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Brainteasers/Math problems - Page 6

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azn_dude1
Profile Joined October 2010
162 Posts
April 09 2011 19:46 GMT
#101
14:
+ Show Spoiler +
2 lines gives 1 intersections
3 lines gives 3 intersections
4 lines gives 6 intersections
For n lines, the number of intersections is equal to the n-1th triangle number. So for 10 lines, you can have 9(10)/2 or 45 intersections.
PepperoniPiZZa
Profile Blog Joined October 2010
Sierra Leone1660 Posts
Last Edited: 2011-04-09 19:53:00
April 09 2011 19:50 GMT
#102
On April 10 2011 04:34 shadowy wrote:
Show nested quote +
On April 10 2011 01:35 Slunk wrote:
On April 10 2011 01:21 Starfox wrote:
On April 10 2011 01:08 Slunk wrote:
Question to #6
Is it correct to assume that the guru is completely useless and only built in as a distraction?

Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people


]But the guru only speaks once and tells something that everybody knows allready.


6. + Show Spoiler +

All of them leave on the first night!

The solution is pretty simple and the guru actually is pretty useless, ah all he does is to confirm, that the solution is possible.

I really don't want to spoil it, so click it, only if you are hundred percent sure you want to know it.

+ Show Spoiler [solution] +

Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.

1. First guy in the column will go to the first guys.
1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right.
1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.

2. Second guy and so forth, save the last one will repeat step 1.

This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.

3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.

By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.

By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.

Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.

You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.



Do i get a virtual present now?


Interaction is not allowed in any way or form. There's a real solution
+ Show Spoiler +

It starts with the assumption that if only 1 person has blue eyes, he will leave on the first night because he doesn't see anybody else with blue eyes. You can work your way up from there on. If there are 2 people with blue eyes, they will both see 1 guy with blue eyes. Assuming that the 1 guy with blue eyes would leave if he wouldn't see any other guy with blue eyes around, it has to be assumed that both of them have blue eyes, making them leave on day 2. Note that in the case of 2 blue eyes people, every brown eyed guy will see 2 blue eyed people and the rest brown eyed while the blue eyed guys will see 1 blue eyed and the rest brown eyed.


I'm sorry, my english isn't as good and I'm a little drunk^_^..
Quote?
Kazius
Profile Blog Joined August 2009
Israel1456 Posts
April 09 2011 19:54 GMT
#103
On April 10 2011 04:44 gyth wrote:
15.+ Show Spoiler +
If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members.
+ Show Spoiler +
Why +2? Wouldn't you only need a 99 member starting group?

+ Show Spoiler +
True, but 100 is of the form 7*n+2, so I proved it for that.
Friendship is like peeing yourself. Anyone can see it, but only you get that warm feeling.
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
April 09 2011 20:02 GMT
#104
On April 10 2011 04:50 PepperoniPiZZa wrote:
Show nested quote +
On April 10 2011 04:34 shadowy wrote:
On April 10 2011 01:35 Slunk wrote:
On April 10 2011 01:21 Starfox wrote:
On April 10 2011 01:08 Slunk wrote:
Question to #6
Is it correct to assume that the guru is completely useless and only built in as a distraction?

Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people


]But the guru only speaks once and tells something that everybody knows allready.


6. + Show Spoiler +

All of them leave on the first night!

The solution is pretty simple and the guru actually is pretty useless, ah all he does is to confirm, that the solution is possible.

I really don't want to spoil it, so click it, only if you are hundred percent sure you want to know it.

+ Show Spoiler [solution] +

Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.

1. First guy in the column will go to the first guys.
1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right.
1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.

2. Second guy and so forth, save the last one will repeat step 1.

This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.

3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.

By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.

By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.

Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.

You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.



Do i get a virtual present now?


Interaction is not allowed in any way or form. There's a real solution
+ Show Spoiler +

It starts with the assumption that if only 1 person has blue eyes, he will leave on the first night because he doesn't see anybody else with blue eyes. You can work your way up from there on. If there are 2 people with blue eyes, they will both see 1 guy with blue eyes. Assuming that the 1 guy with blue eyes would leave if he wouldn't see any other guy with blue eyes around, it has to be assumed that both of them have blue eyes, making them leave on day 2. Note that all the other guy will see only brown-eyed people.


I'm sorry, my english isn't as good and I'm a little drunk^_^..


My version is still OK.

+ Show Spoiler +
Yeah, that works. So all the blue eyes will leave on n+1 night, all the browns (plus the guru) on n+2 night, where n is the number of people with blue eyes.

