Impossible Brain Teezer

[Daveway]

My teacher put a brain teezer out a few days ago and my class thinks it is impossible to figure out.

>>There are 3 crates holding apples and oranges. 1 crate is holding apples. 1 crate is holding oranges. Another crate is holding apples and oranges. The labels for the crates have been put on the wrong crates. You are allowed to open one crate. Which crate do you open to figure out what the other crates have in them? You must only open 1.<<

I'm convinced this is impossible and my teacher is just making a joke out of it.

[mkrishnan]

So you have three crates:

1) marked apples - contains ??
2) marked oranges - contains ??
3) marked A&O - contains ??

You open the one marked A&O. It has to be wrong, so it contains either apples or oranges. Suppose it contains apples:

1) marked oranges - cannot be oranges, cannot be apples, so it must be A&O
2) marked apples - cannot be apples, cannot be A&O (because the one marked oranges is), must be oranges.
3) marked A&O - contains apples

Suppose it contains oranges:

1) marked apples - cannot be apples, cannot be oranges (cuz the one you opened is), so it must be A&O.
2) marked oranges - cannot be oranges, cannot be A&O, so it must be apples.
3) marked A&O - oranges.

Doesn't this work?

[mkrishnan]

Hmmm, now that I think about it, can't you open any of the three crates and follow the same logic? If you open the oranges crate, which must contain apples or apples & oranges (since the label is wrong), then you can open the crate labelled with whatever is actually in the oranges crate, next, and it cannot contain its own label or the actual contents of the oranges crate, leaving only one option. Etc. So I'm not sure. It seems to me like you can open any of the three crates you want, and you'll get the answer.

[angelneo]

This is identical to the question about the 3 light bulbs.

3 light bulbs are located in a room without windows. there are 3 unmarked switches outside the room which supposedly switches on the light bulbs. You are allowed to walk out the room and go back in once and only once and you should be able figure out which switch is linked to which light bulb.

This is relatively easy once you figure it out

EDIT: you cannot see the light bulbs once you are out of the room or even the light from the bulbs

BTW, don't reveal the answer. let other people have the fun of answering

[atszyman]

My teacher put a brain teezer out a few days ago and my class thinks it is impossible to figure out.

>>There are 3 crates holding apples and oranges. 1 crate is holding apples. 1 crate is holding oranges. Another crate is holding apples and oranges. The labels for the crates have been put on the wrong crates. You are allowed to open one crate. Which crate do you open to figure out what the other crates have in them? You must only open 1.<<

I'm convinced this is impossible and my teacher is just making a joke out of it.

Are the labels guaranteed to be wrong? or could they be right? Might the apple crate label be on the apple crate while the oranges and apples/oranges crates are switched?

If they are guaranteed to be on the wrong crates it doesn't matter which one you open first.

Say you open crate 1 that is labeled apples/oranges. If they are guaranteed to be wrong you will find either apples or oranges.

Say crate 2 is labeled apples and crate 3 is labeled oranges. If you found oranges in crate 1. Crate 3 has to contain apples or crate 2 is labeled correctly. You can swap any of the labels and come up with the same logic.

If the crates aren't all labeled wrong then I don't know where to start.

Edit: Beat to the punch and as always more eloquently than I can put it.

[Sun Baked]

If all 3 crates are marked wrong, then they can contain anything but what is marked on them.

Open the first crate and find out what is in it. You know it'll be the label from one of the other two crates.

So you know it'll be one of the "other" two labels.

When you open the crate you'll be taking off a label and grabbing one from another crate and placing it on the crate you open.

The label in your hand is what is in the unopened crate with the label.

[angelneo]

Since we are on the subject of brain teaser, here's another one. (not sure if you guys have heard this one)

A father who recently died left a will to his 4 sons. the oldest son should get half of his asset, the second son should get a quarter of his asset, the third should get an 1/8 of the asset and the last should get a tenth of the asset. Unfortunately, the father left 39 sheeps to his sons, and the sons brought forth their case to you. How would you solve this?


EDIT: For those who want to have a shot at this, do not scroll down as the answer is a few post away

[mkrishnan]

3 light bulbs are located in a room without windows. there are 3 unmarked switches outside the room which supposedly switches on the light bulbs. You are allowed to walk out the room and go back in once and only once and you should be able figure out which switch is linked to which light bulb.

Hmmm...then you must be thinking of a different solution to this than the one I thought of, and saw elsewhere on the net.... The solution I know doesn't seem applicable to me in the case of the fruit crates.

[angelneo]

Hmmm...then you must be thinking of a different solution to this than the one I thought of, and saw elsewhere on the net.... The solution I know doesn't seem applicable to me in the case of the fruit crates.
Come to think of it, you are right, its somewhat different, i guess. Oops

[mkrishnan]

A father who recently died left a will to his 4 sons. the oldest son should get half of his asset, the second son should get a quarter of his asset, the third should get an 1/8 of the asset and the last should get a tenth of the asset. Unfortunately, the father left 39 sheeps to his sons, and the sons brought forth their case to you. How would you solve this?

