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JohnMann
Charter Member
1 posts |
May-23-01, 09:56 AM (EST) |
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"Puzzle:"Then, I know your sum","Then, I know your product""
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I have been trying to solve the following puzzle: Take two integers n and m, selected from the range of integers 2 thru 100. Give the SUM of the two (n+m) to person S, and the PRODUCT of the two (n*m) to person P. Neither S or P know the values of m or n. S says to P: "There is no way you can tell what my sum is" P says to S: "Then, I know your sum" S says to P: "Then, I know your product"
What are the values of n and m such that the above statements are true? The appeal of this puzzle is, of course, that it appears to have insufficient information for the solution. Also, an intriguing point is that the first statement is true -- UNTIL S articulates it. Then it becomes false.
The trouble is, I am not sure if it is possible to solve it. The first statement appears to be true for at least two answers, 11 and 17: Suppose S has the sum as 11 the possible numbers are 2+9 product=18, 3x6 or 2x9 3+8 product=24, 12x2 or 3x8 4+7 product=28, 14x2 or 4x7 5+6 product=30 10x3 or 4x5 so S can say "there is no way you can tell what my sum is" however for 17 this is also the case: 2+15=17 6x5=30,10x3=30,15x2=30 3+14=17 7x6=42,14x3=42,21x2=42 4+13=17 13x4=52,26x2=52 5+12=17 10x6=60,12x5=60,15x4=60,20x3=60,30x2=60 6+11=17 11x6=66,22x3=66,33x2=66 7+10=17 10x7=70,14x5=70,35x2=70 8+9=17 9x8=72,12x6=72,18x4=72,24x3=72,36x2=72 Am I missing something here, or is this really not possible to solve. (I *don't* have the answer!) John Mann jon.mann@btinternet.com
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alexb
Charter Member
672 posts |
May-23-01, 01:33 PM (EST) |
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1. "RE: Puzzle:"Then, I know your sum","Then, I know your product""
In response to message #0
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>What are the values of n >and m such that the >above statements are true? As you rightly note below the first statement, once uttered, becomes false, so that you have to rethink your formulation. But I understand what you mean. >The appeal of this puzzle is, >of course, that it appears >to have insufficient information for >the solution. Also, an intriguing >point is that the first >statement is true -- UNTIL >S articulates it. Then it >becomes false. > >The trouble is, I am not >sure if it is possible >to solve it. The first >statement appears to be true >for at least two answers, >11 and 17: It's true for 11, 17, 23, 27, 29, 35, 37, 41, 47, 53, 59, 65, 67, 71, 77, 79, 83, 89, 95, 97. Call these numbers good sums. >Suppose S has the sum as >11 >the possible numbers are >2+9 product=18, 3x6 or 2x9 >3+8 product=24, 12x2 or 3x8 >4+7 product=28, 14x2 or 4x7 >5+6 product=30 10x3 or 4x5 > > >so S can say "there is >no way you can tell >what my sum is" > >however for 17 this is also >the case: > >2+15=17 6x5=30,10x3=30,15x2=30 >3+14=17 7x6=42,14x3=42,21x2=42 >4+13=17 13x4=52,26x2=52 >5+12=17 10x6=60,12x5=60,15x4=60,20x3=60,30x2=60 >6+11=17 11x6=66,22x3=66,33x2=66 >7+10=17 10x7=70,14x5=70,35x2=70 >8+9=17 9x8=72,12x6=72,18x4=72,24x3=72,36x2=72 > >Am I missing something here, or >is this really not possible >to solve. (I *don't* have >the answer!) If you look at the products in your list and more generally for the products of two numbers that add up to a good sum, you may notice that some of them, like, e.g. 42, may be split into the product of two factors that add up to a good sum in more than 1 way: 42 = 14x3, 14+3=17, and 42 = 21x2, 21+2=23. Or 72 = 24x3, 24+3=27, and 72 = 9x8, 9+8=17. Call the products other than such good products. Far as I can see, there are only few good products: 18, 24, 28, 52 corresponding to the good sums 11, 11, 11, 17, respectively. The first statement makes sense if S holds a good sum. The second statement is true only of P holds a good product. The third statement may only be true if the good sum held by S is associated with a single good product. The only possibility that I can see is S = 17, P = 52, or m = 4, n = 13. In all honesty, I have not verified the fact that there are only four good products. If it's not the case, the problem is unsolvable. |
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alexb
Charter Member
672 posts |
May-23-01, 06:15 PM (EST) |
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2. "RE: Puzzle:"Then, I know your sum","Then, I know your product""
