Puzzles,conundrums, riddles and thoughts?

[Sidhe]

We know that Truthers always tell the truth but Liars will always speak falsely.

An island comprises entirely of Liars and Truthers..

Precisely 45 people attended a dinner in the island. Each one of them sat at a big round table.

After the dinner was over, each of the attendees was asked about their neighbours, and each stated that they were seated between one Liar and one Truther.

It transpired later that precisely two of the Truthers were mistaken in their statements.

Determine the number of Liars And Truthers attending the dinner.

Well since this thread has kind of lapsed here's a logic puzzle to be getting on with, there is only one answer I can prove it if anyone gets it.

 

[azzaman333]

Uh, so 2 truthers lied?

 

[Erik Mesoy]

I think the answer is two, and the seating arrangement was ...LLLLTLTLLLL....

From IWC:

A hundred prisoners are each locked in a room with three pirates, one of whom will walk the plank in the morning. Each prisoner has 10 bottles of wine, one of which has been poisoned; and each pirate has 12 coins, one of which is counterfeit and weighs either more or less than a genuine coin. In the room is a single switch, which the prisoner may either leave as it is, or flip. Before being led into the rooms, the prisoners are all made to wear either a red hat or a blue hat; they can see all the other prisoners' hats, but not their own. Meanwhile, a six-digit prime number of monkeys multiply until their digits reverse, then all have to get across a river using a canoe that can hold at most two monkeys at a time. But half the monkeys always lie and the other half always tell the truth. Given that the Nth prisoner knows that one of the monkeys doesn't know that a pirate doesn't know the product of two numbers between 1 and 100 without knowing that the N+1th prisoner has flipped the switch in his room or not after having determined which bottle of wine was poisoned and what colour his hat is, what is the solution to this puzzle?

Possible answers:

Spoiler :

The 64th square would have more rice than the entire kingdom.
The surgeon is his MOTHER.
You should change your choice to the other door.
The seventh philosopher starves to death.
He commited suicide with an icicle.
16 miles per hour.
Only if the missionary is also the nun's uncle.
The first cannibal on the 29th night at midnight.
Ask him what the other farmer would say is the correct road.
He adds his own horse, then it's left over at the end.
He's too short to reach any button above the 10th floor.

 

[Sidhe]

I think the answer is two, and the seating arrangement was ...LLLLTLTLLLL....

From IWC:


Possible answers:

Spoiler :

The 64th square would have more rice than the entire kingdom.
The surgeon is his MOTHER.
You should change your choice to the other door.
The seventh philosopher starves to death.
He commited suicide with an icicle.
16 miles per hour.
Only if the missionary is also the nun's uncle.
The first cannibal on the 29th night at midnight.
Ask him what the other farmer would say is the correct road.
He adds his own horse, then it's left over at the end.
He's too short to reach any button above the 10th floor.

I'm afraid that's wrong. But keep your riddle on hold.

The liars always lie.

Two liars have told the truth.

Uh, so 2 truthers lied?

No two truthers were mistaken.

 

[Erik Mesoy]

...LLLLLLLTLTLLLLLL....
^In the above, there are two truthers; they will be mistaken if they say that they were sitting between a liar and a truther, and the two liars marked in red will be telling the truth if they say that they were between a truther and a liar.

Two liars have told the truth.

[azzaman's post]
No two truthers were mistaken.

What gives?

 

[Catharsis]

I think he means that in your answer, the Ls underlined in red have told the truth, which they cannot do. So, it is a wrong answer.

EDIT: By this I mean that the red Ls are between an L and a T.

EDIT AGAIN: Beat you Sidhe! I r00l!!!!1

 

[Sidhe]

...LLLLLLLTLTLLLLLL....
^In the above, there are two truthers; they will be mistaken if they say that they were sitting between a liar and a truther, and the two liars marked in red will be telling the truth if they say that they were between a truther and a liar.


What gives?

Liars always lie, the two liars have said that they are seated between a truth teller and a liar, this is the truth yes? So they haven't lied.

Think on

xpost with Catharsis.

You rule C consider me beaten

 

[Erik Mesoy]

#204:

Two liars have told the truth.

Is this the case or not?

Liars always lie, the two liars have said that they are seated between a truth teller and a liar, this is the truth yes? So they haven't lied.

I am confused. Here is how I understand the puzzle at the moment:

"Of 45 people, some are Liars and some are Truthers, and each one knows what each of the others are. They were seated in a ring, and each was asked (by a neutral observer?) who they were sitting between. Each replied that they were sitting between a Truther and a Liar. Two of the Truthers were mistaken (they lied), and two of the Liars told the truth. How many of the 45 are Liars, and how many are Truthers?"

