Best...Puzzle...Ever....

[Tuzenbach]

Perhaps a clock, but in order for it to work, you need to refine the answer:

It's a clock tower, a really big one, filled with a thousand people who have all suffered a horrible accident -- or, rather, all of them but three. All the rest are quadriplegics, and of the three who aren't, one of them lost an arm in the accident that robbed the rest of their limbs.

Honestly, it's a pretty easy riddle once you know the answer.

Daniel

No and also no (to the regular "clock" answer).

OK. For those of you devoid of "The Hobbit", I'll post the riddle from it that inspired my own:

"No-legs lay on one-leg,
Two-legs sat near on three-legs,
Four-legs got some."

And if you tell me you don't know the answer to this Tolkien riddle, I'll break into your abode when you're sleeping and paint you all orange!

 

[Pielorinho]

Oh, God, I just realized what it is:

Spoiler

A vase, painted with various portraits on the outside, into which a psychopathic murderer has thrown the body parts of his myriad victims

.

That's horrifying!
Daniel

 

[Tuzenbach]

Oh, God, I just realized what it is:

Spoiler

A vase, painted with various portraits on the outside, into which a psychopathic murderer has thrown the body parts of his myriad victims

.

That's horrifying!
Daniel

No, but that sounds like a really neat idea for a riddle!

 

[Pielorinho]

Well, I know the Tolkien riddle you're talking about, but other than similarly silly ideas, I don't know what the one you're talking about would be. Wanna post the solution in spoiler tags?

Daniel

 

[Iron Sheep]

I post this one earlier but made a BIG mistake, and while I edited to correct perhpas everyone read it and did not catch the edit. So, here it is in it's correct version:

You are blindfolded and handed a deck of 52 cards. (Alternatively you are not blindfolded and handed a deck of 52 cards but can not tell those upside down from those right side up)

Exactly 13 are upside down randomly scattered throughout the deck.

You must manipulate the cards into 2 piles so that there are an equal number of upside down cards in the two piles.

Nice problem. Barring the Gordian knot strategies people have been suggesting, how about the following:

Spoiler

Count out 13 cards from the deck. You then have a pile of 39 cards with n upside-down cards in it, and a pile of 13 cards with 13-n upside-down cards in it, and n right-side-up cards. Now turn the pile of 13 cards over, so it has 13-n right-side-up cards in it, and n upside-down cards. So both piles have the same number of upside-down cards in them.

I'm currently looking for a good air- and fire-related puzzles (not riddle) for my current campaign. The rope one in this thread was good, and I might use that, but I'd be interested in any other ones people may know.

Corran

 

[Vyper]

Here's a real tough one.

There are two persons, A and B. The evil villain (tm) has imprisoned them, and tells them:

"Tomorrow you will be executed, unless you are able to solve this riddle: I am thinking of two numbers, both greater than 1 and smaller than 100. I will tell A the sum of those numbers (X+Y) and B the product (X*Y). If anyone of you finds out the numbers, he is free. But don't give each other any hints!"

After the villain has told A the sum and B the product, both are silent. Some moments of careful thinking later, B confesses: "Sorry, I have no idea, what the numbers were." A says: "Yes, I already knew that you couldn't find them out." - "Did you", B says, "In this case, I know them now." - A replies: "Okay, now I know them, too."

And indeed, after this conversation (which didn't seem to contain any hints - or so the villain thought), both knew the right numbers.

How did they do that, and what were the numbers?

 

[Pielorinho]

Vyper, trial and error gives me 2 and 9 as the numbers. Here's my thinking:

A knows that however you divide the total of the number he's given, you get two numbers whose product can be reached in multiple ways. A number whose product can only be reached in one way keeps B from knowing the solution.

For example, if A was given the number 8, that could be the total of the villain's secret numbers 5 and 3. But if that were the case, then their product would be 15, and if B were given 15, then B would know the villain's numbers were 5 and three.

A knows that B can't know the answer, which tells us that the number A was given can't be the sum of any two such numbers.

Eleven, however, can be broken down into 2 and 9 (product 18, reachable also by numbers 3 and 6); 3 and 8 (product 24, reachable also by numbers 4 and 6); 4 and 7 (product 28, reachable also by numbers 2 and 14); and 5 and 6 (product 30, reachable also by numbers 2 and 15). So if A were given the sum of 11, he would know that B couldn't possibly solve the problem.

B has to work from that: once he finds out that A knows he couldn't solve it, then he has to look at the sums of the different factor-pairs he was given. We know that he was given either 18, 24, 28, or 30; now we have to see which of numbers can be factored in a way as to tell A that the problem is intractable.

...and here I'm getting stuck to explain the rest of my reasoning. Am I on the right track, at least?

Daniel

 

[Zhaneel]

Here's a real tough one.

There are two persons, A and B. The evil villain (tm) has imprisoned them, and tells them:

"Tomorrow you will be executed, unless you are able to solve this riddle: I am thinking of two numbers, both greater than 1 and smaller than 100. I will tell A the sum of those numbers (X+Y) and B the product (X*Y). If anyone of you finds out the numbers, he is free. But don't give each other any hints!"

After the villain has told A the sum and B the product, both are silent. Some moments of careful thinking later, B confesses: "Sorry, I have no idea, what the numbers were." A says: "Yes, I already knew that you couldn't find them out." - "Did you", B says, "In this case, I know them now." - A replies: "Okay, now I know them, too."

And indeed, after this conversation (which didn't seem to contain any hints - or so the villain thought), both knew the right numbers.

How did they do that, and what were the numbers?

I've seen this one, but not gotten the answer. My math friends are torturous when they care to be.

Zhaneel

 

[Pielorinho]

Iron Sheep, that's the intended answer. Good on ya!
Daniel

 

[orsal]

100 prisoners puzzle

Concerning the 1000 prisoners puzzle -- I got the same answer as Impeesa, and I also did it in my head (in about 2 minutes)

Here's a little hint that might help you formulate a general rule for solving the problem with any number of prisoners: Represent all numbers in binary. Then every time you go through the group, you fix the value of one more bit, and kill everyone whose number doesn't have the right bit. The binary representation of the number of prisoners will tell you which half of the remaining prisoners to kill.