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.

"Recreations in the theory of numbers", by Albert H. Beiler, page 297, n. 27. I was thinking in submit it.

Edited to say:

The funniest thing is that looking again the book, there there is only the answer, not the whole solution, and a month ago I solved it entirely... but canŽt remember how. IŽll try again.

Edited on September 18, 2008, 2:19 pm

Posted by pcbouhid on 2008-09-18 13:39:04