6-5 = 8-7 = 12-11 = 1 (Difference is constant)
Find the LCM of 6, 8, 12 = 24
The smallest number which when divided by 6,8 and 12 leaving 5,7 and 11 as remainder = 24 -1 = 23
The required number will be of the form 23 + 24k (LCM of 6, 8, 24), which is also a multiple of 13.
This means (-3) + (-2)k should give a remainder of 0 when divided by 13.
The first integral value of k which satisfies this condition is k = 5.
Number = 23+120 = 143, Sum of digits = 8.