Lost solution

I'm afraid you simply don't understand how Dale's clock actually works, which is probably why you've been so dismissive of the clever solution associated with it.

To read Dale's clock, merely observe the number pointed at by the hour hand, interpretting this as the current hour. Then observe the number (plus fifths) pointed at by the minute hand, interpretting this as 1/5 of the current minute ("3  3/5" means "18 minutes"). Finally, observe the number (plus fifths) pointed at by the second hand, interpretting this as 1/5 of the current second ("8  1/5" means "41 seconds"). Of course, for the minute and second hands, "12" is read as "0".

So at 3:55:35, the hour hand points at "3", the minute hand points at "11", and the second hand points at "7". Then at 4:00:40, the hour hand points at "4", the minute hand points at "12", and the second hand points at "8".

@ n9531l Yes I jumped on that too soon without really thinking.

@ cobra91 1hour 5 min 5 seconds makes sense if thats is what he meant,5min 5seconds on a normal clock does not. 

If you get to the time you propose by letting the clock run instead of by rotating the face, all the hands cannot be precisely a unit of 5 further along. The hour hand moves exactly 1/12 of a complete rotation in exactly one hour. At your proposed time, the second hand will be 1/12 of a rotation ahead of the starting time, but the hour hand will be more than 1/12 of a rotation ahead, since more than one hour has elapsed. So the angular separation of the second and hour hands will be smaller than at the starting time.

In other words, rotating the face is not a valid way of predicting exactly where the hands will be at some future time, since it ignores the different rates at which the hands move.

It's true that at your proposed time, most people would agree that the hands appear equally spaced. But it's not likely to be the solution, because there's an earlier time at which the hands are even closer to being equally spaced, namely at a time close to 3:37:58.

I said "close to" because I'm assuming the second hand moves continuously and not in one second ticks, and the optimum time for most nearly equal spacing will not occur at a whole number of seconds. For the starting position, the optimum time to the nearest millisecond is 2:54:34.548. This is probably the time Loyd intended to represent in his picture. For the time you have proposed, the optimum is 3:59:39.945, and for the earlier time I mentioned the optimum is 3:37:57.966. This earlier time gives more nearly equal spacing than your proposed time, whether you use whole or fractional seconds.

(Note: All the above assumes a watch whose hands all point straight up at every noontime and all of whose hands move continuously.)

 As a duchman myself with a fair command of the english language I believe there are similar ambiguities in both languages regarding the word "appear". I'd say that one of its translations "lijken" refers to "what something looks like upon superficial inspection" which allows for inaccuracy. The problem with the "show up" interpretation is that it suggests a "disappearance before" as in "the sun appears above the horizon". But the clock hands were always in sight!

Whatever everyone thinks the puzzle means, it has failed by not explaining itself clearly. Nobody can be expected to work through all the thought packages in this thread before deciding on a route to solve it.

And yet, it has succeeded by giving us so much to think about.

The way the puzzle was worded requires the solver to make some assumptions. Different solvers may make different assumptions, and if they wish to, they can argue for years about whose assumptions make more sense. Rather than do that, I think the best approach is to include with any proposed solution the assumptions that were made, and not be bothered by the different solutions that will result from different assumptions. We really don't have to worry about any watches actually being awarded based on our proposed solutions.

@ n9531l I started this thread to find out how someone else approached the puzzle.I must say again that I could not have hoped for a more  thorough set of ingenius solution approaches.Seeing these has solidified my opinion that you can't really be 100% sure because in any solution there is that shadow of a doubt caused by other very logical arguments. (please excuse my 95% emotion 5% logic post of 30 seconds)....So I agree now and have in past posts with both these opinions about the puzzle. 

The same distance between clock hands will be repeated after 1 hour 5 minutes and 25 seconds.

Sorry,i am too hasty.

The actual clock shows time at 02:54:34. It needs 5 minutes and 26 seconds to headed to 3 o'clock.And it needs one hour next to headed to 4 o'clock.At this moment the distance between hour and minute hands to be equal with initial distance.The rest is it needs 40 seconds to separate second hand from hour hand.The total time is 1 hour 6 minutes and 6 seconds.I thought the order of hands should be in consideration.

