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danger_to_walker said:
So, one that's done the rounds recently:

Albert and Bernard just become friends with Cheryl, and they want to know when her Birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19
June 17 June 18
July 14 July 16
August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month, and the day of her Birthday respectively.

Albert: I don't know when Cheryl's Birthday is, but I know that Bernard does not know either.
Bernard: At first I didn't know when Cheryl's Birthday is, but I know now.
Albert: Then I also know when Cheryl's Birthday is.

So, when is Cheryl's Birthday?
Click to expand...

Nice one, never heard this before

July 16th yeah??? assuming alberts clue tells you it cant be in may or june, then bernard now knows the birthday, so the date can''t have been the 14th. so from alberts perspective the only way bernard would be able to figure it out it from the clue he gave him would be if the date was july 16
 
danger_to_walker said:
So, one that's done the rounds recently:

Albert and Bernard just become friends with Cheryl, and they want to know when her Birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19
June 17 June 18
July 14 July 16
August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month, and the day of her Birthday respectively.

Albert: I don't know when Cheryl's Birthday is, but I know that Bernard does not know either.
Bernard: At first I didn't know when Cheryl's Birthday is, but I know now.
Albert: Then I also know when Cheryl's Birthday is.

So, when is Cheryl's Birthday?
Click to expand...


Supposedly it has an answer. And is in fact published, but it's in-answerable without the answer.

June 18 and May 19 ruled out. Obvious.

Albert been given the month. So either May, June, July and August is correct if you don't know month. He does. He knows Bernard doesn't know because he has 4 days and 4 month options. So who would know? Pretty big odds.

Bernard given date. So either 14, 15, 16, 17 x 2 is correct to him and me because he says he doesn't know. And I don't know.But now he knows after hearing Albert doesn't know. So all he knows is Albert doesn't know which of the 4 months it could be in otherwise he'd know.

So far I know that it's either May 15,16. June 17. July 14 & 16. And August 14, 15 & 17. As does Bernard. Albert knows it's only one of those months. Some say you could rule out June, but why? 17 is still active because it appears twice. Bernard knows this and if given 17, it's either June or August. If Albert given June, and knows Bernard doesn't know, which means then it's not June 17 or 18.

I need to take a breath.
 
Ok, now I know it's not June at all.

So either :

May 15,16
July 14 and 16
August 14 and 15.

August 17 of course also ruled out because June 17 gone.

So Albert doesn't know but knows month. Which month? It's now only May, July or August he's been given.

Bernard given date. So either 14, 15 and 16 which now appear twice each.

So if Bernard given 14, rule out May and leaves July and August.
So if Bernard given 15 he now would rule out July. So leaves May or August.
So if Bernard given 16, rule out August. So May and July left.

But Bernard now knows after hearing Albert doesn't know. What does he know? He knows as much as I do which is I now know nobody actually knows.
 

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Fire Storm said:
Supposedly it has an answer. And is in fact published, but it's in-answerable without the answer.

June 18 and May 19 ruled out. Obvious.

Albert been given the month. So either May, June, July and August is correct if you don't know month. He does. He knows Bernard doesn't know because he has 4 days and 4 month options. So who would know? Pretty big odds.

Bernard given date. So either 14, 15, 16, 17 x 2 is correct to him and me because he says he doesn't know. And I don't know.But now he knows after hearing Albert doesn't know. So all he knows is Albert doesn't know which of the 4 months it could be in otherwise he'd know.

So far I know that it's either May 15,16. June 17. July 14 & 16. And August 14, 15 & 17. As does Bernard. Albert knows it's only one of those months. Some say you could rule out June, but why? 17 is still active because it appears twice. Bernard knows this and if given 17, it's either June or August. If Albert given June, and knows Bernard doesn't know, which means then it's not June 17 or 18.

I need to take a breath.
Click to expand...

It's definitely answerable without the answer, dennis kermit is spot on. The comments of Albert and Bernard are important. The answer and working in spoilers below.

Albert: I don't know when Cheryl's Birthday is, but I know that Bernard doesn't know either.
For Albert to know that Bernard doesn't know, the month he is given must be July/August. If he were given May/June, they each have a unique date, which Bernard could conceivably have been given.

Bernard: At first, I don't know when Cheryl's Birthday is, but now I know.
We are left with July 14 & 16, and August 14, 15 & 17. Bernard now knows that May and June are ruled out, given what Albert has just said. Obviously it cannot be the 14th, as Bernard would still not know the actual Birthday date, so we have 3 possible dates remaining. Bernard was given the date, so he now knows the exact date.

