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Old 05-09-2003, 02:14 PM   #31
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Quote:
Originally posted by Majestyk
A multiple of 10? Divisible by 10 and 5. A sum adequate to limit the possible factors. I’m thinking 6, 40.
In that case Sam's sum would be 46. He doesn't know how this number was formed, but one possibility is that it was 41 plus 5. In that case Polly would have a product of 205, and the only way to factor that is 41*5 since 41 and 5 are primes. So, in that case Polly would know from the beginning what the two numbers are. Since Sam is confident there is no way she could know this at the beginning, his sum couldn't be 46.
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Old 05-09-2003, 02:18 PM   #32
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I now agree that the problem is messed up. There is no situation where the sum person can figure it out in the end.
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Old 05-09-2003, 02:27 PM   #33
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I tried to solve this and came up with the exact same results as Jesse.

In order for Polly to know the answer after Sam says he know's she doesn't, the numbers must be one of the following combinations:

Sum = 13
3, 10
4, 9
5, 8
6, 7

Sum = 19
3, 16
4, 15
5, 14
8, 11
9, 10

Sum = 25
3, 22
6, 19
7, 18
8, 17
9, 16
12, 13

Sum = 29
5, 24
6, 23
9, 20
10, 19
12, 17
13, 16

Sum = 31
4, 27
5, 26
8, 23
11, 20
12, 19
13, 18
14, 17
15, 16

None of these allow Sam to know the answer.

I tried adding 2 and 50 to the list of possible numbers and there still is not a solution that allows Sam to figure out the answer.
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Old 05-09-2003, 02:29 PM   #34
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Yeah, there's either no answer or I'm utterly stumped. I'm at the same spot Jesse is:

13: 30 36 40 42
19: 48 60 70 88 90
25: 66 114 126 136 144 156
29: 120 138 180 190 204 208
31: 108 130 184 220 228 234 238 240

The sums (13, 19, 25, 29, 31) are the only sums that one can exclude the possibility of the product in and of itself giving away the answer.

Then we take all of the products that can be made based on numbers adding to that sum as possible products.

Then we remove those that are duplicated in the above list, as those duplicates (78, 84, 100, 150, 154, 168, 198, 210) don't let Polly determine the correct answer.

The last bit is Sam should, theoretically, be able to determine an answer based on the knowledge that Polly was able to determine an answer with the given information. But that bit of information only clears out a few from the list, and doesn't give a single sum that has exactly one remaining possible product.
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Old 05-09-2003, 02:44 PM   #35
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Assuming the problem is solvable, there must be something inherent in the sum that dictates the character of the product. Could restricting the sum to odd numbers, provide such a quality?

*edit* nevermind. I'm out of my league here. I'll just lurk.
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Old 05-09-2003, 02:47 PM   #36
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There is a problem using numbers greater than 1 and a sum of less than 100 which does have a single solution.
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Old 05-09-2003, 03:20 PM   #37
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Quote:
Originally posted by dolem98
There is a problem using numbers greater than 1 and a sum of less than 100 which does have a single solution.
13 and 4?
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Old 05-09-2003, 03:26 PM   #38
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Quote:
Originally posted by dolem98
There is a problem using numbers greater than 1 and a sum of less than 100 which does have a single solution.
Would each individual number have to be smaller than or equal to 50, or can you have combinations like (10, 89)? If each is smaller than or equal to 50, is it important that 99 be a possible sum, or can you restrict it to numbers smaller than but not equal to 50?
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Old 05-09-2003, 03:37 PM   #39
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OK, here's the whole problem.
Quote:
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."
It's on Ask Dr. Math, and it's solvable.
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Old 05-09-2003, 03:41 PM   #40
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Quote:
Originally posted by dolem98
I now agree that the problem is messed up. There is no situation where the sum person can figure it out in the end.
The problem is uniquely solvable as I posted it.




For those who want it, look below for a hint on where to proceed (or at least where I proceeded--there could be a better way). Tf you have trouble reading, either hilight the text or just quote my message:

(1) From Sam's first statement, you (and, in the same manner, Polly) can figure out all of the possile sums he might have been given. It seems many of you have done this so far.

(2) Sam's statement tells us that her product must be even and must have more than one pair of factors (where we neglect the factor pair including the number 2 in all of our calculations).

(3) Ok, so now Polly knows all of Sam's possible sums and she knows the product. Her statement tells us (and Sam) that her her product must factorize such that exactly one factor pair sums to one of Sam's allowed sums. If this is the case, then she can uniquely know the two numbers from Sam's information.

(4) So now Sam knows something about Polly's factor pairs. Let's say he takes his sum and computes all possible products that Polly could have. From this list he can determine whether any of these products meets criteria (3). Sam will only be able to determine the answer if exactly one of these possible products meets criteria (3). Only one of the allowed sums will satisfy this, thereby allowing Sam to know the numbers as well. It is this that allows us as outside observers to figure out what the numbers are as well.


At any rate, both times I solved this (four years ago and last night) I used math to get part way and then wrote a computer program to brute-force it the rest of the way, implementing the filters listed above. I did, however, once witness a professor solve the problem completely on the blackboard, so I know it's possible to do without the aid of a computer.
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