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From: Pucc-H:Pucc-I:Pucc-K:ags@CS-Mordred.UUCP
Newsgroups: net.puzzle
Subject: Re: S and P (REPEAT AGAIN)
Message-ID: <116@pucc-k>
Date: Thu, 17-Nov-83 08:51:22 EST
Article-I.D.: pucc-k.116
Posted: Thu Nov 17 08:51:22 1983
Date-Received: Sat, 19-Nov-83 04:01:08 EST
References: <557@alberta.UUCP>
Organization: Purdue University Computing Center
Lines: 62

The S and P puzzle as stated has no solution.  In fact, the terms are 
contradictory.  I will point out the contradiction and state a revised
version of the puzzle which DOES have a solution.  I will post an answer
to the revised puzzle in a few days if no one gets it.

Here are the conditions on S and P as stated in the original puzzle:


     The judge of the contest chooses two integers tells the mathematicians
         "Both integers are greater than one"
         "The sum of the two integers is less than forty"
         "The product of the two integers is less than four hundred"

First of all, the third condition adds no new information.  If the sum
of the two integers is less than 40 and both integers are greater than one,
we can already deduce that the product is not greater than 20*19 = 380.

Here is my revised version of the judge's instructions:

	"Both integers are greater than one"
	"Neither is greater than one hundred"

As before, the judge gives the sum to S and the product to P and asks them
to deduce both sum and product.

(1) S announces that P does not know the sum.
(2) P says he now knows the sum.
(3) S says he now knows the product.

There is enough information for the reader to determine a unique answer to the 
revised problem.  What is more surprising is that the answer to the revised 
puzzle also satisfies the conditions of the original (namely, the sum is less 
than forty).  From this it does not follow that the same solution works for the
original problem.  Here is why:

(From this point on I assume the original judge's instructions were given)

(1) When S announces that P can not determine the sum, we may deduce that
    the sum is either 11 or 17.  The products corresponding to 11 are
    18, 24, 28 and 30.  The products corresponding to 17 are 30, 42, 52,
    60, 66, 70 and 72.  Each of these products can be factored in more than
    one way, yielding different sums.  For all other sums between 4 and 39,
    there is at least one product that can be factored uniquely.

(2) P has likewise determined that the sum is either 11 or 17.  He has a 
    product that can correspond to only one of these, and therefore deduces 
    the sum.  From this we can determine that the product is not 30, since 
    30 can be factored as 6 * 5 or as 15 * 2, yielding either 11 or 17 for
    the sum.  We are left with the following possibilities:

    sum = 11, product in {18, 24, 28}
    sum = 17, product in {42, 52, 60, 66, 70, 72}

(3) S now claims he knows the product.  He must have miscalculated.  Whether
    his sum is 11 or 17, he cannot know the product.

If you have managed to read this far, your challenge now is to determine
why this reasoning no longer applies if the judge's original instructions
are changed as I have indicated.

				Dave Seaman
				..!pur-ee!pucc-k!ags