View Question
 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?```
 Subject: Re: arithmetic problem Answered By: till-ga on 28 May 2004 05:35 PDT Rated:
 ```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 I´ve 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: and gave an additional tip of: \$5.00 `Thorough, professional and speedy...now if I just were as smart as you!`

 ```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'.```
 `http://mcraefamily.com/MathHelp/PuzzleGuessFromProductAndSum2To99.htm`