@Mr_Edt:  Still too hasty. At 4 o'clock the minute hand is straight up, but 40 seconds later it no longer is. After the starting time, there are a couple of earlier times at which the hands are closer to equal spacing than at your proposed time.

@Mr_Edt read through some of the earlier posts,theoretically the hands will make that seams to or appears to be an equal distance apart two times in that 1 hour 6 min time frame you propose. 

@n95311. It is assuming the clock is a ticking clock type which have 60 ticks in one clock cycle.The minute hand won't be moved until the second hand have reached 60 ticks.

Yes there is earlier time where the hands have separated evenly in order of hour,minute,second of hands in clock wise.If the order of hands are equal with the initial position,it need more than one hour period.It depend on the actual question.

@sameez1. Your question seems to be implied by @n95311 question above.

This is, almost without question, the kind of reasoning that's needed to resolve the apparent contradiction between Loyd's own diagram and the conventional assumptions most people want to make.

How often does the hour hand of a "ticking clock" move? I've never heard of such a clock type, but if the way it works is consistent with the puzzle's given information (namely, that the 3 hands are equally spaced at the time shown in the illustration), then you just may have cracked this thing once and for all! I didn't think a sounder explanation than Dale's could exist (realistically speaking, anyway), but an established clock type producing a clean and nontrivial solution would definitely do the trick!  

The amount of ticks betwen number symbols are 5 ticks.So,there are 60 ticks in one clock cycle (because the numbers are from 1 to 12).Twelve ticks of minute hand will make hour hand to move by one tick.So 5 ticks of hour hand will produce 60 ticks of minute hand and one hour period have been completed. 

@ Mr_Edt please read post #34

As I discover other Sam Loyd puzzles I see he presented as Arisktotle might say a puzzle without the convention of an accurate description to be solved and then left out the solution.LOL there is some devil in the man that makes his puzzles even more endearing (after you get over the frustration it causes)I post this as a sort of proof of my opinion.The stupendously brilliant Sam Loyd’s Cyclopedia of 5000 Puzzles, Tricks and Conundrums, with Answers, originally published in 1914, is now available online.

The riddles are pathetic (“What vine does beef grow on? The bo-vine”), but the rest is mostly terrific. One problem: Loyd withheld the solutions to some puzzles, offering a cash prize. He never followed up with the solutions, so they’ve become stumpers. Here’s one, called “The Trader’s Profit”:

A dealer sold a bicycle for $50, and then bought it back for $40, thereby clearly making $10, as he had the same bicycle back and $10 besides. Now having bought it for $40, he resold it for $45, and made $5 more, or $15 in all.

“But,” says a bookkeeper, “the man starts off with a wheel worth $50, and at the end of the second sale has just $55! How then could he make more than $5? You see the selling of the wheel at $50 is a mere exchange, which shows neither profit nor loss, but when he buys at $50 and sells at $45, he makes $5, and that is all there is to it.”

“I claim,” says an accountant, “that when he sells at $50 and buys back at $40, he has clearly and positively made $10, because he has the same wheel and $10, but when he now sells at $45 he makes that mere exchange referred to, which shows neither profit nor loss, and does not affect his first profit, and has made exactly $10.”

“It is a simple transaction, which any scholar in the primary class should be able to figure out mentally, and yet we are confronted by three different answers,” Loyd says. “The first shows a profit of $15, such as any bicycle dealer would; while the bookkeeper is clearly able to demonstrate that more than $5 could not be made, and yet the President of the New York Stock Exchange was bold enough to maintain over his own signature that the correct profit should be $10.”

I’m thinking the accountant’s right, but then I was a journalism major.

I think this watch puzzle is one of those puzzles.I am sure I will get an opposing opinion,but that is it exactly.Opinion

Unlike the "clock puzzle", this particular question has a relatively straightforward answer:

The dealer ends up with one bicycle less, but 55 dollars more, than what he started with. So the "profit" depends on how we define the true value of the bicycle. If we define that value based on the first sale, then the profit is 5 dollars. If we instead base the value on the cost of buying back the bicycle, then the profit is 15 dollars. Finally, if the bicycle's true value is defined by the second sale, then the profit is 10 dollars.

In summary, it all boils down to one's choice of formal definitions.