Lost solution

[reposted after extensive edits]

My last watch (30 years ago) had one knob (I think they call it a crown!?) which I could pull out in 2 phases; at the first stop I could rotate the hrs hand, at the second stop the minutes hand. In some earlier watches I had 2 knobs for these functions. I assume the hand setting on 1905 watches was similar to one of the above [possibly this is diverse now or old-fashioned; modern instructions say that the minutes and hours hand usually rotate together].

Let's assume we succeeded in setting all hands 120  degrees apart with the help of the knobs (as in the picture, a real time or not), then we start the clock and let it run until the next time the hands are 120 degrees apart. Then you calculate the time interval by comparing the current hours hand to the previous hours hand, the current minutes hand to the previous minutes hand and the current seconds hand to the previous seconds hand (actually, comparing the hours hands is enough). The only condition is that the hands run at the correct speeds! Even on a clock that never points to a real time you can calculate time intervals this way, similar to measuring height differences without ever measuring heights. You shouldn't even think of it as a real clock!

I didn't look at James Coleman's Loyd puzzle (yet). Too busy writing posts

Note: As Cobra said, all hands will certainly be 120 degrees apart again after twelve hours. Whether or not this happens earlier as well I am not sure of but probably at most 1 time where the positions of the hands are switched. And any solution is certainly at an "impossible time" though it won't be far away from a real time just as in the picture. I haven't seriously tried to solve the problem, only to clarify it.

I have never seen a watch or clock that could manipulate the hands that way. How would you communicate a measurement without defining its points?Wrong or right To communicate my answer to you I must define what I measured.Using two men i define how I came to the measurement  Point A top short guys head to point B top of tall guys head I

 don't care how tall they are but I do have to show the points that I am measuring.You could prove it to yourself in your own thoughts without it but now to explain your reasoning you need a point A that everyone agrees to (I can hardly believe a problem there using the time).I did come up with an answer which I really think is wrong.Cobras very first post is what I was thinking when I presented the puzzle.I mean you could have glib answers it still appears equal in 1 second,if it is a mathamatical impossibility then it will always be mathamatically impossible so never.I guess I have to pay for the watch.   BTW I figured out the missing king I never looked at the solution for it but I looked at the solution on another one of those puzzles that used the en-passant move,without seeing that i don't think I would have ever got it.

Well done on solving the missing king puzzle! Did you see the chess puzzle I made to help you (Puzzle voor sameez1)? Then you will understand the relation with the missing king!

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Note on Cobra's first post: What Cobra says is sensible but he ends with the conclusion that the puzzle is wrong. My aim is to suggest which understanding of the puzzle is reasonable and makes it solvable by precise calculation.

Please understand that "the manual setting of the hands by knobs" is irrelevant for solving the clock puzzle. It is only to explain that a watch can display an "impossible time".

I honestly do not understand what you do not understand about my explanation of calculating time difference. I will therefore give an example which uses an "impossible time":

Let's forget about the seconds hand and use only the hrs and minutes hands. I set the hrs hand on 2 and the minutes hand on 6 - impossible as the hours hand ought to be in the middle of 2 and 3 (or 1 and 2). The two hands are now 120 degrees apart.

After some time has elapsed, the hours hand is on 10 and the minutes hand is again on 6. This is also an illegal time on the clock. How much time did pass between the first and the second measurement?

Answer: the hrs hand went from 2 to 10 and meanwhile the minutes hand made 8 rotations. It means that the minutes hand ran 12 times faster than the hours hand (correct) and 8 hours elapsed in between both measurements.

The puzzle did not ask at what time the 120 degree angle between the hands repeats but only "how soon". Though starting and ending with illegal time settings, we could still calculate exactly how much time had passed between them and that was the point of my story.

Conclusion: In order to calculate, we must assume the 3 hands in the original puzzle are exactly 120 degrees apart, but we need not assume that the pictured clock shows a legal time! Even with an illegal starting time we can solve the puzzle as I just did in the simplified example - by calculating the time difference!

