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Logic Puzzles

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I'd still like to know what was wrong with my solution. I've been over it a dozen times and it still works for me...

-Q

I might have misunderstood you, but there is one line in your solution that DOES NOT identify whether the coin is heavier or lighter. You MUST be able to tell which it is. Go through it again and help me if i've simply misunderstood your explanation (on that one point) and i'll apologise.

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I'd still like to know what was wrong with my solution. I've been over it a dozen times and it still works for me...

As blackdiamond said, you need to be able to tell if the counterfeit is too light or too heavy in EVERY case. There is one case in your solution where you cannot tell it. (Black has forwarded your solution to me ... after I sent him mine. ;))

BTW, your solution is almost the same as mine, so you're on the right track; there is only one branch of the decision tree that you'll need to correct. :glare:

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Oh yeah I knew about that. He cc'ed that to me, too, and I thought he meant it was completely wrong. I can't figure out how to tell in that one case, so I wouldn't mind being told. I just thought that maybe blackdiamond didn't think my answer worked at all.

-Q

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I can't figure out how to tell in that one case, so I wouldn't mind being told.

Come on, don't give it up so easily! You're almost there!

in that one case

Just to make sure you don't misunderstand, it is true that you cannot say whether the false coin is too light or too heavy in the specific case we are talking about. The solution is to never arrive at that case, i.e. to do something differently in 2a.

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Qwertz. did you solve the (easier) 27 balls one i gave to Aleph (above) for "practice"? This is not a terribly big clue, but i think if you solved that one, you can identify where your small mistake was for this one. (Sorry for giving you the impression that you were completely off track).

if you still want the answer, PM me or CF.

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  • 10 months later...

Here's another brain teaser (let's see if I can revive my thread):

What is black when you buy it, red when you use it, and gray when you dispose of it?

PM me the answer if you have it, anyone (it's not a trick question; when I asked a friend of mine, his answer was "what." No, it's a real thing.)

If anyone else has an interesting puzzle or brain teaser, please share it here and we'll PM the answer to you.

Thanks,

Black.

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MISLEIGH's finally got it too!!! The answer just CAME later, I'm told, without any "intentional thinking," after struggling with it a bit.

I'm always interested in finding out how the brain solves problems like this one, especially what happened in MISLEIGH's case - and my case too.

(Those who submitted the wrong answer, or have not got any answer yet but want to know it, you can PM me or the other people above who have got it so far, when you give up.)

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ANOTHER INTERESTING ONE:

On a wall are 3 standard on/off switches. One (and only one) controls a light bulb inside a light-tight, well-insulated closet. The other two switches do nothing. You can only open the closet door once, and cannot touch/change any switches after the door is open (or re-closed, for that matter). Damaging or disassembling the door, walls, or switches is against the rules.

Within these constraints, can you determine with certainty which switch controls the light bulb?

[PM Blackdiamond if you have the answer. That includes you too, Meta! :) ]

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Here's a puzzle:

There are three prisoners in a prison, and they are given the chance to be released. Two or three prisoners will be released, but only if they can correctly guess what color hat they're wearing. The prisoners are given these hats according to this rule: There are either two white hats and one red, or two red hats and one white.

They are then lined up so that one prisoner stands behind another. Each prisoner cannot see his own hat, but he can see the hats of those who are standing in front of him. Therefore, the first prisoner can't see anyone wearing hats. The second can see the first, and the third can see the first and the second.

The prisoners have unlimited time to answer, and they do so by raising a hand and the providing the correct answer. If any prisoner offers an incorrect answer, all three prisoners are shot. If a prisoner doesn't want to answer, he stays in jail. Assuming that a prisoner will raise his hand to offer an answer as soon as he knows it, and they are completely rational, how can any two prisoners save themselves in any situation? In which cases is it possible that all three prisoners will save themselves?

Edited by source
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Not exactly a logic pussle, but a fun problem anyway:

Three guys check into a hotel. Desk clerk tells them it is $30 for the night. After they go up to their room he realizes he's overcharged them, the room is only $25. So the desk clerk calls the bellhop over, tells him he's overcharged these three guys, gives him five $1 bills and sends him up to give them a refund.

On the way up the bellhop gets thinking: There are three of them and five dollars. You cant divide three into five evenly. So he decides to keep two dollars, gives them three and goes on his way.

So, the guys originally paid $10 each for the room. Then the bellhop gave them each a dollar back. Now they have paid $9 for the room. There are three of them: Three times 9 is 27 plus the two dollars in the bellhop's pocket is $29. What happened to the other dollar?

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