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?