Another conclusion: I did not say the puzzle has a solution or two or ten. I only explained what the real puzzle is! I explained it has nothing to do with the time on the clock and everything with the 120 degree angles. I can even prove that it doesn't matter where the hands are, if only they are 120 degrees apart. From any such position it will always take the same amount of time to reach the next position with the same angles!

I hope this clarifies things. I'm almost desperate .

Just to be clear, your interpretation of the puzzle is something I considered almost immediately after proving the hands of an ideal clock could never perfectly align as depicted in the illustration (now quite a few days ago). Also, I did not merely propose "12 hours" as an obvious potential solution. Rather, I proved it was the only possible solution under the rudimentary assumptions being made.

Here is the relevant quote from post #3: "...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."

It's too bad my capacity for creative thought is so small - with the work I put into those proofs, it really would've been nice to be the 1st known person to solve a century-old puzzle. But alas, the honor goes to somebody far shrewder than I could ever hope to be. I guess there's just no substitute for a naturally inventive and brilliant mind.  Congrats once more to NM Dale! Outwitted everyone but the great Sam Loyd himself, it seems!! 

Not sure if this helps, but I used a solution from a similar Loyd puzzle to figure out exactly what time is showing on the clock.

Because the hour hand moves 5 times per hour and the minute hand moves 60 times per hour, we can create the following equation:

x/5 = 55-x/60

So now...

60x-275-5x

Then...

65x=275

Isolate the x...

x = 275/65 = 55/13

55/13 becomes 1+ 42/13

Which becomes...

(55 - 1) 42/13

So the time is 2:54:34 and 41/13th of a second

This means that all hands on the dial are separated by 19 minutes, 34 and 41/13th of a second.

I was aware of your "solution" but also noticed you went on to throw doubt on it yourself. Rather than agreeing that the puzzle is wrong, I concentrated on showing why this particular interpretation of the puzzle is much more sensible than any other while ignoring its solutions. My perception was that for some solvers (including myself) your solution could only take hold  after they could mentally let go of the idea that the puzzle clock had to display an actual time! So it was all about identifying the puzzle. Math ain't that difficult , but modeling reality into it, is. And of course you are the greatest of all times by solving it first, even greater than the great Houdini not to mention the even greater Mohammed Ali!

Hm, 3 times that number ain't quite 60, so one or two of the hand distances must be greater than 20 minutes. It doesn't matter a great deal for the current picture but how do you translate these "acceptable errors" into the errors you will accept in the next occurrence of equal angular hand positions? You can't say that won't happen before 12 hours because how often it happens depends completely on the errors you find acceptable (Cobra only proofed it for precise 20 minute distances). Would you reject a solution where one hands distance is 20 minutes and 18 seconds? Or 25 minutes? And after you made up your mind about it, how could all the people solving the puzzle have agreed on the "acceptable error range"? It's not given anywhere in the puzzle!

Good point. It doesn't add up to 60. But then again, is there ever a time on the clock where every hand is exactly 20 minutes apart? There's not. You can get close though.

How about when the time reaches 10:32:13?

The hour hand is in between the 10 and 11

The minute hand is in between the 6 and 7

The second hand is in between the 2 and 3

That's probably about as close as you'll get without waiting a full 12 hours.

That is one of the other 23 nearby solutions. But the questions remain: "why accept this one and not the remaining 22?", "if there is a criterion for accepting nearby solutions, then why wasn't it in the puzzle?", "isn't it more natural to treat it as a geometrical puzzle which does have one exact solution (see Cobra) which respects the 20 minutes distances in full?".

The geometrical solution assumes the clock is a bit off a real clock time but it matches with the clock picture and the only time fact given, namely that the hands are at equal distances. Don't you think the puzzle would have factually stated the clock time as well if required to make it mathematically solvable? The truth is that the time isn't required because it doesn't matter where the hands are, and it doesn't matter that it is a real time. Knowing the angles is sufficient.

@Arisktotle - If knowing only the angles is sufficient to solve, then my original answer of 1 hour, 5 minutes and 5 seconds would be the correct one. The question is "How soon will the hour, minute and second hands again appear equal distances apart?"

According to Cobra (and I think he is right) the first time the angles will be exactly the same again is after 12 hours - kind of an obvious solution.

