Your Favorite Brain Teaser

[2xDouble]

here's one I made up: 1 is 2; 2, 3, 5, and 9 are 5; 4 is 4; 6 is 6; 7 is 3; and 8 is 7.
What is 0? (Hint: it's NOT a math problem)

 

[2xDouble]

My fav riddle:
The billiard balls. There's 12 of them and one is either slightly lighter than the others or slightly heavier. You have a set of scales (ones that strike one way or the other, not indicate weight) and you get to weigh three times. By the end you have to be able to determine which ball is the odd one out and whether it is heavier or lighter.

Obviously the difference is too small for you to tell by guessing manually.

1st weigh in: 6 balls on one scale - 6 balls on the other, whichever bunch are heaviest keep those 6 for the next weigh in
2nd weigh in: 3 balls on one scale - 3 balls on the other, whichever bunch are heaviest keep those 3 for the next weigh in
3rd weigh in: 1 ball on each scale, keep the third on the table, scales will tell you which is the heavier ball, if both balls are the same than the one left on the table must be the heaviest.

Not correct, the odd one out can also be lighter, in which case this doesn't work.

It could, but requires an extra weigh-in.
A better solution would be to weigh all at once, adding a ball to both sides until the scale becomes unbalanced. then, take the two you added last and weigh one against another ball. if it unbalances, the relative weight will be revealed. if it does not, weigh the other ball against the known. and voila you have which ball and it's weight relative to the others.

OR if you recognize them as a billiard set, simply remember the trivial fact that the white cue ball is always slightly smaller than the others and therefore slightly lighter. (of course that would require the set to include all 16 balls...) True fact, that's how coin-operated pool tables work.

 

[skystryke]

Nvm, misread the question ignore this.

 

[BlueTomfoolery]

OK, a turtle walks towards a door 2 metres away at a speed of 1cm a minute. After 5 minutes, a man tries to overtake him. He covers half the distance between him and the turtle, but in that time the turtle has moved forward a small amount.
The man breaks into a run, covering half the distance again, but the turtle has moved.
Now sprinting, he covers half the distance again, but it's still in front.

How can the man ever overtake the turtle?

Is the man in question Achilles by any chance?

 

[RedDeadFred]

here's one I made up: 1 is 2; 2, 3, 5, and 9 are 5; 4 is 4; 6 is 6; 7 is 3; and 8 is 7.
What is 0? (Hint: it's NOT a math problem)

Is 0 zero?
or
0 is a number. Either of those the answer?

 

[TazTheTerrible]

My fav riddle:
The billiard balls. There's 12 of them and one is either slightly lighter than the others or slightly heavier. You have a set of scales (ones that strike one way or the other, not indicate weight) and you get to weigh three times. By the end you have to be able to determine which ball is the odd one out and whether it is heavier or lighter.

Obviously the difference is too small for you to tell by guessing manually.

1st weigh in: 6 balls on one scale - 6 balls on the other, whichever bunch are heaviest keep those 6 for the next weigh in
2nd weigh in: 3 balls on one scale - 3 balls on the other, whichever bunch are heaviest keep those 3 for the next weigh in
3rd weigh in: 1 ball on each scale, keep the third on the table, scales will tell you which is the heavier ball, if both balls are the same than the one left on the table must be the heaviest.

Not correct, the odd one out can also be lighter, in which case this doesn't work.

It could, but requires an extra weigh-in.
A better solution would be to weigh all at once, adding a ball to both sides until the scale becomes unbalanced. then, take the two you added last and weigh one against another ball. if it unbalances, the relative weight will be revealed. if it does not, weigh the other ball against the known. and voila you have which ball and it's weight relative to the others.

OR if you recognize them as a billiard set, simply remember the trivial fact that the white cue ball is always slightly smaller than the others and therefore slightly lighter. (of course that would require the set to include all 16 balls...) True fact, that's how coin-operated pool tables work.

It's not a specific set, the odd one out is determined only by weight and the "all in one go" thing, each time you weigh a different amount would count as a new turn. It's possible though, but complicated.

