Mon Oct 13, 2008 11:00 am by dgarrett@astindale.co.uk 


HELP !!
Can anybody help me with this  it's destroyed my weekend and I seem to have conclusively proved it's not solvable.
Can't eat, can't sleep, can't get on with my life!
Is anybody willing to HELP ME?! PLEASE say yes......
David 




Mon Dec 22, 2008 9:50 pm by alexonfyre 


You could start by posting the riddle...though this seems a bit late probably. 




Tue Dec 23, 2008 10:54 am by dgarrett@astindale.co.uk 


This was the Brain Teaser:
A pack of 9 cards is numbered 1 to 9. One card is placed on the forehead of each of four logicians. Each of them can see the other three numbers, but not their own. In turn, each states whether or not he knows his own number. If not, he announces the sum of two of the numbers he can see.
One game went as follows. Alf: “No, 14”. Bert: “Yes”. Chris: “No, 7. Dave: “No” but before Dave could continue, Alf had worked out his own number.
What was Dave’s number?
And this was my proof that it can't be done:
Definitions:
There are two (and only two) possible “pairs” that make up the 14 that A can see:
68 and 59
These will be called the pairs, and 5,6,8,9 will be referred to as pair numbers. All the other numbers (i.e: 1,2,3,4,7) will be called “anonymous numbers”.
A preliminary general observation:
It is not possible that all four pair cards (5,6,8,9) are on the four foreheads, for there would clearly be no way that C could nominate a 2card total of 7.
Logic:
A can see a 14pair somewhere on B, C & D.
B states that he knows his own number.
So B must have half of a pair – if a whole pair was on C & D then B couldn’t possibly state that he knew his own number unless he was clairvoyant, it could be anything.
So the other half of the pair will be on C or D, and, importantly, the person who doesn’t have the other half must have an anonymous number. For if they had one of the other pair then B couldn’t make his definitive statement. (Unless A conveniently had a third pair card, in which case B would know that he had to have the missing fourth one, but this is precluded – see general observation above).
Just to recap by way of example – say B can see C 8 and D 3. He knows his own number, it must be a 6, because it’s the only way to make the14 that A has called. But if what he saw was C 8 and D 9, he wouldn’t have known his own number – it could be a 6 or a 5.
So we have established: B has one of a pair; C or D has the other half of that pair, and the other has an anonymous number.
So where is the other half of B’s pair? – is it on C or D?
Lets say it’s on C. So C is looking at half a pair on B and an anonymous number on D, and would therefore immediately know his own number, which makes a 14pair with B. So that’s not it, because C was unable to nominate his own number.
So is it on D? No, for the same reason. D is looking at half a pair on B and an anonymous number on C, and would therefore know his own number. But he passed, so that’s not it.
So the whole thing is impossible 




Tue Aug 11, 2009 1:10 pm by mdbbarilla 


they only state the sum of the two numbers they can see. So A can announce the sum of C & D rather than B&C or B&D
A sees 14 ( 6/8 or 5/9)
C sees 7 ( 6/1 or 5/2 or 4/3)
Let say they saw 6.
A can see a 14pair somewhere on B, C & D.
B states that he knows his own number.
So B must have half of a pair B has 6. B = 6
So the only pair that could have a sum of 7 is 6 and 1
If C had 1, C can't say that he sees a 7. C =/= 1
If D had the 1, A wouldn't be able to know his number.
Because A doesn't know anything about his number. D =/= 1
If D didn't have the 1, A would now he has the number 1!
Let say they saw 5.
A can see a 14pair somewhere on B, C & D.
B states that he knows his own number.
So B must have half of a pair. B has 5.
So the only pair that has a sum of 7 is 5 and 2.
If C had 2, C can't say he sees a 7
If D had 2, A won't know his number.
But if D didn't get the 2, A would know his number is 2.
So Dave's number would 3,4 or 7. Doesn't really matter if you use this approach. 




