Riddles

1. The brother in Spain didn't know about the other brother?
2. A hole is empty. There's no dirt in a hole, so you can't remove any.

Number one was wrong but number two was right.

1) A beggar had a brother who lived in Spain but the brother in Spain said that he had no brother. How could this be? 2) How much dirt can be removed from a hole 3 feet long 4 feet wide and 7 feet deep?

1. The beggar is female.
2. There's nothing in a hole…so you can't remove any.

Both answers are right.

You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of root beer you were planning to open for the celebration, but you find out that one of them is poisoned.

The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.

You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.

You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.

What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?

999, right? Or is that too simple?

1.

HUH?!

1.

How came you by this solution?

Actually, you know what, the solution requires the use of binary numbers, which even I, though somewhat fluent in computer programming, barely understand myself. So I'm gonna scrap that one. Unless someone really wants to try it.

Try this one instead:

A stark raving mad king tells his 100 wisest men he is about to line them up and that he will place either a red or blue hat on each of their heads. Once lined up, they must not communicate amongst themselves. Nor may they attempt to look behind them or remove their own hat.

The king tells the wise men that they will be able to see all the hats in front of them. They will not be able to see the color of their own hat or the hats behind them, although they will be able to hear the answers from all those behind them.

The king will then start with the wise man in the back and ask "what color is your hat?" The wise man will only be allowed to answer "red" or "blue," nothing more. If the answer is incorrect then the wise man will be silently killed. If the answer is correct then the wise man may live but must remain absolutely silent.

The king will then move on to the next wise man and repeat the question.

The king makes it clear that if anyone breaks the rules then all the wise men will die, then allows the wise men to consult before lining them up. The king listens in while the wise men consult each other to make sure they don't devise a plan to cheat. To communicate anything more than their guess of red or blue by coughing or shuffling would be breaking the rules.

What is the maximum number of men they can be guaranteed to save?

I've heard this one before.

Hint: To solve this problem, you need to presume that each wise man can count the total number of red hats in front of them without error, that all the wise men have great attention to detail and that all the wise men care about the greater good.

Yeah. It was about lightbulbs though. And I don't remember the answer. facewall

I kinda do… But not very well.

Gooooood night. I'm not even sure if I want to strain my brain to figure this one out… :P

Gooooood night. I'm not even sure if I want to strain my brain to figure this one out... :P

Exactly what I thinking. I mean, you could go with the easy answer and everyone guess red, and hope that he doesn't just put all blue hats except one. xP I don't think I can figure this out…

EDIT - Let's put my comment more plainly - there is no way I'd figure this out. And the clue didn't help either.

Ugh. That doesn't work… keeps thinking

Gooooood night. I'm not even sure if I want to strain my brain to figure this one out... :P

Exactly what I thinking. I mean, you could go with the easy answer and everyone guess red, and hope that he doesn't just put all blue hats except one. xP I don't think I can figure this out... EDIT - Let's put my comment more plainly - there is no way I'd figure this out. And the clue didn't help either.

I could figure it out… but it would take a while… :P

Gooooood night. I'm not even sure if I want to strain my brain to figure this one out... :P

Exactly what I thinking. I mean, you could go with the easy answer and everyone guess red, and hope that he doesn't just put all blue hats except one. xP I don't think I can figure this out... EDIT - Let's put my comment more plainly - there is no way I'd figure this out. And the clue didn't help either.

I could figure it out... but it would take a while... :P

No, that wouldn't work either… keeps thinking

Did you figure this one out, Christian?

Did you figure this one out, Christian?

Not by a longshot. =P

Did you figure this one out, Christian?

Not by a longshot. =P

lol. So, that means you can post the answer, since we're never going to get it? JK. xP

Solution: 99.

You can save about 50% by having everyone guess randomly.

You can save 50% or more if every even person agrees to call out the color of the hat in front of them. That way the person in front knows what color their hat is, and if the person behind also has the same colored hat then both will survive.

