**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.