perplexus.info :: Just Math : ABC's Of Greatest Divisor

ABC's Of Greatest Divisor (Posted on 2008-09-18)

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.

Submitted by K Sengupta

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Solution: (Hide)

(A, B, C) = (3, 1, 5) or (8, 6, 2)

For an explanation, refer to the respective solutions submitted by pcbouhid and Dej Mar in the comments.

From: The Sphinx Collection, No.54 (Sphinx - Feb 1934 #10a - Page 25)

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
solution	Dej Mar	2008-09-19 00:53:30
Remembered!! I´m still in the game?	pcbouhid	2008-09-18 15:56:33
No merit in cheating...	pcbouhid	2008-09-18 13:39:04

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