So how can 99 people be saved? The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y).

If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red.

If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

I was going to say the second one. :P

Solution: 99. You can save about 50% by having everyone guess randomly. You can save 50% or more if every even person agrees to call out the color of the hat in front of them. That way the person in front knows what color their hat is, and if the person behind also has the same colored hat then both will survive. So how can 99 people be saved? The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y). If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red. If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

I don't understand about the second answer about 50% being saved. If everyone called out the color in front of them… 99 could be saved, right? Or am I missing something? Possibly 100, if the first guy calls out the color for the guy in front of him is his own color.

Solution: 99. You can save about 50% by having everyone guess randomly. You can save 50% or more if every even person agrees to call out the color of the hat in front of them. That way the person in front knows what color their hat is, and if the person behind also has the same colored hat then both will survive. So how can 99 people be saved? The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y). If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red. If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

I don't understand about the second answer about 50% being saved. If everyone called out the color in front of them... 99 could be saved, right? Or am I missing something? Possibly 100, if the first guy calls out the color for the guy in front of him is his own color.

You're missing something.

Solution: 99. You can save about 50% by having everyone guess randomly. You can save 50% or more if every even person agrees to call out the color of the hat in front of them. That way the person in front knows what color their hat is, and if the person behind also has the same colored hat then both will survive. So how can 99 people be saved? The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y). If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red. If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

I don't understand about the second answer about 50% being saved. If everyone called out the color in front of them... 99 could be saved, right? Or am I missing something? Possibly 100, if the color he calls out for the guy in front of him is his own color.

But each one has to call the color of the guy in front of him, not his own color. So, say the guy behind you says "blue." You know yours is blue. But the color of the guy in front of you is red. So you have to say red, resulting in your death. But if the guy in front of the guy in front of you is red, then the guy in front of you gets to live, because he must say "red," which is also his color. Make sense?

Solution: 99. You can save about 50% by having everyone guess randomly. You can save 50% or more if every even person agrees to call out the color of the hat in front of them. That way the person in front knows what color their hat is, and if the person behind also has the same colored hat then both will survive. So how can 99 people be saved? The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y). If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red. If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

I don't understand about the second answer about 50% being saved. If everyone called out the color in front of them... 99 could be saved, right? Or am I missing something? Possibly 100, if the color he calls out for the guy in front of him is his own color.

But each one has to call the color of the guy in front of him, not his own color. So, say the guy behind you says "blue." You know yours is blue. But the color of the guy in front of you is red. So you have to say red, resulting in your death. But if the guy in front of the guy in front of you is red, then the guy in front of you gets to live, because he must say "red," which is also his color. Make sense?

Oooooooooooooohhhhhhhhhhhh. Yeah. Thanks for explaining. That does.

You could post another one, though I won't get it. xP

Solution: 99. You can save about 50% by having everyone guess randomly. You can save 50% or more if every even person agrees to call out the color of the hat in front of them. That way the person in front knows what color their hat is, and if the person behind also has the same colored hat then both will survive. So how can 99 people be saved? The first wise man counts all the red hats he can see (Q) and then answers "blue" if the number is odd or "red" if the number is even. Each subsequent wise man keeps track of the number of red hats known to have been saved from behind (X), and counts the number of red hats in front (Y). If Q was even, and if X&Y are either both even or are both odd, then the wise man would answer blue. Otherwise the wise man would answer red. If Q was odd, and if X&Y are either both even or are both odd, then the wise man would answer red. Otherwise the wise man would answer blue.

I don't understand about the second answer about 50% being saved. If everyone called out the color in front of them... 99 could be saved, right? Or am I missing something? Possibly 100, if the color he calls out for the guy in front of him is his own color.

But each one has to call the color of the guy in front of him, not his own color. So, say the guy behind you says "blue." You know yours is blue. But the color of the guy in front of you is red. So you have to say red, resulting in your death. But if the guy in front of the guy in front of you is red, then the guy in front of you gets to live, because he must say "red," which is also his color. Make sense?