In response to message #1
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Last Saturday, I drove to Rockville, MD to visit my parents. As usual, I stopped by the huge Kamkin bookstore of Russian literature. For $4, I bought there a Russian translation of Jeux et casse-tête a programmer by Jacques Arsac, Dunod, 1985. Problem 15, Ch. 1 is exactly your problem with insignificant variations. In a free translation, the conversion appears as: P: I can't find the two numbers. S: I know you can't. P: If so, I can. S: Then I also can. No solution is given, except for suggestions how to write a program that sieves out unwanted numbers. The claim is that, as the result, only one pair of numbers remains. It's then bound to be 4, 13. |
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mark (Guest)

guest
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Jun-28-01, 09:07 PM (EST) |
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3. "RE: Puzzle:"Then, I know your sum","Then, I know your product""
In response to message #2
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i'm no regular at these boards and i'm no math wizz, but someone pointed me here cuz i spent some time trying to solve this problem and i wanted a place to check my solution. so, here goes (all comments appreciated) Knowing only the sum, S claims there is no way P can know what the two numbers are, based solely on the product. That means that S knows that P can not have a product of two primes - for if P had a product of 2 primes, he would be able to tell the factors right away (if P has 35, he knows it's 5 and 7 (primes); if P has 40, he can't possibly tell which the two numbers are (possibilites 2-20, 4-10, 8-5). How can S know this? S can only know that P can not know the two numbers if S holds a number WHICH CAN'T POSSIBLY BE WRITTEN AS A SUM OF TWO PRIMES. That means S has an odd number, for all even numbers can be written as a sum of two primes (this is called Goldbach's theorem i believe). If S holds an odd number, then the two original numbers must be an even one (A) and an odd one (B). Small remark to this - the even number may not be 2, for then too P can find the solution. On to the next step - P says 'then i know your sum'. This means that the knowledge that A is even and B is odd is NECESSARY AND SUFFICIENT to determine A and B from their product. How can this be? It's necessary - there is more then one way to factor the product. It's sufficient - of all possible ways to factor out the product, only one splits it in factors with 1 even and the other one odd (only one factorisation with an odd sum). This can only be the case if the even number is a whole power of 2 (but not 2 itself), while the odd number is a prime. Since the integers must be smaller then 100, this means A can be 4, 8, 16, 32, or 64. B is a prime between 2 and 100 (there are 25 of them i believe, from 3 to 97). Of these couples, all which have a sum that can be written as 2 + a prime must be removed. For example, 8+23 is also 2+29 so it doesn't count (if S held 31 as sum, it couldn't possibly claim P couldn't know the solution because it would be possible for P to hold 2*29 = 58 as sum, in which case P would have the solution immediately). Remarkably, there are a lot of these couples - 62 of them to be exact. This leaves 63 couples of numbers. Finally, S says 'then i know your sum'. This means that, based on the fact that P can find the numbers from S's first claim, S can also find the numbers. This means that the sum S is only the sum of one candidate couple. For example, 11 is the sum of 4 and 7, but also of 8 and 3. In both cases S knows P can't find the factors, and in both cases P can find the factors when S says so, but there is no way for S to know which to choose. Crossing out the couples not matching this final criterium does, however, not narrow the solution down to 1. The previously mentioned solution 4 and 13 is the only solution if you not only suppose A and B are smaller then 100, but P and S also. Then the product is 72 and the sum is 17. However, if you allow P and S to be greater then 100, there are more solutions. These are the couples i found acceptable solutions: 4 - 13, 37, 61, 79 8 - 89 16 - 13, 37, 43, 97 32 - 87, 89 64 - 43, 59, 61, 67, 71, 73, 79, 83, 97 A total of 20 solutions. Hope i didn't make too many mistakes - but if i did i'd love to hear about them and discuss or explain if needed. Mark |
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