Exactly Paulsalomon, and that is what you must realize to solve the riddle.
@flashman, it was used in an interview and I don't think they knew it, but maybe it was unwise of them to do so...

davebishop- are you going to answer my 67.5 minute 3 rope version?

I believe so...
You start one end of the 3rd rope burning at 30 mins (when the first rope's burnt), and light the second end when after 45mins (after the second rope has burnt out).

When the 3rd rope has burnt fully, 67.5 minutes has passed.

sweet. im trying to figure out a hard question. i think im close.

That one wasn't easy... and if you had to solve it without knowing the first one it would be very difficult... nice extension.

Not too hard once you've got the idea: but then, nothing is going to be, I don't think (might be long, but not hard)

Here's one:

With an arbitrarily large number of ropes, which times under 4 hours can you do?
With n ropes, which times under 4 hours can you do?

that's two, actually....

can you set a rope on fire one both ends and in the middle?

GUys-how about a really good live game now?
http://webdiplomacy.net/board.php?gameID=18127

@otto: No

i'm sure someone above has the answer... but i liked the question so I'll give it a try...

assuming you dont have a watch you light one rope on both ends and one on only one end... when the first is finished burning up (could be more or less than 30 min) then you light the 2nd end of the 2nd rope... which should finish burning completely in 45 minutes.

I think thats it.

oh wow - lots of ppl got it... and here i was thinking I'm being all smart =P

i got it after i read the answer! :)

Babak: try my two questions, nobody's done them yet.

Umm, just to be a pedant, dave_bishop's second post in this thread said:

"No tools but a lighter I'm afraid.

(and a stopwatch)"

So if you've got a stopwatch, what you do is this:

1. Start the stopwatch.
2. Wait until the stopwatch tells you that exactly 45 minutes have elapsed.

/end.

Why spend all this time messing with burning ropes if you've got a stopwatch?

Seriously unfunny there.

Ghost, are your ropes still 30mins burning?

With as many as you want and preparation time you could do almost any time you wanted. Just burn down most of the ropes (using method stated above) until you have as smaller time interval as needed. (ie burning one from one end and another from both ends, then after 30mins you can blow out the half-rope, and you have a 30min rope. similarly you can do this for any time 60*2^-n for any natural n. To prepare these ropes you will need n+1 ropes.

but that only gives you a 30 min rope, then a 15 min rope, 7.5 min, 3.75, etc, etc... how would you get 31 mins? I mean it won't be exact

(sorry, 1 hr not 30mins).

If you don't have preparation time, then you could theoretically do virtually any time above 1 hour, having used the first hour to prepare for it. However this would clearly be completely impractical as you would have hundreds of ropes burning at a time and all world need to be stopped and started at the right time.

If you can't blow the ropes out then you're slightly limited, but not much. you can also light a rope in 3 places - each ends and the centre. Then, when either section burns away completely you light the centre of the other part. As such the rope is always burning at 4 places, and so will take 15mins to burn. Expanding this theory you could again make any time 60/(2^n) for natural n, requiring only 1 rope. It should be lit at the end for n=0 , both ends for n=1 and for n=k at both ends and k-1 along it.

Again this could make almost any times as each requires only one rope. I'll get back to you soon for the n ropes question.

No it won't be General, but supposing you had an infinite number of ropes, I could give you some that lasted for such minute fractions of a second. Then it just comes down to expressing 31/60 in the form A + B/2 + C/4 ...

right, using the multiple burnings method (and still assuming I can blow out the ropes) I can now make 60,30,15 - requiring one rope for each of these.
If I then put these ropes in a line, sorted by length, I can choose to light or not light each rope individually. Treating lit ropes as 1 and unlit ropes as 0, this can be thought of as a binary representation. ie,
1 hour: 1000..... (with (n-1) 0s)
45mins:0110... etc
Thus I can make up 2^n different times with n ropes.

(sorry to post so many times, I just walked off then realised this last bit:)
As this second method just requires you to light the ropes at the appropriate time then you dont need to be able to blow them out, so all that matters is you light a section of unburnt rope whenever flaming section burns away.

@ figle: "Seriously unfunny there"

Who's trying to be funny? I just don't understand why he said you had a stopwatch. If you have a stopwatch you don't need the other stuff.

Ah sorry, fair enough. My mistake and was rather harsh response even then - I assumed it was an (attempted) joke.

Also I've realised that with 1 rope you can make an infinite different number of times, simply by lighting it multiple times, which can be used to make any unit fraction of 60. For a fraction of the form 1/[2x] light at x places along the rope, for fractions 1/[2x+1] light at x places along and one end. As above, each time a section burns up completely relight the rope at another location. These times will be less and less useful, rather than the options given above, but could still be done.
Eg, to get 0.9375minutes you have to light the rope at 32 places

Figle:
You cannot set fire to a rope in more than two places.
You cannot prepare.
You are in a mathematical problem where getting all the ropes on fire isn't an issue.
It is an arbitrarily large number of ropes, which is not synonymous with an infinite number, so you cannot do anything that isn't n/2^m

Dealing with just the arbitrarily large number:
You can make, in the first hour: 30, 45, 52.5 etc. mins and the full hour. That much is obvious equally obvious is that you can use the hour to create any number of ropes of length 1/2^A hours where A is arbitrarily large, and therefore can do any time beyond an hour which is of the form n/2^m hours, where n, m are positive integers.

More interesting are the following:
Arbitrarily large without blowing out.
n with blowing out
n without blowing out.

Also, if we did allow multiple burn points, that still only allows you to do rational numbers.

Using the Ghost as my unit of time, where 30mins=1Ghost

You can do following times in Ghosts (all variables are positive integers):
0
1
2-1/2^a
2
3-1/2^a -1/2^b
3
4-1/2^a - 1/2^b -1/2^c
4
etc. (up to 8 Ghosts=4 hours)