Timeline for Divisible by seventeen
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2016 at 16:41 | vote | accept | Alexis | ||
| Feb 16, 2016 at 13:16 | comment | added | Michael Seifert | Good point. In fact, I think any number that's relatively prime to 9 (i.e., not a multiple of 3) will be a divisor of 9m - 1 for some m. | |
| Feb 16, 2016 at 6:58 | comment | added | Sleafar | @MichaelSeifert Actually all numbers equal to 8 (mod 9) and their divisors will have a solution. The numbers which definitely won't have a solution are multiples of 9, because 9m-1 will be never divisible by 9. | |
| Feb 15, 2016 at 22:14 | comment | added | Michael Seifert | This method would generalize to a version of the puzzle where instead of multiples of 17, you wanted multiples of $9m - 1$ for integer $m$ (8, 17, 26, 35, ...) I don't think the puzzle would have solutions for multiples of any numbers that were equal to 8 (mod 9). | |
| Feb 14, 2016 at 18:03 | history | edited | Sleafar | CC BY-SA 3.0 |
added 2 characters in body
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| Feb 14, 2016 at 17:54 | history | answered | Sleafar | CC BY-SA 3.0 |