Oooooooooooooohhhhhhhhhhhh. Yeah. Thanks for explaining. That does.

Okay, good. But yeah, I'd have to read the actual solution probably 10 times before really getting what it's saying.

Next one. This one is slightly less complex, but no less confusing:

I ask Alex to pick any 5 cards out of a deck with no Jokers.

He can inspect then shuffle the deck before picking any five cards. He picks out 5 cards then hands them to me (Peter can't see any of this). I look at the cards and I pick 1 card out and give it back to Alex. I then arrange the other four cards in a special way, and give those 4 cards all face down, and in a neat pile, to Peter.

Peter looks at the 4 cards i gave him, and says out loud which card Alex is holding (suit and number). How?

The solution uses pure logic, not sleight of hand. All Peter needs to know is the order of the cards and what is on their face, nothing more.

Hint: There are only 4 suits, so there will be at least two cards of one suit, one higher and another lower. By careful selection and placement the cards can be used to encode the exact number and suit of the selected card.

Next one. This one is slightly less complex, but no less confusing: _I ask Alex to pick any 5 cards out of a deck with no Jokers._ _He can inspect then shuffle the deck before picking any five cards. He picks out 5 cards then hands them to me (Peter can't see any of this). I look at the cards and I pick 1 card out and give it back to Alex. I then arrange the other four cards in a special way, and give those 4 cards all face down, and in a neat pile, to Peter._ _Peter looks at the 4 cards i gave him, and says out loud which card Alex is holding (suit and number). How?_ _The solution uses pure logic, not sleight of hand. All Peter needs to know is the order of the cards and what is on their face, nothing more._

Wowowow. Let's just say… it's late, and I'm not that smart. =p

Next one. This one is slightly less complex, but no less confusing: _I ask Alex to pick any 5 cards out of a deck with no Jokers._ _He can inspect then shuffle the deck before picking any five cards. He picks out 5 cards then hands them to me (Peter can't see any of this). I look at the cards and I pick 1 card out and give it back to Alex. I then arrange the other four cards in a special way, and give those 4 cards all face down, and in a neat pile, to Peter._ _Peter looks at the 4 cards i gave him, and says out loud which card Alex is holding (suit and number). How?_ _The solution uses pure logic, not sleight of hand. All Peter needs to know is the order of the cards and what is on their face, nothing more._

Wowowow. Let's just say... it's late, and I'm not that smart. =p

Neither am I. This is another one where even the answer is difficult to understand. xP

Hang onnnnnnnnnnnnnn. I might just have gotten it. I was confused. Because I had an idea but thought it was classified as sleight of hand, but Carissa say it isn't… so I dunno.

What's the definition of sleight of hand? xP

What's the definition of sleight of hand? xP

Ummm… a trick where you physically manipulate the cards in some way – by hiding them, switching them around, etc. But this is a purely logical answer.

What's the definition of sleight of hand? xP

Ummm... a trick where you physically manipulate the cards in some way -- by hiding them, switching them around, etc. But this is a purely logical answer.

Exactly. Which is why I won't post my answer because I don't want to make a fool of myself. xp

What's the definition of sleight of hand? xP

Ummm... a trick where you physically manipulate the cards in some way -- by hiding them, switching them around, etc. But this is a purely logical answer.

Exactly. Which is why I won't post my answer because I don't want to make a fool of myself. xp

Post it! You can't possibly make a fool of yourself in answering a question that is next to impossible to answer. xp

What's the definition of sleight of hand? xP

Ummm... a trick where you physically manipulate the cards in some way -- by hiding them, switching them around, etc. But this is a purely logical answer.

