Lost solution

NM Dale's answer must be incorrect.

The minute and second hands would have advanced 5 units, but the hour hand would be in virtually the same position. Distances between the hands would be approximately 16 minutes, 25 minutes and 20 minutes.

It will get back to 40 and beyond. I was wrong when I said it won't go above 40. I didn't think much about the highly unequally spaced case and didn't consider what would happen when one hand is on the low side and the other two are a little below 60. M has a local maximum of 50.3 at 4:54:55. After 12 noon, M first gets back to 40 at 3:43:43.69.

Of course this is unrelated to the solution, which depends on finding the times when M is close to zero.

For fun I checked all of M's minima over the full 12 hour span. Besides Loyd's starting time, I found three other times with M less than 0.1, corresponding to a deviation from exactly equal spacing of less than half a percent.

5:49:09.096  M=0.056
6:10:50.903  M=0.057
9:05:25.450  M=0.028

For me the value of this puzzle was the challenge of finding a mathematical relation between the current time and a reasonable measure of spacing equality. For others, it was the challenge of figuring out what kind of watch Loyd must have based the puzzle on.

You seem to have ignored the fact that, in the context of Sam Loyd's puzzle, the "measure of spacing equality" is not a real number, but rather a boolean value; either the hands are equally spaced, or they are not. So graphs are of no use... besides serving as an indirect and relatively inefficient method of proving that an ideal clock with 3 uniformly rotating hands can never satisfy the puzzle's requirements. Note that I'd already proven this as early as post #3 (~10 days ago, now, I believe).

What I failed to realize (though it became evident to me almost immediately after seeing NM Dale's spectacular solution) is that such a proof also constitutes an indirect proof that the hands of Sam Loyd's clock do not rotate uniformly. I made the mistake of initially assuming the clock's hands moved uniformly, as I considered this to be a necessary property of any clock which keeps perfect time. NM Dale's post #15 demonstrates it is NOT a necessary property, however:

Let h be the number the hour hand points to (rounded to the nearest whole number), m the number the minute hand points to (rounded to the nearest fifth), and s the number the second hand points to (rounded to the nearest fifth). Then the current time displayed by Dale's clock is always  t = h:5m:5s  

Dale's clock also produces the elegant answer of "5 minutes and 5 seconds", making it extremely difficult for any other interpretations to match its soundness.

Not convinced by any of the analyses so far, including my own. The cleanest is the one by NMDale, championed by cobra91, on which the hour hand jumps in one breath from 3 to 4. But I agree with many of the others that this departs too much from actual clock behaviour (especially in Loyd's day) to be an acceptable solution. Loyd was a trickster, but this trick seems to me to be below the belt. 

I agree with cobra91 however that, whatever solution Loyd had in mind, it would have been a pretty "clean" one. The other analyses are all too messy to have been intended.

Can I throw in another observation? The puzzle speaks of the hands being "equal distances apart". Has anyone noticed that when all three hands line up together, they will be equal distances apart, namely a distance of zero? The obvious case here is noon, but there may be other cases too. E.g., slightly after 3:15, all three hands will line up. This is a terrible trick, but this one doesn't seem to me to be below the belt. We should keep this possibility in mind maybe?

Rather than ignoring that "fact", I arrived at my own context which made possible some activity that I enjoyed. So from my perspective Loyd succeeded as a puzzle maker.

That would be another creative solution worthy of Sam Loyd. The only issue I have with it is that if the original hands are really equally spaced on the puzzle time then the clock has an error and they will not be at noon! And if you accept an error in the initial spacing then the other answers (like n9531l's) are valid as well even when imprecise in comparison to your suggestion.

Agreed, the zero spacing option introduces new possibilities without making things any clearer, it seems. I'm close to giving up, as sameez1 says for sake of my health!

I thought keju's suggestion was a good one, and it does provide a counterexample to cobra91's proof that there is never even a single instant when the angular distances are all equal (comment #3). My objection to it as the solution is that there are sooner times when the hands will appear to be equal distances apart, and the puzzle asks how soon it will happen.

@ cobra91 That you would dismiss the word appears but accept the invention of an analog clock that the hour hand jumps an hour all at once.NM Dale even invents the second that the hand jumps on this morphed digital analog clock,wouldn't with this digital thinking hand movement the hour hand won't be ready to do its jump until the second hand reaches the 12.Even with the invention of a clock that don't exist now and certainly didn't then NMDale solution is not one I would choose.

@n9531l I was trying to see if your graph would point to a time other than 12 when all three hands would be exactly together I am wondering if each relation of the three hands together exactly are unique for the whole 12 hour period there will be times when they appear together but they will be off won't they.The time I was asking just puts the minute and hour hand together.A Sam Loyd unanswered puzzle He is laughing right now,look at where some of the arguments are going.Gotta love it.

 I bet it sold a lot of watches. I am kind of new to actually paying attention to puzzles since I have some free time,now just discovering Sam Loyd what adjective can I use?Unbelievable genius with a streak of mischief.

I just only rotate the clock and keeps all the arms on the same place.

So the diferences must keep the same then.

I get  it is just seconds before it is 5 minutes before 4 O clock.

No wait, it is almost 4 O clock. So after one hour plus 5 minutes and 5 seconds. 

Or straight up.

About 1 hour 5 minutes and 5 seconds is my answer.

@ oolalimk1 yes it doesn't say appear at the same angles as shown again, it says How soon will they again appear an equal distance apart.I do believe you caught the Sam Loyd trick.We are all looking only to an even 120 degrees LOL.I do believe that is the type of diversion he uses in his puzzles.I think you are right. I am curious how many agree.

Arisktotle already said it (post #89, almost directly above yours), but it bears repeating since you seem to have a habit of ignoring details which are slightly "inconvenient": keju's suggestion does nothing to address the contradiction one arrives at when assuming ideally conventional behavior for all three hands. Specifically, one can still conclude that either the assumption of ideally conventional clock behavior is incorrect, or Sam Loyd's own illustration is incorrect.

So yes, "noon" could be interpretted as a counterexample to my earlier statement (depending on one's choice of formal definitions), but it's not a relevant counterexample with respect to Loyd's puzzle.

@Sameez1:  Thirty seconds after the starting time, the minute and second hands will be about 60 degrees apart, the second and hour hands will be about 60 degrees apart, and the minute and hour hands will be about 120 degrees apart. Hardly what I would call equal distances apart.