Bogarts Riddle Thread (goto 1st post for details)

[Bogart]

Originally Posted by Phoenix

I suck with numbers

no comment

Originally Posted by Auron

Is there a correct way of knowing? Because it seems impossible to solve. There is no correct way of knowing how many he/she recieved?

Is there?!

As stated in the first post and several times throughot the thread, all of my riddles are solveable by pure logic. no guessing is needed. so there is of course a way of knowing!

[Demi]

I think he got 4

[Bogart]

Originally Posted by Demi

I think he got 4

Unfortunatelly, a simple guess is never enough for winning. You would have to base your result on some chain of logical statements

[Demi]

gosh darn.

[Auron]

Before I say anything, I want to apologise if this makes no sense what so ever.

Let's assign numbers from 1 to 10 for all the people. Bob will be number 1 and his partner 2. The rest we'll assign 3 & 4, 5 & 6, 7 & 8, 9 & 10.

Now, assume that the conference is being held at pair 3 & 4's company, Bob's partner wouldn't need a business card from them as he/she can, most likely, collect one from a desk or something. He would receive one from number 5, another from number 6. However he can say to number 6 that he already has a business card from his partner, so he doesn't need another. Now he has 1 business card. He can do this with 7 & 8 and 9 & 10.
He has 3 business cards.

With that said, the conference could be held at his, and Bob's, company. He would do everything from the last one. That collect one business card from a pair, but since the conference is being held at his company, he now requires a card from pair 3 & 4.
He has 4 business cards.

How far off am I?

[vintage]

Originally Posted by Bogart

.~°") NEW RIDDLE ("°~.
(_/*"~ . _ . ~"*\_)



Busy Businessmen

At an anti-stress conference meeting 10 businessmen from 5 different companies are teamed up pairwise (2 businessmen/company). During the cheerful meeting, some shake hands and exchange business cards, others don't. Ofcourse businessmen from the same company dont exchange business cards.

At the wrap-up dinner, one businessman named Bob asks all the others about how many times they exchanged business cards. He gets an answer from all the other businessmen, but no two answers are the same.

How many business cards did Bob's partner collect during the conference meeting?

Note: Every businessman exchanges business cards at max once per businessman, and businessmen from the same company dont exchange business cards.
.~°") GOOD LUCK ("°~.
(_/*"~. _ .~"*\_)


!!! R E A D__T H E__R U L E S__I N__T H E__F I R S T__P O S T__O F__T H I S__T H R E A D !!!

Originally Posted by Auron

Before I say anything, I want to apologise if this makes no sense what so ever.

Let's assign numbers from 1 to 10 for all the people. Bob will be number 1 and his partner 2. The rest we'll assign 3 & 4, 5 & 6, 7 & 8, 9 & 10.

Now, assume that the conference is being held at pair 3 & 4's company, Bob's partner wouldn't need a business card from them as he/she can, most likely, collect one from a desk or something. He would receive one from number 5, another from number 6. However he can say to number 6 that he already has a business card from his partner, so he doesn't need another. Now he has 1 business card. He can do this with 7 & 8 and 9 & 10.
He has 3 business cards.

With that said, the conference could be held at his, and Bob's, company. He would do everything from the last one. That collect one business card from a pair, but since the conference is being held at his company, he now requires a card from pair 3 & 4.
He has 4 business cards.

How far off am I?

Even though the person before me got the correct answer of 4, I don't follow his reasoning at all, so I'm just gonna post my answer and you can decide who wins.

So, as shatterstar said, no business man can have more than 8 cards and no fewer than 0. And since everyone has a different number of cards, someone must have 0, 1, ... , 8.

Let's call the Bob A and his partner B, then the other business men are C&D, E&F, G&H, and I&J.

Let's note that Bob's Partner can not have 8 cards. If he had 8 cards then no one else (besides Bob) could have 0 cards. He can also not have 0 cards because then no one else (besides Bob) could have 8.

Let's also note that whoever has 8 cards, must have a partner with 0 cards, because everyone else now has 1 card and there must be someone with 0 cards.

So, let's start with partners I&J (although it doesn't matter who we pick) one must have 8 cards and the other 0 cards. Everyone else now has 1 card.

