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        riddles >> medium >> Deduce Their Numbers
        
(Message started by: ThudanBlunder on Dec 28th, 2009, 12:35pm)
Title: Deduce Their Numbers
Post by ThudanBlunder on Dec 28th, 2009, 12:35pm
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?
Title: Re: Deduce Their Numbers
Post by Aryabhatta on Dec 28th, 2009, 1:33pm
[hide]Didn't think it through completely, but seems like 50/3 and 100/3 is one possibility [/hide]
Title: Re: Deduce Their Numbers
Post by R on Dec 28th, 2009, 11:17pm

on 12/28/09 at 13:33:21, Aryabhatta wrote:
[hide]but seems like 50/3 and 100/3 is one possibility [/hide]

It is not, if we are talking about the positive numbers (0 excluded).


on 12/28/09 at 12:35:17, ThudanBlunder wrote:
positive number is printed.

You mean positive integers? or positive real numbers?
Title: Re: Deduce Their Numbers
Post by ThudanBlunder on Dec 29th, 2009, 5:30am

on 12/28/09 at 23:17:19, R wrote:
You mean positive integers? or positive real numbers?

Positive reals.
Title: Re: Deduce Their Numbers
Post by R on Dec 29th, 2009, 6:01am
Aryabhatta is right!! Isn't he? :D
Title: Re: Deduce Their Numbers
Post by SMQ on Dec 29th, 2009, 6:26am

on 12/29/09 at 05:30:42, ThudanBlunder wrote:
Positive reals.

You sure of that? ;) [hide]I find four solutions in positive rationals, with a unique solution in positive integers.[/hide]

--SMQ
Title: Re: Deduce Their Numbers
Post by Hippo on Dec 30th, 2009, 12:17am
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 xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gify. Possibilities are A1=x+y or A2=x-y.
His answer means xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gify.
B: can see x,y where xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gify. Possibilities are B1=x+y or B2=x-y.
His answer means xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gify and from A: x-yhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gify. (xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif2y)
C: can see x,y where xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gify. Possibilities are C1=x+y or C2=x-y.
His answer means xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gify and from A: x-yhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gify (xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif2y) and from B: if y was on C's head x-yhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif2y (xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif3y) and 2(x-y)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gify (xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif3/2y).
A: can see x,y where xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gify. 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 ... {50http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif2y,y} is in {w,2w}, {w,3w}, or {w,3/2w} form.
Hmm, I should probably take paper and pencil to finish it :)
Title: Re: Deduce Their Numbers
Post by ThudanBlunder on Dec 30th, 2009, 4:58am

on 12/29/09 at 06:26:11, SMQ wrote:
You sure of that? ;) [hide]I find four solutions in positive rationals, with a unique solution in positive integers.[/hide]

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. :-/
Title: Re: Deduce Their Numbers
Post by SMQ on Dec 30th, 2009, 5:05am
Alright, here's what I have:

[hideb]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.
[/hideb]
--SMQ
Title: Re: Deduce Their Numbers
Post by Hippo on Dec 30th, 2009, 5:43am
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.
[/edit]
Title: Re: Deduce Their Numbers
Post by cgspam on Jan 28th, 2010, 9:52am
How do you know from A's answer that (2x,x,x) are forbidden?
Title: Re: Deduce Their Numbers
Post by SMQ on Jan 28th, 2010, 9:57am

on 01/28/10 at 09:52:48, cgspam wrote:
How do you know from A's answer that (2x,x,x) are forbidden?

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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