Albert: Then I also know when Cheryl's Birthday is.
For Albert to know what the date is, having been only told the month, it must be July 16, as Bernard has just indirectly confirmed that the date is not 14th. August would still have 2 dates remaining from Albert's point of view, he therefore must have been told July.
 
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Easy to work backwards when you have the answer. :rolleyes:

Because the answer to the 'answer' is found in your reply.

Conceivably. So pure speculation. Awesome logic. :)
 
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Fire Storm said:
Easy to work backwards when you have the answer. :rolleyes:

Because the answer to the 'answer' is found in your reply.

Conceivably. So pure speculation. Awesome logic. :)
Click to expand...

The fact that you can't work it out does not mean everyone else is as dense. The logic is sound, no speculation required. Take your bat and ball and jog on home fella.
 
Albert knows that Bernard doesn't know. How? May and June must be ruled out, because they each have a unique date. If Albert were told one of these months, he could not state with certainty that he knows Bernard doesn't know the full date. Where's the speculation?
 
Because if you take out obvious right at start 18 and 19 you have to then take out 17 because thats obvious that someone would know. So June all gone. And 17 from aug.
so leaves :

M. 15 16
July 14 16
Aug. 14. 15

Albert says Bernard doesn't know. Of course. He's been given a number that at this point appears twice. They both have reached the same conclusion above which is common sense to knock out the obvious.

Bernard has one number that will now appear twice. Albert knows this so why he says bernard doesn't know. As I would. You couldn't know. So he COULD say with certainty that he knows that Bernard doesn't know. Because he'd still have a 50/50 chance. So Albert DOES make the correct statement about Bernard's lack of accuracy.

Bernard now says he didn't know but now does. To me, why? If you have the above left you wouldn't. Impossible. Because why take out all of May? No reason too. None. It COULD be M 15. Could be 16. Why wouldn't it be? If I was Bernard and had been given #16, why would I discount May 16th? I wouldn't. Nobody would.


If Albert WAS given May, that still leaves two same dates for bernard to choose according to Albert. July 16 and August 15. Which is why he said Bernard doesn't know at start. They're both at this same point. Bernard says he now knows. Why would Albert now know too? With that many option above, you wouldn't even if he'd been given May. It's still is either May 15 or 16. Because there hasn't been a reason to knock it out.

Somebody has assumed you'd take out May. If was bernie and given #16, May is still in play. Has to be. Yes it has a unique date. But it also has two other dates that appear again. So from Bernard's standpoint, May is just as much an option as July and August at this point.
 
danger_to_walker said:
So, one that's done the rounds recently:

Albert and Bernard just become friends with Cheryl, and they want to know when her Birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19
June 17 June 18
July 14 July 16
August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month, and the day of her Birthday respectively.

Albert: I don't know when Cheryl's Birthday is, but I know that Bernard does not know either.
Bernard: At first I didn't know when Cheryl's Birthday is, but I know now.
Albert: Then I also know when Cheryl's Birthday is.

So, when is Cheryl's Birthday?
Click to expand...

Maybe I got it wrong - but here is my reasoning:

Albert knows the month - that in itself is not enough info to eliminate any option.
Bernard knows the date - IF it was 18 or 19 (the only single occurrence dates) he would know the birthday. But he doesn't - his options are limited now to 14, 15, 16, 17.
Albert announces that he knows that Bernard does not know. From this, Albert too can eliminate the 18 and 19, leaving:

May 15 May 16
June 17
July 14 July 16
August 14 August 15 August 17

Now we can also eliminate June 17. Albert knew Bernard didn't know, eliminating June 18. But he still says he does not know - so it can't be June as it has only June 17 left as an option - in which case he would know.

This announcement of Albert's changes Bernard's understanding of the situation. Before, all Bernard knew was the numerical date. But now Bernard is privy to the fact that Albert knew that Bernard's initial info (the number) was insufficient. Bernard can then do the same processing as above in his head. He thus concludes that given Albert has not come out with the answer, it can't be June 17, and then (somehow) he says, "Ahh, now I know".

The removal of June 17 presents Bernard with these options (along with the precise knowledge of the numerical part):

May 15 May 16
July 14 July 16
August 14 August 15 August 17

If Bernard was told 14,15 or 16 - he would still not have enough information to be able to conclude "aha - now I know", as each of those numbers has two possibilities. But if he was given 17, he would see that the only possibility would be August 17.
 
