Scott
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How many cups to a gallon?

DiMono
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Man, you stole my question.

Oneiromancer: I wasn't implying any sort of fowl play, just awe.

Anyway, here's the solution:

Clearly, we need to acchieve either 1, 2, 3, or 4 gallons in one of the jugs. If we fill the 5 with water and then pour 3 of it out to the 3 jug, we have 3 and 2. Then we pour out the 3 jug and empty the 2 left in the 5 jug in to it. Refill the 5 jug and top up the 3 jug, leaving 4 gallons in the 5 gallon jug. Empty the 3 gallon jug and fill it from the 5 gallon jug, leaving 1 gallon in that jug. We've now had 1, 2 and 4 gallons in the 5 gallon jug, and 3 gallons in the 3 gallon jug, so whichever happens to be 100 cups is where you pour the coffee mix.

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bibelot
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Oops, meant 3- and 5-cup jugs. Silly me.

DiMono
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...arrive at 100 cups using a 3 cup jug and a 5 cup jug? I think you're going to need to clarify, my friend.

Never mind, I know what you mean, but I don't have time to work out the math.

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TripleM
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Nobody said you needed 100 cups, just to get it to 1%.. I've got it down to about 4%, but I'm stuck. nice problem though.

zex20913
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I suppose that this solution is a bit cheap, but it does work.

Fill 5. (0,5)
Fill 3 from 5. (3,2)
Dump 3. (0,2)
Fill 3 from 5. (2,0)
Fill 5. (2,5)
Fill 3 from 5. (3,4)
Dump 3. (0,4)
Fill 3 from 5. (3,1)
Put coffee into 5. (3,1 (with coffee))
Dump 3. (0,1)

Repeat these 33 times.
Fill 3. (3,1)
Pour 3 into 5. (0, 1+3n)
Hold 5 over never-ending-water supply for overflow.

After 33, drink coffee.

Like I said, it's cheap, but effective.

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TripleM
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I'm pretty sure you won't be allowed to do that.. I don't think filling an already full jug would be allowed as a way to change the concentration..

bibelot
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zex20913 wrote:
Repeat these 33 times.
Fill 3. (3,1)
Pour 3 into 5. (0, 1+3n)
Hold 5 over never-ending-water supply for overflow.

This isn't allowed, really. Also I don't think you'd get the right concentration since you're adding water in three cup units rather than all at once, so the coffee won't be equally distributed in the water.

[Edited by bibelot on 02-22-2004 at 12:35 AM GMT]

TripleM
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Yes! Finally solved it.. this is such a nice problem.


I'll post another puzzle soon..

TripleM
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Heres a nice puzzle.


One time in history the Russian postal system was corrupt. Any letter, package, or box that could be easily opened would be opened and anything inside would be removed (no matter its value), before it would be posted to its target. However, they would never open anything which was locked.

Boris and Natasha each have a large supply of boxes of different sizes, each capable of being locked by padlocks. Also, Boris and Natasha each has a large supply of padlocks with matching keys. The padlocks have unique keys. Finally, Boris has a ring that he would like to send to Natasha. How can Boris send the ring to Natasha so that she can wear it (without either of them destroying any locks or boxes)?

[Edited by TripleM on 02-22-2004 at 08:00 AM GMT]

eytanz
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He should send a box with a combination lock on it.

A week or so later, he should send a letter, with the combination written on the envelope.

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TripleM
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Hmm.. combination locks didn't exist then
Na, that wasn't meant to be the solution. I've looked at another source and rephrased it above.

Koro
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Is this what you're looking for?

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TripleM
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Woah, nope but thats good enough you're up.

other solution below:

The_Red_Hawk
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I've heard this puzzle before, except it was a rich Indian rajah who wanted to send a great big emerald to his brother.

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Koro
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Okay, Pirate Jim must cross a bridge over some troubled water. The bridge has weight capacity of up to 80 kg. Pirate Jim can weigh 78 kg when traveling at his lightest. However, he needs to take with him 3 large gold doubloons, each weighing 1 kg. In these parts, a pirate can be none too careful and can't leave his gold lying around. How does he cross the bridge?
The next bridge has a capacity of 79 kg. How does he cross that one?

[Edited by Koro on 02-23-2004 at 02:14 AM GMT]

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The_Red_Hawk
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If the bridge is short enough...he jumps over it.




Does he have anything with him or anything he can use?