Can I try? I haven't heard this one before. If you pretend that he had forty sheep, then:

Son 1 -> 20 sheep (1/2), 19 left
Son 2 -> 10 sheep (1/4), 9 left
Son 3 -> 5 sheep (1/8), 4 left
Son 4 -> 4 sheep (1/10), none left.

The secret lies in the fact that 1/2 + 1/4 + 1/8 + 1/10 < 1.

Is that correct?

[mkrishnan]

Come to think of it, you are right, its somewhat different, i guess. Oops

Aww, that's too bad. I was hoping for some really neat answer. :D

[angelneo]

Can I try? I haven't heard this one before. If you pretend that he had forty sheep, then:

Son 1 -> 20 sheep (1/2), 19 left
Son 2 -> 10 sheep (1/4), 9 left
Son 3 -> 5 sheep (1/8), 4 left
Son 4 -> 4 sheep (1/10), none left.

The secret lies in the fact that 1/2 + 1/4 + 1/8 + 1/10 < 1.

Is that correct?
yelp. I remember going through this quite sometime ago when I was about 12 and have a hard time solving.

[Daveway]

The boxes had labels identifying what was in them. But, the labels were them scrambled an put on different boxes.

[angelneo]

The boxes had labels identifying what was in them. But, the labels were them scrambled an put on different boxes. I think the idea is that the statement "The labels for the crates have been put on the wrong crates." must be correct in order to solve this.

[blackfox]

Since we are on the subject of brain teaser, here's another one. (not sure if you guys have heard this one)

A father who recently died left a will to his 4 sons. the oldest son should get half of his asset, the second son should get a quarter of his asset, the third should get an 1/8 of the asset and the last should get a tenth of the asset. Unfortunately, the father left 39 sheeps to his sons, and the sons brought forth their case to you. How would you solve this?
I can't be very elegant with my solution, I'm afraid..but :

last son - 4 sheep (10% of 40)
3rd son - 5 sheep (1/8 of 40)
2nd son - 10 sheep (25% of 40)
1st son - 20 sheep. (50% of 40)

involves phantom sheep, but otherwise works. Has something to do with 1/8th me thinks, but haven't messed around with math in a while.

[Sun Baked]

I think the idea is that the statement "The labels for the crates have been put on the wrong crates." must be correct in order to solve this.Only breaks down when the labels are "randomly" placed on the crates -- meaning you don't know if the label is right or wrong until you open them.

[angelneo]

I can't be very elegant with my solution, I'm afraid..but :

last son - 4 sheep (10% of 40)
3rd son - 5 sheep (1/8 of 40)
2nd son - 10 sheep (25% of 40)
1st son - 20 sheep. (50% of 40)

involves phantom sheep, but otherwise works. Has something to do with 1/8th me thinks, but haven't messed around with math in a while.
Oh man, this is way too easy for macrumors members

[angelneo]

Only breaks down when the labels are "randomly" placed on the crates -- meaning you don't know if the label is right or wrong until you open them. But if the labels are random. What if you open a crate which had the label correct, in spite of the randomness, you wouldn't be able to determine the other 2.

[Daveway]

OK. Lets say that the crates were accurately labeled with what was in the box. Could it be done then?

[mkrishnan]

OK. Lets say that the crates were accurately labeled with what was in the box. Could it be done then?

I think that if you assume that it is possible that any less than all of the crates are mislabelled (i.e. either one of the crates is correctly labelled, or all of them are), that this is not possible.

[Sun Baked]

But if the labels are random. What if you open a crate which had the label correct, in spite of the randomness, you wouldn't be able to determine the other 2.That's why I said it "breaks down" only under the randomness example.

OK. Lets say that the crates were accurately labeled with what was in the box. Could it be done then?Then why would you open a crate?

Unless you wanted something to eat.

[angelneo]

That's why I said it "breaks down" only under the randomness example.

Then why would you open a crate?

Unless you wanted something to eat. Oops, sorry for my misunderstanding. :)
talking about fruits and eating is making me hungry for my lunch.

[mkrishnan]

Oops, sorry for my misunderstanding. :)
talking about fruits and eating is making me hungry for my lunch.

Hey, pass me one of those bananas! Don't hog! :mad:

:p:D

[broken_keyboard]

My teacher put a brain teezer out a few days ago and my class thinks it is impossible to figure out.

>>There are 3 crates holding apples and oranges. 1 crate is holding apples. 1 crate is holding oranges. Another crate is holding apples and oranges. The labels for the crates have been put on the wrong crates. You are allowed to open one crate. Which crate do you open to figure out what the other crates have in them? You must only open 1.<<

I'm convinced this is impossible and my teacher is just making a joke out of it.


Open the one with apples and oranges. Then see which is the heavier fruit and you can use that to work out which is in the other crates.

I don't know if that is the offical or clever answer to the puzzle or not but I think it would work in reality.