 

[Sidhe]

#204:Is this the case or not?



I am confused. Here is how I understand the puzzle at the moment:

"Of 45 people, some are Liars and some are Truthers, and each one knows what each of the others are. They were seated in a ring, and each was asked (by a neutral observer?) who they were sitting between. Each replied that they were sitting between a Truther and a Liar. Two of the Truthers were mistaken (they lied), and two of the Liars told the truth. How many of the 45 are Liars, and how many are Truthers?"

OK I'll explain it simply if I can(this is me we're talking about), two truthers were mistaken, ie they did not lie they genuinely believed they were telling the truth. But the two liars know they have told the truth and they always lie, and thus your answer is wrong.

Honestly there is only one answer and yours isn't it, I s**t you not.

Two of the liars cannot tell the truth.

K. My bad it's not clear, two liars have told the truth in your situation, logically they cannot do this thus the answer is wrong.

 

[Catharsis]

When he said 'Two liars have told the truth', I think he was pointing out your mistake, Erik.

EDIT: DANG! One-all, Sidhe.

 

[Sidhe]

When he said 'Two liars have told the truth', I think he was pointing out your mistake, Erik.

EDIT: DANG! One-all, Sidhe.

In your face!

Er ok. anyway, thanks for helping out. It's not a contest, or is it?

 

[Catharsis]

In your face!

Er ok. anyway, thanks for helping out. It's not a contest, or is it?

Of course not. *finger poised on the refresh button*

 

[Sidhe]

Of course not. *finger poised on the refresh button*

I'm of to Bedfordshire soon, you may well win yet, God damn it!

I'd wait it took me about 15 minutes to figure this one out. I'm sure the maths genius Eric will come back sooner, but this one isn't maths as such it's more logical trial and error.

Big clue Erik by the way if your watching logic triumphs.

 

[Renata]

15 liars and 30 truthers.

There can't ever be two or more liars in a row if there are any truthers present, because you always wind up with ...LLT... at the end of the group of liars, which would mean some of them would have been telling the truth when asked who their neighbors were. And the truthers (minus the mistaken ones) have to have exactly one truther as a neighbor. So you wind up with a pattern of ...LTTLTTLTTLTT..., which would make for 15 liars and 30 truthers. Adding in the requirement that two truthers were mistaken doesn't change the numbers (it only takes the all-liars solution out of play); you just change one of the LTT groups to LTTT and another to LT. One of the mistaken truthers was between two other truthers and the other was between two liars.

Edit: And I love the "had" one; it's one of my favorites.

John, while Jim had had "had", had had "had had"; "had had" had had a better effect on the teacher.

 

[Sidhe]

15 liars and 30 truthers.

There can't ever be two or more liars in a row if there are any truthers present, because you always wind up with ...LLT... at the end of the group of liars, which would mean some of them would have been telling the truth when asked who their neighbors were. And the truthers (minus the mistaken ones) have to have exactly one truther as a neighbor. So you wind up with a pattern of ...LTTLTTLTTLTT..., which would make for 15 liars and 30 truthers. Adding in the requirement that two truthers were mistaken doesn't change the numbers (it only takes the all-liars solution out of play); you just change one of the LTT groups to LTTT and another to LT. One of the mistaken truthers was between two other truthers and the other was between two liars.

Edit: And I love the "had" one; it's one of my favorites.

John, while Jim had had "had", had had "had had"; "had had" had had a better effect on the teacher.

Spot on and here's the proof that it is the only answer for anyone who's interested:

We must state a couple assumptions:
1) The liars all told falsehoods (NONE of the liars were next to both a truth teller and a liar)
2) Exactly two of the truth tellers were mistaken, the rest were correct in stating that they sat next to both a liar and a truth teller.

Now, there's *one* possibility that's really easy, which is:

Solution A) FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

But because we know that there are at least two truth tellers who attended the dinner, this answer is wrong.

We can also easily deduce that no two liars sat next to each other. The possibilities would be as follows:

...TFFT... <-- this would make both middle liars correct, and is disqualified
...TFFF... <-- this would make the leftmost liar correct, disqualified
...FFFT... <-- this would make the rightmost liar correct, disqualified
...FFFF... <-- this is plausible, but leads ONLY to Solution A, which is wrong.

Therefore, no two liars sat next to each other.

From that, we know that there are at MOST 22 liars, which means at LEAST 23 truth tellers. And that means that there are at least 21 truth tellers who were correct in their statements. As a result, we know that there are a minimum of 11 liars, because the 21 truth tellers all correctly reference at least one liar who was known to be at the dinner, but half of the accurate truth tellers may have been referencing the same liars.

