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This one I read but is so good I wanted to post it reworded. Two groups of people live on this planet, pure liars and pure truth tellers. I went to a philosophy party(never mind that it's impossible) and met 3 people. The first says something but I don't catch it, the second one replies, "he said he was a liar." The third says to the second, "You are lying!" Is the third person a liar or truth teller?
Hello, jammieg!
A good problem . . .
Two groups of people live on this planet, pure liars and pure truth tellers.
I went to a philosophy party (never mind that it's impossible) and met 3 people.
The first says something but I don't catch it,
the second one replies, "He said he was a liar."
The third says to the second, "You are lying!"
Is the third person a liar or truth teller?
Consider what the first person must have said.
If he were a Truth Teller, he'd say, "I'm a Truth Teller."
If he were a Liar, he'd lie and say, "I'm a Truth Teller."
That is, NO ONE would ever say, "I'm a Liar."
Hence, the second person lied.
Therefore, the third person spoke the truth.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A similar problem . . .
An archaeologist discovered a temple with three gigantic statues
guarding the entrance. To be allowed into the temple, he must
identify each of the three gods.
The God of Truth always tells the truth.
The God of Falsehood always lies.
The God of Diplomacy can tell the truth or lie.
He asked the god on the left, "Who stands next to you?"
"The God of Truth," was the reply.
He asked the god in the middle, "Who are you?"
"The God of Diplomacy," was the reply.
He asked the god on the right, "Who stands next to you?"
"The God of Falsehood," was the reply.
The archeologist immediately identitied the three gods.
Can you?
left=diplomacy
middle=false
right=truth
Truth is obviously not on the left, or he'd be lying about who is next to him.
Truth is not in the middle, or he'd be lying about who he is.
Truth is on the right, therefore, the god next to truth is falsehood, as stated by truth.
That leaves only the left for diplomacy.
Njorl
A man at a table is shot for having 53 bicycles. Why?
Njorl
Originally posted by Njorl
A man at a table is shot for having 53 bicycles. Why?
Njorl
He didn't pay for the parking!
Originally posted by Njorl
A man at a table is shot for having 53 bicycles. Why?
Ha! . . . Took me a few moments.
The "Bicycles" (TM) are playing cards.
(He probably had an Ace up his sleeve.)
theriddler876
Oct7-03, 08:53 PM
here's a simple one
you have 9 coins one is counterfeit and a balance, using only two weighs, find the counterfeit coin (the counterfeit coin is lighter)
hypnagogue
Oct8-03, 04:15 PM
er how about finding the counterfeit coin without any weights at all? Simply try out all permutations of 4 coins on one side of the balance against 4 coins on the other side. Once you have found a permutation such that the two groups of 4 balance perfectly, you know that the unused coin is counterfeit.
edit: or, for a more efficient solution, pick out any two groups of 4 and place them on each side of the balance. If they balance perfectly, then the unmeasured coin is counterfeit. If they do not balance, replace one coin on the lighter side with the unused coin, and continue replacing thus-far-unreplaced coins until you have a perfect balance. This takes at most 5 separate measurements.
Actually, it technically only takes at most 4 measurements since if you have already replaced 3 of the coins on the lighter side of the balance, you can infer the 4th is counterfeit. Of course this only holds if you are correct in assuming that one of the coins is counterfeit in the first place. Reminds me of the superbowl episode of the Simpsons where Chief Wiggum is in jail and declares "one bar in a jail cell is always weak." He goes on to test all the bars except the leftmost one, and infers that it is the weak one, so he rans head-on into it only to fall flat on his back. [:D]
Weigh 3 coins on each side of the balance. If one side is lighter, the fake is in that group of 3. If not, it is in the set of 3 unweighed coins.
Discard the 6 coins you know are true, and balance 2 of the remaining coins. If one is lighter, it is fake. If they weigh the same, the unweighed coin is fake.
You could find one fake in 27 coins with 3 weighings, 1 in 81 with 4 weighings etc.
njorl
jammieg
Oct13-03, 09:17 PM
I think there is one of those annyoing creative solution ways to find the fake coin in 2 measures, with 2 to 2000 or more coins, but I would rather throw it scales and all into the sea.
Among all the various two typings of the people of this world there are those who are apt and preoccupied with finding and percieving the limitations of things and those who are always trying to find ways around those limits.
Originally posted by Njorl
You could find one fake in 27 coins with 3 weighings
Hmm... How's that exactly ?! [b(]
I remmember doing this riddle quite some time ago and
the top number I could solve it for was either 15 or 16
for 3 weighings (I'm a bit tired right now so I won't
check which one it was, for now). But 27 ?! No way ! [o)]
Peace and long life.
Oops... I'm sory Njorl, never mind. I was reffering to
the riddle when you don't know weather the fake is lighter
or heavier. If you do know it's either of these options
it's another story of course. [t)]
Originally posted by jammieg
This one I read but is so good I wanted to post it reworded. Two groups of people live on this planet, pure liars and pure truth tellers. I went to a philosophy party(never mind that it's impossible) and met 3 people. The first says something but I don't catch it, the second one replies, "he said he was a liar." The third says to the second, "You are lying!" Is the third person a liar or truth teller?
I have a variation on this idea that is equally as difficult. It goes along these lines:
You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?
MathematicalPhysicist
Nov29-03, 04:29 AM
Originally posted by Raven
I have a variation on this idea that is equally as difficult. It goes along these lines:
You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?
from which city are you?
if it's a lier than he points you to the wrong way of his city which is the liers city, he points you to the truth tellers city because it's not his city and hes lier than it fits, now if it's truth teller than he points you to the truth tellers city because he tells than truth when asked.
either way you go where the person tells you.
wasteofo2
Dec1-03, 09:14 PM
There are 3 cannibals and 3 missionaries on one side of a river. If ever there are more cannibals than missionaries on one side of the river, the cannibals will eat the missionaries. The boat which will be used to go to the other side of the river seats 2, and at least one person must be in it in order for it to go from one side to the other. In what way can they travel to the other side of the river so that no missionaries are eaten?
-------- ---------
~~~~~~~~~~~~~~~~
Legend: | | = river, < > = boat, C = cannibal, M = missionary
| |< > CCCMMM
|<CM>| CC MM
C M < >| | CC MM
C |< M>| CC MM
C | |< > CC MMM
C |<CC>| MMM
CCC < >| | MMM
CC |<C >| MMM
CC | |< > C MMM
CC |<MM>| C M
CC MM < >| | C M
C M |<CM>| C M
By symmetry, I've reduced the problem to one I already know how to solve, so I'm done. [:D]
Here's a good crossings problem:
Chroot, Monique, Gale, and Hurkyl (four randomly chosen names) are fleeing from the law through a dark tunnel with only one flashlight. They are a good 17 minutes ahead of their pursuers. They find their path blocked by an underground ravine with only a long rickety bridge to permit them to cross. The bridge is surely too unstable to allow more than two of our heroes to cross at any one time, and it's too dangerous to attempt a crossing in the dark. Monique estimates she can cross the bridge in one minute, Chroot estimates that he can do it in two minutes. Gale thinks she needs five minutes to do it carefully, and Hurkyl needs to take ten minutes, just to be safe. Can these fantastic four cross the bridge before the law catches them?
wasteofo2
Dec1-03, 10:11 PM
1 and 2 go across. 15 minutes left
1 goes back. 14 minutes left
5 and 10 go across. 4 minutes left
2 goes back. 2 minutes left
2 and 1 go across. done with no time to spare.
Ok, now someone explain the one with the 3 people in a hotel.
the one where they pay $10 each for the room, but the hotel clerk finds out the room was onlt $25, so he gives $5 to the bellboy to return. The bellboy can't figure out how to divide it evenly so he gives them $1 each and takes the other 2 for himself. They each are now out $9, which means they lost $27 in paying for the room, which leaves $3 left, but the bellboy only kept 2.
phoenixthoth
Dec2-03, 12:10 AM
how 'bout this logic puzzle?
either God exists or God does not exist.
which is it and provide proof.
If you had one question and an all knowing robot will answer it for you but will kill you when it answers you what will you ask him? Assuming he was indestructble and can not be destroyed.
phoenixthoth
Dec2-03, 06:09 AM
is that a logic puzzle? maybe i'd ask it a loaded question like, "why will you never kill me?"
on the other hand, if such a robot existed...
It is the categorical formulation of the simultaneous, situational, instantiated contradiction, where deductive invalidity is the product of the utmost categorical truth of the assumption that if the antecedent of a true conditional is false, then the consequent of the conditional is true or false indifferently, and of the categorical falsehood of the conclusion consequently predicates that if it be not the case that the consequent of a true conditional is true or false indifferently, then, it is not the case that the antecedent of the conditional is false. To pronounce the consequent of a true conditional as being true or false indifferently is tantamount to saying modally that where the antecedent of a true conditional is notoriously false, then the consequent can, or could be, or is possibly true or false. But it may be worthwhile to see that the definitive, simultaneous equality of both true, and false, can be formulated without explicitly including modal terms, which become the predicating operators, which, for the sake of showing that the consequent paradoxical conundrum is not straightforwardly resolvable by appealing to concrete philosophical scruples concerning the intensionality of predicated modal contexts.
and so the natural question i'd gladly give my life for is this: what must one appeal to to resolve the paradoxical conundrum?
(the answer relates to my logic puzzle.)
wasteofo2
Dec2-03, 04:05 PM
Originally posted by THANOS
If you had one question and an all knowing robot will answer it for you but will kill you when it answers you what will you ask him? Assuming he was indestructble and can not be destroyed.
I wouldn't ask it a question.
