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2 pandigital equations (Posted on 2014-09-05) Difficulty: 3 of 5

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.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts First thoughts | Comment 3 of 4 |
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.

  Posted by Steve Herman on 2014-09-05 16:33:57
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