Since 13 divides a+11b , we see that 13 divides a – 2b and hence it also divides 6a –12b. This in turn implies that 13|(6a + 13b).
Similarly 11|(a+ 2b) ⇒11|(6a + 12b) ⇒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.
6a + 6b = 143k + 5b = 144k + 6b - (k + b).
This shows that 6 divides k + b and hence k + b ≥ 6.
We therefore obtain
6(a + b) = 143k + 5b = 138k + 5(k + b) ≥ 138 + 5 x 6 =168.
It follows that a + b ≥ 28. Taking a = 23 and b = 5, we see that the condition of the problem are satisfied. Thus the minimum value of a + b is 28.