Edit: if you open a pure (only apples or only oranges) one first then the other pure one will be the one most different in weight, and the mixed one will have the middle weight.

[mkrishnan]

Open the one with apples and oranges. Then see which is the heavier fruit and you can use that to work out which is in the other crates.

Hmmm...interesting. Assuming that the boxes are packed in a similar enough way, so that the amount of empty space in each box is approximately the same, then this would work. Nice! :)

[Daveway]

Edit: if you open a pure (only apples or only oranges) one first then the other pure one will be the one most different in weight, and the mixed one will have the middle weight.

This would be a logical way of doing it.

[mkrishnan]

Erm, wait a minute. The concerns about empty space aside, you don't know which one contains apples and oranges, so you can't assume that you'll get that one. And this solution doesn't work unless you happen to open that one first.

[puckhead193]

It is a crate can't u peer into it and see inside :confused:

[urban1985_]

It seems like the question is incomplete. Like what ppl said here about the randomness factor, it screws things up.

It's too easy if the labels are ALWAYS wrong though.

[broken_keyboard]

Erm, wait a minute. The concerns about empty space aside, you don't know which one contains apples and oranges, so you can't assume that you'll get that one. And this solution doesn't work unless you happen to open that one first.

If you get one with all apples, then the one with the most different weight must be the oranges and the in-between weight must be the mix. Likewise if you get the oranges.

[jsw]

I'd smell the crates....

Edit: seriously.

[jsw]

BTW: weight-related answers ("weigh them...") assume not only that apples and oranges weigh differently when packed at maximum density (possibly a good assumption), but that all crates are full - which wasn't specified.

[broken_keyboard]

I'd smell the crates....

Edit: seriously.

Wow, I like that answer even more than mine.

[Jon'sLightBulbs]

Since we are on the subject of brain teaser, here's another one. (not sure if you guys have heard this one)

A father who recently died left a will to his 4 sons. the oldest son should get half of his asset, the second son should get a quarter of his asset, the third should get an 1/8 of the asset and the last should get a tenth of the asset. Unfortunately, the father left 39 sheeps to his sons, and the sons brought forth their case to you. How would you solve this?



50% of 39 sheeps will unavoidably involve fractions of sheeps, which i assume is an unacceptable answer. It's clear that the total of the assetts the father has left his sons equals less than 100%. However, this is irrelevant given that 39 sheeps (it seems) constitute 100% of the assetts. So some other person is left with (1 - 1/2 - 1/4 - 1/8 - 1/10) of the father's 39 sheep. This does not avoid the conclusion that 50% of 39 sheep is an uneven number.

Has anyone reconciled this?

[mkrishnan]

Has anyone reconciled this?

Yeah, I guess everyone's solution assumes that giving everyone more sheep than they're entitled to is acceptable. :rolleyes:

[Sun Baked]

50% of 39 sheeps will unavoidably involve fractions of sheeps, which i assume is an unacceptable answer. It's clear that the total of the assetts the father has left his sons equals less than 100%. However, this is irrelevant given that 39 sheeps (it seems) constitute 100% of the assetts. So some other person is left with (1 - 1/2 - 1/4 - 1/8 - 1/10) of the father's 39 sheep. This does not avoid the conclusion that 50% of 39 sheep is an uneven number.

Has anyone reconciled this?Since it all adds up to less than 100% -- everybody can get more than what was in the will.

Otherwise they'll end with a bloody mess for the executor of the will, and he'd still have nearly an entire sheep leftover.

[jsw]

Can I try? I haven't heard this one before. If you pretend that he had forty sheep, then:

Son 1 -> 20 sheep (1/2), 19 left
Son 2 -> 10 sheep (1/4), 9 left
Son 3 -> 5 sheep (1/8), 4 left
Son 4 -> 4 sheep (1/10), none left.

The secret lies in the fact that 1/2 + 1/4 + 1/8 + 1/10 < 1.

Is that correct?
No.

The 1st son gets 19.5 sheep (one is slaughtered).
The 2nd son gets 9.75 sheep (oops - looks like two get slaughtered).
The 3rd son gets 4.875 sheep (darn, three sheep bite the dust).
The 4th son gets 3.9 sheep (there goes sheep number four).

35 sheep make it through alive. Four get butchered and divided.

You get to keep 0.975 sheep for your efforts. Hopefully, at least some of what they give you is edible.

This way, the old man's will is faithfully executed, the sons learn a lesson about being more freaking precise, and you make off with quite a bit to eat.

;)

[jsw]

Since it all adds up to less than 100% -- everybody can get more than what was in the will.

Otherwise they'll end with a bloody mess for the executor of the will, and he'd still have nearly an entire sheep leftover.
You say bloody mess, I say feast. To-may-to, to-mah-to. ;)

[mkrishnan]

This way, the old man's will is faithfully executed, the sons learn a lesson about being more freaking precise, and you make off with quite a bit to eat.

