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Topic: Numbered Hats (Read 2875 times) |
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ThudnBlunder
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Numbered Hats
« on: Jun 19th, 2008, 5:49pm » |
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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?
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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wonderful
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Re: Numbered Hats
« Reply #1 on: Jun 20th, 2008, 1:49am » |
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This is a very nice puzzle.
| hidden: | | 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. |
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| « Last Edit: Jun 20th, 2008, 1:52am by wonderful » |
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ThudnBlunder
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Re: Numbered Hats
« Reply #2 on: Jun 20th, 2008, 3:12pm » |
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on Jun 20th, 2008, 1:49am, 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. |
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When A sees x and y (where x > y) he knows his number is x + y or x - y, not 10 or 50
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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rmsgrey
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Re: Numbered Hats
« Reply #3 on: Jun 21st, 2008, 6:47am » |
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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.
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tohuvabohu
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Re: Numbered Hats
« Reply #4 on: Jun 21st, 2008, 8:40am » |
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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.
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| « Last Edit: Jun 21st, 2008, 2:06pm by tohuvabohu » |
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rmsgrey
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Re: Numbered Hats
« Reply #5 on: Jun 22nd, 2008, 6:48am » |
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on Jun 21st, 2008, 8:40am, tohuvabohu wrote:On A's second turn:
Since:
[...]
C<>2A, if B=1.5C then A=2.5C
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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.
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Hippo
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Re: Numbered Hats
« Reply #6 on: Jun 24th, 2008, 4:55am » |
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on Jun 22nd, 2008, 6:48am, 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. |
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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.
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