I’ve been following this one because the puzzler is often my favorite part of the show. It seems to me that no one is addressing the OP’s question: is there a generic formulation of and solution to the problem of which the puzzler is a particular instance.
∀ m>2, ∃ n = FLOOR( (m-1)/2 ) just says that for any number m greater than 2 there is a number n that is either equal to or one integer to the left on the number line from half of m-1. It’s really just saying that for any number m>2 there is a number that is smaller. I don’t see how this addresses the OP’s post.
I think that Hokiedad is right that it works with an even number of hours. It doesn’t seem to work with an odd number of hours (5,7, etc.).
Mccolvin’s method of solution is the one I used, but would be time-consuming with a 100-hour clock.
Seems to me that cool in mathematics is proving the Riemann Hypothesis and elegant is giving the solution in one short line.