1. A/B=3
2. C/D=9 In both problems ( 1. and 2.) the union of the digits of numbers
A & B (or C & D) is the full set of 9 non-zero decimal digits.
List all existing solutions.
Thanks for the clarification, Ady! In that case:
The sum of all 9 digits are divisible by 9. Therefore, A (mod 9) + B (mod 9) = 0 (mod 9). But A (mod 9) = 3B (mod 9), so 4B (mod 9) = 0. This can only be the case if B is divisible by 9, so A is divisible by 27.
Similarly, C (mod 9) + D (mod 9) = 0. But C (mod 9) = 0, so D is divisible by 9. So C is divisible by 81.