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
