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Q: arithmetic problem ( Answered 5 out of 5 stars,   2 Comments )
Question  
Subject: arithmetic problem
Category: Science
Asked by: applegal-ga
List Price: $30.00
Posted: 28 May 2004 04:50 PDT
Expires: 27 Jun 2004 04:50 PDT
Question ID: 353106
There are 2 integers greater than 1 and less than 50.They may or may
not be equal.S knows the sum of the two numbers, and P knows their
product. S says"I don't know what the two numbers are,and it's
impossible for you to know.(meaning without further info) P says,"In
that case I know what they are."(and he does) S then says,"In that
case, so do I." (and he does) What are the 2 numbers?
Answer  
Subject: Re: arithmetic problem
Answered By: till-ga on 28 May 2004 05:35 PDT
Rated:5 out of 5 stars
 
The solution is 4 and 13. 
This is a tough one, but I found a reference with a great explanation
for the solution. The upper limit (100 here) does not play a role for
the correct solution:

"Two Mathematicians Problem

Date: 05/18/98 at 04:46:56
From: Yusuf Kursat TUNCEL
Subject: 2 Mathematicians Problem

Dear Dr. Math,

We are given that X and Y are two integers, greater than 1, are not 
equal, and their sum is less than 100. A and B are two talented 
mathematicians; A is given the sum, and B is given the product of 
these numbers:

   B says "I cannot find these numbers."
   A says "I was sure that you could not find them."
   B says "Then, I found these numbers."
   A says "If you could find them, then I also found them."

What are these numbers?

Here is my approach:

The reason that B cannot find the numbers from their products is that 
the product has more than 2 prime factors that are not the same (like 
2 * 3 * 5, which results in products like 10 * 3, 2 * 15, and 5 * 6) 
So, for A to be sure that B could not find the numbers, A must have a 
sum that does not contain a two prime solution pair such as 
2 + 13 = 15. A cannot have this because he wouldn't be sure if the 
product is given 2 * 13 = 26 or not.

I have written all of the prime numbers from 2 to 100, and added them
with each other, I have eliminated the the results from the set of 
numbers 2 - 100 (also 3, 4, 5, 6, 7, 8, 9, and 10 can not be sum and 
are eliminated automatically) so that I have a set, which contains 
probable sum of the numbers:

   {11, 17, 23, 27, 29, 33, 35, 37, 41, 47, 51, 53, 57,  59,  65, 67, 
    71, 75, 77, 79, 83, 87, 89, 93, 95, 97, 98}

Any of the numbers in this set cannot be constructed just by using two 
prime numbers. But I couldn't go far. Would you help me?


--------------------------------------------------------------------------------


Date: 05/18/98 at 17:29:46
From: Doctor Rob
Subject: Re: 2 Mathematicians Problem

Your approach is the right one, but you must be careful.

First of all:

   2 + 2 = 4 <= X + Y <= 99   and
   2 * 2 = 4 <= X * Y <= 2450 = 49 * 50

B says, "I can not find these numbers."

Then X * Y cannot be prime, since it is the product of two numbers 
greater than 1. It also cannot be the square of a prime number, 
because X and Y are not equal. If X * Y had exactly two proper 
divisors, then B would know the two numbers. This eliminates the 
product of two distinct primes, and the cube of any prime.

A says, "I was sure that you could not find them."

Yes, X + Y cannot be the sum of two distinct primes, or the sum 
of a prime and its square. This forces X + Y to be one of the 
following numbers:

   11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 
   71, 77, 79, 83, 87, 89, 93, 95, 97

Since all the remaining sums are odd, one of the two numbers must be 
odd and the other even. X * Y must then be even.

B says, "Then, I found these numbers."

B can throw away any factorization of X * Y in which the sum of the 
two factors is not one of these numbers. There must be exactly one 
such possibility left for him to know the numbers. This eliminates
X * Y = 12, for example, because even though it has factorizations  
2 * 6 and 3 * 4, the sums of the two factors, 8 and 7 respectively, 
are not in the set of possible sums, so there are no possibilities 
left. It also throws out X * Y = 120, because 120 = 5 * 24 = 15 * 8, 
and 5 + 24 = 29 and 15 + 8 = 23, which are both on the list, so there 
are two possibilities left. This limits the possibilities for the 
product to:

    18 = 9 * 2, 9 + 2 =     11
    24 = 8 * 3, 8 + 3 =     11
    28 = 7 * 4, 7 + 4 =     11
    50 = 25 * 2, 25 + 2 =   27
    52 = 13 * 4, 13 + 4 =   17
    54 = 27 * 2, 27 + 2 =   29
    76 = 19 * 4, 19 + 4 =   23
    92 = 23 * 4, 23 + 4 =   27
    96 = 32 * 3, 32 + 3 =   35
    98 = 49 * 2, 49 + 2 =   51
   100 = 25 * 4, 25 + 4 =   29
   112 = 16 * 7, 16 + 7 =   23
   124 = 31 * 4, 31 + 4 =   35
   140 = 35 * 2, 35 + 2 =   37
   144 = 48 * 3, 48 + 3 =   51
   148 = 37 * 4, 37 + 4 =   41
   152 = 19 * 8, 19 + 8 =   27
   160 = 32 * 5, 32 + 5 =   37
   172 = 43 * 4, 43 + 4 =   47
   176 = 16 * 11, 16 + 11 = 27
   188 = 47 * 4, 47 + 4 =   51
   192                      67
   198                      29
   208                      29
   212                      57
   216                      35
   220                      59
   222                      77
   228                      79
   230                      51
   232                      37
   234                      35
   238                      41
   244                      65
   246                      47
   250                      35
   ...                      ...
 
A says, "If you could find them, then I also found them."

This means that the sum in the last preceding list must occur only 
once. That eliminates X + Y = 11, 27, 29, 23, 35, 51, 37, 41, 47, ..., 
leaving only a single sum that occurs only once in the above table.  
This tells you X and Y and how A and B figured them out.

from:
( http://mathforum.org/library/drmath/view/55655.html )

Maybe you can read German, then Ive  found another fine reference for you:
( http://knobeln.wiegels.net/1999.phtml?20 )


till-ga

Search strategy: None. My wife is a mathematician and knew the link.
applegal-ga rated this answer:5 out of 5 stars and gave an additional tip of: $5.00
Thorough, professional and speedy...now if I just were as smart as you!

Comments  
Subject: Re: arithmetic problem
From: probonopublico-ga on 28 May 2004 06:59 PDT
 
I'm not sure.

The Question says that the two numbers 'may or may not be equal' but
the Answer assumes that 'It also cannot be the square of a prime
number,
because X and Y are not equal'.
Subject: Re: arithmetic problem
From: graememcrae-ga on 28 May 2004 13:50 PDT
 
http://mcraefamily.com/MathHelp/PuzzleGuessFromProductAndSum2To99.htm

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