Fri Jun 03, 2011 7:22 pm by Unni 


What about placing the card (6 or 9) upside down? 




Sat Jun 06, 2015 6:20 pm by Klasus 


The thing you forgot is that Dave was interrupted by Alf.
The whole procedure would be:
A sees 14, so 14 = B + C / B + D / C + D
14 is either 6+8 or 5+9 we can first assume it's 6+8
B knows his number so he must either be 6 or 8
C sees 7, so 7 = A + B / A + D / B + D we then know that C =/= 6 and C =/= 1 therefore B = 6 and 1 = A / D
D is about to answer (he knows his number) but is interrupted by A who knows his number.
A knows that he must be part of the combination 7 that C saw and therefore he must be 1 because he sees that B = 6.
Finally the only possibility in this case is that D = 8. In the other case when 14 = 5+9 then D = 9.
The only player that does not know his number is C because he is not part of any combination.
A poor illustration:
14 YES
A B
1 6
? 8
C D
7 




Mon Jul 06, 2015 12:57 pm by dk 


Forget whether 14 is 6+8 or 5+9. It seems to me Dave mustnt even say "No" before being interrupted by Alf. Dave knows his number. There's a flaw in the question.
Let's start:
1st step: A says "No, 14"
Thus, all of them know that only one of the following is true. More than one cannot be true since there's only one of each number. If, for example, B+C=14 AND C+D=14, then B and D turn out to have the same value, which cannot happen.
B+C=14
B+D=14
C+D=14
2nd: B says Yes
As its suggested C+D isnt 14. So everyone knows that only either one of the two are true
B+C=14
B+D=14
3rd: C says "No, 7"
Now C knows what B+D is. If it wasnt equal to 14, then he would deduce that B+C was 14 and thereby would have answered Yes.
Therefore,
B+D=14
4th: Problematic step: Dave says "No"
But Dave must know his number since he sees B's number and knows that B+D=14. 




Fri Oct 16, 2015 4:53 pm by king29 


There seems to be a flaw in the question. Either C or D should have known thier number. Here is my explanation:
First let's go in the order the people had answered their questions..
First A: No, and the sum he gives is 14. This means A sees a combination of 8+6 or 9+5 (because 7, 4 and below numbers are not possible for this sum combination). At this point we are not sure what is on whose head but we can say that he sees two of these 4 numbers.
Second B: Yes. If B sees the same combination as A sees (for sum 14) on C&D, then B will not be able to know his number. But B says he knows his number, that means his number belongs to the combination (sum 14). Now let's look at the combinations:
First combination (8+6) : If B sees 8 on C or D then his number would be 6 (or) If B sees 6 on C or D then his number would be 8.
Second combination (9+5) : If B sees 9 on C or D then his number would be 5 (or) If B sees 5 on C or D then his number would be 9.
Third C: No, and the sum he gives is 7. Now let's sit in C's chair and think :)
If A gives a sum 14 and B is part of the sum then either C or D should be part of the combination.
If D is part of the combination then C would not know his number.
If D is not part of the combination then C would be other part of the combination of sum 14 (i.e., C = 14  B)
Here C says, he doesn't know his number that makes D part of the combination of sum 14.
Before going to the flaw, let's continue with A's thoughts by ignoring D's answer:
C says he doesn't know the number that makes D part of the combination of sum 14. In this case D = 14  B.
C gives a sum of 7 that means A and B are the combination for 7 because B and D are already combined for 14. So, A = 7  B.
So, even before D answers Yes/No, A should be able to figure out his number. Now let us stop here and look at the flaw:
C says he doesn't know his number that makes D part of the combination of sum 14. In this case D should have known his number.
D says he doesn't know his number that makes C part of the combination of sum 14. In this case C should have known his number (14  B) and C's answer was wrong.
:roll: :roll: :roll:
Cheers!! 




Tue Jul 12, 2016 7:36 am by Dzallen 


The problem is flawed unless you ignore Dave's 'no' at the end, probably a typo. 






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