Problem Solving (PS) | Re: In the correctly worked computation above A, B, C and D repr

It can be done in a more simpler way as follows:

1. AB + BA = (10A+B)+(10B+A) = 11(A+B) , i.e the sum is always a multiple of 11
2. Since on the other side CDC = AB+BA , therefore CDC is also a multiple of 11
3. The only 3 digit multiple of 11 in the form of CDC is 121
4. Now for 11(A+B) = 121, A+B = 11 therefore A and B can be 6 and 5

So the sum is A+B+C+D= 6+5+1+2= 14

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