Smallest value of $(a+b)$ What can be the smallest value of $(a+b)$ , $a>0$ and $b>0$ where $(a+13b)$ is divisible by $11$ and $(a+11b)$ is divisible by $13$
This is what I have done so far.
We have 
1) $a+ 13b =  11m$
2) $a + 11b = 13n$  
$b = \dfrac{11m - 13n}{2}$ , $a = \dfrac{169n - 121m}{2}$  
Since $a,b > 0$, we have that
$11m - 13n > 0$ and $169n - 121m > 0$
$\implies 11\cdot13m - 13^2n > 0$ and  $13^2n - 11^2m > 0$
$\implies 11\cdot13m > 13^2n > 11^2m$
$\implies (\frac{11}{13}) m > n > (\frac{11}{13})^2m$
How to proceed from here?

Equivalently $11 \vert (a+2b)$ and $13 \vert (a-2b)$.
Hence, $a+2b = 11m$ and $a-2b = 13n$.
This gives us that
$$a = \dfrac{11m+13n}{2} = 6(m+n) + \dfrac{n-m}{2}$$ $$b = \dfrac{11m-13n}{4} = 3(m-n) - \dfrac{m+n}4 $$
This means $2 \vert (n-m)$ and $4 \vert (m+n)$. Hence, $m+n = 4 k_1$ and $n-m = 2k_2$. Hence, $$a = 24k_1 + k_2 > 0$$ and $$b = -6k_2-k_1 > 0$$
Hence, $k_2 > -24 k_1$ and $k_2< - \dfrac{k_1}6$.
$a+b = 23k_1 - 5k_2 \in \left(23 \dfrac56 k_1,143k_1\right)$.
Hence, for $a+b$ to be minimum, our first choice would be $k_1 = 1$.
If $k_1 = 1$, then $-24 < k_2 < -\dfrac16$.
Since $a+b$ goes as $-5k_2$, we want $k_2$ to take a small negative value i.e. $k_2$ should be close to zero and negative. Hence, $k_2 = -1$.
This gives that $$a=23, b=5 \text{ and }a+b = 28$$
