The Constraints Are Reversed

$m+13n \equiv 0 \pmod {11}$ , $m+11n \equiv 0 \pmod {13}$ . (We add $11n$ and $13n$ to the first and second equations respectively to get)

$m+13n+11n = m+24n \equiv 11n \equiv 0 \pmod{11}$ , $m+13n+11n = m+24n \equiv 13n \equiv 0 \pmod{13}$

Since $m+24n \equiv 0 \pmod {11,13}$ , we can deduce that $m+24n \equiv 0 \pmod {143}$ .

Therefore $m+24n = 143k$ for some integer $k$ .

But since $m,n$ are positive, then to minimize $m+n$ , $k$ must be positive but as small as possible, and the smallest positive integer is $k=1$

Therefore, $m+24n = 143(1) = 143$ . We then maximize $n$ such that $m = 143-24n$ is still positive.

The largest satisfying $n$ is $5$ , and the corresponding $m$ is $23$ , so our answer is $23+5 = \boxed{28}$ .

Since $13$ divides $6\left( m+11n \right)=\left( 6m+n \right)+65n$ , $13$ divides $6m+n$ .

Since $11$ divides $6\left( m+13n \right)=6m+n+77n$ , $11$ also divides $6m+n$ .

Hence $11\times 13=143$ divides $6m+n$ , so that $6m+n=143k$ for some positive integer $k$ .

Since $6\left( m+n \right)=143k+5n=6\left( 24k+n \right)-\left( k+n \right)$ , $6$ divides $k+n$ so that $k+n\ge 6$ .

Now $6\left( m+n \right)=143k+5n=138k+5\left( k+n \right)\ge 168$ .

Consequently $m+n\ge 28$ , and this is attained if $m=23$ and $n=5$ .

$m + 13n \equiv 0$ (mod 11)

$m + 11n \equiv 0$ (mod 13)

Starting with the second equation we can deduce that:

$m + 11n \equiv 0$ (mod 13)

$m \equiv 2n$ (mod 13)

$m = 13k + 2n$

This can then substituted into the first equation as follows:

$13k + 2n + 13n \equiv 0$ (mod 11)

$13k + 15n \equiv 0$ (mod 11)

$2k + 4n \equiv 0$ (mod 11)

$k + 2n \equiv 0$ (mod 11)

$9n \equiv k$ (mod 11)

$n \equiv 5k$ (mod 11)

$n = 11j + 5k$

Now we can re-write the sum $m + n$ as :

$28k + 33j$

Now since both $m$ and $n$ are both positive the sum must be more than 0. Hence setting $k = 1$ and $j = 0$ minimizes the sum to $\boxed{28}$

11p=m+13n
13q=m+11n
can be written as,

11y=m+2n
13z=m-2n
solving,

11y+13z=2m
11y-13z=4n

so by substituting for y,z we can know that either y,z both are even or odd. And 11y>13z.

first, possible values for y and z are y=3,z=1 for which m=23 and n=5.

And the next possible values for y and z are y=5,z=3 for which m=47 and n=4 and so on but greater than 28.

so the least possible sum can be 28.