 

[2xDouble]

here's one I made up: 1 is 2; 2, 3, 5, and 9 are 5; 4 is 4; 6 is 6; 7 is 3; and 8 is 7.
What is 0? (Hint: it's NOT a math problem)

Is 0 zero?
or
0 is a number. Either of those the answer?

they are both answers. they are both correct in different contexts. but not for this riddle.

the hard part is figuring out what the heck I'm talking about. once you got that, it's easy.

 

[senorcromas]

My fav riddle:
The billiard balls. There's 12 of them and one is either slightly lighter than the others or slightly heavier. You have a set of scales (ones that strike one way or the other, not indicate weight) and you get to weigh three times. By the end you have to be able to determine which ball is the odd one out and whether it is heavier or lighter.

Obviously the difference is too small for you to tell by guessing manually.

It'd be the cue ball, and it'd be heavier, because it's ever-so-slightly larger than the rest.

I hope?


Anyway, I can't remember the exact wording, but my favorite brain teaser goes something like:


A professor has been murdered in his own room. There is one suspect, and he says to the police:

"I was walking by, and I wanted to check in on the professor. I unfogged the window, and when I did, I saw the professor was collapsed on his desk. I went to investigate, but was already dead, so I called you."

The police knew he was lying, and arrested him on the spot. How did they know?

 

[SeaCalMaster]

There's no chance to discuss? That's the entire point of the exercise, otherwise you can only guess randomly as you have NO reason to suspect a pattern and there's no guarantee that there will be one.
If you CAN discuss the plan then this method is a perfect as you can get. 99 guaranteed, 1 50/50 chance.

No, there is no chance for discussion, because:

I should expand a bit: you're not allowed to talk after the hats are placed on your heads, but you can hear what the people behind you say and whether they were right.

Oh, and I'm aware that I'm a terrible person.

Again, you're not looking at the entire pattern... simply because there is an odd or even number of either color ahead of you has no effect on which color your hat is. and it's NOT 99% guaranteed. It is all still 100 50/50's. your pattern, if you could coordinate it somehow, might raise the overall odds of survival but does nothing for your individual odds except give you confidence in your guess. What happens if you're the first white hat when everyone behind you says "black"? you die. Try it with 10 of your friends (you play the madman, but please don't actually kill them) you'll see what I'm talking about. It's logical probability, and logic has to start with an assumption somewhere.

So, StevieWonderMk2 has the right answer here, and you can, of course, discuss your strategy before hand (are you familiar with what the word "after" means?). Fun fact: This type of strategy works with any finite number of hats and any finite number of hat colors. Now for something really sadistic: suppose you have an infinite number of people. Is there a way to answer such that you are guaranteed to save all but a finite number of people?

 

[2xDouble]

So, StevieWonderMk2 has the right answer here, and you can, of course, discuss your strategy before hand (are you familiar with what the word "after" means?). Fun fact: This type of strategy works with any finite number of hats and any finite number of hat colors. Now for something really sadistic: suppose you have an infinite number of people. Is there a way to answer such that you are guaranteed to save all but a finite number of people?

First off, "after" has no bearing on the other operative word, "madman". Once you toss insanity into the mix, logic pretty much goes out the window.
Second, there could never be infinite people. Even if there were somehow, there's no way a finite number of guards/henchmen/whatever could contain or imprison them all in such a way that they'd be able to see each other. (not even The Matrix was that complex)
Thirdly, yes there is. Simply divide into finite groups and apply the same strategy. In fact, the person in the "front" of each finite-numbered group would also be the "back" of the next finite-numbered group. He or she would know their own color from the person behind him or her so he or she would be "safe" and could start the next line.

This, of course assumes the system is closed (no other outside interference i.e. rescue attempts, coups or deals with the guards, starvation and dehydration from being trapped for eternity while an infinite number of people answer one at a time, etc.), there is ample time for discussion beforehand, and the "madman" keeps his word to spare the correct answerers.

