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Topic: Deduce Their Numbers (Read 998 times) |
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ThudnBlunder
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Deduce Their Numbers
« on: Dec 28th, 2009, 12:35pm » |
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Each of three people is wearing a hat on which a positive number is printed. 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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Aryabhatta
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Re: Deduce Their Numbers
« Reply #1 on: Dec 28th, 2009, 1:33pm » |
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Didn't think it through completely, but seems like 50/3 and 100/3 is one possibility
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R
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Re: Deduce Their Numbers
« Reply #2 on: Dec 28th, 2009, 11:17pm » |
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on Dec 28th, 2009, 1:33pm, Aryabhatta wrote:| but seems like 50/3 and 100/3 is one possibility |
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It is not, if we are talking about the positive numbers (0 excluded).
on Dec 28th, 2009, 12:35pm, ThudanBlunder wrote:| positive number is printed. |
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You mean positive integers? or positive real numbers?
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| « Last Edit: Dec 29th, 2009, 5:58am by R » |
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ThudnBlunder
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Re: Deduce Their Numbers
« Reply #3 on: Dec 29th, 2009, 5:30am » |
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on Dec 28th, 2009, 11:17pm, R wrote:
You mean positive integers? or positive real numbers?
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Positive reals.
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R
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Re: Deduce Their Numbers
« Reply #4 on: Dec 29th, 2009, 6:01am » |
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Aryabhatta is right!! Isn't he? 
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SMQ
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Re: Deduce Their Numbers
« Reply #5 on: Dec 29th, 2009, 6:26am » |
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on Dec 29th, 2009, 5:30am, ThudanBlunder wrote:
You sure of that?  I find four solutions in positive rationals, with a unique solution in positive integers.
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Hippo
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Re: Deduce Their Numbers
« Reply #6 on: Dec 30th, 2009, 12:17am » |
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I have read only the first post ... is there some limit for used numbers?
Say positive integers?
... Please reply editting the first post 
Ohh, ignore it ... there were written positive, integer is implicit.
OK
A: can see x,y where x y. Possibilities are A1=x+y or A2=x-y.
His answer means x y.
B: can see x,y where xy. Possibilities are B1=x+y or B2=x-y.
His answer means xy and from A: x-yy. (x2y)
C: can see x,y where xy. Possibilities are C1=x+y or C2=x-y.
His answer means xy and from A: x-yy (x2y) and from B: if y was on C's head x-y2y (x3y) and 2(x-y)y (x3/2y).
A: can see x,y where xy. Possibilities are A1=x+y or A2=x-y. One of the pairs {Ai,x} or {Ai,y} is of form {w,2w}, {w,3w} or {w,3/2w}.
The other A3-i is equal 50.
We already know {50,50}, {50,25}, {50,100} were excluded both by B and C. {50,75} and {50,150} is also sometimes excluded.
Suppose it's {Ai,y} pair ... {50 2y,y} is in {w,2w}, {w,3w}, or {w,3/2w} form.
Hmm, I should probably take paper and pencil to finish it
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| « Last Edit: Dec 30th, 2009, 3:29am by Hippo » |
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ThudnBlunder
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Re: Deduce Their Numbers
« Reply #7 on: Dec 30th, 2009, 4:58am » |
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on Dec 29th, 2009, 6:26am, SMQ wrote:
You sure of that? I find four solutions in positive rationals, with a unique solution in positive integers.
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I haven't cracked this one yet, and so cannot give you a definite answer. But the question does say 'positive number'.
However, in the next (similar) puzzle it stipulates 'natural numbers'. So perhaps that is what is meant. 
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| « Last Edit: Dec 30th, 2009, 5:07am by ThudnBlunder » |
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SMQ
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Re: Deduce Their Numbers
« Reply #8 on: Dec 30th, 2009, 5:05am » |
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Alright, here's what I have:
| hidden: | Assume you're person A. You see numbers b and c. You know your number a must be either b + c or |b - c|. The only way to deduce which is if one of those two possibilities is ruled out. Initially, the only result which can be ruled out is zero.
a != 0
A can deduce iff |b - c| = 0
A cannot deduce --> b != c
b != 0 AND b != c
B can deduce iff |a - c| = 0 OR a - c = c
B cannot deduce --> c != a AND c != a/2
c != 0 AND c != b AND c != a AND c != a/2
C can deduce iff |a - b| = 0 OR a - b = b OR b - a = a OR |a - b| = a/2
C cannot deduce --> a != b AND a != 2b AND 2a != b AND 3a != 2b
a != c AND a != 2c AND a != b AND a != 2b AND a != b/2 AND a != 2b/3
A can deduce iff b - c = c OR b - c = 2c OR c - b = b OR c - b = 2b OR |b - c| = b/2 OR |b - c| = 2b/3
c - b = b --> 2b = c, a = 3b; a = 50 --> b = 50/3, c = 100/3
c - b = 2b --> 3b = c, a = 4b; a = 50 --> b = 25/2, c = 75/2
c - b = b/2 --> 3b = 2c, a = 5b/2; a = 50 --> b = 20, c = 30
c - b = 2b/3 --> 5b = 3c, a = 8b/3; a = 50 --> b = 75/4, c = 125/4
(b - c = c --> b = 2c, a = 3c; a = 50 --> b = 100/3, c = 50/3)
(b - c = 2c --> b = 3c, a = 4c; a = 50 --> b = 75/2, c = 25/2)
(b - c = b/2 --> b = 2c, a = 3c; a = 50 --> b = 100/3, c = 50/3)
(b - c = 2b/3 --> b = 3c, a = 4c; a = 50 --> b = 75/2, c = 25/2)
So the possible numbers on the other hats are {20, 30}; {25/2, 75/2}; {50/3, 100/3}; or {75/4, 125/4} with {20, 30} being the only integral solution.
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Hippo
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Re: Deduce Their Numbers
« Reply #9 on: Dec 30th, 2009, 5:43am » |
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OK, here I am:
Each person thinks he has sum or abs of difference of seen numbers.
If both signs are allowed and we are to guess only absolute value the riddle would be same.
The only forbidden triples are of form (0,x,z),(x,0,y), and (x,y,0).
Answer A: (?,x,y) gives (x+y,x,y), (|x-y|,x,y) possibilities. Only (0,x,x) may be forbidden therefore (2x,x,x) is the only other forbidden one.
Answer B: similarly (x,2x,x) is forbidden, morever (2x,3x,x) is.
Answer C: similarly (x,x,2x), (x,2x,3x), and (2x,x,3x) morever (2x,3x,5x).
Answer A: Special values which can be determined are except (2x,x,x) and (0,x,y) following:
(3x,2x,x), (4x,3x,x), (3x,x,2x), (5x,2x,3x), (4x,x,3x), (8x,3x,5x) so we can compute all 6 possibilities.
The only (50,20,30) is intergral.
[edit]
I wondered why SMQ has 2 more solutions, but the last two were repetitions.
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| « Last Edit: Dec 30th, 2009, 11:50pm by Hippo » |
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cgspam
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Re: Deduce Their Numbers
« Reply #10 on: Jan 28th, 2010, 9:52am » |
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How do you know from A's answer that (2x,x,x) are forbidden?
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SMQ
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Re: Deduce Their Numbers
« Reply #11 on: Jan 28th, 2010, 9:57am » |
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on Jan 28th, 2010, 9:52am, cgspam wrote:| How do you know from A's answer that (2x,x,x) are forbidden? |
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Because if A saw x and x he would know he had the sum (2x), since if either of B or C had the sum he would have to have 0 (so that 0 + x = x), and zero is not a positive number.
--SMQ
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