First, lets prove something about the system of congruencies: If (A,B) is a solution then (A+1,B-6) is also a solution.
(A+1)+13*(B-6) mod 11
= A+1+13*B-78 mod 11
= A+13*B-77 mod 11
= A+13*B mod 11
(A+1)+11*(B-6) mod 13
= A+1+11*B-66 mod 13
= A+11*B-65 mod 13
= A+11*B mod 13
Next compare (A,B) and (A+1,B-6). Obviously the latter has a smaller sum. So for any known solution to the system we can get a smaller one by increasing A by 1 and decreasing B by 6. We can do this for as long as B>6 to keep things positive, as required by the problem statement.
So now all we need is some (A,B) to start from. (0,0) is the trivial solution, but that is not positive integers. 143 is a multiple of both 11 and 13 so either of A or B can be increased by 143. Take (A,B)=(0,143). 143/6=23.833. So we can apply (A,B)->(A+1,B-6) 23 times. This leaves us at (A,B)=(23,5). B is now too small to reduce further so this is the minimum. A+B=23+5=28 is the final answer,
Analytic Solution