Fire Storm said:
Because if you take out obvious right at start 18 and 19 you have to then take out 17 because thats obvious that someone would know. So June all gone. And 17 from aug.
so leaves :

M. 15 16
July 14 16
Aug. 14. 15

Albert says Bernard doesn't know. Of course. He's been given a number that at this point appears twice. They both have reached the same conclusion above which is common sense to knock out the obvious.

Bernard has one number that will now appear twice. Albert knows this so why he says bernard doesn't know. As I would. You couldn't know. So he COULD say with certainty that he knows that Bernard doesn't know. Because he'd still have a 50/50 chance. So Albert DOES make the correct statement about Bernard's lack of accuracy.

Bernard now says he didn't know but now does. To me, why? If you have the above left you wouldn't. Impossible. Because why take out all of May? No reason too. None. It COULD be M 15. Could be 16. Why wouldn't it be? If I was Bernard and had been given #16, why would I discount May 16th? I wouldn't. Nobody would.


If Albert WAS given May, that still leaves two same dates for bernard to choose according to Albert. July 16 and August 15. Which is why he said Bernard doesn't know at start. They're both at this same point. Bernard says he now knows. Why would Albert now know too? With that many option above, you wouldn't even if he'd been given May. It's still is either May 15 or 16. Because there hasn't been a reason to knock it out.

Somebody has assumed you'd take out May. If was bernie and given #16, May is still in play. Has to be. Yes it has a unique date. But it also has two other dates that appear again. So from Bernard's standpoint, May is just as much an option as July and August at this point.
Click to expand...

Not sure I can help you there mate. You're starting from the wrong position, and once you do that, you can't work it out. The only way that Albert can make the first statement at the beginning, with certainty, that he knows Bernard doesn't know the exact date, is if he was given July or August, because neither of these months contain a unique date. Otherwise he cannot claim this with certainty. This removes both May and June.
 
Up until that statement by Albert is made, they would both have this :

M 15 16
Jy 14 16
A 14 15 17

Logic removes the obvious from beginning. Rule number 1. So Unique dates gone.

When Albert then makes the statement, A17 is gone. Half guess before then it's most likely not it. Statement confirms it. So now this :

M 15 16
Jy 14 16
A 14 15


So even if Albert had May, he confidently knows Bernard doesn't know. He can't with 50/50. Because again, going though step by step logically, both would have the above 'picture'. So his statement is correct with the above picture. Which is the only one that could possibly be obtained up to this point. Nothing else.

To take out May is simply twisting the 'sum' to achieve something somebody wants. Because Alberts statement isn't the beginning. And there's the problem with the 'sum'. It's wrong.
 
I've got a course at Uni that deals with all sorts of problems like this! :)

Two personal favourites:

Q1) Find a number consisting of six digits, that, when multiplied by the numbers 2,3,4,5,6 generates an answer consisting of the same digits, but in a different order.

Q2) There are 31 logicians at a conference. To test if all members were logicians, the conference's host devised a test. Each logician had a dot of some colour placed on their forehead. Each logician could see the dots of the other 30 participants, but not their own. There was to be no communication. Every five minutes, the host would ring a bell, at which point, anyone who had deduced the colour of their dot was to leave the room.

One participant asked if it was possible to deduce the colour of their dot. The host confirmed that it was possible.

At the first ring of the bell: 4 people left the room
At the second ring of the bell: Everyone with red dots left the room
At the third ring of the bell: No-one moved
At the fourth ring of the bell: At least one participant left the room
At the fifth ring of the bell: The man who asked the question, his sister, and some other people left the room. The man and his sister had dots of different colour.

Following the fifth ring of the bell, there were still some participants in the room.

If all 31 participants were logicians (and thus were able to solve the problem), how many times was the bell rung?
 
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Hank Scorpion said:
I've got a course at Uni that deals with all sorts of problems like this! :)

Two personal favourites:

Q1) Find a number consisting of six digits, that, when multiplied by the numbers 2,3,4,5,6 generates an answer consisting of the same digits, but in a different order.

Q2) There are 31 logicians at a conference. To test if all members were logicians, the conference's host devised a test. Each logician had a dot of some colour placed on their forehead. Each logician could see the dots of the other 30 participants, but not their own. There was to be no communication. Every five minutes, the host would ring a bell, at which point, anyone who had deduced the colour of their for was to leave the room.

One participant asked if it was possible to deduce the colour of their dot. The host confirmed that it was possible.

