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Topic: Variation to the sum  product riddle (Read 4369 times) 

Altamira_64
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Variation to the sum  product riddle
« on: Jun 12^{th}, 2012, 4:38am » 
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In our case the numbers are integers from 3 to 100. One of the two friends (A) knows the sum and the second knows the product. Their dialog is as follows: A: I know u can't find the numbers. Unfortunately, neither can I. B: But now I can! A: So do I!


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Grimbal
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Re: Variation to the sum  product riddle
« Reply #1 on: Jun 15^{th}, 2012, 3:59pm » 
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Can the numbers be equal?


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Altamira_64
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Re: Variation to the sum  product riddle
« Reply #2 on: Jun 17^{th}, 2012, 1:53am » 
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Yes


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Grimbal
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Re: Variation to the sum  product riddle
« Reply #3 on: Jun 17^{th}, 2012, 3:12pm » 
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I find 2 solutions: (5,8) and (13,16) But I am not 100% confident with my program.

« Last Edit: Jun 17^{th}, 2012, 3:12pm by Grimbal » 
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0.999...
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Re: Variation to the sum  product riddle
« Reply #4 on: Jun 18^{th}, 2012, 8:25pm » 
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I believe (13,16) is the only solution. (Verified arduously by hand ) That (5,8 ) is not a solution follows from the following reasoning: Suppose A makes the first statement seeing the sum 13. The statement is indeed correct. If B sees 40, then clearly we get a true second statement. I claim that if B sees 30, the second statement is still true! 30 = 5*6 = 3*10 If the pair was (5,6), then A's first statement would be false with 11 = 4+7 ruled out since 4*7 uniquely factors into integers greater than 3. Thus, B determines that (3,10) is the only possible pair. On the third statement, therefore, A cannot determine whether it is (3,10) or (5,8 ).

« Last Edit: Jun 18^{th}, 2012, 8:45pm by 0.999... » 
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Altamira_64
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Re: Variation to the sum  product riddle
« Reply #5 on: Jun 19^{th}, 2012, 3:29pm » 
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Can anyone briefly explain the reasoning?? How did you get to this result?


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0.999...
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Re: Variation to the sum  product riddle
« Reply #6 on: Jun 19^{th}, 2012, 9:26pm » 
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Let S be the set of all sums we would rule out due to A's first statement. S contains 7 and 199 because A would know their summands. S contains all even numbers 6200 because they can be written as a sum of two primes and so A would not be sure that B cannot know their factors. S contains all sums p+4 where p is odd and prime, because p*4 is the only way to factor that product into integers greater than or equal to 3. Now, for any remaining sum r: If r = a+b = x+y where all other factorizations (a',b') of a*b and (x',y') of x*y sum to elements of S, then B's statement will be true, but A cannot deduce the pair, falsifying A's second statement. What we have left are the sums such that there is exactly one pair (a,b) with all other factorizations of a*b summing to elements of S. For instance, 13 = 3+10 = 5+8; 30 = 3*10 = 6*5 (= 15*2) where 6+5 = 11 = 7+4 an element of S. 40 = 5*8 = 10*4 (= 20*2) where 10+4 is even and so in S. So 13 can be ruled out. On the other hand, 29 = 3+26 = 4+25 = 5+24 = 6+23 = 7+22 = 8+21 = 9+20 = 10+19 = 11+18 = 12+17 = 13+16 = 14+15 78 = 3*26 = 13*6; 29 and 19 are both not in S. 100 = 4*25 = 20*5 120 = 5*24 = 3*40 138 = 6*23 = 3*46 154 = 7*22 = 11*14 168 = 8*21 = 7*24 180 = 9*20 = 45*4 190 = 10*19 = 5*38 198 = 11*18 = 9*22 204 = 12*17 = 51*4 (*)208 = 13*16 = even*even 210 = 14*15 = 7*30

« Last Edit: Jun 19^{th}, 2012, 9:55pm by 0.999... » 
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Altamira_64
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Re: Variation to the sum  product riddle
« Reply #7 on: Jun 20^{th}, 2012, 2:42am » 
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Great, many thanks to all of you!


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Grimbal
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Re: Variation to the sum  product riddle
« Reply #8 on: Jun 20^{th}, 2012, 2:03pm » 
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on Jun 18^{th}, 2012, 8:25pm, 0.999... wrote:I believe (13,16) is the only solution. (Verified arduously by hand ) 
 Yo are correct, indeed. And my program was wrong. I corrected the program and made it run a 100 times as a punishment.


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0.999...
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Re: Variation to the sum  product riddle
« Reply #9 on: Jun 22^{nd}, 2012, 9:08am » 
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on Jun 20^{th}, 2012, 2:03pm, Grimbal wrote: Yo are correct, indeed. And my program was wrong. I corrected the program and made it run a 100 times as a punishment. 
 Oh no, you must write that program in 100 different (programming) languages.

« Last Edit: Jun 22^{nd}, 2012, 9:18am by 0.999... » 
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Altamira_64
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Re: Variation to the sum  product riddle
« Reply #10 on: Jun 24^{th}, 2012, 12:39pm » 
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on Jun 19^{th}, 2012, 9:26pm, 0.999... wrote:78 = 3*26 = 13*6; 29 and 19 are both not in S. 
 What do you mean by this? 29 is the sum of the two numbers 13 + 16; how can they not be in S?


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Altamira_64
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Re: Variation to the sum  product riddle
« Reply #11 on: Jun 25^{th}, 2012, 4:24am » 
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Sorry, my mistake, I just realized that S contains the ones we rule out, not the ones we keep!


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