Freethought & Rationalism Archive

Jesse · 05-09-2003, 04:41 PM

Quote:

Originally posted by Lobstrosity
Well, if you allow all numbers from 3 to 50 inclusive then there's a different solution. I have never worked through this problem allowing the number 2, so I cannot comment at this time on how that would change things.

You mean 3 to 49 inclusive, right? (it probably wouldn't make a difference if you include 50, but you originally stated it as between 2 and 50) Like Silent Acorns, I'd like to hear what your solution is, because I don't think there is one.

Silent Acorns · 05-09-2003, 04:45 PM

I found a slight problem in my code, but it only applies to cases where the minimum number can be 1. If the numbers can be 1 to 50 inclusive then the unique solution is:
1 and 10

Silent Acorns · 05-09-2003, 04:49 PM

Quote:

Originally posted by Jesse
with that many sums it seems like the only way to solve the problem would be to write a computer program to do it for you

That's what I was forced to do in order to look for cases that allow for unique solutions.

Principia · 05-09-2003, 05:11 PM

Nevermind. These problems hurt me.

Lobstrosity · 05-09-2003, 05:36 PM

You know, I think I fucked up with regards to putting the cap on the numbers instead of on the sum. I really apologize for that. If you say that the sum must be less than 100 rather than each number less than 50, you can get a unique answer. If I'm correct, this increases the number of allowed sums to:

13, 19, 25, 29, 31, 37, 43, 49, 53, 55, 59, 61, 67, 73, 79, 81, 85, 89, 91, 95, 97, 99

Having a larger set of allowed sums then ends up working to your advantage. Damn, I'm sorry for messing that part up--I was trying to remember this from four years ago.

Lex Talionis · 05-09-2003, 05:59 PM

Ahh, yes, if the limitation is numbers > 2 and sum < 100, then the answer is: 13 + 4

Running it through my program, I get a similar but larger table as posted above. Possible sums: 11 17 23 27 29 35 37 41 47 53.

A lot of duplicates this time -- the sum 17 has 7 possible products (30 42 52 60 66 70 72), all but 52 are duplicated by products availible from the good set of sums. And the two numbers that have a sum of 17 and a product of 52 are 4 and 13.

Lobstrosity · 05-09-2003, 06:06 PM

It can't be 4 and 13. It can't be 4 with any prime.

4 x 13 = 2 x 2 x 13 = 52

the only factors of this are (2, 26) and (4, 13). Since 2 is not allowed, should Polly see a product of 52, she would instantly know the numbers were 4 and 13.

Lex Talionis · 05-09-2003, 06:10 PM

Oops. got a bug somewhere. hrm.

Lex Talionis · 05-09-2003, 06:20 PM

Ok, the last one I was solving for >= 2, duplicate numbers not allowed. That's what I get for writing code after 9 pm.

For > 2, duplicate numbers allowed, I get 13 and 16

Lobstrosity · 05-09-2003, 06:37 PM

And we have a winner!

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05-09-2003, 04:45 PM	#52
Silent Acorns Veteran Member Join Date: Oct 2002 Location: SW 31 52 24W4 Posts: 1,508	I found a slight problem in my code, but it only applies to cases where the minimum number can be 1. If the numbers can be 1 to 50 inclusive then the unique solution is: 1 and 10

05-09-2003, 05:11 PM	#54
Principia Veteran Member Join Date: Mar 2002 Location: anywhere Posts: 1,976	Nevermind. These problems hurt me.

05-09-2003, 05:36 PM	#55
Lobstrosity Senior Member Join Date: Feb 2003 Location: San Diego, California Posts: 719	You know, I think I fucked up with regards to putting the cap on the numbers instead of on the sum. I really apologize for that. If you say that the sum must be less than 100 rather than each number less than 50, you can get a unique answer. If I'm correct, this increases the number of allowed sums to: 13, 19, 25, 29, 31, 37, 43, 49, 53, 55, 59, 61, 67, 73, 79, 81, 85, 89, 91, 95, 97, 99 Having a larger set of allowed sums then ends up working to your advantage. Damn, I'm sorry for messing that part up--I was trying to remember this from four years ago.

05-09-2003, 05:59 PM	#56
Lex Talionis Regular Member Join Date: Aug 2001 Location: Indeterminate Posts: 447	Ahh, yes, if the limitation is numbers > 2 and sum < 100, then the answer is: 13 + 4 Running it through my program, I get a similar but larger table as posted above. Possible sums: 11 17 23 27 29 35 37 41 47 53. A lot of duplicates this time -- the sum 17 has 7 possible products (30 42 52 60 66 70 72), all but 52 are duplicated by products availible from the good set of sums. And the two numbers that have a sum of 17 and a product of 52 are 4 and 13.

05-09-2003, 06:06 PM	#57
Lobstrosity Senior Member Join Date: Feb 2003 Location: San Diego, California Posts: 719	It can't be 4 and 13. It can't be 4 with any prime. 4 x 13 = 2 x 2 x 13 = 52 the only factors of this are (2, 26) and (4, 13). Since 2 is not allowed, should Polly see a product of 52, she would instantly know the numbers were 4 and 13.

05-09-2003, 06:10 PM	#58
Lex Talionis Regular Member Join Date: Aug 2001 Location: Indeterminate Posts: 447	Oops. got a bug somewhere. hrm.

05-09-2003, 06:20 PM	#59
Lex Talionis Regular Member Join Date: Aug 2001 Location: Indeterminate Posts: 447	Ok, the last one I was solving for >= 2, duplicate numbers not allowed. That's what I get for writing code after 9 pm. For > 2, duplicate numbers allowed, I get 13 and 16

05-09-2003, 06:37 PM	#60
Lobstrosity Senior Member Join Date: Feb 2003 Location: San Diego, California Posts: 719	And we have a winner!

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