Go back to previous page
Forum URL: http://www.cut-the-knot.org/cgi-bin/dcforum/forumctk.cgi
Forum Name: This and that
Topic ID: 907
Message ID: 2
#2, RE: Chinese Remainder Theorem
Posted by jprice2 on Aug-22-09 at 10:04 AM
In response to message #1
i see. So, t(m1/gcd(m1, m2)) - s(m2/gcd(m1, m2)) = n0 can be written as (t/n0)(m1/gcd(m1, m2)) - (s/n0)(m2/gcd(m1, m2)) = 1.
And, therefore (t/n0)(m1/gcd(m1, m2)) = 1 (mod(m2/gcd(m1, m2))) has a solution which implies that t(m1/gcd(m1, m2)) = n0(mod(m2/gcd(m1, m2))) has a solution. I new that as + bt = 1 is true for integer s and t when a and b are coprime; so, I should have seen this.
Thanks for the insight.