Lost solution

@ keju Your long post worked out to the milli-second is  fantastic, well appreciated and something for me to digest.Yes that pesky second hand is the brunt of the problem. I had wondered before if i watched from 12:00:00 to the first 120 degree appearance of the three hands 12:21:41 thats 21 min 41 seconds if I doubled it,thats 43 min22 seconds if that time would stay constant  but as I tested that it would steadily get short as I went to the next closest appearance so I scrapped that idea.Now with this fraction of a second added to it I am thinking you are right.Well done ...I am going to submit all the answers i have from my friends on this thread  and get my free watch for sure.

With any luck, you will get all 10,500 of them!

Not confident that my answer is right though! There are signs that it is still wrong . Back if I have something more decisive. Still can't put the thing down.

Yes I don't think any solution is 100% without a logical argument  (from yourself) that it is accurate. I wanted to see if the puzzle would draw the interest of any one else besides myself,you could almost say it haunts you.

No confident analysis yet but I noticed something curious. 

The diagram shows a time, just before 3, when all three hands are equally spaced. Almost everyone here mentioned that, if the clock is perfect, there is no such time, and I wanted to verify this for myself.

I first ignored the second hand and looked for the exact time (just before 3) when the minute and hour hands would be exactly 20 minutes apart. I used this diagram:

Here, the two hands are meant to be separated by exactly 20 units, counting 60 such units around the clock. The minute hand falls short of 12 by t units. (Fat blue triangle.) So the hour hand must be short of 3 by t/12 units. (Skinny blue triangle.) Cos the minute hand moves twelve times faster around the clock than the hour hand. So we can see from the diagram that:

 t + 15 - t/12 = 20

Solving this, t = 5 and 5/11. So the time shown is 2 hours, 54 and 6/11 minutes. This is the exact time, just before 3, when the minute and hour hands are exactly 20 units apart.

I draw it like this:

giving the exact position of the hands in units of 60 around the clock. The hour hand must be at 14 6/11 because that's exactly 20 units away from 54 6/11.

And then I worked out (for this exact time) where the second hand must be. For Loyd's equidistance, it has to be at 34 6/11 units (shown in red below), but unfortunately its correct location at this time can be shown to be 32 8/11 units.

We get this correct location by noticing that there is a balance of 6/11 minutes on the minute hand, which translates to 32 8/11 seconds.

In other words, slightly before 3, when the minute and hour hands are exactly 20 units apart, then (if the clock is perfect) the second hand cannot be in the (ideal) red position shown, but must be in the (less than ideal) black position.

My observation is that if you look at Loyd's diagram, the minute, hour and second hands seem to be placed very carefully at 54 6/11, 14 6/11 and 34 6/11 respectively! The minute hand for example is drawn between 54 and 55, just a little closer to 55. Indeed, if I had been asked to portray the minute hand as showing 54 6/11, I would have drawn it exactly where Loyd drew it. Likewise for the hour and second hands. Here is the diagram again:

It seems to me that Loyd was drawing a very precise diagram here. He calculated the time (just before 3) when the minute and hour hands would be 20 units apart and drew them into the diagram very carefully. Minute hand at 54 6/11 and hour hand at 14 6/11. 

If he knew exactly where the minute and hour hands were, he must also have known that the second hand would then have to be at 32 8/11. I can only conclude that he nevertheless deliberately misdrew it at 34 6/11, the ideal red position where it would be equidistant from the other two hands.

This seems to me a clue of some sort. Loyd is telling us: I know that the second hand belongs at 32 8/11 but I'm placing it at 34 6/11 anyway. Can you see what that means?

I considered the possibility that he was trying to tell us that the second hand was simply broken! It just happens to be stuck at 34 6/11 the whole day, even though the minute and hour hands are working perfectly. (Obviously not an Ingersoll clock.) If so, then we simply have to look for the next time at which the three hands are equally spaced, given that the second hand is permanently frozen at 34 6/11. Unfortunately, I looked carefully and there is no such time, apart from the trivial one exactly 12 hours later.

So whatever Loyd was trying to tell us, it's not that the second hand is broken.

