Article 596 of sci.physics:

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:

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3,4 3,5 3,6

4,5

(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.

or (1,6)

By saying "I know that P doesn't know", S informs P that the sum is not 5.

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:

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 <898@sri-arpa.ARPA> Richard Pavelle

>writes:

>>

>> P: I don't know what the numbers are.

>> S: I knew you didn't. Neither do I.

>> P: Oh! Now I know.

>> S: Oh! So do I.

>>

>>What are the two integers?

>

>1 and 4

>

>1=>product not prime or 1

>2a=>sum odd

>2b=>sum > 3

>3=>product is product of 2 primes since only two ways of getting product

>4=>sum < 7 since only 2 ways of getting sum

>

than two ways of getting the product are allowed as long as all but one

are eliminated by the requirement that the sum be odd. Any product of

two or more primes will be odd unless one (or more) of them is 2.

Thus, unless a 2 is involved, the sum of 1 plus the product and the sum

of any two numbers derived by taking subproducts will always be even

and 2a would not be satisfied. So, at least one of the prime factors

must be a 2. In this case, the sum of 1 plus the product will be odd.

But, unless all of the prime factors are twos, at least one pair of

numbers derived by taking subproducts of the prime factors will also

namely 1 and the product itself. Thus, 2a gives P the answer. From

the knowledge that P then knows the two numbers, S will be able to

Steve Stewart