|
||
Title: find remainder of the sum when divided by 81 Post by gkwal on Jul 16th, 2007, 9:42pm what is the remainder when 13^99 + 19^93 is divided by 81 |
||
Title: Re: find remainder of the sum when divided by 81 Post by SMQ on Jul 17th, 2007, 7:24am 13^n (mod 81): 13, 7, 10, 49, 70, 19, 4, 52, 28, 40, 34, 37, 76, 16, 46, 31, 79, 55, 67, 61, 64, 22, 43, 73, 58, 25, 1, 13, ... (period of 27) 19^n (mod 81): 19, 37, 55, 73, 10, 28, 46, 64, 1, 19, ... (period of 9) So 13^99 + 19^93 (mod 81) = 13^(99 mod 27) + 19^(93 mod 9) (mod 81) = 13^18 + 19^3 (mod 81) = 55 + 55 (mod 81) = 29. --SMQ |
||
Powered by YaBB 1 Gold - SP 1.4! Forum software copyright © 2000-2004 Yet another Bulletin Board |