least positive value

let a+13b = 11k (k belongs to N)
or a+2b= 11(k-b) = 11c

let a+11b=13l (l belongs to n)
or a-2b = 13(l-b) = 13 d

now a-2b < a+2b (since a,b are +ve)
hence 13d < 11c
hence min. (d)=0
min (c)=1
therefore a-2b=0
and a +2b =11 or a=5.5 (not possible)

hence let us take the next case ,
d=0
c=2
there we get b=5.5(again not possible)

similarly going on we find that for c=4 and d=0 ,
a=22 and b=11

hence min (a+b) =33

why we will n"t increase the min value of d.

I will give you some steps , I think you can complete it .

1 . 13 divides " a - 2 b " .

2 . 13 divides " 6 a + b " .

3 . 11 divides " a + 2 b " .

4 . 11 divides " 6 a + b " .

5 . gcd ( 11 , 13 ) = 1 .

6 . 6 a + b = 143 n . . . . . . . . ( n E N ) .

7 . n + b ≥ 6 .

8 . " a = 23 , b = 5 " works .

By the way , this is an old RMO problem from what I can gather from my memory .

Aditya - " min. (d)=0 "

How is it true ?

" b " can be greater than " l " .