Mod m, m-1, and m-2 residues

We try to first find x and m:

When x has the residue 1 mod m, the for some integer $k_{1}$ :

$\frac{x-1}{m}$ = $k_{1}$

When x has the residue 2 mod m-1, the for some integer $k_{2}$ :

$\frac{x-2}{m-1}$ = $k_{2}$

When x has the residue 3 mod m-2, the for some integer $k_{3}$ :

$\frac{x-3}{m-2}$ = $k_{3}$

Now we can minimize m first.

m has to be greater than 1, so the most minimized value of m is 2. But 2-2=0, and you can't divide by 0, so we try m=3.

Plugging in m=3 we get:

$\frac{x-1}{3}$

and:

$\frac{x-2}{2}$

and:

$\frac{x-3}{1}$

Using guess and check, we find that the minimum value of x is 4.

So the answer is 4+3= $\boxed{7}$