Okay, but my way is more vegetarian-friendly. :p

[KingSleaze]

I can't be very elegant with my solution, I'm afraid..but :

last son - 4 sheep (10% of 40)
3rd son - 5 sheep (1/8 of 40)
2nd son - 10 sheep (25% of 40)
1st son - 20 sheep. (50% of 40)

involves phantom sheep, but otherwise works. Has something to do with 1/8th me thinks, but haven't messed around with math in a while.

Here, let me add my 1 sheep to the flock, temporarily. The sons each get their cut, and I get my 1 sheep back.
10% +12.5% + 25% + 50% = 97.5%

[angelneo]

No.

The 1st son gets 19.5 sheep (one is slaughtered).
The 2nd son gets 9.75 sheep (oops - looks like two get slaughtered).
The 3rd son gets 4.875 sheep (darn, three sheep bite the dust).
The 4th son gets 3.9 sheep (there goes sheep number four).

35 sheep make it through alive. Four get butchered and divided.

You get to keep 0.975 sheep for your efforts. Hopefully, at least some of what they give you is edible.

This way, the old man's will is faithfully executed, the sons learn a lesson about being more freaking precise, and you make off with quite a bit to eat.

;) LOL... hahaha... this is the most funny answer I have seen. Crap, I am laughing uncontrollably with my colleagues staring at me.

[puckhead193]

can we have the answer to the apples one.... grr so annoying :eek:

[cyanide]

maybe im missing it but you guys are using forty for the sheep thing. there are only thirty nine sheep as stated in the original post. what did i miss?

[MOFS]

maybe im missing it but you guys are using forty for the sheep thing. there are only thirty nine sheep as stated in the original post. what did i miss?

some dodgy maths involving a lot of people allowing 1/2 of 39 to equal 20 - and some realists who just like lamb dinners! :p

[Veldek]

maybe im missing it but you guys are using forty for the sheep thing. there are only thirty nine sheep as stated in the original post. what did i miss?What you're missing is that this is a famous story where it is solved by using another animal which will be left over in the end. The original story is about camels though.

The solution comes down to the father being a math idiot because his will doesn't sum up to 1.

The fruit thing was already solved in the first posts, as one can safely assume that ALL boxes are labelled wrong, because that's what the riddle said.

Unless this was the case, the riddle would be indeed unsolvable. The very nice answers about weighing or smelling are nice indeed, but I think you are only allowed to open a crate, neither to weigh them nor to smell on them.

[AmigoMac]

A father who recently died left a will to his 4 sons. the oldest son should get half of his asset, the second son should get a quarter of his asset, the third should get an 1/8 of the asset and the last should get a tenth of the asset. Unfortunately, the father left 39 sheeps to his sons, and the sons brought forth their case to you. How would you solve this?

I'll include one more sheep from myself

1st son: 20
2 : 10
3 : 5
4: 4

I take back my sheep and everyone is happy... :mad:

Why should this world discuss for a sheep? ;)

[AmigoMac]

What you're missing is that this is a famous story where it is solved by using another animal which will be left over in the end. The original story is about camels though.


I guess the name of the book is

"The man who calculates(d)"

nevermind, should be something like that... yeah, Camels...

[angelneo]

Since the answers are all out, you guys might as well post the answer for the 3 light bulbs. Any takers?

[jsw]

Unless this was the case, the riddle would be indeed unsolvable. The very nice answers about weighing or smelling are nice indeed, but I think you are only allowed to open a crate, neither to weigh them nor to smell on them.
Actually, you're allowed to open only one crate, but you aren't forbidden to do anything else, and the original puzzle never stated that all the labels were wrong.

However, I'd imagine that the puzzle was poorly worded and implied that they were, indeed, all mislabeled.

[mad jew]

From my rural experience, if you leave 39 sheep in a good paddock for long enough there'll be 40 sooner rather than later... ;)

[jsw]

Since the answers are all out, you guys might as well post the answer for the 3 light bulbs. Any takers?
This has been posted before.

Turn one switch on for a while, turn it off. Turn another one on. Walk into the room. The light which is on is obviously controlled by the switch you left on. The light which is still warm is controlled by the other one you turned on. The cold light is the one controlled by the final switch.

[Veldek]

Actually, you're allowed to open only one crate, but you aren't forbidden to do anything else, and the original puzzle never stated that all the labels were wrong.

However, I'd imagine that the puzzle was poorly worded and implied that they were, indeed, all mislabeled."You are allowed to open one crate". This means you aren't allowed to do anything else. These riddles have to be followed word by word.

"The labels for the crates have been put on the wrong crates". Again you have to take it exactly as it was said, which means the labels ARE on the wrong crates, no matter which label you take.

I agree that he probably wrote it from memory and it was better asked before he restated it, but that's how these things work. It's fun to find possibilites to circumvent the "right" solution like smelling or weighing or telling that these are crates so you should be able to see the contents, but let's be honest: that's not what it's all about as long as it isn't a joke.