But I wanted a better solution, so i read it quite carefully. It does NOT say that interaction in way or form is forbidden. It says the communication is not allowed, hence my version is allowed and everybody can figure out the color of their eyes in one night.

Also, the last person's eye's color can be determined by some switching around and does not obligatorily lies on counting numbers.
[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
Last Edited: 2011-04-09 20:16:38
April 09 2011 20:08 GMT
#105
On April 10 2011 01:54 Kazius wrote:
Ohh! I love these.
1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):

12. + Show Spoiler +
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
.


Wrong logic
+ Show Spoiler +

Once of the doors is open, the chances now are 50/50. It doesn't matter if you switch or not - there is no better or worse chance to get it right.

Edit: To explain it better: There are now 2 closed doors, 1 car and 1 goat. It's 1/2 chance to get the car...or a goat.


[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
Murderotica
Profile Blog Joined December 2009
Vatican City State2594 Posts
April 09 2011 20:14 GMT
#106
On April 10 2011 04:34 shadowy wrote:
Show nested quote +
On April 10 2011 01:35 Slunk wrote:
On April 10 2011 01:21 Starfox wrote:
On April 10 2011 01:08 Slunk wrote:
Question to #6
Is it correct to assume that the guru is completely useless and only built in as a distraction?

Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people


]But the guru only speaks once and tells something that everybody knows allready.


6. + Show Spoiler +

All of them leave on the first night!

The solution is pretty simple and the guru actually is pretty useless, ah all he does is to confirm, that the solution is possible.

I really don't want to spoil it, so click it, only if you are hundred percent sure you want to know it.

+ Show Spoiler [solution] +

Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.

1. First guy in the column will go to the first guys.
1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right.
1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.

2. Second guy and so forth, save the last one will repeat step 1.

This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.

3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.

By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.

By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.

Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.

You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.



Do i get a virtual present now?

No, you are incredibly wrong.
ǝsnoɥ ssɐlƃ ɐ uı sǝuoʇs ʍoɹɥʇ ʇ,uop || sıʇɹoɟ ɹǝdɯǝs
cmpcmp
Profile Joined March 2010
84 Posts
April 09 2011 20:16 GMT
#107
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.

As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.

The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister?
Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.


are these being added to the OP? I think that would be a good idea.
If you are not confused, then you are not paying attention.
Talanthalos
Profile Joined September 2010
Germany153 Posts
April 09 2011 20:20 GMT
#108
On April 10 2011 04:04 Manit0u wrote:
New one:

People are seated around the round table in regular intervals (spaced equally to fill entire table). If 9th person is sitting directly across 22nd, how many people are seated at the table?

+ Show Spoiler +
26 people are at the table.

from 9th to 22th there sit 12 people + 12 from 22th to 9th + these 2
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
April 09 2011 20:20 GMT
#109
On April 10 2011 05:14 Murderotica wrote:
Show nested quote +
On April 10 2011 04:34 shadowy wrote:
On April 10 2011 01:35 Slunk wrote:
On April 10 2011 01:21 Starfox wrote:
On April 10 2011 01:08 Slunk wrote:
Question to #6
Is it correct to assume that the guru is completely useless and only built in as a distraction?

Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people


]But the guru only speaks once and tells something that everybody knows allready.


6. + Show Spoiler +

All of them leave on the first night!

The solution is pretty simple and the guru actually is pretty useless, ah all he does is to confirm, that the solution is possible.

I really don't want to spoil it, so click it, only if you are hundred percent sure you want to know it.

+ Show Spoiler [solution] +

Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.

1. First guy in the column will go to the first guys.
1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right.
1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.

2. Second guy and so forth, save the last one will repeat step 1.

This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.

3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.

By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.

By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.

Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.

You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.



Do i get a virtual present now?

No, you are incredibly wrong.


Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.

+ Show Spoiler +

If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
Kazius
Profile Blog Joined August 2009
Israel1456 Posts
April 09 2011 20:23 GMT
#110
[QUOTE]On April 10 2011 05:08 shadowy wrote:
[QUOTE]On April 10 2011 01:54 Kazius wrote:
Ohh! I love these.
1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):

12. + Show Spoiler +
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
.

Wrong logic
+ Show Spoiler +

Once of the doors is open, the chances now are 50/50. It doesn't matter if you switch or not - there is no better or worse chance to get it right.


