Let the "carry" number be x, which is either 0 or 1

   x
   CB
  +AA
---------
  CCC

Then we have this system:



or




The second equation becoms:



C cannot be 0, and x can only be no larger than 1,
therefore C can be no larger than 1, since A is less
than 20.  So C must be 1 and x must be 1 since A must
be less than 10. So C =1, x=1 and A = 10(1) - 1 = 9

Substituting in

B + A = 10x + C
B + 9 = 10(1) + 1
B + 9 = 10 + 1
B + 9 = 11
    B = 2

Thus the solution is:
 1
 12
+99
---
111

Edwin