A twenty-one handed clock Calculating when clock hands will align is a well-known problem. Here's a similar but much more complicated one, based on an actual device.
Instead of just hour and minute hands imagine 21 hands, all driven from a central shaft. Each rotates at a different rate. The slowest has a ratio of 0.15 (54 degrees per full shaft rotation of 360 degrees). Each intermediate gear ratio increases by 0.025 (9 degrees), up to the fastest at 0.65 (234 degrees per shaft 360).
Assuming the hands start straight up, how many shaft rotations will it take until the hands are all once again at 'noon'? How many times (if any) will the hands also align during the grand tour?
I'm interested in an approach to solving. The real device has cams instead of hands and will only complete one full shaft rotation, then reverse -- this is strictly an academic exercise.
 A: Let's let the unit of time be one full rotation of the central shaft. Also, we'll label the hands from $0$ to $20$, in increasing order of speed.
Then hand $0$ (the slowest hand) rotates once every $\frac{360}{54}=\frac{20}{3}$ time units; hand $1$ (the next-fastest once) every $\frac{360}{63}=\frac{40}{7}$ time units, and so on. In general, hand $n$ will rotate $54+9n$ degrees in a single time unit, so it will take $\frac{360}{54+9n}=\frac{40}{n+6}$ time units to make a full rotation. Our goal is then to find the amount of time before all of the hands have completed an integer rotation: that is, the least common integer multiple of the numbers
$$
\frac{40}{6},\frac{40}{7},\dots,\frac{40}{26}
$$
Simply by staring at all these numbers, you can see that $40$ is an integer multiple of all of them. So their actual least common multiple must be some fraction $\frac{40}{k}$. Also, it must be at least as large as $\frac{40}{6}$, since that is one of the numbers in question. But none of the numbers $\frac{40}{2},\frac{40}{3},\dots,\frac{40}{6}$ are integer multiples of $\frac{40}{7}$, and so $40$ is in fact their least common integer multiple. That is, the first time the hands will all simultaneously return to their starting place is after $40$ units of time (central shaft rotations).
You can figure out when the hands will all be aligned in much the same way. Imagine that you are a small ant who feels no gravity riding on hand $0$. Then from your perspective, hand $n$ rotates by $9n$ degrees every time unit, and you want to know when the positive-numbered hands will have all — from your perspective — completed an integer number of full rotations. Since hand $n$ takes $\frac{360}{9n}=\frac{40}{n}$ time units to complete a full rotation, this amounts to finding the least common multiple of
$$
\frac{40}{1},\frac{40}{2},\dots,\frac{40}{20}
$$
which is clearly $40$. So the hands will only all align when they are all in their initial position.
