Author |
Topic: Numbered Hats (Read 2890 times) |
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
Numbered Hats
« on: Jun 19th, 2008, 5:49pm » |
 Quote  Modify
|
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?
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
wonderful
Full Member
Posts: 203
|
|
Re: Numbered Hats
« Reply #1 on: Jun 20th, 2008, 1:49am » |
Quote Modify
|
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. |
|
| « Last Edit: Jun 20th, 2008, 1:52am by wonderful » |
IP Logged |
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: Numbered Hats
« Reply #2 on: Jun 20th, 2008, 3:12pm » |
Quote Modify
|
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. |
|
When A sees x and y (where x > y) he knows his number is x + y or x - y, not 10 or 50
|
|
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
rmsgrey
Uberpuzzler
    
Gender:
Posts: 2873
|
|
Re: Numbered Hats
« Reply #3 on: Jun 21st, 2008, 6:47am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
tohuvabohu
Junior Member
Gender:
Posts: 102
|
|
Re: Numbered Hats
« Reply #4 on: Jun 21st, 2008, 8:40am » |
Quote Modify
|
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.
|
| « Last Edit: Jun 21st, 2008, 2:06pm by tohuvabohu » |
IP Logged |
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: Numbered Hats
« Reply #5 on: Jun 22nd, 2008, 6:48am » |
Quote Modify
|
on Jun 21st, 2008, 8:40am, 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.
|
|
IP Logged |
|
|
|
Hippo
Uberpuzzler
Gender:
Posts: 919
|
|
Re: Numbered Hats
« Reply #6 on: Jun 24th, 2008, 4:55am » |
Quote Modify
|
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. |
|
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print