A number theory problem by Anish Roy

We have

$\displaystyle \begin{aligned} a + 13b & \equiv 0 \pmod{11} \\ a + 11b & \equiv 0 \pmod{13} \end{aligned}$

or

$\displaystyle \begin{aligned} a + 2b & \equiv 0 \pmod{11} \\ a - 2b & \equiv 0 \pmod{13} \end{aligned}$

which can be rewritten as a system of equations

$\displaystyle \begin{aligned} 2a & = 11p + 13q \\ 4b & = 11p - 13q \end{aligned}$

Thus, $a = 23$ and $b = 5$ , with $p = 3$ and $q = 1$ .

Since $13$ divides $a+11b$ , we see that $13$ divides $a-2b$ and hence it also divides $6a-12b$ . This in implies that $13|(6a+b)$ . Similarly $11|(a+13b) \Rightarrow 11|(a+2b) \Rightarrow 11|(6a+12b) \Rightarrow 11|(6a+b)$ . Since $\gcd(11,13) = 1$ , we conclude that $143|(6a+b)$ . Thus we may write $6a+b=143k$ for some natural number $k$ . Hence, $6a+6b=143k+5b=144k+6b-(k+b)$ . This shows that $6$ divides $k+b$ and hence $k+b \geq 6$ . We therefore obtain $6(a+b)=143k+5b=138k+5(k+b) \geq 138k+5 \times 6=168$ . It follows that $a+b \geq 28$ . Taking $a=28$ and $b=5$ , we see that the conditions of the problem are satisfied. Thus the minimum value of $a+b= \boxed{28}$