At the first ring of the bell: 4 people left the room
At the second ring of the bell: Everyone with red dots left the room
At the third ring of the bell: No-one moved
At the fourth ring of the bell: At least one participant left the room
At the fifth ring of the bell: The man who asked the question, his sister, and some other people left the room. The man and his sister had dots of different colour.

Following the fifth ring of the bell, there were still some participants in the room.

If all 31 participants were logicians (and thus were able to solve the problem), how many times was the bell rung?
Click to expand...
A1
142857

working on A2 now
 
okay so two sets of two colors leave on the first bell.

On the second bell 3 reds leave

On the third bell nobody leaves

On the fourth bell 5 people leave

On the fifth bell 12people leave, 6 of each color

That leaves 7 people so they will all leave on the 6th bell ring?
 
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Both correct

Solution to Q1:

Let the number be x = abcdef, where a, b, c, d, e, f are the digits of the original number.

a = 1, as 6*x would have 7 digits for any other value of a.


If a = 1, there must be 1 in the sixth column for some multiple of x (as the digits are in a different order)

f = 7, as 3*x generates a number a 1 in the sixth column. This cannot be achieved for any other value of f:

Multiplying x by each number, we find the sixth digits in each case
1*x = . . . . . 7
2*x = . . . . . 4
3*x = . . . . . 1
4*x = . . . . . 8
5*x = . . . . . 5
6*x = . . . . . 2

So, the digits are 1,2,4,5,7,8



The sum of these digits its 27. Each column must hold each digit exactly once**.

So the sum of: x + 2x + 3x + 4x + 5x + 6x = 21x = 2999997:

x = 1 . . . . 7
2x = 2 . . . . 4
3x = 4 . . . . 1
4x = 5 . . . . 8
5x = 7 . . . . 5
6x = 8 . . . . 2
___ _________
21x 2999997

x = 2999997/21
x = 142857



**Proof: If two multiples, say y and z, of x had the same digit in a respective column y*x-z*x would have have a zero or nine in the column. But (y*x-z*x) is a multiple of x , so this is impossible.

Hence, each multiple of x has a unique digit in its columns


Solution to Q2:

We know the problem can be solved, so it is known that all logicians shared their colour with another participant. If there were 31 different colours, for example, there would be no way to solve the puzzle.

Each logician has figured this out, hence they know that there must be at least two dots of each colour in the room. If a logician looks around and sees only one other logician with his colour, say blue, he knows that he must have the other blue dot.

So, the colours that have been distributed twice are the first to leave. As four participants left, two pairs of coloured dots left after the first bell. There are 27 participants remaining in the room.

From the point of view of someone with a red dot, he has witnessed the pairs of participants leave. He also observes two red-dot people, who did not leave. Hence, he must have a red dot, or else they'd have left. This is the though process of all (three) red-dotted people, they all leave on the next bell. 24 people remain.

No-one leaves on the next bell. Using the same process, there is clearly no colour that marks 4 participants. 24 participants remain

Next bell, 5 people leave, same thought process. 19 participants remain.

Next bell, any dot colour that marks 6 people leave. As a man and his sister both have different colour dots, at least two groups leave. It must be exactly two groups that leave, or else they'd be a lone remaining participant, which is invalid at this stage. Hence 12 people total leave at this bell, 7 people remain.

7 people leave on the next bell

Answer: 6 bell rings
 
The puzzle of enlightenment.

On your journey to the monastery you come upon a junction with three roads leading out, the left road, the middle road and the right road.

There is a monk standing at the junction, who will truthfully answer any yes-no question, if he can do so without contradicting himself; otherwise he will immediately sit and meditate for eternity. The monk will only answer one question per traveler. What question shall you ask of the monk to find out which road is the monastery road?

Hint: A monk saw a turtle in the monastery garden and asked his mentor, "All beings cover their bones with flesh and skin, so why does this being cover its flesh and skin with bones?". The mentor took off one of his sandals and covered the turtle with it.
 
Moshie The Mook said:
The puzzle of enlightenment.

On your journey to the monastery you come upon a junction with three roads leading out, the left road, the middle road and the right road.

There is a monk standing at the junction, who will truthfully answer any yes-no question, if he can do so without contradicting himself; otherwise he will immediately sit and meditate for eternity. The monk will only answer one question per traveler. What question shall you ask of the monk to find out which road is the monastery road?

Hint: A monk saw a turtle in the monastery garden and asked his mentor, "All beings cover their bones with flesh and skin, so why does this being cover its flesh and skin with bones?". The mentor took off one of his sandals and covered the turtle with it.
Click to expand...

