Lost solution

Yes,the solution is 1 hour 5 minutes and 5 seconds.It is because there is a moment when the time pointing out at 03:59:39 which is the other exact evenly hands separation.

Well,that is turn out to be the impossibility for the puzzle to exist because the accuracy depend on demand.

I found a solution that satisfies me!

Suppose that the clock hands are designed to move continuously and also that Loyd has depicted them exactly 120 degrees apart. As we know, this would be impossible if the clock was accurate, so at least one of the hands must be malfunctioning, e.g., it is frozen in place, not moving. (Loyd expects us to deduce this.)

I considered the possibilities and the only sensible culprit is the hour hand. The hour hand is indeed not moving -- it slipped off its shaft (these things do happen) and has been frozen in the same position all day. However, the other two hands are working fine and are correctly shown, exactly 120 degrees apart. Their exact positions are in fact these, consistent with Loyd's diagram:

Note that the balance of 34/59 on the minute hand equals 34 and 34/59 seconds, confirming that the hands are exactly 20 units apart. 

The true time shown by the clock is unknown because we don't know where the hour hand really should be. It could be in many different places - twelve different places, if I'm not mistaken. Just by luck, however, it happens to be frozen in the place (shown in red below) where all three hands are exactly 120 degrees apart:

So we have a clock with a broken hour hand and we don't know what the true time is. This does not matter, because in exactly one hour, the minute and second hands will appear again in the same positions and -- the hour hand being frozen -- all three hands will again be exactly 120 degrees apart.

Answer: Exactly one hour!

I like this solution because it means that Loyd's diagram is drawn very accurately (e.g. the minute hand is not pointing to 54 or 55 but in between, slightly closer to 55). Also, the clock hands move continuously and are separated by exactly 120 degrees, which is the kind of precision I think Loyd would have liked. Finally we have this funny frozen hour hand which is just the kind of mischievous trick Loyd was known for.

What would you guess is the probablility that a watch manufacturer would publish a puzzle to promote its new watch whose solution depends on the watch being broken?

@ Keju ...A frozen hour hand,does this really leave no doubt for you.

I agree it's not ideal but it seems to be a minor issue. Besides, having already commissioned Loyd to produce a puzzle, they would have had to accept whatever he came up with. Too late to commission someone else!

Look at the original news clip:

At the bottom it says, "Insist on an Ingersoll - the name is on the dial". I bet they added that after seeing Loyd's solution. Loyd's watch doesn't say "Ingersoll" on the dial!

By the way, sameez1 anticipated this possibility way back in post 4: 

Quite right I think. That's an acceptable ad gimmick. Besides, it's just a puzzle. Nobody would seriously confuse the watch in Loyd's puzzle with an Ingersoll watch.

There's always doubt  but I'll explain why this solution satisfies me. It's because equally spaced hands is an outright impossibility on an accurate clock. Loyd surely knew this and yet (we can imagine) he wanted a puzzle with equally spaced hands anyway. Why? Well, because the idea is just too pretty to ignore!

So what to do? If you have equally spaced hands, the clock must be inaccurate. So why not construct a puzzle around this? Loyd could reasonably also have expected his readers to deduce all this. A clock can of course be inaccurate in many different ways, but a stuck hand is a pretty common one. So he could reasonably have expected us to consider it also. I certainly did.

Loyd had other puzzles with funny, misbehaving clocks. One of them was this:

It's six o'clock and the time is shown correctly. Unfortunately this clock has its minute and hour hands wrongly mounted on each other's pinions. When will it next show the correct time? 

Loyd was not averse to weird clocks!

On the other hand, Loyd was also rather precise. In this puzzle, previously mentioned on this thread:

the minute and hour hands are equally spaced around the 6. What is the time? The accepted answer is 8 hours, 18 and 6/13 minutes. So Loyd was thinking of a clock with continuously moving hands. The answer is correct to the decimal.

Combining such precision with a potentially misbehaving clock, I'm quite satisfied with the solution I suggested. I remain open to a better one though!

We could divide a circle with subtle spacing as many as possible.And still the accuracy of hour hand is only describing the minutes part left into one hour completion.The accuracy of minute hand is only describing the second part left into one minute completion.Thus the subtle dividing on around a circle is solely enhancing the second part accuracy of a digital clock.And the deviation is not more than one second.In practical clock,the time in diagram could be achieved.But the second part accuracy depend on demand.

But the puzzle doesn't ask for equally spaced hands. It asks for hands that have the appearance of being equally spaced, without specifying how to judge that.

But I don't object to your being satisfied. That's entirely up to you.

@ Keju I think the judges would have to send you a watch after veiwing all of your entries as a most persistant contestant prize.I thank you for that post with the entire original ad. Now I have the crazy clock of Zurich to think about too.

