#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.
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