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Well, I have a system of congruences it is : $$n\equiv13\pmod{19}$$ $$n\equiv6\pmod{12}$$ I'm trying to prove that for any pair of integers $(u,v)$ the number $N=13\times12v+6\times19u$ is a solution to the system of equations above.
actually i don't know even how to begin :(.
Oh i forgot : $(u,v)$ satisfy $19u+12v=1$.
i got it it's very easy actually :
well, we have $19u+12v=1$ and that implies $19u=-12v+1$.
$$19u\equiv1\pmod{-12}\implies 19u\equiv1\pmod{12}\implies19u\times6\equiv6\pmod{12}$$
$$12\times13v\equiv0\pmod{12}$$
adding both of the two congruences give us :
$$19u\times6+12\times13v=N\equiv6\pmod{12}$$.
and the same way to prove that $N\equiv13\pmod{19}$.
