Freethought & Rationalism ArchiveThe archives are read only. |
05-09-2003, 01:53 PM | #21 |
Regular Member
Join Date: Nov 2000
Location: Chch, NZ
Posts: 234
|
Since Sam knows that Polly doesn't know, and then Polly says she knows, then Sam knows it must be a power of two and a prime, otherwise Polly still wouldn't know!! I explained this in my post above. Since the only way to express 13 as 2^n + p is 8+5, Sam knows it is 8+5.
Edit: of course it could be 2+11 but 2 is illegal Scrambles |
05-09-2003, 01:53 PM | #22 |
Junior Member
Join Date: Oct 2000
Location: Boise, Idaho
Posts: 9
|
Ok, yep, 5 sums, 13, 19, 25, 29, 31... writing a program for the multiples...
|
05-09-2003, 01:58 PM | #23 | |
Moderator - Science Discussions
Join Date: Feb 2001
Location: Providence, RI, USA
Posts: 9,908
|
Quote:
|
|
05-09-2003, 02:00 PM | #24 | |
Regular Member
Join Date: Aug 2001
Location: Indeterminate
Posts: 447
|
Quote:
How does Sam know it's not 4+9 instead of 8+5? If it's 4+9, the product is 36. This could be 4x9 or 3x12 instead, but since there are multiple products, Polly still doesn't know which one it is. |
|
05-09-2003, 02:03 PM | #25 |
Regular Member
Join Date: Nov 2000
Location: Chch, NZ
Posts: 234
|
Jesse,
3 and 10: sam knows polly doesn't know. Polly knows the product is 30 and polly still has 5*6 or 3*10...so Polly still doesn't know!! Polly isn't told that the sum is 13 remember 4 and 9: Sam knows Polly doesn't know and Polly is still left with 4*9 or 3*12 so Polly still doesn't know!! 6 and 7: Sam knows Polly doesn't know and Polly is still left with 6*7 or 3*14. Polly still doesn't know!! Polly is not told the sum remember. All he knows is that Sam knows that Polly doesn't know. Scrambles |
05-09-2003, 02:06 PM | #26 | |
Regular Member
Join Date: Aug 2001
Location: Indeterminate
Posts: 447
|
Quote:
|
|
05-09-2003, 02:08 PM | #27 |
Veteran Member
Join Date: Mar 2003
Location: Northeast Ohio
Posts: 2,846
|
The sum must be less than 98 (96) and yet encompass 2 integers who’s product will yield a non-prime number. Sam knows that Polly doesn’t.
This then eliminates all but one set of factors possible in the product. Polly knows. Which allows the numbers comprising the Sum to be determined. Polly’s knowing, allows Sam to know. A multiple of 10? Divisible by 10 and 5. A sum adequate to limit the possible factors. I’m thinking 6, 40. *edit* maybe, 7, 40. |
05-09-2003, 02:09 PM | #28 |
Regular Member
Join Date: Nov 2000
Location: Chch, NZ
Posts: 234
|
Lex,
(1) Sam states that he knows Polly doesn't know...Hence one even number one odd number (reasons stated in my first post) (2) Poly then states that she knows the number. This means that after putting all of the 2's in one number there is no freedom to choose. E.g. if it were 4 and 9 then the product is 36: 36 = 2*2*9 = (2*2)*9 or (2*2*3)*3 So since Poly stated that she knows the numbers, it can't be 4 and 9. Similarly you can reject all other possibilities except 5 and 8. Scrambles |
05-09-2003, 02:09 PM | #29 |
Moderator - Science Discussions
Join Date: Feb 2001
Location: Providence, RI, USA
Posts: 9,908
|
Scrambles:
3 and 10: sam knows polly doesn't know. Polly knows the product is 30 and polly still has 5*6 or 3*10...so Polly still doesn't know!! Polly isn't told that the sum is 13 remember Yes, but Polly knows that if it was 5 and 6 then Sam's sum would be 11, in which case it would be impossible for him to know from the beginning that Polly didn't know the numbers, since the two numbers could then be 4 and 7, and then Polly's product would be 28 and she'd know the numbers were 4 and 7. So, in this case Polly would know, based only on her product and Sam's claim that she can't know, that the two numbers must be 3 and 10. Scrambles: 4 and 9: Sam knows Polly doesn't know and Polly is still left with 4*9 or 3*12 so Polly still doesn't know!! Polly knows it can't be 3 and 12, since if it was Sam's sum would be 15, so from his point of view the sum could 4+11, and if it were Polly would have a product of 44 and would thus know the two numbers right away. Scrambles: 6 and 7: Sam knows Polly doesn't know and Polly is still left with 6*7 or 3*14. Polly still doesn't know!! Polly knows it can't be 3 and 14, since in that case Sam's sum would have been 17, and he would have no way of being sure the numbers weren't 4 and 13 (if they were, then again, Polly would be able to deduce them right away). |
05-09-2003, 02:13 PM | #30 |
Regular Member
Join Date: Nov 2000
Location: Chch, NZ
Posts: 234
|
Jesse,
Ah yes...recursion. Doh!! Scrambles |
Thread Tools | Search this Thread |
|