@ kesie  sorry I missed your post before.Your suggested 12:00:00 is without a doubt a time they will actually be equally spaced,but a sooner time keeps putting a doubt.I have just used your observation at an earlier time around 3:16 thank you I am going to post this.

@ all The hands will be equally distant at 3:16:21:81 they will all be exactly together.And they will certainly appear equally distant.                                   I was suspecting a diversion of some sort in the puzzle they don't have to be 120 degrees apart they just need to be an equal distance apart I am confident of an answer for the first time.

This one's easy, though. Let a_m and a_h (angular displacements relative to angular positions when t = 0) be measured in radians, while t (time elapsed relative to 6:00:00) is measured in minutes. Then we just need the lowest positive value of t satisfying:

a_m = a_h + 2πn  for some integer n

After substitution, we get:

πt/30 = πt/360 + 2πn --> 11t/360 = 2n --> t = 720n/11

So the lowest positive value of t satisfying the equation occurs when n = 1. Therefore, the correct value for t is:

t = 720/11 minutes = 65 5/11 minutes = 1 hour, 5 minutes, 27 3/11 seconds

Finally, we conclude from the above that 7:05:27 is the next time the clock will show the correct time.

@ cobra91 great solution to the crazy clock,curiously the same 1 hour 5min 27.27 seconds it takes for the hands to come all exactly back together again.

I hadn't seen it prior to finding it here, if that's what you mean.

Is it worth posting under "More Puzzles"? Such decisions can only be made on an individual basis (I can't decide the matter for you!), but it's not a puzzle that I (personally) would create a new and separate forum topic for. Instead, I'd probably post it in a related thread which already exists.

Sorry - this doesn't work at all (at least not for an ideal clock with 3 uniformly rotating hands). At 3:16:21.81, the hour and minute hands are between the '3' and '4' (and do line up exactly as you intended), but the second hand is between the '4' and '5'. I already proved (a long time ago) that the 3 hands will only line up at noon or midnight.

Note also that it's irrelevant whether noon/midnight is considered a valid instance of equal spacing, because the diagram itself would still present the same contradiction (the hands still cannot be equally spaced at any time near the one shown, unless the clock's mechanics differ from what is being assumed).

Perhaps someone would like to consider the case of a real wristwatch, for which the second hand typically moves in discrete jumps. I have two of those, and for both, the second hand jumps once per tick. The Timex has five ticks per second, and the Bulova has six ticks per second. The minute and hour hands appear to move continuously, but presumably also jump once per tick in steps that are too small to see.

Assume all hands are straight up at midnight, and take as the start time the time close to Loyd's hand position that gives the least possible deviation from equal spacing. Is there ever a time when the deviation is less than it is just one tick (1/5 or 1/6 second) after the start time? If so, what is the first such time?

To solve this puzzle, you will have to define a measure for deviation from equal spacing, as I did previously, and include that definition as part of your solution.

Approaching this logically, we have to conclude that:

a) this is a depiction of a real watch, not a mathematical or digital interpretation.  If it were otherwise, Loyd would have to have stated it.  Since he didn't, we have to assume this is a real watch from that time period, which is analog and moves as such.  The time, as depicted, is obviously 2:54:34.something.

b) as n9531l and others have carefully shown, there is no time where the hands are all spaced exactly 120 degrees apart.

c) Loyd would have known this, and being the genius that he was, would not have carefully crafted the word "appear" into the puzzle, had he not known there was no exact solution.

Therefore the best solution is the one of approximation that n9531l has laid out on pages 4 and 5 of this thread.  Either that or noon/midnight.

@Platzerwasel:  For my previous comments to be valid, there is one assumption I neglected to mention, which is that the proposed solution position must appear different from the starting position. Perhaps Loyd thought this would be understood. Without it, a time one microsecond after the start time would be a potential solution.

Back in post 34, sameez1 considered a clock whose face was divided into 60 units, where every unit represents one tick. The second hand ticks once a second, the minute hand ticks once a minute, and the hour hand ticks once every 12 minutes.

At 2:54:34, the second hand is at 34, the minute hand at 54, and the hour hand at 14. Exactly 120 degrees apart. This is Loyd's diagram:

At 3:38:58, the second hand will be at 58, the minute hand at 38, and the hour hand at 18. Exactly 120 degrees apart again. This is 44 minutes and 24 seconds away. This is the answer sameez1 recommended.

We might also consider 3:16:16, when all three hands point to 16. This is only 21 minutes and 42 seconds away.

I don't like this clock because I always thought Loyd would have gone for continuously moving hands without making us guess the clock mechanism. Also, it means that Loyd's diagram was not completely precise. (Loyd's second hand is not at 34, but is clearly between 34 and 35. Why did he not draw it exactly at 34? Was it so hard?)

But almost everyone else thinks these points are minor! If so, then I would endorse this clock and maybe Loyd's trick was all three hands lining up at 3:16:16?