Originally posted by wasteofo2
Ok, now someone explain the one with the 3 people in a hotel.
the one where they pay $10 each for the room, but the hotel clerk finds out the room was onlt $25, so he gives $5 to the bellboy to return. The bellboy can't figure out how to divide it evenly so he gives them $1 each and takes the other 2 for himself. They each are now out $9, which means they lost $27 in paying for the room, which leaves $3 left, but the bellboy only kept 2.
It's adding and subtracting the wrong numbers -- the 27 spent is 2 for the bellboy and 25 for the inn. The $3 is the amount that they got back. It makes just as much sense to ask how the inkeeper could have kept $25 when there are only $3 left.
D'oh, forgot to mention that if you don't ask a question the robot will kill you anyway.
Questions that will take more than my natural life to answer:
What are the first 10^10^10 digits of π in order?
Questions that require resolution of the dilemma to be meaningful:
Why did you self-distruct?
Questions that are paradoxical:
What happens when an immovable object meets an irresistable force?
Questions for a solution to the dilemma:
How do I avoid getting dying after you answer this question?
Questions that have no correct answer:
What is an effective method for destroying you?
(Robot is indistructible by setting.)
Questoins that provide meta-abuse:
What is the text of questions and answers that a comprehensive list of questions and corresponding answers that would allow me to avoid this dilemma would contain?
wasteofo2
Dec2-03, 09:34 PM
Originally posted by NateTG
Questions that require resolution of the dilemma to be meaningful:
Why did you self-distruct?
The robot could say "I did not self destruct, you are confused" then kill you
Originally posted by NateTG
Questions that are paradoxical:
What happens when an immovable object meets an irresistable force?
The robot could say "I don't know" then kill you.
Originally posted by NateTG
Questions for a solution to the dilemma:
How do I avoid getting dying after you answer this question?
The robot could say "You cannot" and then kill you
Originally posted by NateTG
Questions that have no correct answer:
What is an effective method for destroying you?
"There is no effective method" then you're killed
Originally posted by NateTG
Questoins that provide meta-abuse:
What is the text of questions and answers that a comprehensive list of questions and corresponding answers that would allow me to avoid this dilemma would contain?
"There are no questions with corresponding answers that would allow you to avoid this dilemma" and then you're killed.
I'll just post the question that was meant for the question i asked.
the question is "What is the one question you can not answer?"
MathematicalPhysicist
Dec3-03, 10:44 AM
if i could answer your question then it wouldnt be the right question if i havent answered to your question then it would make the criteria therefore this statement is the same as saying "im a lier".
Originally posted by wasteofo2
"There are no questions with corresponding answers that would allow you to avoid this dilemma" and then you're killed.
Well, unless you want to debunk the first question (which is fairily solid), there are questions, and corresponding answers that resolve the dilemma.
Thanos: The robot can reply, "There is no such single question." and blow your head off because you include 'one'.
"What are all the questions you cannot answer?" can be answered as well.
In fact, you did not stipulate that the robot answer truthfully, so I suppose it could just respond with '42' before ventilating your skull, regardless of how clever the qestion was.
I suppose you could work around the problem by asking a question that takes longer than your own natural life to complete asking.
Ask the robot to say aloud all of the digits of pi, hehe. Hopefully by the time the gets around to blowing my head off I would have already found a way out, hacked the internal circuitry of the robot, saved all the (useful) information on some sort of storage device, and change the robot's OS to Microsoft, thus effectivly dooming the robot.
jammieg
Jan31-04, 12:19 PM
Here's another one some strange guy said to me once:
So your employers give you more work than you don't want?
What does he mean?
This might seem really simple but for some reason it confused me greatly and I'm wondering if there is more to it.
where is the missing square from the attachement bellow?
The red and green triangles have different slopes, the appearance that the (macroscopic) figures are triangles is misleading.
You can see this by, for example, looking at the height of each of them 5 units from the left point. The lower shape has a height of two, but the upper shape has a height of 15/8 which, even in the illustration, is less than 2.
We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's.
Every man has his own, single, different pet, and smokes his own, type of cigarettes.We know that:
1.The englishman lives in the red house
2.The sweedish has a dog
3.The dutch drinks tea
4.The green house is situated at the left of the white house
5.The person who lives in the green house drinks coffee
6.The person who smokes PallMall has a bird
7.The person who lives in the yellow house smokes Dunhill (like I do[6)] )
8.The person who lives in the 3rd house drinks milk
9.The norwegian lives in the 1st house
10.The Blend smoker lives in the house near the house where lives a cat
11.The person who owns a horse lives near the Dunhill smoker
12.The person who smokes Blue Master drinks beer
13.The german smokes Prince
14.The norwegian lives next to the blue house
15.The Blend smoker is the neighbour of the one who drinks only watter
The question is : Which one of them has a fish?
Pergatory
Feb10-04, 07:45 PM
Originally posted by marqq
We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's.
Every man has his own, single, different pet, and smokes his own, type of cigarettes.
I loved this one! This was created by Einstein originally, claiming that only 2% of the world's population could solve it. It would be interesting to find the percentage of people on this forum who can solve it.
jammieg
Feb11-04, 08:58 AM
Why doesn't a bird put a roof over it's nest, and yet they will live in birdhouses?
My guess is that anyone can solve it given enough time and effort.
We have 5 houses, everyone of them painted in a different colour.In every house lives a person with a different nationality, which preferes his own drink, different then other's. Every man has his own, single, different pet, and smokes his own, type of cigarettes.
It is the German, right? That was extremely hard for me; these puzzles are great! This thread has kept me entertained today. [:)]
Yes,it is the German...I delayed the answer, hoping that i'll find other answers :D guess not...leto I think we r among the 2% [6)](but i doubt that only 2% could solve this problem)[6)]
I think the solution to the puzzle depended on how
you interpreted the word "near"- the term "next to"
was used elsewhere so the usage of the word "near"
in other descriptions becomes ambiguous.
jammieg
Feb14-04, 08:05 AM
I mentioned this in another thread and thought it worth repeating, there is a way to solve this type of problem in one's head without pencil and paper, even if you don't have a good imagination although it helps to visualize it, the key is to read through each factual and try to memeorize the logical functions of each factual a bit more, by this I mean to think along the lines of, "well it could go here or here but not here unless that other could go there, ok move on to the next one" and so reading through the list over and over and focusing only on one at a time the functions of those factuals start to engrain in the memory and things may start to snap together seemingly without effort, although it took me way over an hour to do it this way I'm sure most anyone could do it this way with practice, it's the same principle as memorizing a long poem except one is trying to memorize the possibilities and limitations of each factual you would be amazed at how long a poem a person can memorize if they want to.
Let me know if it doesn't work, that means I probably don't know what I'm talking about.
It's a trick question; the 5th pet is really a salamander!
1. Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.
2. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.
3. When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.
How old are they now?
[o)]
Nice coder
Feb17-04, 04:09 AM
for that robot
"Calculate 1/3 until there is no remainder ,the formula 14700/44100 = the answer is true and using only 2 decimal places (without rounding)"
The robot will never finish that because
1/3 to 2 decimal places (no rounding) is 0.33
0.33 < 0.33333333333 (to infinity)
14700/44100 = 1/3
another one would be,
Find the largest posible numerator in a fraction that would equal pie
That would never work, because pie is a never-ending number, and the highest possible numerator would be infinity!
Those questions would be great to combine and ask the robot, you don't have one by any chance?
Nice coder for the 1st question it can answer you 0.(3) :D i think "pi" is the question who may save us :)
pedestrian
Feb17-04, 11:11 PM
sorry nvm
Originally posted by marqq
1. Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.
2. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.
3. When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.
How old are they now?
[o)]
I don't think this problem is solvable. 2 requires Tim's age to be
an odd number while 1 requires Tim's age to be an evem number and 3 requires alot of tylenol.
I don't think this problem is solvable. 2 requires Tim's age to be
an odd number while 1 requires Tim's age to be an evem number and 3 requires alot of tylenol.
Eyesaw would think that, trapped as he is in his private twilight zone where time has 3 dimensions. But actually, it turns out that it's just algebra. Let N = the current year ("Now"), let T = the year Tim was born, M = the year Mary was born, J = the year Jane was born. And x1 through x10 can represent the various other unknown years in the clues.
N + 10 - T = 2(x1 - J)
x1 - M = 9(x1 - T)
N - 8 - M = (1/2)(x2 - J)
x2 - J = 1 + x3 - T
x3 - M = 5(N + 2 - T)
x4 - T = 1
x4 - M = 3 + x5 - T
x5 - J = 3(x6 - M)
x6 = x7 - 6
x7 - J = (1/2)(x8 - T)
x8 - M = 10 + x9 - M
x9 - J = (1/3)(x10 - T)
x10 - M = 3(J - M)
Now, work from the bottom up, substituting out the x's as you go. After eliminating the "x" years, all of part 3 boils down to
-4T -3M + 7J = 1, and the other two equations are
13T - M - 8J = 4N + 40
-6T + 3M = -3N - 27
Solve them in terms of N to get the ages (T-N, M-N and J-N).
[Edit: sorry about that; I meant to write N-T, N-M and N-J]
Tim is 3
Mary is 15
Jane is 8
we have 10 bags with 10 gold ingots each...Each ingot has 10kg, but we have a bag full of fake ingots (9.900 kg each).we have a coin scale, and a single coin (for a single use of the scale), and we have to find the bag with the fake ingots!
Good Luck![;)]
Thought so. That makes it too easy. Just number the bags 1 to 10. Take 1 ingot from bag 1, 2 from bag 2, ..., 10 from bag 10 & weigh those 55 ingots together.
Need I say more?
