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        riddles >> medium >> Numbered Hats
        
(Message started by: ThudanBlunder on Jun 19^th, 2008, 5:49pm)

Title: Numbered Hats
Post by ThudanBlunder on Jun 19^th, 2008, 5:49pm

A, B, and C are each wearing a hat on which is printed a positive number. Each can see the numbers on the others' hats but not their own number. All are told that one of the numbers is the sum of the other two. The following statements are made in the hearing of all:

A: I cannot deduce what my number is.

B: I cannot deduce what my number is.

C: I cannot deduce what my number is.

A: I can deduce that my number is 50.

What are the numbers on the other two hats?

Title: Re: Numbered Hats
Post by wonderful on Jun 20^th, 2008, 1:49am

This is a very nice puzzle.

[hideb]I will show that the other two are B=20 and C=30. Seeing B and C, A derives that A=10 or A = 50. If A = 10, then C knew C=10 or 30; C then derived that C # 10 otherwise B would announce B= 20 given A=C=10; thus C would announce C= 30. So, A can know A =50 given B=20, C=30.[/hideb]

Title: Re: Numbered Hats
Post by ThudanBlunder on Jun 20^th, 2008, 3:12pm

on 06/20/08 at 01:49:59, wonderful wrote:

I will show that the other two are B=20 and C=30. Seeing B and C, A derives that A=10 or A = 50.

When A sees x and y (where x > y) he knows his number is x + y or x - y, not 10 or 50

Title: Re: Numbered Hats
Post by rmsgrey on Jun 21^st, 2008, 6:47am

Wonderful has verified a possible answer - that is to say, he's shown that, for his values of B and C, events would play out as described in the problem statement.

What remains to be shown is that those values of B and C are (or are not) unique.

Title: Re: Numbered Hats
Post by tohuvabohu on Jun 21^st, 2008, 8:40am

On his first turn A could answer if B=C. THen A=2B

Bcould answer if A=C. THen B=2A.
Since B<>C, if A=2C. THen B=3C.

C could answer if A=B. THen C=2A.
Since:
B<>C, if A=2B. THen C=3B.
A<>C, if B=2A. Then C=3A.
A<>2C, if B=1.5A. Then C=2.5A

On A's second turn:
Since:
A<>B, if B=2C, A=3C
A<>C, if C=2B then A=3B
A<>2B, if C=3B then A=4B
A<>2C, if B=3C then A=4C
C<>2A, if B=1.5C then A=2.5C
B<>2A, if C=1.5B then A=2.5B

Only the last two result in integers when A=50.
C=20, B=30 and B=20, C=30 both seem like valid answers to me. Which is unique if you are not required to say who is wearing which hat.

Title: Re: Numbered Hats
Post by rmsgrey on Jun 22^nd, 2008, 6:48am

on 06/21/08 at 08:40:45, tohuvabohu wrote:

On A's second turn:
Since:
[...]
C<>2A, if B=1.5C then A=2.5C

I don't see where this line comes from - on B's turn, he doesn't yet know that B<>A so wouldn't be able to deduce his number from C=2A...

Removing that case leaves a unique answer.

Title: Re: Numbered Hats
Post by Hippo on Jun 24^th, 2008, 4:55am

on 06/22/08 at 06:48:34, rmsgrey wrote:

I don't see where this line comes from - on B's turn, he doesn't yet know that B<>A so wouldn't be able to deduce his number from C=2A...

Removing that case leaves a unique answer.

Yes, there should be C=5/3B case instead leading to A=8/3B=8/5C.
BTW: .... there was not stated the numbers are whole so all 6 solutions are OK for me.