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Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

.....the king of the skies.....

RoboBob3000
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Hmm... can he toss the dubloons ahead of him one at a time, or is that not allowed?

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The_Red_Hawk
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I thought of that, but it's the same as leaving them unguarded on the other side.

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Slashing, whirling, diving, twirling,
Snapping, turning, rising, swirling,
Screeching, flipping, gliding, sliding,
The red hawk's dance of death.

.....the king of the skies.....

Koro
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If the bridge is short enough...he jumps over it.

Both bridges are too long.

Does he have anything with him or anything he can use?

Afraid not.

Hmm... can he toss the dubloons ahead of him one at a time, or is that not allowed?

Not sure what you're after. If he does this, all the doubloons are still on the bridge at some time with him if it is of sufficient length, so it would break.

[Edited by Koro on 02-23-2004 at 12:42 AM GMT: Oh and Red Hawk is right if you meant throwing them all the way over]

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eytanz
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If the answer you are looking for is juggling, could one of the physicists on this board explain why that won't work or will I have to look up a detailed explanation?

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Koro
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eytanz wrote:
If the answer you are looking for is juggling, could one of the physicists on this board explain why that won't work or will I have to look up a detailed explanation?

Are you making a guess? Or are you guessing that it is unsolvable? And if you are making a guess that you don't actually believe for the sake of solving the puzzle, what about the second bridge? Need some elaboration.

Also, a note on capacity that I thought about putting in the original post. If the bridge has a capacity of 80 kg, it would have a reasonable fudge factor and all weights portrayed aren't accurate to infinite sig figs. If we were that stingy, Pirate Jim would have to do his own compensation and wouldn't dare set foot an the bridge if he equalled its capacity because a light breeze or the mere act of walking would kill it. I don't want to say much more on anything...

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eytanz
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I'm familiar with this puzzle (or a very similar one), in which the answer is supposed to be that the pirate juggles the dubloons. Unfortunately, that's phyisically impossible - when you're juggling, you're exerting force to keep the balls up, which means that due to some sort of Newtonian thing, an equal force is operating on you in the other direction.

In other words, if you weight 68kg and are juggling 3 1kg dubloons, you effectively weigh 71kgs - exactly the same as if you were carrying these dubloons, if not more (depending on how high the dubloons go), even if at any given time you aren't physically holding the dubloons.

If you have another solution, then this is irrelevent.

[Edited by eytanz on 02-23-2004 at 01:46 AM GMT]

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Koro
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You are right about the first bridge, although you admitted to having seen the problem before
The physics dilemma can be cleared up, although I wasn't going to require it, but maybe I should now. The second bridge is a corallary, and needs a corallary-ish answer.

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DiMono
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Can Pirate Jim phone a friend? Someone who weighs less, or has a trampoline, or something?

Actually, I'm going to say he juggles them with one hand. Here's my explanation to remove the physics problems:

You're juggling your dubloons with two hands. At any time, you are holding one dubloon, and throwing another in the air. As long as you don't throw your dubloons too high, the extra weight caused by the juggling is negligable.

Now, if you're juggling with one hand, two are in the air instead of one, so that's one more kilo of weight not being applied, saving you for the bridge.

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Scott
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Whats all this juggling about. The mans a pirate after all. What he will do is dig a hole and put the extra coins in that then place a big fat X over it.

Koro
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DiMono is right for the second bridge. Either one handed juggling or a shower pattern. So both bridges are solved. We can argue about physics or what pirates should or should not do... or DiMono can post the next question.

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eytanz
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It doesn't matter how many hands you use, or how many dubloons are in the air at any given time. If you're juggling dubloons, the force you're exerting on the bridge (i.e. your effective weight) will not increase (actually, it will increase and decrease constantly, but the average weight will be 71kg - at times it will be below 70kg, but at times it will be even higher, and the bridge will collapse before you have a chance to continue juggling).

There's a nice explanation of this somewhere on the web - I can find it if you want.

Of course, this is nitpicking - and I definitely think that DiMono should have the question.

If you want to test this for yourself, by the way, try juggling while standing on a scale. You'll see you don't weight less than you'd do if you just carry the balls.

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DiMono
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Alright, time for a new type of puzzle to this thread. The bolded text is a puzzle to be solved, where the solution is a seven letter word. What is the word, and how do you arrive at it?

Detectives and drunken lushes around edge of tavern

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bibelot
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