We move on to partners G&H (again any partners would do accept for Bob and his patner who are special) one must have 7 and the other 1. The same logic applies here, once one has 7 everyone else has 2 so the patner must have 1.

We continue with the other partners 6-2, 5-3, and finally we reach Bob and his partner who both have 4. However, this is OK because Bob asked the question and did not have to answer it, so he is allowed to have the same number of cards as someone else.

As far as I can tell it doesn't matter who hosts the conference.

[Bogart]

Originally Posted by Auron

Before I say anything, I want to apologise if this makes no sense what so ever.

Let's assign numbers from 1 to 10 for all the people. Bob will be number 1 and his partner 2. The rest we'll assign 3 & 4, 5 & 6, 7 & 8, 9 & 10.

Now, assume that the conference is being held at pair 3 & 4's company, Bob's partner wouldn't need a business card from them as he/she can, most likely, collect one from a desk or something. He would receive one from number 5, another from number 6. However he can say to number 6 that he already has a business card from his partner, so he doesn't need another. Now he has 1 business card. He can do this with 7 & 8 and 9 & 10.
He has 3 business cards.

With that said, the conference could be held at his, and Bob's, company. He would do everything from the last one. That collect one business card from a pair, but since the conference is being held at his company, he now requires a card from pair 3 & 4.
He has 4 business cards.

How far off am I?

uhm... Noth pole <-> South pole? Yeah that quite tells how far But thanks for the try!

Originally Posted by vintage

Even though the person before me got the correct answer of 4, I don't follow his reasoning at all, so I'm just gonna post my answer and you can decide who wins.

So, as shatterstar said, no business man can have more than 8 cards and no fewer than 0. And since everyone has a different number of cards, someone must have 0, 1, ... , 8.

Let's call the Bob A and his partner B, then the other business men are C&D, E&F, G&H, and I&J.

Let's note that Bob's Partner can not have 8 cards. If he had 8 cards then no one else (besides Bob) could have 0 cards. He can also not have 0 cards because then no one else (besides Bob) could have 8.

Let's also note that whoever has 8 cards, must have a partner with 0 cards, because everyone else now has 1 card and there must be someone with 0 cards.

So, let's start with partners I&J (although it doesn't matter who we pick) one must have 8 cards and the other 0 cards. Everyone else now has 1 card.

We move on to partners G&H (again any partners would do accept for Bob and his patner who are special) one must have 7 and the other 1. The same logic applies here, once one has 7 everyone else has 2 so the patner must have 1.

We continue with the other partners 6-2, 5-3, and finally we reach Bob and his partner who both have 4. However, this is OK because Bob asked the question and did not have to answer it, so he is allowed to have the same number of cards as someone else.

As far as I can tell it doesn't matter who hosts the conference.

Congratulations, this is the exact correct result! Here is your prize:

Be proud to find yourself in the winners list of the first post!

Next riddle will follow at monday supposedly. Thanks to all who participated so far!

[vintage]

Originally Posted by Bogart

.~°") NEW RIDDLE ("°~.
(_/*"~ . _ . ~"*\_)



Freaky foreheads


Two math-freaks (X and Y) with unfailable logic sit face to face, each staring at the others forehead. A blind hatter has written a number on each freaks forehead and told them to find out what number is on their own forehead. The blind hatter explains the quest:
"You may only look at the other persons forehead, and you are not allowed to look at you own forehead in any way. You task is to find out which number is written on your own forehead. I wrote integers that are greater than zero. The only thing you are allowed to do is to ask the other person the following question:
'Do you know what your number is?'
That person may only answer the qustion with either 'yes', or 'no'! One thing I can tell you is, that the sum of you numbers is either 24 or 27. Now as you both have understood the rules, I am curious how many times you must ask each other before you will know..."

The blind hatter wrote "12" on both foreheads. The two freaks stare at each other for a short duration, then one starts to ask:

Q/X: "Do you know what your number is?" - A/Y: "no!"
Q/Y: "Do you know what your number is?" - A/X: "no!"
Q/X: "Do you know what your number is?" - A/Y: "no!"
...
...


