How can 'cyclical alignment' problems be solved? E.g.

Or 

I used to believe that problems like these could be solved simply by finding the LCM of the lengths of revolution for the each entity.
This method works for the first problem. The answer is 10, which is clearly the lowest common multiple of 2.5 and 2. However, it does not work for the second problem.
Using prime factor decomposition:
Othello:

20 = 2*2*5

Hamlet:

45 = 3*3*5

Romeo:

120 = 2*2*2*3*5

Therefore, the lowest common multiple of all three is

2*2*2*3*3*5 = 360

...which isn't an option.
The actual answer is 72. Why is this the case, especially when 72 is not a multiple of any of the lengths of revolution for any of the moons. It's worth noting that
this is because dividing 360 by 5, the only factor common to all of the revolution lengths, gives 72.
Also, what are these types of problem actually called? It might be useful to know because it might enable an understanding of the theory behind how to solve them.
 A: The reason you got the wrong answer in the second problem is that you solved for the wrong thing.
The least common multiple of the circuit times tells you the amount of time it takes for all three moons to wind up back in their starting positions. But that's not the only way that they could wind up in the right configuration! An easy way to visualize this is to think about the minute hand and the hour hand on a clock: with each starting at the $12$ position, it will take $12$ hours until they both return to that position, but they'll be pointing in the same direction long before that (indeed, shortly after $1$ hour).
Now as a reality check for your specific problem, note that ${72\over 20}$, ${72\over 45}$, and ${72\over 120}$ all have the same fractional part, namely ${3/5}$; this means that after $72$ hours the three moons are each ${3\over 5}$ of the way around their orbits, and so in particular they're lying in a line as desired.
Of course this does not imply that $72$ is the right answer - we also need to know that no earlier time makes the three moons wind up in the right configuration. To be completely sure that the answer is $72$ we have to check that the only other possibility, $36$, does not work. But this is easy: we have $${36\over 20}=1.8, {36\over 45}=0.8, {36\over 120}=0.3,$$ so after $36$ hours the first two moons are pointing in the same direction but the third isn't.

So the issue is that the "coincidence" you're trying to time is more complicated than the one corresponding to merely the least common multiples of the circuit times. Always think carefully about what it is that you're measuring!
