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April 14th, 2010, 05:38 PM  #1 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  application to ax=b (mod m)
describe all solutions of the given congruence. 22x = 5 (mod 15) proof: the gcd(22,15) =1 and 1 is a divisor of 15. Thus by a corollary, there is solutions. the multiplicative inverse of 22 in Z_15 is 13. So (22*13)x = 15 *13 (mod 15) => x = 65 mod 15 => x = 5 mod 15 Hence the reminder is 5. Solutions in Z_15 is 5+15Z? thanks 
April 18th, 2010, 09:42 PM  #2  
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: application to ax=b (mod m)
Hello tinynerdi, Quote:
Hoempa  
April 27th, 2010, 09:08 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: application to ax=b (mod m)
Typo; should have been 5 not 15.


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