These riddles are all about logic and I knew some of them. I'll try to find one of them and post it. Sometimes you really think it's totally stupid and it can't be solved, but when you hear the solution you are stunned.

[ExoticFish]

My teacher put a brain teezer out a few days ago and my class thinks it is impossible to figure out.

>>There are 3 crates holding apples and oranges. 1 crate is holding apples. 1 crate is holding oranges. Another crate is holding apples and oranges. The labels for the crates have been put on the wrong crates. You are allowed to open one crate. Which crate do you open to figure out what the other crates have in them? You must only open 1.<<

I'm convinced this is impossible and my teacher is just making a joke out of it.

ok here's my bet:

"There are 3 crates holding apples and oranges." - all 3 crates have both apples and oranges. then they tell you what's in each box and they would all be correct. it doesn't say "1 crate is holding JUST apples."

so you don't have to open any of the crates to know what's in them, but who says the labels that were messed up are even for these crates?

sounds like the best solution to me.

[Veldek]

A student visits his Math professor at home. They chit-chat a while until the professor mentions his three daughters. The students gets interested at once and wants to know how old the daughters are. The professor doesn't want to make it easy for his student and says:

"If you multiply their ages, you get 36."
"If you add their ages, you get my house number."

The student thinks about it and says: "That's not enough information." And so the professor says:

"The younger ones are twins".

Now, the student knows the answer. And you?

Ok, it might be easy for the math cracks here, but we'll see. I'll see if I can find another one.

[atszyman]

If all crates are full the weighing option becomes even easier. Weigh all three crates and you'll end up with x>y>z. Open x or z, y is the mixed crate and x and z contain only apples or oranges.

[jsw]

A student visits his Math professor at home. They chit-chat a while until the professor mentions his three daughters. The students gets interested at once and wants to know how old the daughters are. The professor doesn't want to make it easy for his student and says:

"If you multiply their ages, you get 36."
"If you add their ages, you get my house number."

The student thinks about it and says: "That's not enough information." And so the professor says:

"The younger ones are twins".

Now, the student knows the answer. And you?

Ok, it might be easy for the math cracks here, but we'll see. I'll see if I can find another one.
(1, 1, 36), (3, 3, 4) or (2, 2, 9) meet the "twins" criteria.

Possible solutions and their sums (ignoring twin constraint) are:

1, 1, 36 [38]
1, 2, 18 [21]
1, 3, 12 [16]
1, 4, 9 [14]
1, 6, 6 [13]
2, 2, 9 [13]
2, 3, 6 [11]
3, 3, 4 [10]

Thus, the daughters must be 2, 2, and 9, as that's the only solution that matches a non-younger-twin option.

I've yet to have coffee this morning, so perhaps I missed something.

[Veldek]

ok here's my bet:

"There are 3 crates holding apples and oranges." - all 3 crates have both apples and oranges. then they tell you what's in each box and they would all be correct. it doesn't say "1 crate is holding JUST apples."

so you don't have to open any of the crates to know what's in them, but who says the labels that were messed up are even for these crates?

sounds like the best solution to me.Although I really like your solution, you draw the wrong conclusion, when you say that you don't need to open any to know what's in them, because you still don't know if a crate contains only apples or apples and oranges.

I think, the original story was better worded and as has been said before, this is essential for these kind of problems.

[Veldek]

Thus, the daughters must be 2, 2, and 9, as that's the only solution that matches a non-younger-twin option.

I've yet to have coffee this morning, so perhaps I missed something.You didn't say anything why 3, 3 and 4 couldn't be a solution.

Edit: Well, you did in a way that I don't understand. What do you mean with "non-younger-twin option"?

[jsw]

Although I really like your solution, you draw the wrong conclusion, when you say that you don't need to open any to know what's in them, because you still don't know if a crate contains only apples or apples and oranges.

I think, the original story was better worded and as has been said before, this is essential for these kind of problems.
Yes, I agree - the original puzzle must have been worded more precisely.

And, instead of smelling them (my original solution), I'd just X-ray them. Why open any crates if you don't have to? ;) Seed distribution would clearly identify apples vs oranges.

[jsw]

You didn't say anything why 3, 3 and 4 couldn't be a solution.
Yes I did (well, indirectly) - it doesn't match any other sum, so therefore, if that were the prof's street address, the student would already have known the answer. There had to be at least two sums that were the same, one of which used two younger twins.

That's what I meant with my poorly worded "that's the only solution that matches a non-younger-twin option".

[Veldek]

Yes I did (well, indirectly) - it doesn't match any other sum, so therefore, if that were the prof's street address, the student would already have known the answer. There had to be at least two sums that were the same, one of which used two younger twins.

That's what I meant with my poorly worded "that's the only solution that matches a non-younger-twin option".Yes, I noticed it too late, sorry. You seem to be good in these things, perhaps I should get you another one?