[/QUOTE]
+ Show Spoiler +
You are wrong. The probability of success after choosing does not magically change to 50%. Are you arguing that it was 50% to begin with?
Friendship is like peeing yourself. Anyone can see it, but only you get that warm feeling.
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
April 09 2011 20:27 GMT
#111
[QUOTE]On April 10 2011 05:23 Kazius wrote:
[QUOTE]On April 10 2011 05:08 shadowy wrote:
[QUOTE]On April 10 2011 01:54 Kazius wrote:
Ohh! I love these.
1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):

12. + Show Spoiler +
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
.

Wrong logic
+ Show Spoiler +

Once of the doors is open, the chances now are 50/50. It doesn't matter if you switch or not - there is no better or worse chance to get it right.


[/QUOTE]
+ Show Spoiler +
You are wrong. The probability of success after choosing does not magically change to 50%. Are you arguing that it was 50% to begin with?
[/QUOTE]

+ Show Spoiler +

In the begining there were 3 doors. So the chance to get the car was 1/3 and a goat 2/3.
Once one of the doors is open a goat is revealed, there is a new cituation at hand.
Although, not by magic, but by the actions of the person opening that door, now you have one-in-two chance to get a car. 1/3 or 2/3 chance can not exist anymore, as there are no more 3 choices. You get one try, from two choices, hence 1/2 chance or 50/50.
[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
Talanthalos
Profile Joined September 2010
Germany153 Posts
April 09 2011 20:30 GMT
#112
[QUOTE]On April 10 2011 05:27 shadowy wrote:
[QUOTE]On April 10 2011 05:23 Kazius wrote:
[QUOTE]On April 10 2011 05:08 shadowy wrote:
[QUOTE]On April 10 2011 01:54 Kazius wrote:
Ohh! I love these.
1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):

12. + Show Spoiler +
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
.

Wrong logic
+ Show Spoiler +

Once of the doors is open, the chances now are 50/50. It doesn't matter if you switch or not - there is no better or worse chance to get it right.


[/QUOTE]
+ Show Spoiler +
You are wrong. The probability of success after choosing does not magically change to 50%. Are you arguing that it was 50% to begin with?
[/QUOTE]

+ Show Spoiler +

In the begining there were 3 doors. So the chance to get the car was 1/3 and a goat 2/3.
Once one of the doors is open a goat is revealed, there is a new cituation at hand.
Although, not by magic, but by the actions of the person opening that door, now you have one-in-two chance to get a car. 1/3 or 2/3 chance can not exist anymore, as there are no more 3 choices. You get one try, from two choices, hence 1/2 chance or 50/50.
[/QUOTE]
+ Show Spoiler +
your are wrong. read the wikipage to that riddle http://en.wikipedia.org/wiki/Monty_Hall_problem
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
April 09 2011 20:31 GMT
#113
On April 10 2011 05:16 cmpcmp wrote:
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.

As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.

The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister?
Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.


are these being added to the OP? I think that would be a good idea.


I give up. I don't see how to do it with only one question to only one of the sisters and not stated, but provided that I can tell the age of the princess by visually looking at them. I will be very glad if post ( or PM) with a solution.
[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
cmpcmp
Profile Joined March 2010
84 Posts
April 09 2011 20:34 GMT
#114
Solution to the Monty Hall / 3 doors problem

The Answer is:
+ Show Spoiler +
there is a 2/3 chance of getting the car if you switch (if you don't believe me read the explanation in the next spoiler, or try to figure it out before you do that if you like)


The Explanation is:
+ Show Spoiler +
Step 1: Put most simply: your odds of having chosen a wrong door on the first guess is clearly 2/3 because there are three doors and you have no useful information about any of them.
Step 2: When the host opens another door for you, assuming you chose wrong on the first guess, if you switch you will end up with the car because it will be the other door.

Essentially, because you know that your odds of picking wrong on the first guess were 2/3 you have information that helps you with your second guess.
If you are not confused, then you are not paying attention.
Talanthalos
Profile Joined September 2010
Germany153 Posts
April 09 2011 20:38 GMT
#115
On April 10 2011 05:31 shadowy wrote:
Show nested quote +
On April 10 2011 05:16 cmpcmp wrote:
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.

As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.

The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister?
Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.


are these being added to the OP? I think that would be a good idea.


I give up. I don't see how to do it with only one question to only one of the sisters and not stated, but provided that I can tell the age of the princess by visually looking at them. I will be very glad if post ( or PM) with a solution.

i dont see any possible solution here, because the middle sister could in a very unlucky coincidence completly behave like one of her other sisters. so its logically impossible to seperate them
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
April 09 2011 20:42 GMT
#116
[QUOTE]On April 10 2011 05:30 Talanthalos wrote:
[QUOTE]On April 10 2011 05:27 shadowy wrote:
[QUOTE]On April 10 2011 05:23 Kazius wrote:
[QUOTE]On April 10 2011 05:08 shadowy wrote:
[QUOTE]On April 10 2011 01:54 Kazius wrote:
Ohh! I love these.
1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):

12. + Show Spoiler +
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
.