I also considered that he was trying to tell us that the second hand was simply running fast by 1 1/9 seconds, this being the difference between 32 8/11 and 34 6/11. Perhaps the clock took a knock previously and the second hand was displaced forward by 1 1/9 seconds. The clock otherwise works perfectly. So, again, when is the next time that the hands are equidistant, given that the second hand is running fast by 1 1/9 seconds?

But this led nowhere either. I couldn't find any such time, except the trivial one exactly 12 hours later.

Maybe I'm just barking up the wrong tree but it seems very curious that Loyd's diagram shows the hands at pretty much the exact positions, 54 6/11, 14 6/11 and 34 6/11. The first two numbers are precisely where the minute and hour hands would be if they were exactly 20 units apart shortly before 3 o'clock.

This guy is seriously keeping me awake at night ..

..@ Keju Without the absolute precise math you have so elegantly displayed I was in a similar frame of mind as you are when I posted this, wondering if I would be alone in my lack of complete confidence of the solution.I could not have hoped for any better a response.Without question Sam Loyd was  a unique genius with a little bit of rascal in him knowing what havoc this puzzle could produce in the obsessive mind (most people would look a few minutes come up with an answer and forget about it). I had mistakenly wondered out loud if  I could find out if Ingersol had actually given any free watches out,which would prove to me there was a correct solution. My wife answered you know this borders on crazy when you know you still won't have the answer your stuck in an endless circle quit.(easy for her) but with all the good answers on here I have shelved this puzzle.I suggest you do the same for your sanity.Not saying I am deserting you if you need to discuss it.Just worried about your health LOL.

Has anyone tried simulating the problem for an empirical solution?

@Rubenoganhout If you look close the big hand is not on the 11 or whole 5minute mark it is slightly before, the second hand is slightly before the 7 mark,and the small hand is slightly before the 3 mark 2:54:34 would be a more accurate non fraction time,but this time as shown is questionable to the mechanical structure of the watch,as shown in keju post,so wouldnt the answer be never.Then again we have the word appears brings up another perfectly logical argument for doubt,and more as pointed out on this thread.Of course a person could devise a solution and know in his heart he is right,I can't. I hope our language differences don't create any problem or misunderstanding.

Ok I didn t see it so precise. My eyes are not so good lol. But I downloaded the foto.

When I zoom in I estimate that it must be about 25.4 seconds before it is exactly 5 minutes before three O clock.  ( yes in Dutch we say 5 before Three not 55 past ) and we say ten to half and not 20 past )  If that is the case then the seconde wijzer  ( arm )  is precisely on the twelve and the Big arm is on the 5 minutes to three and the small arm is still before the three. because it is still not three O clock. I this case the second arm must still goes 5 times around till the big arm is on the twelve again and also the second arm is on the twelve again. could the small arm be then on the three? Do I make a thinking error?  I any case what I thought was the big arm and the small arm are depended on each other but you have never the second arm precisely on top on every minute and every hour. In Practice. So I mean I don t know if that is the assumption of the problem if the second arm must be on top too. If not I thought the second arm is then every round about 4.6 minutes past the half hour and it doesnt matter then how the big and small arm are developed then after because the round of the second arm then never change en will be there all the tome every minute and the others arms can have very different positions. The second arm will then never be sinchrone with the other arms. Therefore I thought all the positions could be only the same after 12 hours past and it is again the same time as started. But maybe I am wrong.

I've been out a while and just came across this thread. It gives the kind of math puzzle I like to solve, so I decided to take a crack at it.

All three hands of a real mechanical watch move in steps. The size of the steps is determined by the number of teeth in the gears chosen by the watch designer. Those numbers are different for different watches, so I will assume all three hands move continuously.

