If a number is divisible by 9, the sum of its digits will also be divisible by 9.
The largest a can be, then, is 18162 (that's 2018 '9s'). b is then 18 in this case.
The largest b can be is 36 (one possibility of that is when a is 9999). So, b can only ever be 36, 18, or 9. The sum of all these digits is 9, so...
c must be 9.