Question 198255
Since A+C results in a two digit result, and addition can't result in more than a carry of 10. Thus, A must be 1.

This also applies to B+D. That is, you can only carry a maximum of 1 out of that addition.

{{{1B + CD = 111}}}
In order to have the high order addition result in "11", C must be 9. That way, the sum of 1+9 plus the carry from the low order addition will yield 1+9+1 = 11.

So now we know that B+D must equal 11 too. There are several values that can be used to get that result. Since C is already used, B and D can;t be 9. But any of several pairs of values for B + D can result in a sum of 11 works.
For instance B,D could be 8,3 or 7,4 or 6,5