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ABCD's Of Greatest Divisor (Posted on 2009-09-24) Difficulty: 3 of 5
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.

See The Solution Submitted by K Sengupta    
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Solution Solution | Comment 2 of 3 |

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
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