nm, figured it out.
11k+4 != 17l+3 so no solutions exist
x congruent to 4 (mod 11)
x congruent to 3 (mod 17)
M = 11 * 17 = 187
so M1 = 11, M2 = 17
i'm trying to find the inverse of M1 and M2 but am unsure how because M1M1' congruent to 1 (mod 11) but M1M1' is a factor of 11 so it will always have a remainder of 0.
can this be proven?
If GCD(m1,m2) = 1 then a solution exists. That is the first step.
That's a good 2nd step.trying to find the inverse of M1 and M2
The modular inverse of (m1,m2) and the modular inverse of (m2,m1).
I do not know what you are attempting with M1M1? I can't understand it. It seems as if you are trying to say: 11 will be a factor of m1m2 and 17 will be a factor of m1m2. Not sure. Just wondering.i'm trying to find the inverse of M1 and M2 but am unsure how because M1M1' congruent to 1 (mod 11) but M1M1' is a factor of 11 so it will always have a remainder of 0.
Best advice: use google to get additional info on the Chinese Remaider Theorem (CRT).