Wrong logic
+ Show Spoiler +

Once of the doors is open, the chances now are 50/50. It doesn't matter if you switch or not - there is no better or worse chance to get it right.


[/QUOTE]
+ Show Spoiler +
You are wrong. The probability of success after choosing does not magically change to 50%. Are you arguing that it was 50% to begin with?
[/QUOTE]

+ Show Spoiler +

In the begining there were 3 doors. So the chance to get the car was 1/3 and a goat 2/3.
Once one of the doors is open a goat is revealed, there is a new cituation at hand.
Although, not by magic, but by the actions of the person opening that door, now you have one-in-two chance to get a car. 1/3 or 2/3 chance can not exist anymore, as there are no more 3 choices. You get one try, from two choices, hence 1/2 chance or 50/50.
[/QUOTE]
+ Show Spoiler +
your are wrong. read the wikipage to that riddle http://en.wikipedia.org/wiki/Monty_Hall_problem
[/QUOTE]

I stand corrected. But I have to be honest - that is one hell of a riddle.
I had to read most of the wiki twice to begin to understand why it's so.

For this one I will feel no shame, seeing how many people, professors included do not get the correct answer.
[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
Joe12
Profile Joined March 2010
183 Posts
April 09 2011 20:44 GMT
#117
A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.

During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.

How many drowned?


Its a while since i heard this one, but im pretty sure the wording is correct..

cmpcmp
Profile Joined March 2010
84 Posts
Last Edited: 2011-04-09 20:52:38
April 09 2011 20:46 GMT
#118
Solution to the 3 princesses / most eligible bachelor puzzle

Yes, there is a logical, reasonable, and not stupid solution if that's what ur thinking.

Solution
+ Show Spoiler +
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.


Explanation
+ Show Spoiler +
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest

1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter
2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters
3. She will answer that the youngest daughter is the youngest, and you will pick her.

If you are not confused, then you are not paying attention.
ixi.genocide
Profile Blog Joined June 2010
United States981 Posts
Last Edited: 2011-04-09 20:56:49
April 09 2011 20:49 GMT
#119
+ Show Spoiler +
On April 10 2011 05:16 cmpcmp wrote:
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.

As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.

The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister?
Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.


are these being added to the OP? I think that would be a good idea.



Because of the random elements the middle sister has it is not possible to find out in 1 question which sister is not her (that's all that really matters).

Are you the oldest sister? yes, ?, yes

Are you the youngest sister? No, ?, No

Would you marry the middle sister? No, ?, Yes

Because their is no reasoning behind the middle sisters lies/honesty it is impossible.



+ Show Spoiler +
On April 10 2011 05:46 cmpcmp wrote:
Solution to the 3 princesses / most eligible bachelor puzzle

Yes, there is a logical, reasonable, and not stupid solution if that's what ur thinking.

Solution
+ Show Spoiler +
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.


Explanation
+ Show Spoiler +
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest

1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter
2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters
3. She will answer that the youngest daughter is the youngest, and you will pick her.



Ah ok.... I was under the assumption that you are asking all 3 daughters at once and thus this line of questioning was not possible. Woops
shadowy
Profile Blog Joined August 2010
Bulgaria305 Posts
Last Edited: 2011-04-09 20:51:39
April 09 2011 20:50 GMT
#120
On April 10 2011 05:46 cmpcmp wrote:
Solution to the 3 princesses / most eligible bachelor puzzle

Yes, there is a logical, reasonable, and not stupid solution if that's what ur thinking.

Solution
+ Show Spoiler +
You ask any of the daughters "which of your 2 sisters is the youngest?" Based off of this information, you pick the daughter that is indicated to be younger.


Explanation
+ Show Spoiler +
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest

1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter
2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters
3. She will answer that the youngest daughter is the youngest, and you will pick her.



I am very sorry, but this solution does not corresponds to the question being only YES or NO.

+ Show Spoiler +

You are actually asking one of the sisters to pick, which is quite different from what you tasked us to solve.

Edit: typo
[Fear the leather Gracket!] // ¯\_(ツ)_/¯ // Liquid'Hero hwaiting!
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