For A, 35 = 7 x 5
For B, 27 = 3 x 3 x 3
For C, 84 = 2 x 2 x 3 x 7
For D, 34 = 17 x 2

Therefore the LCM = 2 x 2 x 3 x 3 x 3 x5 x 7 x 17 = 64,260

Divide that by the number of spokes for each to get the number of revolutions for each.

Revs for A = 1836
Revs for B = 2380
Revs for C = 765
Revs for D = 1890

Posted by Lawrence on 2003-08-24 05:59:40