If are four distinct digits, how many possible values does have?
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+ C A D C 0 C B A D C C C B A D C D C B A B
Observe the cryptarithm carefully that the first four terms in the sum list some of the possible arrangements of A , B , C and D , where the thousands digit of C D A is 'invisibly' B . Noting that B = 0 , we have A + C + D ≡ 0 m o d 1 0 Since A + C + D ≥ 1 0 , and A + C + D exists in three remaining decimal places, there are 4 leftmost carry digits of the same value C , which is nonzero. So either C = 1 or C = 2 .
Thus, since the sum depends on the variables A and B , there are 1 ⋅ 1 ⋅ 6 = 6 possible choices to substitute A B C D .
Note: I will check to see if there are other accurate/complete approaches for this problem. I believe my solution seems to be less rigorous because there can be other choices of B that can also work.