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?

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@John0417 The questions says that $a>0$ and $b>0$. – Fly by Night Oct 31 '12 at 19:59

$\rm\begin{eqnarray}{\bf Hint}\rm\quad a+13b &=&\rm 11m &\:\Rightarrow\:&\rm 13a+169b &=&\rm 143m\\ \rm a+11b &=&\rm 13n &\:\Rightarrow\:&\rm 11a+121b &=&\rm 143n\\ && &\:\Rightarrow\:&\rm\ \ 2a+\ \ 48b &=&\rm 143(m-n)\ \ \ by\ subtracting\ prior\ from\ first \end{eqnarray}$

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.

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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 $$a=23, b=5 \text{ and }a+b = 28$$

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