
2. Find the least possible value of $a+b$, where $a, b$ are positive integers such that 11 divides $a+13 b$ and 13 divides $a+11 b$.


Solution:Since 13 divides $a+11 b$, we see that 13 divides $a-2 b$ and hence it also divides $6 a-12 b$. This in turn implies that $13 \mid(6 a+b)$. Similarly $11|(a+13 b) \Longrightarrow 11|(a+2 b) \Longrightarrow$ $11|(6 a+12 b) \Longrightarrow 11|(6 a+b)$. Since $\operatorname{gcd}(11,13)=1$, we conclude that $143 \mid(6 a+b)$. Thus we may write $6 a+b=143 k$ for some natural number $k$. Hence

$$
6 a+6 b=143 k+5 b=144 k+6 b-(k+b)
$$

This shows that 6 divides $k+b$ and hence $k+b \geq 6$. We therefore obtain

$$
6(a+b)=143 k+5 b=138 k+5(k+b) \geq 138+5 \times 6=168
$$

It follows that $a+b \geq 28$. Taking $a=23$ and $b=5$, we see that the conditions of the problem are satisfied. Thus the minimum value of $a+b$ is 28 .