Originally posted by gnome
Eyesaw would think that, trapped as he is in his private twilight zone where time has 3 dimensions. But actually, it turns out that it's just algebra. Let N = the current year ("Now"), let T = the year Tim was born, M = the year Mary was born, J = the year Jane was born. And x1 through x10 can represent the various other unknown years in the clues.
N + 10 - T = 2(x1 - J)
x1 - M = 9(x1 - T)
N - 8 - M = (1/2)(x2 - J)
x2 - J = 1 + x3 - T
x3 - M = 5(N + 2 - T)
x4 - T = 1
x4 - M = 3 + x5 - T
x5 - J = 3(x6 - M)
x6 = x7 - 6
x7 - J = (1/2)(x8 - T)
x8 - M = 10 + x9 - M
x9 - J = (1/3)(x10 - T)
x10 - M = 3(J - M)
Now, work from the bottom up, substituting out the x's as you go. After eliminating the "x" years, all of part 3 boils down to
-4T -3M + 7J = 1, and the other two equations are
13T - M - 8J = 4N + 40
-6T + 3M = -3N - 27
Solve them in terms of N to get the ages (T-N, M-N and J-N).
[Edit: sorry about that; I meant to write N-T, N-M and N-J]
Tim is 3
Mary is 15
Jane is 8
Goofus, if Tim is 3, 10 years from now, he would be 13. What is 13/ 2? If fractions are allowed in the problem, then your final answer should include the month, day and second they were all born.
You're a real piece of work, Eyesore. I guess we were wrong to advise you to re-study relativity.
Based on your comments above, you should first re-study fractions. [:D]
I'm in a charitable mood now, so I'll help you through the first few lines.
Ten years from now Tim will be 13. Half of that is 6½. Now, Jane is 7 years younger than Mary, so when Jane was 6½, Mary was 13½. Mary is 12 years older than Tim. So, when Mary was 13½, Tim was 1½. 13½ is 9 times 1½.
The rest is no more difficult than that. Let us know if you need any more help.
No need to worry about months, days, hours, minutes and seconds. Unless you're a masochist. Since you like to do things the hard way, go ahead. Eventually you'll get the same answer if you don't screw it up.
Originally posted by gnome
You're a real piece of work, Eyesore. I guess we were wrong to advise you to re-study relativity.
Based on your comments above, you should first re-study fractions. [:D]
I'm in a charitable mood now, so I'll help you through the first few lines.
Ten years from now Tim will be 13. Half of that is 6½. Now, Jane is 7 years younger than Mary, so when Jane was 6½, Mary was 13½. Mary is 12 years older than Tim. So, when Mary was 13½, Tim was 1½. 13½ is 9 times 1½.
The rest is no more difficult than that. Let us know if you need any more help.
No need to worry about months, days, hours, minutes and seconds. Unless you're a masochist. Since you like to do things the hard way, go ahead. Eventually you'll get the same answer if you don't screw it up.
Hey, how often do you hear someone say I am 6 years and 6 months older than you? And what are the odds of 3 randomly chosen people being born on the same day, same hour and same second? Based on common practice of excluding all but the year in calculating age differences, this problem is unsolvable.
I agree -- if you don't understand fractions, you can't solve it.
Originally posted by gnome
I agree -- if you don't understand fractions, you can't solve it.
I knew you could do it! I knew you'd see your mistake
sooner or later. Well done.
Eyesaw, you're making a fool of yourself.
Originally posted by Raven
I have a variation on this idea that is equally as difficult. It goes along these lines:
You have reached a fork in the road. One path leads to the city of truth where everyone always tells the truth and the other path leads to the city of lies where everyone always tells a lie. There is a stranger at the fork of the road who is a native of one of the two cities. Given one question, what can you ask him in order to find out which path leads to the city of truth?
i figured it out the second i noticed the relationship to the first one, but then i got lost again when i tried analyzing the relationships...
i nearly lost my mind trying to figure this one out...
i thought it was impossible... but i realized i just lost track...
it is very complicated, can someone break this down into logical variables? maybe some sort of computer code for me to better understand it?
An order of perfectly logical monks lives isolated in their monastery. They are sworn not to communicate with each other in any way. They have no mirrors, and no other means by which a monk can see his own face. They see each other only once each day when they all gather together for afternoon prayers.
There is a demon loose in the land - the relativity demon. If a person becomes possessed by this demon, the demon's sign (e=mc2) appears on his forehead. The monks know that this is a very powerful demon -- one that can never be exorcised -- and so, if a monk would discover that he bore this mark, that evening, in the privacy of his cell, he would commit suicide.
One afternoon, a visitor (one who is not sworn to silence) comes to the prayer meeting. He looks around the room, announces "At least one monk in this room has the demon's mark on his forehead", and immediately leaves. The monks all look around, examining each other's foreheads. Nothing unusual happens that evening.
At the second day's prayer meeting the monks all look at each other, but again that evening nothing unusual occurs.
...and so on for the third, fourth, fifth, and sixth days.
But on the seventh evening, all the monks commit suicide.
What happened? How many monks were there? How many had the demon's mark?
Enjoy. [:)]
There were 7 monks, all 7 of them had the mark from day 1.
Originally posted by gnome
An order of perfectly logical monks lives isolated in their monastery. They are sworn not to communicate with each other in any way. They have no mirrors, and no other means by which a monk can see his own face. They see each other only once each day when they all gather together for afternoon prayers.
There is a demon loose in the land - the relativity demon. If a person becomes possessed by this demon, the demon's sign (e=mc2) appears on his forehead. The monks know that this is a very powerful demon -- one that can never be exorcised -- and so, if a monk would discover that he bore this mark, that evening, in the privacy of his cell, he would commit suicide.
One afternoon, a visitor (one who is not sworn to silence) comes to the prayer meeting. He looks around the room, announces "At least one monk in this room has the demon's mark on his forehead", and immediately leaves. The monks all look around, examining each other's foreheads. Nothing unusual happens that evening.
At the second day's prayer meeting the monks all look at each other, but again that evening nothing unusual occurs.
...and so on for the third, fourth, fifth, and sixth days.
But on the seventh evening, all the monks commit suicide.
What happened? How many monks were there? How many had the demon's mark?
Enjoy. [:)]
is relativity the answer to my former question? it makes sense that it is... the confusion is thru the fact that the equation changes from each persons point of view...
it seems the only way to solve this problem is to evaluate each monks perspective on the first evening that it was announced by the stranger that atleast one of the monks bore the markings, and the situation that caused them all to commit suicide on the 7th day.
the first day there are 2 marked monks.
these two monks see only one marked monk, so they wait a day to see if the monk they saw with the marking commits suicide (they all know that atleast one person has a marking, and if no other monk has the marking then they know it is them who has the marking). the rest wait and do nothing, helpless to aid their fellow monks to realize this, and suspicious of whether they have markings or not.
the second day the reason the two monks with the markings come back shows to these two monks that they also have a mark on their heads. hence causing them to kill themselves on the second day. this doesnt happen, which means a marking has appeared on another monk. the rest of the group see this change.
the monk with the new marking now still see's two marked monks. he waits a day to see if they will kill themselves, since now they should have realized that they are the ones with the markings. since the third day there is no death, the third monk realizes he is the one with the marking due to the fact that the two monks still live, and since noone commits suicide the night of the third day, it is apparent that there is yet another marked monk.
the pattern continues and eventually every monk has the marking->
causing the seven monks to all to commit suicide eventually when there are no monks left without markings (the 7th day).
they all realize they have markings when noone is left without a mark.
i think i have solved it correctly.
thanks for this gnome. solving this has given me better clarity on relativity then directly replying to my question would have.
added later:
also i have noticed that it isnt for sure the amount of monks there were in the group. the incrementation of the amount of monks bearing new marks each day could have been any amount, and the original number of monks having the markings could start at any number....
so i am unsure.
Thats the first conclusion I came to, but it bothers me because it assumes the monks know the demon possesses the same number of monks per night.
yea, but regardless of how many more happen to be possesed each night it also seems that eventually they all will be possesed and will commit suicide... still the amount of monks seems undefinable...
to me and you that is ;)
can anyone seem to understand this?
Originally posted by elibol
yea, but regardless of how many more happen to be possesed each night it also seems that eventually they all will be possesed and will commit suicide... still the amount of monks seems undefinable...
to me and you that is ;)
can anyone seem to understand this?
I think you solved the problem incorrectly.
The problem assumes no more monks are being
possessed after the visitor's announcement, we are merely
asked to find that fixed number.
But even assuming more monks are being possessed each day, the minimum number that was possesed on the visitor's announcement
should be 7. But since they all commited suicide on the same day, there was never more than 7 monks.
eyesaw said:
"I think the problem assumes no more monks are being
possessed after the visitor's announcement, we are merely
asked to find that fixed number."
why would they sit around and look at each other for seven days and then kill themselves then?
all seven monks examine each other and find that the other 6 besides them are possed... all of them are unsure of whether they are possesed or not.
but in the case that one of them were not possesed->
from the perspective of the 6 possesed monks they see 5 possed monks and one that is not, it seems they have no basis on which to predict whether they are possesed or not.
and from the one that is not possesed, he see's six monks that are possesed, he also has no way to predict whether he is possesed or not.
at this point whether they decide to sleep on it they have no pattern to recognize to know if they themselves are possesed. the next day would be the same, maybe the last fellow being possesed as well, but he is unaware of this, and all the rest are just sitting around unsure if they themselves are possesed. it doesnt matter at this point, even if the last monk becomes possesed noone learns anything more than they didnt know before.
basing it on the fact that they would kill themselves if they were not certain of the fact that they may not be possesed leads me to believe that in this scenario, they would all kill themselves the first night.
being the logical monks they are, they should realize there is no true way of knowing whether they are possesed or not. better safe than sorry maybe?
i am still uncertain but i think this disproves your theory.