[jsw]

Yes, I noticed it too late, sorry. You seem to be good in these things, perhaps I should get you another one?
Feel free! :)

I need to do at least some real work today, but I'll check back to see if you post anything new!

[Veldek]

Feel free! :)

I need to do at least some real work today, but I'll check back to see if you post anything new!Three cowboys have been captured by Indians. The chieftain says he will release them when they solve a riddle, otherwise they'll die.

The cowboys have to sit down behind each other, meaning the first can't see the others, while the second can see the first and the third can see both. They mustn't talk or give signs. The chieftain has three (edit: five, of course) feathers, two black and three white ones. Each cowboy gets one feather in his hat. The cowboys can neither see which feathers they have nor which are remaining, only the ones from the cowboys that are sitting before them. One of them must say what feather he has on his hat.

Do they die? Or who knows the correct answer and what is it?

My wife (who isn't a math crack) solved it in about a minute while some friends of mine didn't find out at all.

I found another one which I couldn't solve yet. I'll try again, but if you're interested...

[jsw]

Three cowboys have been captured by Indians. The chieftain says he will release them when they solve a riddle, otherwise they'll die.

The cowboys have to sit down behind each other, meaning the first can't see the others, while the second can see the first and the third can see both. They mustn't talk or give signs. The chieftain has three feathers, two black and three white ones. Each cowboy gets one feather in his hat. The cowboys can neither see which feathers they have nor which are remaining, only the ones from the cowboys that are sitting before them. One of them must say what feather he has on his hat.

Do they die? Or who knows the correct answer and what is it?

My wife (who isn't a math crack) solved it in about a minute while some friends of mine didn't find out at all.

I found another one which I couldn't solve yet. I'll try again, but if you're interested...
OK, here's my guess:

If the 3rd one sees two black feathers, he'll know he has a white one, and they go free.

If the 3rd one sees either two white feathers or a black one and a white one, he doesn't know.

If the 3rd one can't tell, the 2nd one knows he (the 3rd one) sees either a black one and a white one or two white ones. Therefore if the 2nd one sees a black feather, he knows his is white, so they go free.

If the 2nd one can't say, it means he (the 2nd one) sees a white feather.

If the 3rd and 2nd ones can't tell, the 1st one knows he must have white feather on his head, so they all go free.

Close?

[atszyman]

The chieftain has three feathers, two black and three white ones.

Anyone else confused by this?

[Veldek]

Close?Perfect.

Now a VERY easy one:

You are a busdriver. On your tour there are several stops where passenger get in and out.
On the first stop some passengers get in.
On the second stop 3 get out and 4 get in.
On the third stop 7 get out and 2 get in.
On the fourth stop 4 get out and 5 get in.
On the last stop all the passengers get out.

How old is the busdriver?

This is NO joke, just a very simple question. Don't give an explanation so everyone can have fun ;) Just tell me if you found out.

[jsw]

Perfect.

Now a VERY easy one:

You are a busdriver. On your tour there are several stops where passenger get in and out.
On the first stop some passengers get in.
On the second stop 3 get out and 4 get in.
On the third stop 7 get out and 2 get in.
On the fourth stop 4 get out and 5 get in.
On the last stop all the passengers get out.

How old is the busdriver?

This is NO joke, just a very simple question. Don't give an explanation so everyone can have fun ;) Just tell me if you found out.
Gosh, that's an old (and, yes, easy) one!

[Veldek]

Anyone else confused by this?Oops I meant five feathers. Sorry!

[jsw]

Anyone else confused by this?
I read it as a typo: "five feathers, three of which will be used, one per head"

[Veldek]

Here's the one I couldn't solve yet.

In a monastery there are some monks with a vow of silence. One day there's a mysterious illness breaking out. Everyone with it has a green face, but feels fine nevertheless. There are no mirrors in the monastery and the vow of silence is this special that they aren't even allowed to give signs or communicate in any other way (oh man, this is really construed). That's why a monk doesn't know if he's ill. All they know is, that at least one of them is ill.
The ill monks mustn't go to the evening prayer for religious reasons. In the beginning all monks still come there. But one evening, suddenly only the healthy monks appear.

How did the monks find out who's ill and how long did it take to find out?

[Mitthrawnuruodo]

Since the answers are all out, you guys might as well post the answer for the 3 light bulbs. Any takers?
Is there a way you can tell if a lightbulb recently has been on...?

Edit: Apparently answerd in full, hours ago... I'm lagging a bit behind... :o :p

Edit2+: Got the busdriver, not sure about the monks... only thought is that it won't take long because they can see their reflection in the water while drinking... or that frost came and something froze and the green ones could see their reflection...

[jsw]

Here's the one I couldn't solve yet.

In a monastery there are some monks with a vow of silence. One day there's a mysterious illness breaking out. Everyone with it has a green face, but feels fine nevertheless. There are no mirrors in the monastery and the vow of silence is this special that they aren't even allowed to give signs or communicate in any other way (oh man, this is really construed). That's why a monk doesn't know if he's ill. All they know is, that at least one of them is ill.
The ill monks mustn't go to the evening prayer for religious reasons. In the beginning all monks still come there. But one evening, suddenly only the healthy monks appear.