 

[SeaCalMaster]

Tom Cruise has been married three times. How can this be?

I LOL'd.

OK, a turtle walks towards a door 2 metres away at a speed of 1cm a minute. After 5 minutes, a man tries to overtake him. He covers half the distance between him and the turtle, but in that time the turtle has moved forward a small amount.
The man breaks into a run, covering half the distance again, but the turtle has moved.
Now sprinting, he covers half the distance again, but it's still in front.

How can the man ever overtake the turtle?

At each half-way point, he has taken only half the time to reach the next one. Therefore, both the distance traveled and the time taken converge to finite values, allowing the man to overtake the turtle.

 

[2xDouble]

Anyway, I can't remember the exact wording, but my favorite brain teaser goes something like:


A professor has been murdered in his own room. There is one suspect, and he says to the police:

"I was walking by, and I wanted to check in on the professor. I unfogged the window, and when I did, I saw the professor was collapsed on his desk. I went to investigate, but was already dead, so I called you."

The police knew he was lying, and arrested him on the spot. How did they know?

Windows only fog on the inside. There are no windows in the hallway. I'm sure there are a few other possibilities... something along that line.

 

[SeaCalMaster]

So, StevieWonderMk2 has the right answer here, and you can, of course, discuss your strategy before hand (are you familiar with what the word "after" means?). Fun fact: This type of strategy works with any finite number of hats and any finite number of hat colors. Now for something really sadistic: suppose you have an infinite number of people. Is there a way to answer such that you are guaranteed to save all but a finite number of people?

First off, "after" has no bearing on the other operative word, "madman". Once you toss insanity into the mix, logic pretty much goes out the window.
Second, there could never be infinite people. Even if there were somehow, there's no way a finite number of guards/henchmen/whatever could contain or imprison them all in such a way that they'd be able to see each other. (not even The Matrix was that complex)
Thirdly, yes there is. Simply divide into finite groups and apply the same strategy. In fact, the person in the "front" of each finite-numbered group would also be the "back" of the next finite-numbered group. He or she would know their own color from the person behind him or her so he or she would be "safe" and could start the next line.

This, of course assumes the system is closed (no other outside interference i.e. rescue attempts, coups or deals with the guards, starvation and dehydration from being trapped for eternity while an infinite number of people answer one at a time, etc.), there is ample time for discussion beforehand, and the "madman" keeps his word to spare the correct answerers.

You're being way too literal here. Of course you can't actually have an infinite number of people, mostly because there aren't infinitely many people. It's a thought exercise.

Your answer is incorrect, by the way. The person at the front of each finite group (I feel dirty conflicting terms like that, but we'll go with it) might be able to answer their color correctly, but that color might not correspond with the information that they need to give the people in front of them.

 

[The_root_of_all_evil]

At each half-way point, he has taken only half the time to reach the next one. Therefore, both the distance traveled and the time taken converge to finite values, allowing the man to overtake the turtle.

Close enough. What the question is measuring is the whether the man has overtaken the turtle when he's behind it, so the answer there is no. (And the same with the door)
Using finite counting (Any fraction will hit infinity repeatedly), it's a simple matter.

Is the man in question Achilles by any chance?

Indeed

http://en.wikipedia.org/wiki/Zeno%27s_paradoxes

 

[Composer]

Here's a good one:
You have been captured by some organization. They give you a choice of three ways to die. A room filled with poison gas, a room filled with lions that haven't eaten in a year, or a room full of expert assassins.
Which room do you choose and why?

The lions because they would have starved to death?

i remember that one from grammer school.

 

[rokkolpo]

Here's a good one:
You have been captured by some organization. They give you a choice of three ways to die. A room filled with poison gas, a room filled with lions that haven't eaten in a year, or a room full of expert assassins.
Which room do you choose and why?

the lions because they starved

 

[2xDouble]

So, StevieWonderMk2 has the right answer here, and you can, of course, discuss your strategy before hand (are you familiar with what the word "after" means?). Fun fact: This type of strategy works with any finite number of hats and any finite number of hat colors. Now for something really sadistic: suppose you have an infinite number of people. Is there a way to answer such that you are guaranteed to save all but a finite number of people?

