Mishari wrote:

AB

CD+

AAA=

where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

Aha finally got it.

Since AB and CD are two digit numbers and AAA is a three digit number. AAA has to be less than 200. Ex/ 99+99=198. So since the last digit of AAA has to be 1 then all the A's are 1.

So the new problem looks like this:

1B

+

CD
111

To get 1 from B+D, B+D has to either equal 11 or 1 (0+1). In this case only 11 works b/c C can't be 10.

It doesnt matter what the values of B and D are, but just that they equal 11. Since there is a carry over of 1. C must be equal to 9. Since 9+1+1=11.

The numbers could look like this.

15

+96

111

Ans D