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Path: utzoo!linus!decvax!cca!ima!ism780!jim
From: jim@ism780.UUCP (Jim Balter)
Newsgroups: net.math
Subject: Re: divisibility by 7
Message-ID: <26@ism780.UUCP>
Date: Thu, 25-Aug-83 13:44:00 EDT
Article-I.D.: ism780.26
Posted: Thu Aug 25 13:44:00 1983
Date-Received: Sat, 27-Aug-83 16:32:03 EDT
Lines: 18

That's cute!  And it's easy to prove.  It is equivalent to saying that
n mod 7 = 0  iff  (floor(n/10) - 2*(n mod 10)) mod 7 = 0.

Say n = 7*x = 10*a + b (a and b integers).
Then floor(n/10) - 2*(n mod 10) = a - 2*b.
By algebra  a - 2*b = 21*a - 14*x = 7*(3*a-2*x).
So, n mod 7 = 0  iff  x is an integer  iff  a - 2*b is divisible by 7.

I like your sample (but hardly random) number.  Note for instance

	142857*1 =  142857
	142857*3 =  428571
	142857*2 =  285714
	142857*6 =  857142
	142857*4 =  571428
	142857*5 =  714285

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