A gear wheel has 12 teeth on the smaller wheel and 32 teeth on the larger one. Due to a manufacturing defect, one tooth on the small wheel and one gap on the large wheel are faulty, causing them to rattle when they meet. How often will we hear the rattling if the time for one revolution of the large wheel is 3 seconds? How can the rattling be eliminated if we assume that all other tooth-gap pairs are flawless and make contact silently?
Between any two clicks, both wheels make an integer number of revolutions, since the meeting always occurs at the same place. Therefore - measuring time in terms of one tooth's rotation - the time between any two clicks is a multiple of both 12 and 32. The repetition interval is the least common multiple of these, which is 96 teeth rotations. In this time, the large wheel makes 3 revolutions, so the interval length is 9 seconds.

From this, we also see that the faulty notch of the large wheel can only meet with 3 different teeth of the small wheel: the 8th tooth from the fault - because in one revolution of the large wheel, the small wheel makes 2 revolutions and rotates by 8 more teeth - and similarly, the 4th tooth from the fault, because $2 \cdot 32 = 5 \cdot 12 + 4$. Therefore, the click can be eliminated by aligning the faulty notch with any of the 9 additional teeth of the small wheel.

Fekete József (Kisújszállás, Móricz Zs. g. I. o. t.)

Note. Many papers only considered the click eliminable if the faulty notch was aligned with an odd-numbered tooth of the small wheel. It is clear that the number can also be even, but it cannot be divisible by 4.