Who is the liar in this logic puzzle?

[jammieg]

Silly questions

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.

 

[leto]

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.

 

[marqq]

Response

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% (but i doubt that only 2% could solve this problem)

 

[Eyesaw]

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]

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.

 

[Hurkyl]

It's a trick question; the 5th pet is really a salamander!

 

[marqq]

May I ?

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?

 

[Nice coder]

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?

 

[marqq]

Nice coder for the 1st question it can answer you 0.(3) :D i think "pi" is the question who may save us :)

 

[pedestrian]

sorry nvm

 

[Eyesaw]

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?

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 a lot of tylenol.

 

[gnome]

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 a lot 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

 

[marqq]

:D

Bravo !
Well Done !

 

[marqq]

Here we go again

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!

 

[gnome]

A balance or a scale?

 

[gnome]

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?

 

[Eyesaw]

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.

 

[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.

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 6&frac12;. Now, Jane is 7 years younger than Mary, so when Jane was 6&frac12;, Mary was 13&frac12;. Mary is 12 years older than Tim. So, when Mary was 13&frac12;, Tim was 1&frac12;. 13&frac12; is 9 times 1&frac12;.

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.

 

[Eyesaw]

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.

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 6&frac12;. Now, Jane is 7 years younger than Mary, so when Jane was 6&frac12;, Mary was 13&frac12;. Mary is 12 years older than Tim. So, when Mary was 13&frac12;, Tim was 1&frac12;. 13&frac12; is 9 times 1&frac12;.

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.

 

[gnome]

I agree -- if you don't understand fractions, you can't solve it.

 

[Eyesaw]

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.

 

[leto]

Eyesaw, you're making a fool of yourself.

 

[elibol]

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?

 

[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.

 

[Eyesaw]

There were 7 monks, all 7 of them had the mark from day 1.

 

[elibol]

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 doesn't 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 no one 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 no one 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 isn't 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.

 

[leto]

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.

 

[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?

 

[Eyesaw]

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.

 

[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 doesn't matter at this point, even if the last monk becomes possesed no one 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.

 

[Eyesaw]

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 doesn't matter at this point, even if the last monk becomes possesed no one 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.

 

[elibol]

ok, but this isn't 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 doesn't seem possible to me to know the number of monks.

either there is something I am missing, or it is just not possible.

if it is something I am missing please, you have the honor of pointing it out to me =]

 

[Eyesaw]

Originally posted by elibol
ok, but this isn't 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 doesn't seem possible to me to know the number of monks.

either there is something I am missing, or it is just not possible.

if it is something I am 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.

 

[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. 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.

 

[elibol]

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 doesn't 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.