How did the monks find out who's ill and how long did it take to find out?I'll try to figure out the "real" answer. My "trivial" one: they stick out their lips or look at their nose and see if they're green - you can see those things without a mirror.

Edit: other "bad" solution: the disease is fatal as soon as the temperature reaches a certain high or low. The weather changes, and all the green monks die. ;)

[Veldek]

Edit: other "bad" solution: the disease is fatal as soon as the temperature reaches a certain high or low. The weather changes, and all the green monks die. ;)Well, who needs monks with these kinds of vows anyway :D

[jsw]

not sure about the monks... only thought is that it won't take long because they can see their reflection in the water while drinking... or that frost came and something froze and the green ones could see their reflection...
Ah, that's much more humane than my guess that the weather killed them! :D

[mkrishnan]

Ah, that's much more humane than my guess that the weather killed them! :D

Also, can't the monks see each others' reflections in their eyes? So they just stare at each other like lovers, and they know. :D

[Pittsax]

My teacher put a brain teezer out a few days ago and my class thinks it is impossible to figure out.

>>There are 3 crates holding apples and oranges. 1 crate is holding apples. 1 crate is holding oranges. Another crate is holding apples and oranges. The labels for the crates have been put on the wrong crates. You are allowed to open one crate. Which crate do you open to figure out what the other crates have in them? You must only open 1.<<

I'm convinced this is impossible and my teacher is just making a joke out of it.
I think you're all overthinking this....

All of the labels are incorrect according to the riddle. So to solve this you:

1. Open the crate marked "apples and oranges" This crate will either contain apples OR oranges (but not both, since the label is wrong). Whichever fruit is in the crate, you put that label on it (let's say it has apples).

2. Now, you're left with one crate with no label (it used to say "apples" but you put that on the crate that you opened), and one crate that says "oranges")

3. Since the labels are all wrong, the crate that says "oranges" must not contain oranges, and since you already showed the other crate had apples, the oranges must be in the crate with no label.

4. Now you're left with one crate, which contains apples and oranges.

Voila!

[PixelFactory]

Here's the one I couldn't solve yet.

In a monastery there are some monks with a vow of silence. One day there's a mysterious illness breaking out. Everyone with it has a green face, but feels fine nevertheless. There are no mirrors in the monastery and the vow of silence is this special that they aren't even allowed to give signs or communicate in any other way (oh man, this is really construed). That's why a monk doesn't know if he's ill. All they know is, that at least one of them is ill.
The ill monks mustn't go to the evening prayer for religious reasons. In the beginning all monks still come there. But one evening, suddenly only the healthy monks appear.

How did the monks find out who's ill and how long did it take to find out?

Well here is my try. As we dont know the quantity of monks who are sick, I'll start with one sick monk.

On the first day, all the well monks will see the sick monk and return the next day. The sick monk will see all well monks and not return to pray.

2 sick monks.
the first day the well monks will see the two sick monks but the sick monks will see only one sick monk. They will all think they are well. The second day everyone returns. Now the two sick monks realize that there is more than one sick monk and that they only see one sick monk they must be sick too.

For each additional sick monk add a day.

[mkrishnan]

I think you're all overthinking this....

All of the labels are incorrect according to the riddle. So to solve this you:

Ummm, did ya *read* the thread? We posted that solution a loooong time ago. :D We've been arguing about technicalities since then! :p

[jsw]

Well here is my try. As we dont know the quantity of monks who are sick, I'll start with one sick monk.

On the first day, all the well monks will see the sick monk and return the next day. The sick monk will see all well monks and not return to pray.

2 sick monks.
the first day the well monks will see the two sick monks but the sick monks will see only one sick monk. They will all think they are well. The second day everyone returns. Now the two sick monks realize that there is more than one sick monk and that they only see one sick monk they must be sick too.

For each additional sick monk add a day.
I don't get it. We're told that all they know is that at least one monk is sick. Suppose three are. They all come to prayer. The three sick ones each see two sick ones and wonder why they keep coming, but they have no way of knowing that they are the third one.

[PixelFactory]

If there are three sick monks.

The sick monks will see two sick monks. On the first day they will think there are two sick monks. On the second day they will still think there are two sick monks and not leave. (edit: if there were only two sick monks they would only see one sick monk each and realize that they are sick and not return) On the third day when all the monks appear they will realize that there are more than two sick monks. Since they only see two they must be sick too. So they will not return on the fourth day.

[broken_keyboard]

If there are three sick monks.

The sick monks will see two sick monks. On the first day they will think there are two sick monks. On the second day they will still think there are two sick monks and not leave. (edit: if there were only two sick monks they would only see one sick monk each and realize that they are sick and not return) On the third day when all the monks appear they will realize that there are more than two sick monks. Since they only see two they must be sick too. So they will not return on the fourth day.