Is the answer to this one "say no to drugs" by any chance?
 
Fire Storm said:
Up until that statement by Albert is made, they would both have this :

M 15 16
Jy 14 16
A 14 15 17

Logic removes the obvious from beginning. Rule number 1. So Unique dates gone.

When Albert then makes the statement, A17 is gone. Half guess before then it's most likely not it. Statement confirms it. So now this :

M 15 16
Jy 14 16
A 14 15


So even if Albert had May, he confidently knows Bernard doesn't know. He can't with 50/50. Because again, going though step by step logically, both would have the above 'picture'. So his statement is correct with the above picture. Which is the only one that could possibly be obtained up to this point. Nothing else.

To take out May is simply twisting the 'sum' to achieve something somebody wants. Because Alberts statement isn't the beginning. And there's the problem with the 'sum'. It's wrong.
Click to expand...

I got the same thing.
 
Fire Storm said:
To take out May is simply twisting the 'sum' to achieve something somebody wants. Because Alberts statement isn't the beginning. And there's the problem with the 'sum'. It's wrong.
Click to expand...
Remember, Albert knows the month. He looks at the list and sees the 'month' doesn't contain any unique number. Therefore, he is certain Bernard cannot know.

If the 'month' had a unique number in it, Albert cannot make his first statement.

From Albert's statement, we know the month does NOT contain a unique number. You eliminate May and June.
 
  1. 26 L of the A
  2. 7 D of the W
  3. 7 W of the W
  4. 12 S of the Z
  5. 66 B of the B
  6. 52 C in a P (W J)
  7. 13 S in the U S F
  8. 18 H on a G C
  9. 39 B of the O T
  10. 5 T on a F
  11. 90 D in a R A
  12. 3 B M (S H T R)
  13. 32 is the T in D F at which W F
  14. 15 P in a R T
  15. 3 W on a T
  16. 100 C in a D
  17. 11 P in a F (S) T
  18. 12 M in a Y
  19. 13 is U F S
  20. 8 T on an O
  21. 29 D in F in a L Y
  22. 27 B in the N T
  23. 365 D in a Y
  24. 13 L in a B D
  25. 52 W in a Y
  26. 9 L of a C
  27. 60 M in an H
  28. 23 P of C in the H B
  29. 64 S on a C B
  30. 9 P in S A
  31. 6 B to an O in C
  32. 1000 Y in a M
  33. 15 M on a D M C
 
  1. 26 L of the A - 26 letters of the alphabet
    [*]7 D of the W - 7 days of the week
    [*]7 W of the W - 7 wonders of the world
    [*]12 S of the Z - 12 signs of the zodiac
    [*]66 B of the B - 66 books of the bible
    [*]52 C in a P (W J) - 52 cards in a pack (without jokers)
    [*]13 S in the U S F - 13 stripes in the United States Flag
    [*]18 H on a G C - 18 holes on a golf course
    [*]39 B of the O T - 39 books of the old testament
    [*]5 T on a F - 5 toes on a foot
    [*]90 D in a R A - 90 degrees in a right angle
    [*]3 B M (S H T R) - no idea
    [*]32 is the T in D F at which W F
    [*]15 P in a R T - 15 players in a rugby union team
    [*]3 W on a T - 3 wheels on a tricycle
    [*]100 C in a D - 100 cents in a dollar
    [*]11 P in a F (S) T - 11 players on a football (soccer team)
    [*]12 M in a Y - 12 months in a year
    [*]13 is U F S - 13 is unlucky for some
    [*]8 T on an O - 8 tentacles on an octopus
    [*]29 D in F in a L Y - 29 days in February in a leap year
    [*]27 B in the N T - 27 books in the new testament
    [*]365 D in a Y 365 days in a year
    [*]13 L in a B D - 13 loaves in a bakers dozen
    [*]52 W in a Y - 52 weeks in a year
    [*]9 L of a C - 9 lives of a cat
    [*]60 M in an H - 60 minutes in an hour
    [*]23 P of C in the H B - no idea
    [*]64 S on a C B - 64 squares on a chess board
    [*]9 P in S A - 9 pints in a saturday arvo :drunk:
    [*]6 B to an O in C 6 balls in an over in cricket
    [*]1000 Y in a M - 1000 years in a millenia
    [*]15 M on a D M C - 15 months on a December - March Calendar (not sure about this one)
 
nerds1.jpg
 
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