So! Between 11 and 22 liars, proven.

We've shown that no liars can be seated next to each other. This means that *EVERY* liar must be seated in between TWO truth tellers. And, the vast majority of those truth tellers must in turn be seated next to *another* truth teller because the majority of truth tellers are correct in their assesments.

The only thing left are the anomolies. The anomolous truth tellers can either be seated as TTT or FTF. The remainder of people MUST be situated in the pattern: ...TTFTTFTTF... The pattern itself is therefore a multiple of 3.

Hence, we are left with possibilities:
insert a TTT and a TTT
insert a TTT and a FTF
insert a FTF and a FTF

But notice that by inserting a TTT into the pattern, we must NECESSARILY be adding *one* person (a truth teller) to a single phase of the pattern. And conversely, by inserting an FTF into the pattern, we must necessarily be REMOVING one person (a truth teller) from the pattern. So, let's reanalyze our possibilities:

insert a TTT and a TTT - wrong, this would add 2, and our total would not be a multiple of 3.

insert a TTT and a FTF - possible

insert a FTF and a FTF - wrong, this would remove 2, and our total would not be a multiple of 3.

As a result, we must take our base pattern of TTFTTFTTF, and insert a TTT and an FTF, in order to keep the total at 45. Where we add them is irrelevant.

If we start with our base pattern, we get that there must be twice as many truth tellers as liars, or 30 truth tellers and 15 liars. Then, we put in the necessary TTT, and we get 31 truth tellers and 15 liars. Finally, we put in the necessary FTF, and we go back to having 30 truth tellers and 15 liars.

Therefore, the solution of 30 truth tellers and 15 liars is unique.

Yeah I've seen the had had had one before so I wouldn't of put up an answer even if I could, too easy. If you want to find another riddle, up to you otherwise anybody feel free to have a go.

 

[Steph]

I have one.

a and b are two integers, comprised in [2, 40]

Mr P and Mr S both know that, but they don't know a or b

Mr P is given the product of the two numbers ( a * b).

Mr S is given the sum of the two numbers ( a + ).

Mr P knows Mr S knows the sum, but Mr P doesn't know the value of the sum

Mr S knows Mr P knows the product, but Mr S doesn't know the value of the product.

After a while, Mr S calls Mr P, and says "knowing the sum, I'm sure you can't find a and b"

A bit later, Mr P called Mr S back and says "I've found a and b"

Eventually, Mr S called Mr P and says "Me to".

What are a and b?

 

[Sidhe]

I have one.

a and b are two integers, comprised in [2, 40]

Mr P and Mr S both know that, but they don't know a or b

Mr P is given the product of the two numbers ( a * b).

Mr S is given the sum of the two numbers ( a + ).

Mr P knows Mr S knows the sum, but Mr P doesn't know the value of the sum

Mr S knows Mr P knows the product, but Mr S doesn't know the value of the product.

After a while, Mr S calls Mr P, and says "knowing the sum, I'm sure you can't find a and b"

A bit later, Mr P called Mr S back and says "I've found a and b"

Eventually, Mr S called Mr P and says "Me to".

What are a and b?

I presume you mean between 2 and 40 inclusive. Just to make sure it's not some vector quantity or something?

If so I get 4 or 2+2 or 2*2

4 has only one way of producing four and regardless of the sum it is known that 4 will only be made by 2*2

a and b are 2

of course 1 is not a number we can use as it's not between 2-40. So this counts out 1*4 or 1+3 the sum is the only solution 2+2 and the product is also 2*2.

Of course I'm assuming that a and b need not be different and are in the range of 2-40 inclusive.

If they have to be different then 2 and 3 have the same deal. With six and five being the product and sum with only 4+1illegal or 3+2 or 2+3 and 3*2 and 2*3 as product answers the only conclusion is a and b must be 3 and 2. It's obvious the answer must be 3 and 2, although it's an either or deal. Any other numbers it's not certain that by knowing sum or product you must know a and b. And prime numbers obviously can't be used being only divisible by themselves and 1. Or products of 1 and themselves

 

[Steph]

I presume you mean between 2 and 40 inclusive. Just to make sure it's not some vector quantity or something?

inclusive, yes.
Your answer is not correct.
If it was 2 and 2, then the sum would be 4, and Mr S would know that as the sum is 4, the two numbers are 2 and 2. So the product is 4.
But then Mr P, would also find a and b easily.
But Mr S said Mr P couldn't find it.

 

[Abaddon]

Is this not a game?

 

[Steph]

no, it's a number , not a game