No, it only takes 2 days.
At the first prayer session, each sick monk will see two other sick monks.
At the second prayer session, each sick monk again sees two sick monks. But this time they know that each of the sick must have also seen two sick. If they had only seen each other, and the rest healthy they would have disappeared the day before, but they didn't. Since our guy can see everyone else apart from those two is healthy, he knows he must be the third sick one.

[Don't panic]

I don't get it. We're told that all they know is that at least one monk is sick. Suppose three are. They all come to prayer. The three sick ones each see two sick ones and wonder why they keep coming, but they have no way of knowing that they are the third one.

it works if you assume that thee monks are intelligent and that there is a lapse of time that count as one "period" after which every monk re-evaluates the situation. It's exactly the same ase the feathered (but untarred) cowboys.

[PixelFactory]

No, it only takes 2 days.
At the first prayer session, each sick monk will see two other sick monks.
At the second prayer session, each sick monk again sees two sick monks. But this time they know that each of the sick must have also seen two sick. If they had only seen each other, and the rest healthy they would have disappeared the day before, but they didn't. Since our guy can see everyone else apart from those two is healthy, he knows he must be the third sick one.

It will still take 4 prayer sessions.

day one: they will think there are only two sick monks.

day two: they will still think there are only two sick monks. If there are only two they would only see one and realize they are the other sick monk and nit return

day three: Since the two sick monks return they must see two sick monks too. Therefore there must be a third and since they only see two they must be him and they wont retun on day four.

Day four: only healthy monks go to prayer session

[Pittsax]

Ummm, did ya *read* the thread? We posted that solution a loooong time ago. :D We've been arguing about technicalities since then! :p
It's official...I am a dumbass http://forums.operationsports.com/vBulletin/images/smilies/graemlins/brickwall.gif

[mkrishnan]

It's official...I am a dumbass http://forums.operationsports.com/vBulletin/images/smilies/graemlins/brickwall.gif

Naaaahhhhh, now you can come argue about vomiting monks with us. :D

[Doctor Q]

I love brain teaser puzzles, but the thread title says the thread is about impossible puzzles, so obviously they must be unsolvable, which means there's no use in trying to solve them.

But that makes me a bit surprised that you have collectively solved some of them! :rolleyes:

[Pittsax]

Naaaahhhhh, now you can come argue about vomiting monks with us. :D
Speaking of the Monks with green faces, I'm guessing their noses aren't green? Because if I cross my eyes, I can see that my nose isn't green...

[Veldek]

It will still take 4 prayer sessions.

day one: they will think there are only two sick monks.

day two: they will still think there are only two sick monks. If there are only two they would only see one and realize they are the other sick monk and nit return

day three: Since the two sick monks return they must see two sick monks too. Therefore there must be a third and since they only see two they must be him and they wont retun on day four.

Day four: only healthy monks go to prayer sessionYou're right. That's it.

[Nermal]

Here's the one I couldn't solve yet.

In a monastery there are some monks with a vow of silence. One day there's a mysterious illness breaking out. Everyone with it has a green face, but feels fine nevertheless. There are no mirrors in the monastery and the vow of silence is this special that they aren't even allowed to give signs or communicate in any other way (oh man, this is really construed). That's why a monk doesn't know if he's ill. All they know is, that at least one of them is ill.
The ill monks mustn't go to the evening prayer for religious reasons. In the beginning all monks still come there. But one evening, suddenly only the healthy monks appear.

How did the monks find out who's ill and how long did it take to find out?

Just get a bucket of water and see your reflection :)

Edit: Oops, somebody already said that.

[Applespider]

I'm still not seeing the monk answer even after reading Pixelfactory's correct answer.

Do we know for sure that 1 monk gets sick everyday? Otherwise I'm bamboozled as to whether 80% of the monks could suddenly go down with it overnight and although they see lots of other green monks, they'd never know they were green too. The answer must assume that the monks get a daily update of how many monks are sick?

I prefer the practical 'notices the colour of face in the communion chalice' since they're generally pretty shiny!

[Veldek]

I'm still not seeing the monk answer even after reading Pixelfactory's correct answer.

Do we know for sure that 1 monk gets sick everyday? Otherwise I'm bamboozled as to whether 80% of the monks could suddenly go down with it overnight and although they see lots of other green monks, they'd never know they were green too. The answer must assume that the monks get a daily update of how many monks are sick?

I prefer the practical 'notices the colour of face in the communion chalice' since they're generally pretty shiny!There's a given amount of ill monks and this amount doesn't change anymore. All we know for sure is that there's at least one ill monk.

Imagine you being the only ill monk. Then you wouldn't see any other ill monks and could conclude that you must be ill.

If you see one other, there are two possibilities. A) You're healthy and he's the only ill monk or B) You're both ill. When he appears the next day, you know that you have to be ill too, because otherwise he would have known that he's ill and would have stayed away.

You can now go on like this with a given amount of ill monks.