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

 

Sorry about the garbled tables (looked ok before post). I won't cut and paste anymore.

Upper letters uppercase, lower letters lowercase: A=5, B=0, C=1, D=6, a=6, b=1, c=0, d=7

The ones column gives a small set of relationships between A, c, a, and C (Since A=a+/1 likewise for C).

The 100s and 10,000s columsn give a set of relationships between A, C, a, and d (and c, D). Combining with 1's column there are only 16 possibilities (power of 2 - cool)

(Table excluded to avoid UML nightmare)

The 10's column has 4 sets for B and b (B+B=b and abs(B-b)=1):

B: 0  1  8  9

b: 1  2  7  8

Using the 1000's column and combining all tables give 10 possibilities. Don't need to compare anymore because only one combination has 5 different digits. (Plenty more comparisons are available bu 50166107 works so no need to try and eliminate it).

Posted by Rajal on 2004-08-11 17:57:22