Your solution is the second time they are nearby the 20 minutes angles. The first time would be after about 43 minutes and 23 seconds when the order of the hands is switched. But then you will always have the discussion about "what is nearby?".

43-44 minutes and 23 seconds would have to be the correct answer then.

@ Arisktotle To clarify my thoughts on this,as DarkArmy rightfully pointed out no where in the puzzle does it say the hands are exactly evenly spaced,it says how soon will they again appear to be.This is not a mathamatical clock with a non existant time it is a normal clock at the time 2:54:34 which appears to be evenly spaced.Going with that the next nearest normal clock time that the hands will appear evenly spaced will be 3:38:58,that is 44minutes 24 seconds away.A lawyer could argue with the creator of this puzzle that this is correct because of the wording used.Wrong or right I have no other way of conveying that thought to you without mentioning the time.The next time after that that they would appear evenly spaced is 4:00:40.What time will they be perfectly 120degrees apart never.I now think that the real misconception of the puzzle is a mathamatically perfect  solution, no where does it even imply a perfect separation of the three hands.I certainly appreciate the interest in the puzzle from everyone.Thanks .....No I missed the puzzle I will look for it, I seen a puzzle with a solution at that futility closet.    Please check post 34 for my detailed reasoning on this.

@ DarkArmy I missed your post of the actual add somehow fantastic thanks.

That is an interesting understanding of the language. I always understood it as "the hands appear again at equal distance apart" but I can read it your way as well.

The problem with the loose understanding of "appear" is that what appears to one man does not appear to another. There is no predetermined definition of how far out "appears" can be from "exactly". And indeed, a puzzle has to be lawyer-proof (at least).

For instance, I certainly do not share your observation that the seconds hand is on 34 and the minutes hand on 54 while the hours hand is indeed pretty close to 3 (or 15). The truth however is that none of the hands is on any of these numbers exactly unless it is a ticking clock - which is unlikely from the picture. If you were to zoom in on the clock (like they do with hawkweye images in tennis), you will see the seconds hand may be at something like 34.628734719902... and so on. That doesn't mean it is on 34 and it doesn't mean it is on 35; it just is where it is. And in about 44 minutes all the hands will be in very nasty positions again. The answers on how soon the hands will start appearing to be in the same positions all depend on (a) how and in how many decimals you read the original time and (b) how much of an error your eyes will tolerate for hands "appearing" to be in the same positions. And we must assume your visual judgement is flawed or the hands in the picture could never "appear" to be at equal distances when they actually aren't! The word "appear" does therefore not automatically imply that your eye can distinguish 34.97883 from 35.003 or 34.4778821 from 34.5111234. And this simply makes the puzzle unsolvable because it is ambiguous.

If your interpretation is right that the puzzle asks for "when the hands appear to be at equal distances again" then I have to agree with Cobra's earlier evaluation that Loyd went off the rail with this puzzle. Even we disagree on reading the clock on the picture, and we are just 2 people. So I'd rather believe in my own mathematical clock which at least has one decent solution.

Note that none of this is glib. With the numbers you observed for the hands: 34, 54 and 15 (or 3), you (the puzzle) believes they can appear to be at equal angles. So your eyes (the eyes assumed by the puzzle) have an error of at least 6 degrees which you believed (in an earlier post) to be at most one second. But that is fiction! It all depends on the hand where the error occurs. An error of 6 degrees in the hours hand makes a difference of 12 minutes and in the minutes hand of 1 minute. And you would still think the angles appear equal! That is the reason why you couldn't say it is after 42, 43 (the closest) or 44 minutes. And this error comes on top of the errors in reading the original clock time plus differences between observers which could add another 2 minutes to the uncertainty.

@ Arisktotle I am by no means 100% sure my post is the answer,until I read different ideas about the puzzle I was (again not 100%sure) of the same opinion that was in Cobras first post.I understood that solving the puzzle your way of looking at it could be done without stating the time but along the lines I wanted to get my idea across I needed an agreement on the time.I guess if you were looking at a huge clock on a tower or you put the watch under a magnifying glass you could argue again for an exact in between a whole second time which brings us back to exact.I see the original add in DarkArmys post .You think its easy.Lol you gotta love it.

