88 lines
3.7 KiB
Plaintext
88 lines
3.7 KiB
Plaintext
Article 596 of sci.physics:
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Path: puukko!santra!tut!enea!mcvax!uunet!husc6!uwvax!oddjob!ncar!gatech!mcnc!decvax!decwrl!pyramid!ctnews!andrew!TS0014%OHSTVMA.BITNET@CUNYVM.CUNY.EDU
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From: TS0014%OHSTVMA.BITNET@CUNYVM.CUNY.EDU
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Newsgroups: sci.physics
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Subject: Re: Mathematical Puzzle]
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Message-ID: <903@sri-arpa.ARPA>
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Date: 21 Mar 88 18:28:19 GMT
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Lines: 21
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From: Joe Damico <TS0014%OHSTVMA.BITNET@CUNYVM.CUNY.EDU>
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Assuming the integers must be "different", it follows that:
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sum 3 4 5 6 7 8 9
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---------------------------------------------------------------------
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possible 1,2 1,3 1,4 1,5 1,6 1,7 1,8
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pairs 2,3 2,4 2,5 2,6 2,7
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3,4 3,5 3,6
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4,5
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If S doesn't know, then sum>4. If S knows P doesn't know, then sum>6.
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(IF sum=5 then numbers could be 1 and 4, and so P could know the numbers)
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(IF sum=6 then numbers could be 1 and 5, again, P could know the numbers)
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SO the numbers could be 1 and 6.
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P knows the product is 6, but doesn't know whether the factors are (2,3)
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or (1,6)
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By saying "I know that P doesn't know", S informs P that the sum is not 5.
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P says "Now, I know"
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But, by similar argument, the numbers could be 1 and 8.
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Correct me if I'm wrong, but I don't think the problem has a unique solution
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->Joe Damico
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Article 599 of sci.physics:
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Path: puukko!santra!tut!enea!mcvax!uunet!husc6!mailrus!nrl-cmf!ames!ucsd!nosc!cod!stewart
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From: stewart@cod.NOSC.MIL (Stephen E. Stewart)
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Newsgroups: sci.physics
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Subject: Re: Mathematical Puzzle]
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Message-ID: <1039@cod.NOSC.MIL>
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Date: 22 Mar 88 23:53:07 GMT
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References: <898@sri-arpa.ARPA> <5818@watdragon.waterloo.edu>
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Reply-To: stewart@cod.nosc.mil.UUCP (Stephen E. Stewart)
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Organization: Naval Ocean Systems Center, San Diego
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Lines: 41
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In article <5818@watdragon.waterloo.edu> bpdickson@trillium.waterloo.edu (Brian P. Dickson) writes:
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>In article <898@sri-arpa.ARPA> Richard Pavelle <RP%OZ.AI.MIT.EDU@XX.LCS.MIT.EDU>
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>writes:
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>>
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>> P: I don't know what the numbers are.
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>> S: I knew you didn't. Neither do I.
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>> P: Oh! Now I know.
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>> S: Oh! So do I.
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>>
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>>What are the two integers?
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>
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>1 and 4
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>
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>1=>product not prime or 1
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>2a=>sum odd
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>2b=>sum > 3
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>3=>product is product of 2 primes since only two ways of getting product
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>4=>sum < 7 since only 2 ways of getting sum
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>
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I submit that there are exactly 5 possible solutions. The inferences
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which Brian made from 3 and 4 above are not completely rigorous. More
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than two ways of getting the product are allowed as long as all but one
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are eliminated by the requirement that the sum be odd. Any product of
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two or more primes will be odd unless one (or more) of them is 2.
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Thus, unless a 2 is involved, the sum of 1 plus the product and the sum
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of any two numbers derived by taking subproducts will always be even
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and 2a would not be satisfied. So, at least one of the prime factors
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must be a 2. In this case, the sum of 1 plus the product will be odd.
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But, unless all of the prime factors are twos, at least one pair of
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numbers derived by taking subproducts of the prime factors will also
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give an odd sum and P could not determine the numbers from 2a. Only
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when all of the prime factors of the product are twos will all possible
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pairs of numbers resulting in the product have an even sum except one,
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namely 1 and the product itself. Thus, 2a gives P the answer. From
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the knowledge that P then knows the two numbers, S will be able to
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deduce from the foregoing arguments that the only possibilities are:
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1&4, 1&8, 1&16, 1&32, and 1&64. Knowing the sum, he can determine
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which one it is and announce that he knows also.
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Steve Stewart
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