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The detailed process of solving the linear congruence equation103x ≡ 57 (mod 211) in discrete mathematics.
103 x≡57(mod 2 1 1)

be equal to

103x+2 1 1y=57

103(x+2y)+5y=57

Let z=x+2y ①.

103z+5y=57

3z+5(y+20z)=57

Let w=y+20z ②.

3z+5w=57

Obviously, z=9 and w=6 are a set of solutions, or z= 19 and w=0 are also a set of solutions.

(If you can't see it, continue to exchange yuan:

3(z+2w)-w=57

Let u=z+2w.

3u-w=57=3* 19

Obviously, U = 19 and W = 0 are a set of solutions.

)

Then according to ① ②, we get

x = z-2(w-20z)=-2w+4 1z = 357 mod 2 1 1

Calculated by modulus

x= 146 mod 2 1 1

Is the solution of congruence.