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.

N1 = 90ABC17 = 9,000,017 + 100*ABC 

N2 = 79ABC = 79,000 + ABC

N3 = 491ABC4 = 4,910,004 + 10*ABC

(10*N3 - N1) = 40,100,023 = 547*73,309

(N3 - 10*N2) = 4,120,004 

73,309 is not a factor of 4,120,004, but 547 is.

So mdc(N1, N2, N3) = 547

 Multiples of 547 that starts with 79 is 145*547 = 79,315 and  146*547 = 79,862.

So, ABC = 315 or 862 and the mdc is 547.

Posted by pcbouhid on 2008-09-18 15:56:33