In this cryptarithm, each letter above the line represents a digit differing by 1 from the digit represented by the same letter below the line. For example, if A=3 above the line, then A=2 or A=4 below the line. All occurrences of a letter on the same side of a line represent the same digit. It is also known that there are total of five digits in the solution.

 ADABA
+CACBA
------
 DBABC

First, I looked at the second column from the right. I found only two possibilities: either the upper B=9 and the lower B=8 [and there was no carry from the first column] or the upper B=0 and the lower B=1 [and 1 was carried over from the first column].

I decided to try out the first possibility. Looking at the middle column, A+C+1 (the carry) should equal A-1, or A+1, or A-1+10 (and carry 1), or A+1+10 (and carry 1). The first case is impossible. The second one implies C=0, so the lower C=1, but then the first column doesn't work out: A+A should be even. The third one gives upper C=9, so lower C=8, but we already have lower B=8. The fourth case is also impossible. Thus, all cases fail, so upper B isn't 0.

Trying out the second possibility, we need to carry 1 from the first column, so upper A=5, 6, 7, 8, or 9, and lower C=0, 2, 4, 6, or 8. Looking now at the third column, only upper A=5 or upper A=6 do not produce repeated numbers or inconsistent results, and our sum looks like 5D505+15105=D1610 or 6D606+16106=D1712. The former works if D=6; the latter cannot work, so the solution is 56505+15105=71610.

Posted by Oskar on 2004-08-11 13:40:31