Originally posted by elibol
eyesaw said:
"I think the problem assumes no more monks are being
possessed after the visitor's announcement, we are merely
asked to find that fixed number."
why would they sit around and look at each other for seven days and then kill themselves then?
all seven monks examine each other and find that the other 6 besides them are possed... all of them are unsure of whether they are possesed or not.
but in the case that one of them were not possesed->
from the perspective of the 6 possesed monks they see 5 possed monks and one that is not, it seems they have no basis on which to predict whether they are possesed or not.
and from the one that is not possesed, he see's six monks that are possesed, he also has no way to predict whether he is possesed or not.
at this point whether they decide to sleep on it they have no pattern to recognize to know if they themselves are possesed. the next day would be the same, maybe the last fellow being possesed as well, but he is unaware of this, and all the rest are just sitting around unsure if they themselves are possesed. it doesnt matter at this point, even if the last monk becomes possesed noone learns anything more than they didnt know before.
basing it on the fact that they would kill themselves if they were not certain of the fact that they may not be possesed leads me to believe that in this scenario, they would all kill themselves the first night.
being the logical monks they are, they should realize there is no true way of knowing whether they are possesed or not. better safe than sorry maybe?
i am still uncertain but i think this disproves your theory.
Start with 2 monks. If only one is possessed, the possessed monk will commit suicide that night since he will see no mark on the other monk. If both have marks, they won't know until the second day. And they would both commit suicide on the second day.
With 3 monks, one monk sees the other two monks with marks. Since the situation with two monks out of 3 having marks on their head is the same as that when there were only two monks and both have marks, the watching monk would expect the other two to commit suicide on the second day if he had no mark on his own head. The fact that they don't would signify that he himself has a mark on his head. But since this point of view is valid for all 3 monks, all three will come to the same conclusion and commit suicide on the 3rd day.
And the pattern repeats itself. The day of the suicide directly corresponds to the number of marked monks (if there are at least
two monks to start with of course) - i.e., 1 marked monk = suicide day 1, 2 marked monks = suicide day 2, 3 marked monks = suicide day 3. Additionally, all the marked monks commit suicide on the same day.
Applying these facts to the problem, since they all commit suicide on the 7th day, 7 monks were marked on the first day. We know there are only 7 monks since there were no monks left over.
ok, but this isnt what you said in what i quoted you on.
when we start with 2 monks having marks, and each day one other monk gains a marking, then this makes sense that 7 monks will die on the 7th day.
but...
there is no certainty that it started with 2 monks.
and there is no certainty on how many monks gain a marking each passing day.
so with these left undefined it doesnt seem possible to me to know the number of monks.
either there is something im missing, or it is just not possible.
if it is something im missing please, you have the honor of pointing it out to me =]
Originally posted by elibol
ok, but this isnt what you said in what i quoted you on.
when we start with 2 monks having marks, and each day one other monk gains a marking, then this makes sense that 7 monks will die on the 7th day.
but...
there is no certainty that it started with 2 monks.
and there is no certainty on how many monks gain a marking each passing day.
so with these left undefined it doesnt seem possible to me to know the number of monks.
either there is something im missing, or it is just not possible.
if it is something im missing please, you have the honor of pointing it out to me =]
I only used the example of 2 or 3 monks to give you an idea
of how to solve the problem. Again, no where in the original
problem does it say that monks are gaining marks each passing
day. There had to be 7 monks marked to begin with or else it wouldn't take until day 7 for some monk to commit suicide.
Let's assume for a minute there were 200 monks to start
with. Now, if only one monk was marked, he would know
about it on the first day since he sees no marks on the
other 199 monks even though the visitor had announced that
at least one of the monks were marked. So the number
of unmarked monks does not affect the day of the suicide,
only the number of marked monks. Now if we had started
with 200 and we know that on day 7, 7 will commit suicide
if 7 were marked from day 1, then there would be 193 monks
left. But since the problem also states that "ALL" the monks
commited suicide on day 7, this requires that there be
only 7 monks to start with.
Originally posted by Eyesaw
Let's assume for a minute there were 200 monks to start
with. Now, if only one monk was marked, he would know
about it on the first day since he sees no marks on the
other 199 monks even though the visitor had announced that
at least one of the monks were marked.
yes, and he would commit suicide the first night.
Originally posted by Eyesaw
So the number of unmarked monks does not affect the day of the suicide, only the number of marked monks.
what? what was the point in pointing this out?
this has already been apparent thru out our argument.
it is obvious that the amount of marked monks determines the day of suicide.
Originally posted by Eyesaw
Now if we had started
with 200 and we know that on day 7, 7 will commit suicide
if 7 were marked from day 1, then there would be 193 monks
left.
wrong. why would it take them 7 days to notice that 7 were marked?
it states in the puzzle:
"At the second day's prayer meeting the monks all look at each other, but again that evening nothing unusual occurs."
why would they bother to examine each other if all of them had the markings? why would they sit around and wait 7 days? i cannot bother with this anymore. you are missing the fact that there is no logic in waiting 7 days when they all have markings.
Originally posted by Eyesaw
There had to be 7 monks marked to begin with or else it wouldn't take until day 7 for some monk to commit suicide.
what? explain this if you care to. this doesnt make logical sense that they would all wait 7 days.
Originally posted by Eyesaw
But since the problem also states that "ALL" the monks
commited suicide on day 7, this requires that there be
only 7 monks to start with.
yes but you are still assuming a given amount to begin with. you are saying 7... bah! you arent making any sense...
im sorry but until you cease to repeat yourself i will nolonger continue to argue with you. you are not being the least bit rational.
Originally posted by elibol
yes, and he would commit suicide the first night.
what? what was the point in pointing this out?
this has already been apparent thru out our argument.
it is obvious that the amount of marked monks determines the day of suicide.
wrong. why would it take them 7 days to notice that 7 were marked?
it states in the puzzle:
"At the second day's prayer meeting the monks all look at each other, but again that evening nothing unusual occurs."
why would they bother to examine each other if all of them had the markings? why would they sit around and wait 7 days? i cannot bother with this anymore. you are missing the fact that there is no logic in waiting 7 days when they all have markings.
what? explain this if you care to. this doesnt make logical sense that they would all wait 7 days.
yes but you are still assuming a given amount to begin with. you are saying 7... bah! you arent making any sense...
im sorry but until you cease to repeat yourself i will nolonger continue to argue with you. you are not being the least bit rational.
Read the problem again. It says that the monks could not tell each
other who has marks on their heads. So again, look at the simplest scenario where there are only two monks, both with marks on their heads. The first monk sees a mark on the second monks head but he cannot tell whether his own head is marked or not. From the second monk's point of view, it looks the same. So even though both monks are marked and there are only two of them, they cannot know on the first day whether they both have marks.
The only way for one of the monks to know he himself is marked is by the fact that the other monk doesn't commit suicide on the first evening.
If there are 3 possessed monks, it would take them 3 days to figure out they all have marks on their heads. And if there are 4, it would take 4 days.. et al.
i was the first to explain the answer man...
it isnt like i dont understand something. its more like you keep adding these weird things to your explanations.
like some crap about 7 monks having marks when the dude that could speak came and spoke.
why you keep dodging what you said before?
i am done.
Originally posted by elibol
i was the first to explain the answer man...
it isnt like i dont understand something. its more like you keep adding these weird things to your explanations.
like some crap about 7 monks having marks when the dude that could speak came and spoke.
why you keep dodging what you said before?
i am done.
Nowhere in the original problem does it state that monks are being possessed every passing day, you made up that assumption. But even if a million monks were being possessed every passing day, the
only relevant point is how many monks were marked on the first
day. It takes 7 monks 7 days to figure out they all have marks on their heads.
I stand corrected. If elibol was one of the monks,
it would take him quite a few years to commit suicide.
umm, yep. you sure do.
there is no solution without assumption in this puzzle... assuming there are any number of monks anywhere in your answer is an assumption. what the hell are you talking about? all i hear is bla bla bla...
my assumptions make logical sense, in that the conclusion of the puzzle is solved.
all i hear you say is 7, 7, 7... lucky number 7.
you stand corrected?
you arent the least bit rational.
if you are flawed (trust me, everyone is flawed at one point in their life or another), how will you ever know this?
your ignorance is impossible to deal with.
obviously it is an impossible task to try and convince you to be rational.
you will never be taken seriously if you are not rational.
you can stand corrected, in your room, wherever you are, by yourself. you are the only one that agrees that you are correct. it seems your in your own little world trying to prove people wrong. does it feed your selfesteem to prove others wrong? is your life so depressing that you feed off of the frustration of others inpatience to deal with your ignorance? it probably satisfies you that people give up against your ignorance. it is an obvious paradox, but from your perspective this is not obvious. you just love the feeling and dont really care to take a look at yourself for one second and realize your way out of line.
is it so hard for you to prove people wrong in a logical sense that instead of being rational you have to be ignorant enough to keep writing on and on? until the person you argue with gives up?
arguing over the internet is like running the special olympics.
whether you win or not, you are still RETARDED FOR ARGUING.
state your opinion, and sit down. let others decide whether you "stand correct".
