The greatest common divisor of five positive integers ABCD, 1920CD41, 496BC3, 872AB76 and 10A25D8 is ≥ 2, where each of A, B, C and D represents a different base 10 digit from 0 to 9.

Determine all possible quadruplet(s) (A, B, C, D) that satisfy the given conditions.

Given the five values have the same common factor, any integral combination must also have that factor.

1920CD41 + 100*872AB76 - 100*ABCD =

  1920CD41
+872AB7600
-   ABCD00
----------
 891207641

10A25D8 + 10*496BC3 - 10*ABCD

 10A25D8
+496BC30
-  ABCD0
--------
 5962538

GCD(891207641, 5962538) = 2131 (prime)
ABCD must be a multiple of 2131: 2131, 4262, 6393, or 8524 Plugging values into the four larger numbers leaves one solution: ABCD=8524

Posted by Brian Smith on 2009-09-24 12:43:50