How many times did the freaks ask the question?
how many 'no!'s were necessary till one answered with 'yes!' and got the correct number?
Is it at all possible to solve this riddle?


.~°") GOOD LUCK ("°~.
(_/*"~. _ .~"*\_)


!!! R E A D__T H E__R U L E S__I N__T H E__F I R S T__P O S T__O F__T H I S__T H R E A D !!!

** WARNING ** tl;dr


Well, this is clearly an extension of the "hat problems" but definitely a lot trickier. I feel fairly confident that I'm going to give the correct analysis and yet the wrong solution because I miss some simplification trick, but here goes anyway:

X sees a 12, and therefore knows his number must be 12 or 15.
Y sees a 12, and therefore knows his number must be 12 or 15.

So, let's say X begins; he asks Y if he knows his number. Y says NO, because his number could be either 12 or 15.

Y now asks X the same question. X says NO, because his number could be either 12 or 15.

Now, X knows that his number could be 12 or 15 and ALSO knows that Y couldn't figure out his own number. Since Y could only have possibly seen a 12 or a 15, he knows that Y must conclude that Y's number was either 12 (12 + 12 = 24), 15 ( 12 + 15 = 15 + 12 = 27), or 9 (15 + 9 = 24).

Of course, he must have known that Y couldn't have have known his number on the first pass, so perhaps there is a simplification here, but since I don't see it, I'm going to ignore this fact for the rest of my answer.

Note that Y knows the same information as X, so I won't retype it from the opposite perspective.

This information does not proclude either 12 or 15 as the possible number on the freak's forehead, so on the next round of questioning both freaks will again answer NO.

Now the information has changed again. They know their number must be 12 or 15, they know that the other freak must see one of these two numbers, and they know the other freak has not been able to figure out their own number even after hearing them answer NO twice.

So, if they saw a 12 their number could have been 12 or 15 and if they saw a 15 their number could have been 12 or 9. Assuming they saw a 15 and thought their number was a 9 (This assumption is important because when it leads to an impossibility we have solved the problem), this would have to mean that the other freak saw a 9 and answered NO twice. A 9 would lead the other freak to think their own number was either 15 (9 + 15 = 24) or 18 (9 + 18 = 27). So again, both are possible and preclude neither 12 nor 15. So, both freaks must answer NO for a third time.

This logic continues on each pass, always focusing on the "new" number 18 leads to 6 or 9 on the 4th pass ; 6 -> 21 or 18 on the 5th pass ; 21 -> 3 or 6 on the 6th pass ; 3 -> 24 or 21 on the 7th pass. And finally we have important information.

24 is NOT possible, since that would mean the freaks number would be 0 which isn't a possibility since all numbers were non-zero integers. So, that means the assumption was FALSE and the freak can preclude 15 as the possible number on his forehead and he can now know with certainty that his number is 12. You may be wondering, why this proves the assumption false, since 21 IS a possible number. It's true that there is still one number that is valid, but if only one possible number was valid then the previous freak would have answered YES instead of NO. Since he answered NO, both numbers need to be valid.

So... given that logic, let me answer your questions:
I believe each freak asked the question 7 times (ignoring simplifications).
That would mean there were 14 NO answers before freak X answered yes.
And finally, of course the problem is solvable. In fact, I think it's been proven that it can be extended to x number of people and x number of possible "sums".

Again, I think my solution has a good chance of being wrong, but I know that's the logical process of solving this. So, if someone else uses my post to answer the question I want to marked down for the assist.

[Formation]

The strange screen

Quote:
Originally Posted by Bogart

You are in a room with three doors. You dont know why, you dont know wherefrom. Infront of you is a screen displaying some text and a keyboard below. You read the text:

"One of these doors will lead you out, two
doors will lead you to your certain death!
After pressing 'any key' you may enter three
questions, one after the other, which are to
be answered with only 'yes' or 'no' and I will
answer every of them. However, I will only
tell the truth ONCE." followed by a blinking cursor.

What questions will you ask?
Important: You can ask questions depending on previous answers you got. You cannot ask nested questions like "Is it true that door 1 leads to the exit and the answer to this question is a lie?" with basically two question combined with a conjunction. Now get your brains going, after all you do want to be home for dinner.

