Fretful my yes or no questions.
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Do they plan to eat the carrots?
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Do they need the carrots for the dinner party?
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Do they have a pet rabbit?
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Is the answer some sort of pun?
erislover: Nope, sort of right track, though.
ivylass:
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Do they plan to eat the carrots?
No. -
Do they need the carrots for the dinner party?
No. -
Do they have a pet rabbit?
No. -
Is the answer some sort of pun?
No.
String problem
You take both strings - you light one on both ends, and the other on just the one end. When the first string is burned through, you light the second end of the second string.
When the second string is burned - 45 minutes.
Quoth MaxTheVool:
I should imagine that finding the solution for thirteen is difficult, since it is in fact impossible. While three trits is enough information to distinguish between 26 (or even 27) possibilities, there’s no way to implement it in this case. Every weighing must have the same number of coins on each pan, or it’s guaranteed to not balance (I’m assuming that the difference in weight is less than the weight of a genuine coin). If we weigh four vs. four (or 3 vs. 3, 2 vs. 2, or 1 vs 1) on the first weighing, and they balance, then we have ten possibilties left: One of five coins is the counterfeit, and it can be either heavy or light. Since we only have two weighings left, we can only distinguish between 9 possibilities.
If, on the other hand, we put five or more on each side, then we’ve got a problem if they don’t balance, since the counterfeit can be any one of those ten. Again, we only have enough weighings left to distinguish nine possibilities.
Re: the pirates. This seems too simple and I suspect I’m missing something, but here’s my solution anyway.
Three pirates go swimming. Pirate 1 (lowest ranked) proposes to give nothing to himself and pirates 2-5 get 25 gold coins each. Pirate 1 votes yes to the plan because he doesn’t want to go swimming. Pirates 2 and 3 vote yes because 25 is the best they can expect anyway. Pirate 4 votes yes because he does the math and sees that if this keeps up he’s not going to get anything. Pirates 6-10 vote no, because they are getting screwed and pirate 5 votes no, because he knows he can get more if he plays his cards righ. The vote is 6-4 and pirate 1 goes swimming.
Pirate 2 proposes the same thing (3-6 get 25 each) with the same result. 6-3 and pirate two goes swimming. Pirate 3 proposes that he get’s one and pirates 4,5,6 each get 33. 3 votes yes, 4 votes yes (he’ll take anything at this point), but 5-10 vote no. 6-2 and pirate 3 goes swimming.
Pirate 4 says “aw crap, I knew this was going to happen” and proposes that pirates 5 and 6 split it 50-50. Pirate 4 votes yes, pirates 5 and 6 vote yes (having done the math and realizing 50 is the best they can hope for) and pirates 7-10 vote no. The vote is 3-3 and the cap’n says “make it so”.
Here are a handful.[list=1]
[li]I’m going to ask you if there are more than 6.02 x 1023 stars in the universe. Write the answer on a piece of paper. Make sure that everyone will agree you have written the correct answer on the paper.[/li][li]What common English word begins and ends with “he”? What common English word has “adac” in the middle?[/li][li]Let me tell you a story about a miller’s daughter. In a certain town lived a miller, his daughter, and the evil mayor. The miller was in debt to the mayor, and the mayor had his eye on the miller’s daughter. The mayor made a proposition: he would place a black stone and a white stone in a bag, and the miller’s daughter would pick one out in front of the whole village. If she drew the white stone, the mayor would forgive the miller’s debt. If she drew the black stone, the mayor would marry the miller’s daughter and take the mill. The miller had no choice but to agree. Late one night, the miller’s daughter overheard the mayor saying that he would place two black stones in the bag. The miller’s daughter was quite upset, but knew that if she told anyone, the mayor would seize her father’s mill. How can she get out of marrying the mayor and save the mill?[/li][li]I lay four cards on a table. Each one has a back and a front, and these sides are marked. You can see the side of the card which is facing up. The first card has a square on the front. The second card has a circle on the front. The third card has a square on the back. The fourth card has a circle on the back. If I tell you that any card with a square on its front also has a square on its back, which cards must you turn over to verify that statement?[/li][li]In Nintendo’s classic videogame The Legend of Zelda, enemies and items come in two colors, red and blue. Red items are stronger than blue items, but red monsters are weaker than blue monsters. Is this a consistent usage of the colors red and blue?[/li][/list=1]
Addendum to the above: The alternate solution would be that no pirates get thrown into the ocean. After 1 proposes his initial solution, 5 may vote yes bringing the vote to 5-5. Since there is no reason why pirate 4 couldn’t pick pirates 9 and 10 when it’s his turn, 5 may decide to play the probability, and consider a 100% chance of 25 gold coins better than a 40% (2 out of 5) chance of 50 gold coins.