Let's ignore the second hand temporarily. At roughly 2.55

the minute and hour hand are the desired distance apart. The next time this happens will be roughly 3.40

and thereafter at roughly 4.00

That's a large movement of the minute hand (a little more than 40 min), followed by a small movement (a little more than 20 min). This cycle of large movement, small movement will repeat 11 times until the clock returns to the original position, 12 hours later.

So the length of each cycle is 12/11 hours. Within each cycle, I also suspect that the ratio of the large to the small movement of the minute hand is exactly 2 to 1.

If so, then the duration of the large movement will be 2/3 x 12/11 hours, or exactly 43 and 7/11 minutes.

So this is the answer to the puzzle? By the time of the second diagram, exactly 43 and 7/11 minutes will have elapsed.

We can draw the second hand into the diagrams (in the ideal positions) but I don't think it makes much difference because the positions of the minute and hour hands will appear to be the same, wherever the second hand is. So the second hand is a bit of a diversion. The puzzle essentially concerns the minute and hour hands. The answer is still 43 and 7/11 minutes.

Thx keju for the nice clock pictures you made! I think we already approached the puzzle from all angles now  including your calculations. Without seconds hand the angle on an accurate clock would be exactly 20 minutes at 45 different occasions per full day and the thread wouldn't be that long. It's the third hand that throws in the confusion and turns certain solutions into disputable ones. The answers then start varying with the opinions of the posters on where the inaccuracy is and how big it is permitted to be. The only agreed upon answer for all approaches is "after twelve hours" since noone doubts the angles are then the same again! But smaller outcomes can be defended in court!

Update & correction

My answer above was 43 and 7/11 minutes. I realized after a while that this cannot be right! because if we start roughly like this

then after 43 and 7/11 minutes, there is no way we will end up in the following desired position

because while the minute hand has indeed shifted by about 43 minutes, the second hand (in red) has shifted by only about 25 seconds, and 7/11 minutes is obviously way more than 25 seconds. 

I realized at once that my error was the following suspicion:

The ratio can't be 2 to 1 because this gives the wrong result of 43 and 7/11 minutes. The ratio must still be close to 2 to 1, but it cannot be exactly that. 

So I started thinking how to find the exact ratio. I mentioned previously that the minute hand has clearly made the larger movement (and not the smaller one) between the two diagrams above. But I noticed that something similar is true of the second hand. The second hand too has a corresponding "larger" and "smaller" movement, as we cycle through the various possibilities, but it makes the smaller movement (and not the larger one) between the two diagrams above. Combining this info, we can find the ratio in question. 

For the minute hand, the larger + smaller movements must sum to 12/11 of an hour, as before. Correspondingly, for the second hand, they must sum to 12/11 of a minute.

We need 43 and x minutes, where x is about 25 seconds. Also, x must equal the "smaller movement" of the second hand.

A little algebra shows that the ratio we need is exactly 485 to 247, rather than the mistaken 2 to 1 previously supposed. I skip the algebra to keep things short, but it can easily be confirmed "backwards", see below.

So where I previously calculated the minute hand to move 2/3 x 12/11 hours, I now calculate it to move 485/732 x 12/11 hours. This equals 43 and 247/671 minutes, which looks right because the balance of 247/671 minutes is roughly 25 seconds, matching the movement of the second hand in the diagrams. 

To confirm it mathematically "backwards", we check that the second hand indeed makes the smaller movement mentioned above. Given the ratio 485 to 247, this smaller movement must be 247/732 x 12/11 minutes. Since this is exactly 247/671 minutes, it matches the balance of minutes calculated above. So the ratio is exactly right.

So I'm changing my answer to 43 and 247/671 minutes, a.k.a., 43 minutes, 22 and 58/671 seconds. The second hand is not a diversion after all but necessary to work out the answer.

Sorry for the long post but I tried to be as clear as possible. sameez1, I hope this is useful. The puzzle is quite captivating and I couldn't stop thinking about it. I agree that Sam Lloyd was amazing.