Mod m, m+1, and m+2 residues

From the given conditions of the problem, for positive integers $x, m, p, q, r$ we can write

$x=pm+1=q(m+1)+2=r(m+2)+3$

Since we are searching for the minimum of $x$ and since $m>1$ , we take the minimum value of $m=2$ into consideration. Then we have

$x=2p+1=3q+2=4r+3$

$\implies p=\dfrac {3q+1}{2}=2r+1$

So $q$ must be odd and so also $\dfrac {3q+1}{2}$

We can easily see that this is achieved for $q=3$ , when $p=5$ and $r=2$

Then $x=11$ and $m+x=2+11=\boxed {13}$ .

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.

Plugging in m=2 we get:

$\frac{x-1}{2}$

and:

$\frac{x-2}{3}$

and:

$\frac{x-3}{4}$

Using guess and check, we find that the minimum value of x is 11.

So the answer is 11+2= $\boxed{13}$