Originally posted by elibol
umm, yep. you sure do.
there is no solution without assumption in this puzzle... assuming there are any number of monks anywhere in your answer is an assumption. what the hell are you talking about? all i hear is bla bla bla...
my assumptions make logical sense, in that the conclusion of the puzzle is solved.
all i hear you say is 7, 7, 7... lucky number 7.
you stand corrected?
you arent the least bit rational.
if you are flawed (trust me, everyone is flawed at one point in their life or another), how will you ever know this?
your ignorance is impossible to deal with.
obviously it is an impossible task to try and convince you to be rational.
you will never be taken seriously if you are not rational.
you can stand corrected, in your room, wherever you are, by yourself. you are the only one that agrees that you are correct. it seems your in your own little world trying to prove people wrong. does it feed your selfesteem to prove others wrong? is your life so depressing that you feed off of the frustration of others inpatience to deal with your ignorance? it probably satisfies you that people give up against your ignorance. it is an obvious paradox, but from your perspective this is not obvious. you just love the feeling and dont really care to take a look at yourself for one second and realize your way out of line.
is it so hard for you to prove people wrong in a logical sense that instead of being rational you have to be ignorant enough to keep writing on and on? until the person you argue with gives up?
arguing over the internet is like running the special olympics.
whether you win or not, you are still RETARDED FOR ARGUING.
state your opinion, and sit down. let others decide whether you "stand correct".
I think you are overreacting a little here. All I did was
gave my solution to gnome's puzzle and you make me out as if
I were the evil demon in the problem. I don't understand what
your objection is to my solution. Where am I making assumptions
that are not already stated in the problem? We know there is
at least one monk there from what the visitor said about the
marks on the monks.
You made assumptions that weren't stated in the original problem,
so your solution isn't even valid. Find where the problem tells
us to assume each passing day, new monks were being possessed?
The assumption is not even relevant to the problem.
You are the one out of line here for calling me ignorant and all sorts of names when you can't even produce a good reason why my solution is wrong.
one_raven
Feb24-04, 05:28 AM
Eyesaw is correct.
Let me try...
If you are 1 of 2 monks, you will look at the other monk.
If he has no mark, then you will kill yourself that night since you know you are posessed.
If he DOES have a mark, then you may or may not have a mark.
Keep in your mind that he is thinking the EXACT same thing.
The next day, since he did not kill himself, you realize that he was waiting to find out just like you were, so that dictates that you must have the mark as well, that night you both kill yourselves.
See how that works so far?
If you are 1 of 3 monks you will look at the other two monks to see if they are marked.
If neither of them are marked, then you must be, so you kill yourself that night.
If one of them is marked, then you wait till tomorrow since if teh marked one does kill himself then you know the he did not see the mark on the others.
If one of them is marked, and on the second day no one is dead, that means you are marked, so that night you will kill yourself.
That also rings true for the other one since no one was dead on day two and you know that only one of the others are marked.
On day three, the monk that saw that you were both marked is still alive since he did not know if he was marked.
He knew that if he was not marked you would both kill yourselves on day two.
So, if you were not both dead on the morning of day three, then he must be marked as well so all 3 would kill themselves that night.
They all look at each other every day for two reasons.
1. To see who is still alive.
2. To make sure that the markings haven't changed.
Now if you continue that logic to seven days the only way all the monks would be dead is if there were seven of them.
If there were only 6, then they would have all been dead by day 6.
If there were more than 7, then it would have taken more than seven days to figure it out.
Personally, i would have said, "Fucx this vow, am I marked?"
Seems gnome's puzzle was a spin off of one called
the "blue-eyed monks".
Here's a solution given from a website:
http://ai.eecs.umich.edu/people/dreeves/brainteasers/archives/
I wasn't nuts after all.
one_raven
Feb24-04, 06:08 AM
Originally posted by Eyesaw
I wasn't nuts after all.
Now who's making assumptions?[:))]
Well done, Eyesaw. I knew you could do it -- it didn't require any fractions. [;)]
I hope you enjoyed that debate as much as I did reading it. By the way, I didn't get it from the U Mich. site. I guess we both got it from some other source, several times removed. I couldn't find where I originally saw it, so I was writing from memory. I may have embellished it slightly, but my source definitely involved a demon and marks on the forehead.
Oh, and one_raven: nice explanation. [:)]
sorry eyesaw, but when you said:
"I stand corrected. If elibol was one of the monks,
it would take him quite a few years to commit suicide."
that really pissed me off...
i take back my remarks.
also, note: i have taken out a large amount of writing that was once this post, i now realize it is unessecary based on what i have concluded to.
Originally posted by gnome
I hope you enjoyed that debate as much as I did reading it.
what does this mean?
Elibol, Eyesaw didn't explain it very well, but I think you might want to stop and think. You appear to me to be making a fool of yourself much like eyesaw did the other day. [;)] Don't worry about the criticism, everyone is wrong from time to time.
I think one_raven did a good job at explaining it, but I'll take a stab in my own terms since the concept wasn't easy for me to grasp either.
If there were three monks that all had a marking on their heads, each would see two other monks with a marking on their head on day one. At this point, none of the three monks would be certain there was a mark on his head. On the second day, the monks waited for the two they observed to have markings to kill themsevles; the two monks would realize they had a marking on their head if the observing monk did not. When all three monks came back on the 3rd day, all three monks would realize they had a marking on their heads.
i already know this.
this makes perfect sense man...
its much deeper than this though... read my posting on page 7 please. everyone read it, if you care to look into this puzzle further. dont turn your heads the other way, because without what i wrote on page 7, the means of getting to the solution still has holes even though the logic the monks use makes sense...
ok, i was wrong, and i was able to figure this out with the blue-eyed monks version's follow up question.
it is essential to accually understanding the purpose of the announcement, and for me, to understanding how the entire problem makes any sense.
it states:
"It's clear that the visitor was necessary in the
proof in order to establish the base case. However, the only information
that the visitor provides (there exist blue eyes on this island) is
something that (assuming more than 1 blue-eyed monk) every single monk
already knew. The question: What was different after the visitor's
announcement? Ie, as a monk on the island, what do you know after the
visitor's announcement that you didn't know before?"
this is what i am trying to say, what does the monk know now that he didnt know before the announcement?
i will answer this after i have addressed my former flaw:
lets take now the 7 marked monks, i will explain it now the way i wanted it to be explained for me to understand it. i knew there was something i was missing, but the flaw in gnomes version (the pointlessness of the announcement) caused me to come to some alternate possible case scenario with an assumed aspect (the daily incrementation).
if i had read formerly the blue-eyed monks version i wouldnt have had this problem.
there are 7 marked monks. each one see's 6 other marked monks, and each one assumes the other monks only see 5 marked monks based on the assumption that he is not marked. now he must look from the perspective of the 5 monks, to know whether they are marked or not they must assume first that they are not, thus seeing only 4 marked monks. he must now look from the perspective of the 4 monks, to know whether they are marked or not they must assume first that they are not, thus seeing only 3 marked monks. he must now look from the 3 monks point of view, to know whether they themselves is marked they must first assume that they are not, thus seeing only 2 marked monks. he must now look from the two marked monks perspective, to know whether he himself is marked, he must first assume that he is not, thus seeing only one marked monk. when this monk does not commit suicide on the first day, he now knows he is marked. thus the addition to his knowledge base on how many monks know that themselves are marked becomes two.
i think now that the assumption they all have of not being marked causes this logical solution to be possible.
i believe the announcement marked a starting point in which all of them could take advantage of this logical method to figure out whether they were blue-eyed or not.
allowing them to begin their evaluation of the situation at the same time.
an attempt to figure out the color of their eyes anytime before this would be impossible, since they have no way of telling when they can start the countdown to get an accurate count on how many monks knew the color of their own eyes.
it is now obvious that it would be illogical for a monk to just begin counting down days with no starting point to base his information on.
I just finally got the nerve to attempt to explain it with 7, and you go and burst my bubble. [*(]
Edit: Anyway, my bad for not reading what you wrote before I replied to you, but I thought the analogy with 3 made the concept easier to understand. Not much changes between 3 monks/3 days, and 7 monks/7 days. It also takes much less explaining.
Elibol, I really don't know what I did to make you sick. I certainly had no malicious intent in posting this puzzle. I don't see that there is anything wrong with enjoying a debate. If we all agreed all the time, what would be the point in talking?
And I do not agree that my version of the puzzle is flawed, either. In fact, some of your own objections highlight the fact that the U.Mich. version is flawed. That is, blue eyes are obviously a permanent feature (anybody with blue eyes has always had blue eyes), so your point about what information was provided by the visitor is well taken. Also, the U.Mich. version says they will commit suicide "by sunset", thus allowing the possibility of someone committing suicide immediately at prayer meeting, which can screw up the sequentiality of it. My version specifies that suicides only occur in private in the evening, so the information about the suicides is only available the next day. For these reasons, I think that the U.Mich. version is ambiguous.
My version, on the other hand, allows the interpretation that the marks first appeared on "day 1" when the visitor arrived. Yes, you are correct in observing that if 6 out of 7 monks were marked that day, they would all see marks on either 5 or 6 of their fellows. But the puzzle doesn't tell you how many there are, or how many are marked. So the visitor is certainly necessary for the benefit of the puzzle. IF there were only two or three monks, and IF only one monk (out of any total) was marked, the visitor is necessary for the monks as well. As far as the question of what knowledge is provided by the visitor in the case where there are 7 monks, all marked, I haven't heard or thought about that question before now, and I'll admit I don't know the answer. But if the number of monks is unknown, and the number of marked monks is unknown, and you don't know if any of the monks know if any of the monks are marked, there is no puzzle.
Anyway, while the last few posts were going up, I've been busily working on this explanation which I see is now redundant, but since I spent all this time on it, I'll throw it up here anyway. Maybe it will help someone else.
I still believe that a total of 7 monks, all marked beginning at the moment of the visitor's announcement, is the correct answer. I probably can't explain it any better than the others have done, but I'll try. The key to it is the dimensionality. Monk A is trying to deduce what monk B (and every other monk) is thinking, knowing that monk B is trying to deduce what monk C (and every other monk) is thinking, and so on. So it becomes one of those picture in a picture in a picture in a ... situations. Monk A thinks that Monk B thinks that Monk C thinks that ... etc. I can't see 4 dimensions, let alone 7.
