TITLE: Smallest value of $(a+b)$
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: $\rm\begin{eqnarray}{\bf Hint}\rm\quad   a+13b &=&\rm 11m &\:\Rightarrow\:&\rm 13a+169b &=&\rm 143m\\
\rm a+11b &=&\rm 13n &\:\Rightarrow\:&\rm 11a+121b &=&\rm 143n\\
   && &\:\Rightarrow\:&\rm\ \ 2a+\ \ 48b &=&\rm 143(m-n)\ \ \ by\ subtracting\ prior\ from\ first
\end{eqnarray}$ 
Thus $\rm\:a + 24b = 143c.\:$ Its solution with minimal $\rm\:a+b\:$ is $\rm\:(a,b) = (23,5),\:$ since all other solutions arise by adding to $\rm\:(a,b)\:$ nonnegative multiples of $\rm\:(-1,6)\:$ or $\rm\:(24,-1),\:$ increasing $\rm\:a+b,$ viz.
$$\begin{eqnarray}\rm a+24b = 143&\:\Rightarrow\:&\rm (a,b) = (23,\ \ 5),\ (47,\ \ 4),\ (71,\ \ 3),\ \ldots\\
\rm a+24b = 286&\:\Rightarrow\:&\rm (a,b) = (22,11),\ (46,10),\ (70,\ \ 9),\ \ldots\\
\rm a+24b = 429&\:\Rightarrow\:&\rm (a,b) = (21,17),\ (45,16),\ (69,15),\ \ldots\\
                      \cdots && \cdots
\end{eqnarray}$$
In this solution table, adding $\rm\:(24,-1)\:$ moves right, and adding $\rm\:(-1,6)\:$ moves down. But moving left or above the border by subtracting these terms results in a solution with $\rm\:a\:$ or $\rm\:b\:$ negative.