Thread by @ProfDFrancis

This Sam Loyd question is generating lots of very nice equation-based answers, but let me try a method, derived from Zugzwang et al, to see if I can do it without so much algebra, just by thinking.

Suppose we CHANGE the question to this:

"The hour hand and minute hand are SUPERIMPOSED on each other. How long till this superimposition next happens?"

Here is my approach. Suppose it is now just-after-two o'clock. When will this next happen?

First answerer is correct.

And this will continue on in a similar pattern.

When you get to "just-after-11 o'clock", what actually is that time, after 11 o'clock, that the hour and minute hands superimpose?

So at the beginning, the time "between 12 and 1" is 12:00,
"between 1 and 2" is 1-and-a-bit, etc.
but when you get to the end, there isn't really a time "between 11 and 12", because that just turns out to be 12:00.

How many distinct positions are there in the cycle, therefore?

And it is obvious by handwaving, that they are equally spaced in time. Therefore each interval is 1/11 of 12 hours.

NOW let's think about Sam Loyd's question. OPPOSITE hands, not superimposed hands.

The first is at 12:00.

The next is when?

OK! And this keeps happening. Except now the intervals are LESS than an hour, rather than MORE than an hour.

By the time you get to "just before 11 oclock", it is quite a bit before 11 oclock. In fact it is only just past 10 o clock.

How many times in each 12-hour cycle?

If it happens 13 times per 12 hours (counting the 12:00 at the beginning but not at the end), the intervals are exactly 12/13 of an hour.

So the generalised answer to Sam Loyd's puzzle is:
n*(12/13) hours after 12:00, for any integer n.

For the one between 8 and 9, put n=9.