1 of course is trivial. If there is only 1 monk, & the visitor says "at least 1 is marked", the monk commits suicide that evening & the game is over.
2 is not much harder. Monk A tells himself, "only 1 of us is marked, so if I am unmarked, B will see that, deduce that he IS marked, and he will commit suicide on evening 1. When Monk B appears at prayers on day 2, Monk A knows that B sees the mark on A, so A commits suicide on evening 2. You can see that my choice of who is A and who is B is completely arbitrary, so swap the letters, & now BOTH commit suicide on evening 2. Note that this is true only if IN FACT BOTH ARE MARKED. If only Monk B is marked, he sees that A is unmarked and commits suicide on evening 1. This confirms in A's mind that he is unmarked so he does not commit suicide. Period. Either way, the condition that "all of the monks commit suicide on evening 7" does not occur, so 2 monks is not a valid answer.
Let's look at 3.
IF there are 3 and all 3 are marked, on day 1, Monk A sees that his two brethren are marked. Monk A wants to believe that he is unmarked. He knows that Monk B wants to believe that HE is unmarked. Monk A thinks, "If I am unmarked and Monk B believes that he is unmarked, he (B) must assume that C is the only one marked. He (B) would then expect C to commit suicide on evening 1, and therefore he (B) would expect that C would not show up at prayers on day 2. Since C was still alive on day 2 (A reasons), B would have to conclude (on day 2) that C does NOT see two unmarked monks. So, if C is still alive on day 2, he must see at least 1 marked monk. If I (A) am unmarked, B sees that, and therefore he will know on day 2 that HE (B) is that marked monk, so he will commit suicide on evening 2. If B shows up at prayers on day 3, then he must see that I (A) am marked, so on day 3 I commit suicide.
If all of the monks are marked, they all follow the same reasoning, and all commit suicide on evening 3. If only 1 was marked, he would commit suicide on evening 1, and the other two would know that they are unmarked. If 2 were marked, following the same reasoning they would both commit suicide on evening 2, saving the 3rd.
If there are 4...
I don't have time to take this to 4. I'm not sure that it can be described adequately verbally. And even if I did, you'd say "prove it for 5". I believe the inductive proof already described on the U. Mich. site is valid for my scenario, even more than it is for theirs.
If you don't agree, I guess none of us is going to convince you. I hope you - all of you - enjoyed puzzling over it anyway.
Originally posted by leto
I just finally got the nerve to attempt to explain it with 7, and you go and burst my bubble. [*(]
thanks.
i admit i was wrong. but i dont think i ever would have figured it out or understood it as fully without the blue-eyed version's follow up question...
it might sound a bit pathetic to some of you, but the process it took for me to get to the solution to this problem will be a pretty big asset to my problem solving skills...
know that im amateur to these problems, and if im wrong im humbly requesting that if i ask of someone to explain the problem the way i need it to be explained i would greatly appreciate it. otherwise it seems i keep going and going and... yall saw what happened =[
well, i guess the time it took you to post that i had already posted that we have a mutual agreement on how the monks logic work, and fixed alot of what i said before... you dont make me sick, i have corrected this posting as well.
i was utterly frustrated, and i admit, a bit childish.
thank you for the puzzle, it was very enlightening.
one_raven
Feb25-04, 03:15 AM
gnome,
I have a suggestion.
The first time I read this I thought, "That visitor must be a prick! Why wouldn't he just tell which monks are marked, knowing that none of them can communicate with each other?"
Maybe next time you can avoid such confusion (about whether the curse spreads and why the visitor's announcement is important) by changing one small tidbit. The visitor is the one who curses them. Maybe it is a demon or devil of some sort. He visits teh island and says, "I will place a curse on at least one of teh monks on this island. I will not tell you which ones, because I do not want you to know, but every one else will know because there will be a mark on your forehead. You will live the rest of your life not knowing whether or not you are cursed." Or something like that. Then he disappears. Seven days later, all the monks are dead of suicide.
I think that makes it a seamless puzzle.
What do you think?
Originally posted by one_raven
gnome,
I have a suggestion.
The first time I read this I thought, "That visitor must be a prick! Why wouldn't he just tell which monks are marked, knowing that none of them can communicate with each other?"
Maybe next time you can avoid such confusion (about whether the curse spreads and why the visitor's announcement is important) by changing one small tidbit. The visitor is the one who curses them. Maybe it is a demon or devil of some sort. He visits teh island and says, "I will place a curse on at least one of teh monks on this island. I will not tell you which ones, because I do not want you to know, but every one else will know because there will be a mark on your forehead. You will live the rest of your life not knowing whether or not you are cursed." Or something like that. Then he disappears. Seven days later, all the monks are dead of suicide.
I think that makes it a seamless puzzle.
What do you think?
Well, agreed- that visitor was a prick. But I think gnome wrote a clever version of that blue-eyed monk puzzle actually. And I don't think there are any loose ends. The key clue in the puzzle was that all the monks committed suicide on the same day. With this knowledge and also knowing the day of the suicide, it's kind of implicit that the puzzle wasn't talking about monks getting possessed every passing day but about how many monks were marked at the time
of the visitor's announcement.
Let's try to solve the problem the other way to see why it is
not possible. Let's say on the first day, two monks are marked but one isn't. Let's also assume that on the second evening the demon will come to possess the third monk. Well, since the two monks marked on the first day have already concluded by the second day that they both are marked, they will kill themselves that evening, so they won't even live long enough to see the mark on the third monk. And once they are dead, the third monk would have no way of knowing if he is marked, so he will live on.
Even if the third monk was marked by the morning of the second day, it still won't prevent the first two monks from committing suicide
on the second evening since their deductions of marks on their heads only depended the marks they saw on the first day.
So you see, if the monks were getting possessed each passing day,
there will always be at least one monk left alive. So the only way all 7 would be dead if they were all marked by the first day.
BTW, I had never saw this puzzle before and I solved it correctly
the first time so it must have been written appropriately. I only bothered to look for second source for the solution to convince a doubting elibol.
I agree with Eyesaw (of course, since he agrees with me [;)] ) but I'll take this opportunity to be a prick & say, Why should I tell the reader that all the monks become infected simultaneously? Part of the puzzle is to figure that out for yourself.
Eyesaw, you get a gold star for being such a discerning judge of puzzles. [:D]
Oh, and elibol: no sweat. We all get aggravated at times, as I did the other day when Eyesaw was enjoying hassling me about* the birthday puzzle.
*[my correct solution to] [;)]
Originally posted by elibol
sorry eyesaw, but when you said:
"I stand corrected. If elibol was one of the monks,
it would take him quite a few years to commit suicide."
that really pissed me off...
i take back my remarks.
also, note: i have taken out a large amount of writing that was once this post, i now realize it is unessecary based on what i have concluded to.
I thought it was a good joke, my bad. From your first explanation
of the puzzle, I don't doubt you would've solved the problem correctly if the assumptions were made more explicit. I probably just got lucky because the first thought I had when I read the line " all monks committed suicide on day 7" was that they must've
all been marked on the same day. And I also had a strong hunch that 7 was going to be a key number- it just seemed like a great way to complete the puzzle.
Originally posted by gnome
I agree with Eyesaw (of course, since he agrees with me [;)] ) but I'll take this opportunity to be a prick & say, Why should I tell the reader that all the monks become infected simultaneously? Part of the puzzle is to figure that out for yourself.
Eyesaw, you get a gold star for being such a discerning judge of puzzles. [:D]
Oh, and elibol: no sweat. We all get aggravated at times, as I did the other day when Eyesaw was enjoying hassling me about* the birthday puzzle.
*[my correct solution to] [;)]
I found the lazy way to solve that problem.
Originally posted by Eyesaw
I only bothered to look for second source for the solution to convince a doubting elibol.
hehe, my last name is accually elibol. so for you to put it that way it makes alot of sense!
:)
please stop quoting all posts...please do not post off topic....please post some new puzzles
A general is capturing 3 ennemies....being a puzzle fan,he proposes them a possibility to gain their lives
He puts the one behind another , face on the wall(the 1st was seing just the wall, the 2nd was seing the 1st etc)and he brings a bag with 5 hats(3 black and 2 white)
He puts a hat on each head (not the small heads[:)] )and asks the colour of the hat that they are wearing.
The 3rd says that he dosen't know...and he dies
The 2rd says that he dosen't know...and he dies
The 1st says that his hat is black and he's free
Explain his logic
third guys logic-> he see's the second and first person either is wearing a black hat or a white hat. he would have been able to figure out which color he was wearing if both had a white hat on, so one of them must have had atleast one black hat on. he dies.
second guys logic-> since the guy behind him is dead, he knows one of them must have a black hat on or else the guy behind him would have gone free. if he see's a white hat on the guy infront of him he would guess that he had a black hat, and he would have gone free. this isnt the case, so he is dead.
first guys logic -> since he is logical enough to understand the logic the third and second guy went thru, he knows for sure he has a black hat on. he lives.
I think the puzzle with the monks was an offspring
of this hat puzzle. The monk one is harder.
We have a long cable with 10 wires inside....we have a battery and a light ball....we have to find the ends of each wire, but in the mean time we can move only 2 times between the sides of the cable
enjoy[;)]
Pick up end of cable, drag it to the other end, test wires, drag it back. [:D] [:D]
u r joking, right [?] ....
DO NOT ERASE MY POSTS please[!:)]
OK, you don't like that approach, how about this. I'll try to upload a diagram, but in case that doesn't work, maybe you can follow this.