First off, "after" has no bearing on the other operative word, "madman". Once you toss insanity into the mix, logic pretty much goes out the window.
Second, there could never be infinite people. Even if there were somehow, there's no way a finite number of guards/henchmen/whatever could contain or imprison them all in such a way that they'd be able to see each other. (not even The Matrix was that complex)
Thirdly, yes there is. Simply divide into finite groups and apply the same strategy. In fact, the person in the "front" of each finite-numbered group would also be the "back" of the next finite-numbered group. He or she would know their own color from the person behind him or her so he or she would be "safe" and could start the next line.

This, of course assumes the system is closed (no other outside interference i.e. rescue attempts, coups or deals with the guards, starvation and dehydration from being trapped for eternity while an infinite number of people answer one at a time, etc.), there is ample time for discussion beforehand, and the "madman" keeps his word to spare the correct answerers.

You're being way too literal here. Of course you can't actually have an infinite number of people, mostly because there aren't infinitely many people. It's a thought exercise.

Your answer is incorrect, by the way. The person at the front of each finite group (I feel dirty conflicting terms like that, but we'll go with it) might be able to answer their color correctly, but that color might not correspond with the information that they need to give the people in front of them.

yeah, you're right. there would need to be separation between the finite groups. It was just a thought.

Besides (going way off topic), why do thoughts need exercising anyway? Isn't it our thought processes that need exercising? Wouldn't that then become a process exercise, or at least a brain exercise? Or, if it can be applied like regular exercise, wouldn't the problem itself be "thought resistance" and the process of thinking out the problem is the exercise? I could go on, but... I don't care anymore. lol!

 

[sheic99]

Your last good ping-pong ball fell down into a narrow metal pipe embedded in concrete one foot deep.
How can you get it out undamaged, if all the tools you have are your tennis paddle, your shoe-laces, and your plastic water bottle, which does not fit into the pipe?

You pee into the hole.

 

[QuadrAlien]

here's one I made up: 1 is 2; 2, 3, 5, and 9 are 5; 4 is 4; 6 is 6; 7 is 3; and 8 is 7.
What is 0? (Hint: it's NOT a math problem)

Is 0 zero?
or
0 is a number. Either of those the answer?

they are both answers. they are both correct in different contexts. but not for this riddle.

the hard part is figuring out what the heck I'm talking about. once you got that, it's easy.

0 is 6. The second number represents the number of lines that make up the number as they appear on a calculator or digital clock's display.

Your last good ping-pong ball fell down into a narrow metal pipe embedded in concrete one foot deep.
How can you get it out undamaged, if all the tools you have are your tennis paddle, your shoe-laces, and your plastic water bottle, which does not fit into the pipe?

You pee into the hole.

The water bottle presumably was meant to have still contained some water, so there was no need for that.... Tempted as I have been to enter 'URINE' instead of 'WATER' when this puzzle appeared in Professor Layton.

 

[2xDouble]

here's one I made up: 1 is 2; 2, 3, 5, and 9 are 5; 4 is 4; 6 is 6; 7 is 3; and 8 is 7.
What is 0? (Hint: it's NOT a math problem)

Is 0 zero?
or
0 is a number. Either of those the answer?

they are both answers. they are both correct in different contexts. but not for this riddle.

the hard part is figuring out what the heck I'm talking about. once you got that, it's easy.

0 is 6. The second number represents the number of lines that make up the number as they appear on a calculator or digital clock's display.

Your last good ping-pong ball fell down into a narrow metal pipe embedded in concrete one foot deep.
How can you get it out undamaged, if all the tools you have are your tennis paddle, your shoe-laces, and your plastic water bottle, which does not fit into the pipe?

You pee into the hole.

The water bottle presumably was meant to have still contained some water, so there was no need for that.... Tempted as I have been to enter 'URINE' instead of 'WATER' when this puzzle appeared in Professor Layton.

bingo