EDIT: Note that in order to ask questions depending on future answers, you will need an if-clause, which is allowed. In any case you cannot ask if any answer is/was/will be a lie (or the truth ofcourse).

Reserved.

[Demi]

You are in a room with three doors. You dont know why, you dont know wherefrom. Infront of you is a screen displaying some text and a keyboard below. You read the text:

"One of these doors will lead you out, two
doors will lead you to your certain death!
After pressing 'any key' you may enter three
questions, one after the other, which are to
be answered with only 'yes' or 'no' and I will
answer every of them. However, I will only
tell the truth ONCE." followed by a blinking cursor.

What questions will you ask?
Important: You can ask questions depending on previous answers you got. You cannot ask nested questions like "Is it true that door 1 leads to the exit and the answer to this question is a lie?" with basically two question combined with a conjunction. Now get your brains going, after all you do want to be home for dinner.

EDIT: Note that in order to ask questions depending on future answers, you will need an if-clause, which is allowed. In any case you cannot ask if any answer is/was/will be a lie (or the truth ofcourse).

______________

I've spent ALL DAY working on this, Bogart. Instead of doing my homework. xD I really hope it makes sense.

Question #1
"Is the sky blue?"
Should be a question that you definitely know the answer to. Something like, Is the sky blue? That way, if the computer answers "NO", you know that is one false answer, and if it answers "YES", it is one true anser.


If the computer answers "YES" to Question #1...
Questions #2 and 3
"Is door #1/2/3 safe to pass through?"
Since you have already used the one "TRUE" answer, you know that the next two answers will be false. You can really ask about any door you want. Say you ask if door #1 is safe to pas through. If the computer says "NO", you know it is lying, and that door is safe to pass through. If it says "YES", you know it is lying, and the door is NOT safe to pass through. If you get two "YES"'s, you automatically know by process of elimination that the door you did NOT ask about is the safe door.


If the computer answers "NO" to Question #1...
Question #2
"If I go through either of the leftmost doors, will I live?"
So the answer to question #1 was false. That means, you have one false answer and one true answer left. So the question you ask next is "If I go through either of the leftmost doors, will I live?" The computer can either answer "NO" or "YES" to this. You won't know if it's true until the next question you ask.

If the computer answers "YES" to Question #2...
Question #3
"If I go through the middle door, will I live?"
SO. The computer has told you that either of the two leftmost doors will lead you out safely. And now you ask the computer if the middle door, one of the two leftmost doors, will lead you to safety.

Lets say the computer says "YES" to question 3. Okay. So now you know that one of those two yes's is a lie. Let's assume it's the first one. This means that the two leftmost doors will lead to death. Now, that means that question #3 MUST be true. And that means that the middle door will lead you to safety. BEEEEEEP. Question #2 being false has told you that both doors on the left side will lead to death. So, this is impossible. That means that question #2 will be true (the two leftmost doors lead to safety) and the middle door will NOT lead to safety, meaning the door on the left will lead you to safety.

Lets say the computer answers "NO" to question 3. So you know that one of the answers is truth, and one of the answers is a lie. Lets assume that the first statement is a lie. That means that either of the two doors on the left will lead to death. This leaves you with the answer "NO" to question 3 being the truth (you will NOT live if you go through the middle door.) Since you have eliminated one of the left-side doors, you can safetly say that the other is the one to go through.

If the computer answers "NO" to Question #2...
Question #3
"If I go through the middle door, will I live?"
This can basically be looked at the same way as if the computer had said "YES" for question #2.

Lets say that the computer is telling the truth for question #2. This means that you will NOT live if you go through either of the leftmost doors. SO, you can assume that it is on the right. But, this would also mean that the computer would have to answer question #3 with a false answer. If the computer said "NO" to question 3, which would be a false answer, that would mean you would life if you went through the middle door, which we already determined leads to death. WWWWROOONG.

So lets say that the computer is LYING for question #2. This means that you will live through one of the doors on the left. Now, the computer must tell the truth for the last question. It says "NO." And it is telling the truth. And since you already determined that you would live if you went though one of the doors on the left, the door farthest to the left must be the right one.


Wow.
I think I got that totally wrong. :c