Article 596 of sci.physics:
Path: puukko!santra!tut!enea!mcvax!uunet!husc6!uwvax!oddjob!ncar!gatech!mcnc!decvax!decwrl!pyramid!ctnews!andrew!TS0014%OHSTVMA.BITNET@CUNYVM.CUNY.EDU
From: TS0014%OHSTVMA.BITNET@CUNYVM.CUNY.EDU
Newsgroups: sci.physics
Subject: Re: Mathematical Puzzle]
Message-ID: <903@sri-arpa.ARPA>
Date: 21 Mar 88 18:28:19 GMT
Lines: 21
From: Joe Damico
Assuming the integers must be "different", it follows that:
sum 3 4 5 6 7 8 9
---------------------------------------------------------------------
possible 1,2 1,3 1,4 1,5 1,6 1,7 1,8
pairs 2,3 2,4 2,5 2,6 2,7
3,4 3,5 3,6
4,5
If S doesn't know, then sum>4. If S knows P doesn't know, then sum>6.
(IF sum=5 then numbers could be 1 and 4, and so P could know the numbers)
(IF sum=6 then numbers could be 1 and 5, again, P could know the numbers)
SO the numbers could be 1 and 6.
P knows the product is 6, but doesn't know whether the factors are (2,3)
or (1,6)
By saying "I know that P doesn't know", S informs P that the sum is not 5.
P says "Now, I know"
But, by similar argument, the numbers could be 1 and 8.
Correct me if I'm wrong, but I don't think the problem has a unique solution
->Joe Damico
Article 599 of sci.physics:
Path: puukko!santra!tut!enea!mcvax!uunet!husc6!mailrus!nrl-cmf!ames!ucsd!nosc!cod!stewart
From: stewart@cod.NOSC.MIL (Stephen E. Stewart)
Newsgroups: sci.physics
Subject: Re: Mathematical Puzzle]
Message-ID: <1039@cod.NOSC.MIL>
Date: 22 Mar 88 23:53:07 GMT
References: <898@sri-arpa.ARPA> <5818@watdragon.waterloo.edu>
Reply-To: stewart@cod.nosc.mil.UUCP (Stephen E. Stewart)
Organization: Naval Ocean Systems Center, San Diego
Lines: 41
In article <5818@watdragon.waterloo.edu> bpdickson@trillium.waterloo.edu (Brian P. Dickson) writes:
>In article <898@sri-arpa.ARPA> Richard Pavelle