Exactly. Which is why I won't post my answer because I don't want to make a fool of myself. xp

Post it! You can't possibly make a fool of yourself in answering a question that is next to impossible to answer. xp

Well… a shot in the dark… my parents use to play a game like this with Daniel a long time ago, and friends of ours showed us how you can have someone pull a card like that and give the answer by placing the cards in a certain order. Though, there has to have pre-planned talking on what ways the cards are placed. But the way you worded it sounded like he'd just be able to tell the number and picture right away… I dunno!

What's the definition of sleight of hand? xP

Ummm... a trick where you physically manipulate the cards in some way -- by hiding them, switching them around, etc. But this is a purely logical answer.

Exactly. Which is why I won't post my answer because I don't want to make a fool of myself. xp

Post it! You can't possibly make a fool of yourself in answering a question that is next to impossible to answer. xp

Well... a shot in the dark... my parents use to play a game like this with Daniel a long time ago, and friends of ours showed us how you can have someone pull a card like that and give the answer by placing the cards in a certain order. Though, there has to have pre-planned talking on what ways the cards are placed. But the way you worded it sounded like he'd just be able to tell the number and picture right away... I dunno!

No, you're actually on the exact right track. But it's a fairly complex system that would have to be agreed upon by the dealer and Peter before the trick started.

What's the definition of sleight of hand? xP

Ummm... a trick where you physically manipulate the cards in some way -- by hiding them, switching them around, etc. But this is a purely logical answer.

Exactly. Which is why I won't post my answer because I don't want to make a fool of myself. xp

Post it! You can't possibly make a fool of yourself in answering a question that is next to impossible to answer. xp

Well... a shot in the dark... my parents use to play a game like this with Daniel a long time ago, and friends of ours showed us how you can have someone pull a card like that and give the answer by placing the cards in a certain order. Though, there has to have pre-planned talking on what ways the cards are placed. But the way you worded it sounded like he'd just be able to tell the number and picture right away... I dunno!

No, you're actually on the exact right track. But it's a fairly complex system that would have to be agreed upon by the dealer and Peter before the trick started.

Coolness. XD I haven't played stuff that hard of finding the exact picture and number, but very similar which has pre-planned talking about what order things are laid.

I allllmost solved this one but I think I reached a dead end… sigh :P

Unfortunately, I have misplaced the webpage on which I found these riddles…. I shall have to do a backwards Google search for it..

Here we are:

Solution: Pick out two cards of the same suit. Select a card for Alex where adding a number no greater than six will result in the number of the other card of the same suit. Adding one to the Ace would cycle to the beginning again and result in a Two. E.g. if you have a King and a Six of Diamonds, hand the King to Alex. The other three cards will be used to encode a number from 1 through 6. Devise a system with Peter to rank all cards uniquely from 1 to 52 (e.g. the two of hearts is 1, the two of diamonds is fourteen etc…). That will allow you to choose from six combinations, depending on where you put the lowest and highest cards.

Here's an easier, better one:

Six logicians finish dinner. The waitress asks, “Do you all want coffee?”

First logician: “I don’t know.”

Second logician: “I don’t know.”

Third logician: “I don’t know.”

Fourth logician: “I don’t know.”

Fifth logician: “I don’t know.”

Sixth logician: “No.”

Who gets coffee and why?

It can't be the first 5. Probably the 6th? Still trying to figure how his word would take affect in light of the others.

Here's an easier, better one: Six logicians finish dinner. The waitress asks, “Do you all want coffee?” First logician: “I don’t know.” Second logician: “I don’t know.” Third logician: “I don’t know.” Fourth logician: “I don’t know.” Fifth logician: “I don’t know.” Sixth logician: “No.” Who gets coffee and why?

The first five. They all would have answered "no" if they themselves did not want coffee, because the waitress used the word all.
That's my guess, anyway.

It can't be the first 5. Probably the 6th? Still trying to figure how his word would take affect in light of the others.

I'm thinking the sixth wouldn't, because he said he didn't want any. I figure that if any waitress heard all of these responses, though, she'd just shrug and walk away, assuming none of them wanted any. :P