Lost solution

In 1905, the Ingersoll company engaged Sam Loyd to invent a puzzle to promote its new dollar watch. Loyd sent the illustration above and asked:

How soon will the hour, minute and second hands again appear equal distances apart?

The company advertised the puzzle in Scribner’s in June, promising a free watch to the first 10,000 correct respondents. “The full problem is stated above,” the ad ran, “and no further information can be given in fairness to all contestants.” Further, it said, Loyd’s solution “is locked in our safe, inaccessible to any one.”

Perhaps it still is — I can’t find any record of a solution to the puzzle. I offer it here for what it’s worth.

   I know it is not a chess puzzle but I need to see what someone else thinks it is.

As a former math major, I'm afraid Sam Loyd outsmarted himself this time. Such puzzles have always been a hobby of mine, and (unfortunately) I've just proven that, if the clock in the problem is an ideal clock (i.e. one which always shows the correct time with perfect precision), then there will never be even a single instant when all angular distances between its three hands are exactly the same.

To make matters worse, I also proved that if the clock in question is "almost" ideal (that is, if the orientation of hands is not ideal, but they still rotate ideally), then only the trivial answer of "12 hours" is possible. My best guess is that Loyd's solution features a deviation from ideal rotation speeds, but I suspect there must surely be multiple ways of varying the clock's mechanics to produce such solutions. So it is very likely that the intended idea or concept is not well founded.

Yes I was wondering if this particular time was the only instant in the total 12 hour rotation or if the advertising gimmick would be Thats not one of our watches its broke.The first 10,000 thats a lot of watches.Btw I forgot Thanks

I've no idea what cobra said (but he sounds like he knows what he's talking about + my maths is atrocious) - but it took me literally about 10 seconds of googling to find the puzzle + 'solution'. Search for 'a question of time' by Sam Loyd.

^ Are you sure about that? Compare the clock diagrams, and pay close attention to the approximate orientations. I doubt the puzzle you're referring to is identical to the one being discussed.

Yes I noticed that and I agree you're correct but would it actually make any difference to the method of working it out?  I assumed the clock diagram above was just a placeholder.

It makes a vast difference, actually. For one thing, the puzzle in this thread also deals with a second hand, in addition to the hour and minute hands. The question being asked is important, as well: one problem concerns absolute angular positions, while the other is about relative angular positions. Because of this, calculations for the "lost solution" above are somewhat more involved than for "a question of time" (making the latter simpler and easier).

Wouldn't that be a third hand?

Perhaps the answer is

not very.

@ JamesColeman add that it is 1/4 degree movement of the hour hand to a half rotation of the second hand to that formula for this puzzle and it is hard to calculate a time when they will all be exactly 120 degrees apart, the second hand always ends up in the wrong place.

@sameez1 OK. Happy to take your word for that! I've certainly learnt something in this thread...

I think 5 minutes and 5 seconds from now is possibly the answer if we take the current time to be

3:55:35 and imagine the hour hand only moves to the next hour when that hour is reached instead of gradually.

@ NMDale in 5 min the minute hand and the hour hand will be almost 90 degrees apart.the hands have to all be 120 degrees apart, with the different speeds of the three hands I think Cobra91 is right there is never a time a properly fuctioning clock would meet that position.  I think a Sam Loyd trick to advertise the watch without having to give any away. I appreciate your post thanks.

@NM Dale: I've been beaten by the better man - congratulations! The puzzle was obviously well above my pay grade, and I doubt I'd ever have come up with such an ingenious solution, even if I'd thought about this for years. Well done.   *applause*

To clarify for others: In the brilliant solution from post #15, it was omitted (because it goes without saying for any genius who is capable of solving a problem this difficult) that the minute hand behaves in exactly the same way as the hour hand; it only moves to the next minute "all at once" when that minute is reached. So neither the hour hand nor the minute hand exhibit constant rotation.

@sameez1: No, NM Dale was correct. In 5 minutes and 5 seconds, the hour hand will be at '4', the minute hand will be at '12', and the second hand will be at '8'.

@ cobra91 I see what your are saying about the movement of the hands not being constant (the second hand has to make a complete rotation to move the other hands in increments)but the time is 2:54 34 seconds in 5 minutes and 5 seconds it  will be 2:59 and 39 seconds not 4. I Do believe though that 4 would be the 120 degrees as long as the second hand could make it to the 8 without moving the minute hand i didn't think it could.

^ Read post #15 again. The time shown in the illustration is [a little counterintuitively] 3:55:35 (the hour hand won't move until 4:00:00, while the minute hand won't move until 3:56:00), so in 5 minutes and 5 seconds the time will be 4:00:40.

@ cobra91 you lost me the time in this illustration is 2:54:34 it is almost 3:00 not 4:00

Yes, exactly. The time 2:54:34

The question is, "How soon will the hour, minute and second hands again appear equal distances apart?"

Perhaps the key word here is "Appear". As noted by Cobra91, there is not a single instance where the hands will all be a perfect distance from each other. But then again, are we looking for perfection or something that just appears perfect?

The easy answer is 24 hours when the time falls on 2:54:34 the following day.

But can it happen sooner?

In 1 hour, 5 minutes and 5 seconds, all three hands will appear to be equal distances apart when the time is 3:59:39

This is not perfect however because the minute hand will have advanced 5 seconds further into it last minute than the original position. But still, it appears to be equal.

@ Dark_Army if  we are looking for just appear equal then 3:38:58 or you could argue 3:39:59 is the next closest time where the three hands will appear an equal distance apart. I guess you could argue they will still appear equal in 1minute 1second at 2:55:35.and it would be the exact same time in 12hr.   Again I don't think they intended to give out any watches.