Let t be the current time in seconds, with t=0 at noon when the hands are all straight up. Let s, m, and h be the positions of the second, minute, and hour hands, measured in sixtieths of one complete rotation. Then we can calculate their values at the current time as

s=FracPart(t/60)*60
m=FracPart(t/3600)*60
h=FracPart(t/43200)*60

Now we need to define M, a figure of merit that tells how evenly the hands are spaced. Let v1, v2, and v3 equal the values of s, m, and h when they are arranged in numerically increasing order. We can calculate those values as

v1=min(h,m,s)
v3=max(h,m,s)
v2=h+m+s-v1-v3

Ideally we would like the difference of the two larger numbers and difference of the two smaller numbers to both be exactly 20. So we will take the sum of the amounts by which those differences differ from 20 as our figure of merit. That is,

M=abs(v3-v2-20)+abs(v2-v1-20)

At noon, M=40 and it never gets any larger. The smaller M is, the more evenly spaced the hands are. I found an online graphing calculator (meta-calculator.com) and used it to plot M against t. The graph for Loyd's starting position is shown below. M is has a minimum value of 0.03 at a value of t corresponding to a time of 2:54:34.57. If the solution needs to be in whole seconds, the best time is 2:54:35, giving M=0.87. M goes up fast at the second hand moves, and approaches a minimum again just one minute later, at 2:55.35, giving M=0.96 as the minimum. This could be the solution, if you think the hands still appear to be equally spaced.

If you want more equal spacing, you'll have to wait over an hour to the time shown in the second graph. M gets to a minimum of 0.20 at a time of 3:57:37.97. As several people noted, the ambiguity in the problem is the lack of any critera for what counts as an appearance of equal spacing.

@ n9531l glad to see your post Yes it seems that if you want to define appears you get into a problem no matter how accurately you calculate the clocks movement.I thank you for your well thought out graphing. does the graph fit the angles at around 3:37:57  3:38:58.? 8:00:20 should have a low M if I am getting this right?

I like n9531l's clear analysis. I think it is good for at least 7000 of the 10,500 watches that were either never rewarded or incorrectly!

@sameez1:  Here are the values for the times you listed.

3:37:57 (t=13077)  M=1.16

3:38:58 (t=13138)  M=1.68

8:00:20 (t=28820)  M=0.36

@Arisktotle:  I hope you will communicate your opinion to the Ingersoll Company. I will be waiting for some watches in the mail.

@Arisktotle and n9531l  before  I even read the content it was nice to see a post from n9531l. as I was missing his always informative point of view. Now I have something new to ponder over.I ask though how soon is it?One minute?or the more mathematically closer 1hour 3minute 3 seconds.I really think if this contest was in the present I would send in each viable answer individually ,not so much for the watch but the answer,which I also have a shadow of a doubt there is one.

Problem is, if there were still any watches to award when this thread was created, then NM Dale would have already won them all. I'm virtually 100% certain that no puzzle by Sam Loyd would involve any sort of approximation-related ambiguity ("equal distances apart" means "equal distances apart", not "roughly equal distances apart"), nor would such a puzzle feature a trivial solution that could easily be guessed without thought or calculation.

@ cobra91  NMDale answers 5minute 5seconds is how soon the hands will appear equal again.no disrespect to him I would not be entering that answer.You could send it  if you like it.

@ 9531l what time will the M be exactly 40 again will it take 12 hours or does it work out at 1:05:25 approximately.

The tell-tale sign is that nobody found an official solution nor the story of what happened with the prizes. Can only mean the puzzle collapsed due to diverse interpretations. I bet it was withdrawn before its closure date.

Then I fear you'd be wasting your [hypothetical] time. As I see it, the probability that NM Dale's answer is identical to Loyd's is at least 0.9-0.95, while the probability that Loyd's intended solution depends upon arbitrary unspecified definitions (such as an exact mathematical definition for "appear") is precisely zero. The probability that either of my proposed answers from post #3 ("never" or "12 hours") is correct is nonzero, but negligible.

There's a small but significant chance (~0.05-0.1) that Loyd's answer involves some other unconventional (but very logical) mechanism for the pictured clock, which no one has thought of yet. The key, though, is that the clock in NM Dale's solution always displays the correct time with perfect precision (albeit via a different encoding than is typical for analogue clocks), which is an important property of ideal clocks that is tough to replicate for alternative rotation schemes.