-=> Quoting Ron Mcdermott to Charles Beams <=-
 CB>Start with a number, cross off the last digit, subtract twice that 
 CB>deleted last digit from the remaining portion of the number.  Repeat 
 CB>the process.  as needed.  if the remainder is 0 or any other 
 CB>multiple of 7, the number is divisible by 7.
 
 CB>Example...
 
 CB> 2156  Start with this number, cross off the six, leaving...
 
 CB> 215  but now subtract 2 x 6 (12)...
 CB> -12
 CB> ___
 CB> 203  Repeat the process - drop the 3...
 
 CB> 20   and double the 3 (6) to subtract...
 CB> -6
 CB> __
 CB>  14  remainder is 14, a multiple of 7, so 2156 is divisible by 7.
 RM> Ok... Now WHY does this work?
I was wondering that.  I know a more complicated rule for determining
the remainder when dividing by 7 that works because 100-2 is divisible
by 7.  This rule doesn't give the correct remainder for non-divisible
cases, it only tests yes or no for divisibility.  I was going skip
figuring out how it works until I got to your challenge.
The method is equivalent to the assertion that if 10a+b is divisible
by 7, then so is a-2b.  Which is equivalent to saying for integer
variables, 10a+b=7c, a-2b=7d+e, prove e is zero.  adding twice the
first equation to the second equations shows 21a=7*(2a+d)+e, so e must
be divisible by 7 but is in the range 0-6 by construction, so e is 0 and
a-2b=7d is divisible by 7.  QED.  (Not ALL Americans learn math poorly)
 
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