number theory - Find minimal $xinBbb N$ that solves the linear congruence

I need to find minimal $x\in\Bbb N$ that solves the linear congruences:

$6x \equiv 2 \pmod {\!4}$

$3x \equiv 6 \pmod {\!9}$

$x \equiv 15 \pmod {\! 17}$

I divided the first congruence by $2$ and the second congruence by $3$, then used the Chinese remainder theorem.

I got $x\equiv 393 \pmod {2\cdot 3\cdot 17}$

I checked that this solution actually solves the 3 equations.

But is it the smallest one? If so, then why ?

Thanks.

Answer

Your congruences may be reduced to

$x\equiv 1\pmod{2}$

$x\equiv 2 \pmod{3}$

$x\equiv 15 \pmod{17}$

We have $\mod {17}$, $x\equiv 15\equiv 32\equiv 49 \equiv 66 \equiv 83$.

And we note that $83$ satisfies the other two congruences.

The general solution is then $x\equiv 83 \pmod{102}$