Difference between revisions of "KGS math club/solution repunit"
(Created page with "Let x be an integer ending in 7, then 9x is a multiple of 9 which has no common factor with 10, so 10^k = 1 (mod 9x) for k = <a href="https://en.wikipedia.org/wiki/Euler%27s_t...") |
|||
| Line 1: | Line 1: | ||
| − | Let x be an integer ending in 7, then 9x is a multiple of 9 which has no common factor with 10, so 10^k = 1 (mod 9x) for k = | + | Let x be an integer ending in 7, then 9x is a multiple of 9 which has no common factor with 10, so 10^k = 1 (mod 9x) for k = [https://en.wikipedia.org/wiki/Euler%27s_totient_function Euler's totient function] of 9x. then 10^k - 1 is a multiple of 9x, so (10^k - 1)/9 is a multiple of x. (10^k - 1)/9 = 111...111 with k copies of 1. |
Latest revision as of 15:53, 7 May 2016
Let x be an integer ending in 7, then 9x is a multiple of 9 which has no common factor with 10, so 10^k = 1 (mod 9x) for k = Euler's totient function of 9x. then 10^k - 1 is a multiple of 9x, so (10^k - 1)/9 is a multiple of x. (10^k - 1)/9 = 111...111 with k copies of 1.
