The greatest common divisor of three positive integers 90ABC17, 79ABC and 491ABC4 is  ≥ 2, where each of A, B and C represents a different base 10 digit from 0 to 9.

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

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

90ABC17 = 9000017 + ABC x 10²
491ABC4 = 4910004 + ABC x 10¹
  79ABC =   79000 + ABC x 10⁰

491ABC4 x 10¹ = 49100040 + ABC x 10²
(49100040 + ABC x 10²) - (9000017 + ABC x 10²) = 40100023
= (547 x 73309)


79ABC x 10¹ = 790000 + ABC x 10¹
(4910004 + ABC x 10¹) - (790000 + ABC x 10¹) = 4120004
= (547 x 2² x 7 x 269)


79ABC x 10² = 7900000 + ABC x 10²
(9000017 + ABC x 10²) - (7900000 + ABC x 10²) = 1100017
= (547 x 2011)


Therefore the greatest common divisor, the common factor, of 90ABC17, 491ABC4 and 79ABC is 547.


79000 mod 547 = 232
547 - 232 = 315
Thus, 79315 mod 547 = 0.

79315 + 547 = 79862.

79862 + 547 > 79999

Therefore, (A, B, C) is (3, 1, 5) or (8, 6, 2).

Posted by Dej Mar on 2008-09-19 00:53:30