1. ## chinese remainder theorem

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?

2. nm, figured it out.

11k+4 != 17l+3 so no solutions exist

3. Originally Posted by silentbob
11k+4 != 17l+3 so no solutions exist
Why not? $11(3)+4\,=\,37\,=\,17(2)+3.$

One way of solving such a problem is to find a particular solution by trial and error. In this case, it’s 37. The general solution is then of the form (particular solution) + (multiple of LCM of moduli). In this case, the general solution is $x\,=\,37+187n,\quad n\in\mathbb Z.$

4. Originally Posted by silentbob
nm, figured it out.

11k+4 != 17l+3 so no solutions exist
If GCD(m1,m2) = 1 then a solution exists. That is the first step.

trying to find the inverse of M1 and M2
That's a good 2nd step.
The modular inverse of (m1,m2) and the modular inverse of (m2,m1).

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.
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.