Starting at the left end of the cable, connect wires together in a group of 4, a group of 3, and a group of 2. (Leaving 1 wire unattached to anything.)
Going to the right side, you can use the battery and the bulb (I assume that your "light ball" is actually a light bulb) to test and identify the wires that are in the 4-group, the 3-group, the 2-group and the single one. Label them a - j as I show in the diagram. Now connect b to e, c to h, d to j, and f to i. (a and g are left unconnected.)
Now go back to the left side with the battery and bulb. Label the wires in each group so you will remember which ones they were, disconnect them from each other, and test to find:
- a wire from the group of 4 that makes no circuit with any other wire - that one is a.
- a wire from the group of 4 that makes a circuit with a wire from the group of 3 - those are b and e.
- a wire from the group of 4 that makes a circuit with one from the group of 2 - those are c and h.
- a wire from the group of 4 that makes a circuit with the one that was originally unconnected - those are d and j.
- a wire from the group of 3 that makes a circuit with one from the group of 2 - those are f and i.
- a wire from the group of 3 that makes no circuit - that is g.
I SAID "GOOD WORK" DUDE!!! (do not erase my postreplys please )
i found 2 solutions on this problem but you kinda used them both in your answer
so....we have 12 balls.Only one is different(lighter or heavyer)
We also have a balance....and only 3 trys to find the different ball
Enjoy[:D]
I got something nice too.
I am a Gabber with a bold head.
Am I lying or telling the truth? [?]
[o)] Kindest regard Dj You_nis
yes you are [:D] [:D] [:D]
wrong, i aint a skinnhead [!:)] I am the devil :P
Tell me y3nis [:))] where did i said that you were a skinhead ? [:)]
you didn't. lets make this really philosophy. What exectly does skinhead mean? Someone that is bold i'd say. Else my previous post werent correct
Did I really said smth?[:D] [:D]
Pls. quote me if did [;)]
Originally posted by marqq
Did I really said smth?[:D] [:D]
Pls. quote me if did [;)]
Save My Tits Here=smth?
Well i'll tell you the truth. I aint a gabber and i aint bold.
Did i really tell the truth? [g)]
Originally posted by y3nis
Save My Tits Here=smth?
something
Btw...try to puzzle me by offering an corect answer to my 12 balls problem [:D] [:D]
Originally posted by marqq
so....we have 12 balls.Only one is different(lighter or heavyer)
We also have a balance....and only 3 trys to find the different ball
Enjoy[:D]
you didn't forget about that didn't you?
where r the signatures?
:mad:
tribdog
Mar29-04, 03:36 AM
You are the big winner of a television game show. The host says "Grand prize is behind door #1, 2 or 3. pick one." After you make your choice (lets say you pick #1) the host says "You may be right. It's either #1 or #2 but its not number 3. Now I'm going to give you one last chance do you want to stick with door #1 or would you like to pick again?" What should you do?
I just took the time to answer that horribly long balance puzzle, and my post didn't work. I'm not certain I am right, I am tired and can't think anymore. There is more to it than this, but this is all I had in my clipboard and there is no way in hell I'm typing anymore:
I would take 8 of the 12 balls and weigh them with 4 on each side of the balance.
If they were even, then I would know the odd ball was one of the 4 I haven't weighed. If the balance tipped then I would label the balls on the tipped side as l1 l2 l3 and l4. The balls on the other h1 h2 h3 and h4.
I would then take l1, h4, h3 and weigh against h1, h2 and l2. If the balance was even, then I would know the odd ball was either l3 or l4. if the balance tipped to the first side, then i would know the odd ball was either h4, h3 or l2. if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l2.
If it was l3 or l4 then I would use my last balance to measure l3 again any ball other than l4. If the scale was even then l4 is the odd ball. If the scale dips to l3 then l3 is the odd ball.
It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.
If it was h1, h2, or l2, then I would do the same as directly above and find the odd ball.
If my initial weigh was balanced, then I would take 3 of the 4 remaining balls and weigh them against 3 other balls. If the balance stayed level, then I would know the odd ball was the one I didn't weigh. If it did not, then I would know whether the 3 balls were lighter or heavier. I would then take 2 of the 3 remaining balls and weigh them against eachother. If the balance was level then the odd ball is the one I didn't weigh. Otherwise, it is the heavier/lighter one, depending on whether the 3 balls were lighter or heavier.
It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.
If it was h1, h2, or l2, then I would do the same as directly above and find the odd ball.
If my initial weigh was balanced, then I would take 3 of the 4 remaining balls and weigh them against 3 other balls. If the balance stayed level, then I would know the odd ball was the one I didn't weigh. If it did not, then I would know whether the 3 balls were lighter or heavier. I would then take 2 of the 3 remaining balls and weigh them against eachother. If the balance was level then the odd ball is the one I didn't weigh. Otherwise, it is the heavier/lighter one, depending on whether the 3 balls were lighter or heavier.
isn't it cute? BUT IS WRONG!!! :biggrin: :biggrin: remember that u don't know if the ball is lighter or heavyer... It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.
why?
but you had a good start!!
You are the big winner of a television game show. The host says "Grand prize is behind door #1, 2 or 3. pick one." After you make your choice (lets say you pick #1) the host says "You may be right. It's either #1 or #2 but its not number 3. Now I'm going to give you one last chance do you want to stick with door #1 or would you like to pick again?" What should you do?
whatever you want....you have a 33% chance to find the prize....you pick #1...and you find that the door #3 is a fake....Your chances increase to 50%....so either you take #2 or you rest on #1 you heave the same chances to win the prise :)
I had one typo, but the fact that I don't know whether it is heavier or lighter was taken into account. This puzzle would be simple if I knew if it was lighter or heavier.
"if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l2."
should of been:
"if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l1."
I can connects the dots if you still think I am wrong, but I'm too lazy to do that until then :P
whatever you want....you have a 33% chance to find the prize....you pick #1...and you find that the door #3 is a fake....Your chances increase to 50%....so either you take #2 or you rest on #1 you heave the same chances to win the prise :)
no I don't think that is correct. its close, but not right. What if there are a million doors and after you choose Monty Hall/Bob Barker throws out 999998 of the bad doors and says it's either the one you chose or this one? do you think it is still a 50/50 chance then?
tribdog if there were 1milion doors you would have 1/1000000 chances to win the prise...if we-ll take off a fake one, either you change or you keep your door, you'll have a 1/999999 chances to win the prize...so...the answer is the same...you can do whatever you like...you have the same chances
leto ... let me give you a hint...try to use the 1st weigh...make 2 options: 1.what if it tippes ao a 1st side, and 2.what if it tippes on second side... You'll have to work longer, but you have a lot more chances to find the odd ball :biggrin:
I disagree. By removing the 999998 doors, the chance of the door you didn't pick being the door increases. The door you did pick could not be removed, and is therefore less likely to be the prized door.
There are three possibilities with the first weigh. Either it tips to the left, right or stays level. I accounted for all three of these.
Edit:
I think you are assuming I label them l1-l4 and h1 - h4 because I assume I know this. I only label them as such so that I know later that, if one of them is the odd ball, they have to follow that pattern. A ball that was on the heavy side will never be the lighter odd ball.
I disagree. By removing the 999998 doors, the chance of the door you didn't pick being the door increases. The door you did pick could not be removed, and is therefore less likely to be the prized door.
correct. in the million door scenario the chances of the door you picked being the correct door is 1 in a million, therefore wouldn't the chance of the other door being the correct one be 999999 in a million? The point being it is always better odds to make the switch in a situation like this.
another way of looking at it is this: after you make your first choice you are given the option of keeping your door or taking all 999999 of the other doors.(which is the same thing as taking one door that may be right and throwing out 999998 wrong ones) Do you keep your 1 or take all 999999 of the others?
I disagree. By removing the 999998 doors, the chance of the door you didn't pick being the door increases. The door you did pick could not be removed, and is therefore less likely to be the prized door.
The problem allow us to eliminate only one door...From 3 doors or from 100000000 doors....we are able to remove only one door...
I'll reed about probabilyties and i'll give you an answer
There are three possibilities with the first weigh. Either it tips to the left, right or stays level. I accounted for all three of these.
Edit:
I think you are assuming I label them l1-l4 and h1 - h4 because I assume I know this. I only label them as such so that I know later that, if one of them is the odd ball, they have to follow that pattern. A ball that was on the heavy side will never be the lighter odd ball.
sorry ... i missunderstood your "l-f" definition...i was tired and i thought you ment left-right...and swiching the balls, an left becomes right and viceversa.....therefore i missjudged your solution....i am deeply sorry....your answer is as corect as it can be..... i'm really sorry about that
The mother is 21 years older than the son.In 6 years, the son will be 5 times younger then his mother.Question: where is the father now? :rolleyes:
enjoying himself :)
you are closer to the answer than you imagine :biggrin:
you are closer to the answer than you imagine :biggrin:
i got the answer, that's why i wrote that ;)
would you please enlarge your answer for the others? :wink:
vertigo
Apr12-04, 05:35 PM
Here's an easy one. See if you can solve this riddle in under 30 seconds...
Two Russians are walking down a street. One Russian is the father of the other Russian's son. How are they related?
Husband and wife? In second thought, I don't know how things work in Russia so my answer would be "they had sex".
no I don't think that is correct. its close, but not right. What if there are a million doors and after you choose Monty Hall/Bob Barker throws out 999998 of the bad doors and says it's either the one you chose or this one? do you think it is still a 50/50 chance then?