I was refering to the post above ultrafilter’s, of course.
Ultrafilter:
2.)The word is the pronoun “he” 
3.) the only thing I can think of is that she palm a white stone and cheat using slight of hand. The mayor will know she cheated, but can’t do anything because then the town will know he cheated.
4.) the card with the square on the front and the card with the circle on the back. You said specifically “any card with a square on its front also has a square on it’s back”. This does not mean the converse is true and that cards with squares on the back must have squares on the front, so the card with the square on the back doesn’t matter - it fits no matter what’s on it’s front. The card with the circle on the front obviously doesn’t have a square on the front, so you eliminate that one as well. That leaves the two I mentioned.
Beeblebrox: You got 4). For 2), the word is at least four letters long. And for 3), there’s no palming involved.
- the answer is the same for both…headache
- yes…there are more than that many stars. Anyone disagree?
Well, #2 kinda contains the answer to both questions in itself: “headache”.
People who don’t know anything about the size of the universe, or who disagree with scientific conclusions. Think creationists. They should be able to agree that you’ve written the correct answer on the piece of paper.
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headache for both parts
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She takes a stone and drops it on the floor. She then says “Lets look in the bag and see what stone is left. I must have chosen the other one.”
Uh … you draw a thumbs-up sign on the paper, and if there are any creationists present, you hand it to them upside down?
The trick here is to, as quickly as possible, find two blocks which are different. That is, one of them is aluminum and one is duraluminum. Once you have that pair, you can measure that pair against a pair of unknown blocks and easily observe whether that pair has 0, 1, or 2 aluminum.
So, here’s the solution:
(1) Pick two blocks. Weigh them against each other. If they don’t balance, then it’s trivial to solve the whole thing in 10 weighings.
if they do balance, we know that the two blocks we have picked up so far are the same type. But we don’t know what type they are. So we proceed to step 2…
(2) Take our current two blocks and weigh them, together, against two fresh blocks. If these also balance, then all 4 of these blocks you have so far are the same type (but you don’t know what type it is). If so, put 2 of these 4 blocks into a special pile somewhere, and repeat step 2.
(3) At some point, as long as there’s at least one aluminum and one duraluminum, your 2 vs. 2 weighing will not balance. When that happens, you’ll immediately know whether the 2 you already had, and anything in the special pile, are aluminum or duraluminum. Now take the 2 new ones you were weighing against your 2 “control” blocks. Weigh them against each other. If they don’t balance, one is aluminum and one is duraluminum, and it’s easy from there on. If they do balance, then they are both the same, and both the opposite of whatever your control group was. So take one of these, and one from your control group, and proceed.
(I hope that was all clear)
A convincing argument… but not quite correct. In fact, it is possible to determine that a coin is counterfeit without determining whether it is heavy or light. For instance, if I hand you 3 coins, tell you “one of these is counterfeit… which is it?” and you weigh (a) vs (b) and they balance, you will have determined that © is counterfeit without gaining any information about whether it’s light or heavy. However, this is only possible if that coin has never been weighed. And if there are two coins that have never been weighed, there’s obviously no way to distinguish between them. So no more than one coin can possibly end up in this state. Therefore, if we have 5 coins and two weighings, there can be 9 possible outcomes:
(a)-heavy, (a)-light
(b)-heavy, (b)-light
©-heavy, ©-light
(d)-heavy, (d)-light
and (e)-counterfeit.
Regarding the pirates: Suppose pirate 1 proposed to split the loot 50-50 between pirates 5 and 6. Pirate 1 votes yes, pirates 2 and 3 would realize that they die unless they vote yes, and pirates 5 and 6 would not do any better and vote yes.
About the miller’s daughter- she draws a stone, pops it in her mouth and swallows it. Then she tells the mayor to reveal the stone she didn’t select.
About the carrots: I give up.
I’m a creationist and I say there ARE more than 6000 some odd stars. So there. 
As for the stars one, you can never guarantee that everyone will agree. Even if you know every written language in the world, you still have to deal with people who have no knowledge of a written language, blind people who can’t read Braille, babies, the illiterate, insane, irrational, and obtuse, most of whom will never agree that you have written “the correct answer” on that paper. (And if they somehow all do, I won’t. I’m just going to be that way.)
The Zelda red/blue logic : Assuming that ‘all items are good’ and ‘all monsters are bad’, and ‘stronger items are better for you’ and ‘stronger monsters are worse for you’, then red represents ‘better’ and blue represents ‘worse’. (Certainly the colors could have been different and a rationale could have been formed for that too.)
Are the carrots meant to be consumed by an animal?