Times brainteaser from Sun 12-Oct-2008

[beermagnet]

Can't see this one at all:

A pack of nine cards is numbered l 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?

[weecalf]

was it number nine as the only way to get 14 is5&9 ect ect

[Dilbert53]

I can't get to the end, to much work, but I can take it someway.

A must be able to see either 9 and 5 or 8 and 6.
Because B knows what he has either C or D must have only one of 9,5,8 or 6 and B can conclude that he has the other pairing.
Because C can't deduce what he has then he must be able to see the other card making 14 so therefore D must have the other pairing to B.
C can see either 6 and 1, 5 and 2 or 4 and 3.

I will revisist later but this may put someone on the right track to finish.

[beermagnet]

You're thinking the same way as me D63.
I had thought through as follows:
14 can only be made by 6+8 or 5+9
This 14 must be from one of B+C, C+D or B+D

For B to be able to deduce his number he must be able to see a set of numbers A C D where one of C or D cannot be involved in the sum making 14 i.e. 1,2,3,4,or 7, So B can
see that the other, which must be one of 5,6,8,9 was involved in the 14 declared by A and his number B is the other half of the sum making 14.
I can't see any other way that B can deduce his number.

The trouble with this is that it means the other bloke making up the 14, C or D, can now use the same logic to deduce his (C or D) number. For example if D has the 1,2,3,4, or 7 and C has the 5,6,8, or 9, then C can see B's number so can deduce his C number from that.

So how come the game proceeds with "C No" and "D No" ?

There must be some other way B can deduce his number.

These brainteasers simply state the answer 2 weeks later without any hint of how they are derived which in cases like this can be irritating.

[Gizmonster]

I'm also at a loss on this one .......... the puzzle seems flawed in my opinion .......

After A announces that he doesn't know his number and he states the number 14, we now know that either:
B+C=14,
or B+D=14,
or C+D=14

When B announces that he knows his number, then C+D=14 can be ruled out, 'cos if this was the case, he wouldn't be able to deduce his own number.

So we now know that either:
B+C=14,
or B+D=14

When it comes to C and he doesn't know his number, we now know that B+D=14, otherwise he'd be able to deduce his number using the same logic as B.

This is when it breaks down ....... when it comes to D, he should now be able to work his number out, using the same logic as B, but he answers, "no" ...... eh ???

Are you sure you wrote the puzzle down correctly ???

Basically, if B+C=14, then when it comes to C's guess, he'll see that B+D is not 14 and he'll be able to work out his own number.

Or if B+D=14, then when it comes to D's guess, he'll see that B+C is not 14 and he'll be able to work out his own number.

Either way, C or D will be able to work out their own number .....

[beermagnet]

> Are you sure you wrote the puzzle down correctly ???
Yes.

> puzzle seems flawed in my opinion
That's what I'm thinking:
I am wondering if the line that says
> If not, he announces the sum of two of the numbers he can see.
should be
> If not, he announces the sum of two or three of the numbers he can see.
or even
> If not, he announces the sum of one, two or three of the numbers he can see.

If B can work out that A is using the B+C+D = 14 method he could immediately knows his number whereas C and D might be stymied on that score.
This could happen if C+D<14 and neither are 5, 6, 8 or 9 so can't be involved in the two-number 14 sums
e.g. BCD is 941 932 842 743

Take BCD=941
B knows he is 9 as 14 can only be from the B+C+D route but C sees BD 91 but doesn't know A's 14=B+C+D it might've been B+C only!
Plus we could infer from C's "no" that A is not the x of a B+x=14
That would remove B from that two-number possibility.
Same for D

So B must be 9 or 8 or 7
C and D must be any 2 of 1, 2, 3 or 4

What about the 7 that C says
From 3 numbers only 124
From 2 numbers 61 52 43

We know that B can't be involved as it is too big so this must be a two-number sum involving D+A
And having said that we see that after C says 7 D can always work out his number - so we're stuck again ....

So finally what if players can announce the sum two or three of the numbers or just one of the numbers he can see.

Then D is not sure as D=7 is now a possibility

When D says "No" A can see D is not 7 so can know his A number.
Trouble is now, I can't see how *we* can know D's number I now have these possibilities that satisfy this last scheme:
ABCD
6941
3914
4923
5842
3824

06:43 Thu 23rd Oct 2008

[beermagnet]

Anyone following this saga might be interested in today's published answer. It was:


There is no possible answer (with apologies)


!!!!
Makes you wonder what went wrong there.