Ahh yes, why not? Because Bob says "it is either the one you chose or the other", which implies that one of the doors is the right one. He does not ask "would you like to change your pick now", which doesn't necessarily mean one of the doors is the right one. So yes, your chances would be 50/50.
We have a driver who has to make the distance between A and B in his own car...He drives the 1st 3rd of the distance with 40 km/h (he was reall calm) then he becomes angry and drives the 2nd 3rd with 120 km/h...then it calms down and asks himself what is the speed that he should use for haveing a decent average of 90 km/h
I'll ask you to try (before starting to calculate) an aproximation by intuition, and then make the difference to see it it was a small or a huge error :tongue:
Is 120 km/sec a typo or did you really mean that? If it's a typo he should drive at 110 km/h, if it's not... I don't think he can reach an average of 90 km/h after driving at ~400,000 km/h.
If it's a typo he should drive at 110 km/h, if it's not...
sorry
I was thinkin at smth else
120 km/h
And about the 110km/h you are wrong :biggrin: !Hint : make an ecuation
I did...
90 = \frac{40 + 120 + x}{3}
x = 110
:confused:
1/90 = 3(1/40 + 1/120 + 1/x)
You do realize, marqq, that the solution to that equation is negative?
is not negative...is just the infinite x
would you please enlarge your answer for the others? :wink:
He is located inside the wife, at least a certain part of him.. I didn't want to spoil since it's simple but funny :)
As for the driver.. If t is time in which he would have to make the whole trip for his average to be 90 km/h, he spent exactly t on the first two thirds of the trip. Which means he would have to make the remaining 1/3 in 0 hours, going infinitely fast :)
He is located inside the wife, at least a certain part of him.. I didn't want to spoil since it's simple but funny :)
As for the driver.. If t is time in which he would have to make the whole trip for his average to be 90 km/h, he spent exactly t on the first two thirds of the trip. Which means he would have to make the remaining 1/3 in 0 hours, going infinitely fast :)
Well done Pig :smile:
i'm inside of an imaginary cyrcle on which a shark swims inteligently (he can change the direction or he may stop but he can't enter the cyrcle) 4times faster then i do. Can I exit the cyrcle without been cought by the shark?
:eek:
Btw ...who-s the phorum administrator? i have a problem with my avatar and my signature :)
jammieg
Apr18-04, 09:55 AM
True or false, what is your name?
off topic. erase your post :)
Angel Loupe
Apr27-04, 04:18 PM
Taken Hurkyl's legend:
l l=river < >=canoe C=Cannibals M=Missionaries
CCC MMM < > l l
C MMM l<C C>l < > CC
C MMM l< C > l C
CC MMM < > l l C
MMM l<C C>l C
MMM l l < > CCC
MMM l< C > l CC
C MMM < > l l CC
C M l<MM>l CC
C M l l < > CC MM
C M l<CM>l C M
CC MM < > l l C M
CC l<MM>l C M
CC l l < > C MMM
CC l< C >l MMM
CCC < > l l MMM
C l<C C>l MMM
C l l < > CC MMM
C l< C >l C MMM
CC < > l l C MMM
l<C C>l C MMM
l l < > CCC MMM
this problem allready has a solution on this thread...pls stay focused :tongue:
True or false, what is your name?
False. In fact, I have never met anyone named 'what'. Also, since most names are capitalized, I will point out that I have never met anyone named 'What' either.
Njorl
Angel Loupe
Apr28-04, 10:16 AM
My bad. Focus is not my strong point...however, the solution was good? Then it was still to my own benefit, I suppose. And I love solving those kinds of puzzles. I'll remember to read all next time...thanks!
i'm inside of an imaginary cyrcle on which a shark swims inteligently (he can change the direction or he may stop but he can't enter the cyrcle) 4times faster then i do. Can I exit the cyrcle without been cought by the shark?
:eek:
try this one Angel Loupe
Two guys walk into a bar sit down both order scotch on the rocks the first guy slowly sips his and the second guy gulps his first one down then after a bit he orders a second drink and gulps the second one down while the first guy is still slowly sipping his drink. After the second guy finishes his second drink the first guy slumps over dead on the bar.
If both drinks were poisoned then why is the first guy dead and the second guy still alive.
I was recieved this Logic Puzzle. But I havn't been able to answer such puzzle. I think it is flawed. Since there involves three drinks, two of them are poinsoned. One Person drinks one drink which is poisoned and dies. Now that leaves two drinks whereas the other guy drinks both of those drinks. Then one of those drinks must be poisoned therefore he should drop and die just like the other guy. Anyways am I right? Someone help me here.
MathematicalPhysicist
May2-04, 03:11 AM
Two guys walk into a bar sit down both order scotch on the rocks the first guy slowly sips his and the second guy gulps his first one down then after a bit he orders a second drink and gulps the second one down while the first guy is still slowly sipping his drink. After the second guy finishes his second drink the first guy slumps over dead on the bar.
If both drinks were poisoned then why is the first guy dead and the second guy still alive.
I was recieved this Logic Puzzle. But I havn't been able to answer such puzzle. I think it is flawed. Since there involves three drinks, two of them are poinsoned. One Person drinks one drink which is poisoned and dies. Now that leaves two drinks whereas the other guy drinks both of those drinks. Then one of those drinks must be poisoned therefore he should drop and die just like the other guy. Anyways am I right? Someone help me here.
it depends which two drinks were poisoned the one which the first guy drank and one of the two which the second guy drank or two of the drinks which the second man drank were poisoned.
if it's the second option then it could be say that the first man was an old person (and that's why he sip the drink slowly) and he died out of old age and the drinks which the second man drank neutrilized each other.
This is not a logic puzzle...is just a childish question...like What DIANA means? (Died In A Nasty Accident)
We can asume tons of possible answers...but i think that the correct answer is : The poison was on the ice rocks,and it had the time to melt for the 1st drinker :smile:
MathematicalPhysicist
May3-04, 01:21 PM
This is not a logic puzzle...is just a childish question...like What DIANA means? (Died In A Nasty Accident)
We can asume tons of possible answers...but i think that the correct answer is : The poison was on the ice rocks,and it had the time to melt for the 1st drinker :smile:
are you familiar to lateral puzzles, these are those sorts of puzzles.
anyway, your answer seems more appropiate to the puzzle.
In 3 days i'll post the solution of the shark problem...this way I'll be able to post someother easyer problems.
i'm inside of an imaginary cyrcle on which a shark swims inteligently (he can change the direction or he may stop but he can't enter the cyrcle) 4times faster then i do. Can I exit the cyrcle without been cought by the shark?
Yes, you can. Supposing the circle has a diameter of 4 units, the cut of point of where you have the ability to travel more degrees over the circle, and therefore are able to keep the shark to the opposite side of the circle would be in 1/2 units. At this point you are 1.5 units away from the edge of the imaginary circle, so it becomes a simple question of whether you can travel 1.5 units in the time it takes the shark to travel half the circumference of the circle, which you can.
Edit: Realistically, you are cutting it so close that it would depend on how long you were and how big the circle was. I think that is the answer you want though unless you care to give us more details.
leto that was the expected answer.thank you. :smile:
3 mens are going to see a whore. they have only 2 condoms on them (not each).find a solution that allows all 3 of them to have sex with the prostitute without any exchange of liquids between them. :biggrin: :biggrin:
First guy wears both condoms. Second guy wears one that was on top, and third guy wears the one the first guy wore insided out with second guy's on top.
:smile: good job! As allways :wink:
5 men had a whole bunch of coconuts.as they gathered the coconuts all night long, they've decided to share them the next day.As soon as they went to bed, the 1st one wakes up, counts the nuts, and realises that he has to throw away one, in order to split them in 5 equal parts..He throws away the coconut,then he takes one fifth and hides it.30 min after, the 2nd man does the same thing...he counts, and he realises that he has to throw away a coconut in order to have 5 equal shares...he throws the f*ckin nut away, and he takes one fifth of the rest and goes to hide his part...Following the same pattern the third wakes up, then the fourth, then the fifth...The next day, they counted the coconuts, and they finded that they need to throw away a nut in order to share them equaly...
How many nuts were before the 1st stoled his share? :smile:
15621-1=15620
15620/5=3124
15620-3124=12496
12496-1=12495
12495/5=2499
12496-2499=897
897-1=896
896/5=179.2
179.2 dosen't belong to N
But you are really close :smile:
12496-2499=897 :confused:
Pattielli
May24-04, 06:59 PM
He didn't follow the pattern either,
:smile:
1st man 15621-1=15620
15620/5=3124
15620-3124=12496
2nd man 12496-1=12495
12495/5=2499
12495-2499=9996
3rd man 9996-1=9995
9995/5=1999
9995-1999=7996
4th man 7996-1=7995
7995/5=1599
7995-1599=6396
5th man 6396-1=6395
6395/5=1279
6395-1279=5116
2nd day 5116-1=5115
5115/5=1023
That's correct...sorry about that, i was tired, i knew that the number was 15xxx, i started to calculate, and I made a mistake :shy:
Good job!
Pattielli
May25-04, 12:16 PM
You have any other riddles or quizes ?
a b c d e
..........x4
---------
e d c b a
Rogerio
Jan28-05, 08:09 AM
a b c d e
..........x4
---------
e d c b a
abcde = 21978
:-)
scarecrow
Feb2-05, 12:12 PM
A good problem . . .
Consider what the first person must have said.
If he were a Truth Teller, he'd say, "I'm a Truth Teller."
If he were a Liar, he'd lie and say, "I'm a Truth Teller."
That is, NO ONE would ever say, "I'm a Liar."
Hence, the second person lied.
Therefore, the third person spoke the truth.
True, but You forgot about the speaker: He said "he didn't catch it", which means he's part of the liar group, and he really did hear what the guy said.
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