Water clocks puzzle
Mar. 1st, 2006 01:10 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
jackPuzzle: I have two identical cylindrical lemonade bottles I punched small holes into the sides of, just above the bottom1. I find that when I fill them each with water the first empties in 225 seconds and the second more quickly in 75 seconds. Without any other form of measurement, how do I use them to time 25 seconds?
Q. That's just one of those hourglass/numerology puzzles. If one hourglass empties in time p minutes, and the second in time q, then calculate p-1 mod q, and you know pp-1=1 (mod q) so pp-1=1+nq. Run glass Q n times and glass P p-1 times; when the first has finished you have exactly one minute before the second does.
A. If you think that's easy, try it.
Q. It's a trick, isn't it.
A. Probably.
Q. Tilt one of the bottles...
A. If you like. They *are* cylindrical, so you *can* halve the amount of water by tilting them until the water only just covers the base, if you think it'll help.
Q. I could use a...
A. No other equipment is necessary. You can use additional vessels of unknown size if you want to.
Q. How quickly did you solve it?
A. Actually, I made the puzzle up myself last night.
Q. Are you sure your solution works?
A. I may be mistaken. In which case I apologise, and promise to feel very embarassed.
Q. Do I need to use much maths for this?
A. Some. Nothing not taught at A-level IIRC.
Q. Can I find a different solution?
A. I don't think so, but try; it might be better.
[-1] That's not a footnote, that means 'inverse of p'
[1] Really.
Q. That's just one of those hourglass/numerology puzzles. If one hourglass empties in time p minutes, and the second in time q, then calculate p-1 mod q, and you know pp-1=1 (mod q) so pp-1=1+nq. Run glass Q n times and glass P p-1 times; when the first has finished you have exactly one minute before the second does.
A. If you think that's easy, try it.
Q. It's a trick, isn't it.
A. Probably.
Q. Tilt one of the bottles...
A. If you like. They *are* cylindrical, so you *can* halve the amount of water by tilting them until the water only just covers the base, if you think it'll help.
Q. I could use a...
A. No other equipment is necessary. You can use additional vessels of unknown size if you want to.
Q. How quickly did you solve it?
A. Actually, I made the puzzle up myself last night.
Q. Are you sure your solution works?
A. I may be mistaken. In which case I apologise, and promise to feel very embarassed.
Q. Do I need to use much maths for this?
A. Some. Nothing not taught at A-level IIRC.
Q. Can I find a different solution?
A. I don't think so, but try; it might be better.
[-1] That's not a footnote, that means 'inverse of p'
[1] Really.
no subject
Date: 2006-03-01 02:07 pm (UTC)In that case the volume remaining in the slow bottle won't be 1/3 of its total volume; but it will still be the correct volume to empty out of the slow bottle in 75 seconds. Therefore, assuming that the fast bottle empties at an instantaneous rate three times that of the slow bottle when they're equally full (which seems reasonable in spite of the nonlinearity you mention), it should still be the right amount to empty out of the fast bottle in 25 seconds.
no subject
Date: 2006-03-01 02:21 pm (UTC)