RM> Ok... Now WHY does this work?
LP>I was wondering that.  I know a more complicated rule for determining
LP>the remainder when dividing by 7 that works because 100-2 is divisible
LP>by 7.  This rule doesn't give the correct remainder for non-divisible
LP>cases, it only tests yes or no for divisibility.  I was going skip
LP>figuring out how it works until I got to your challenge.
LP>The method is equivalent to the assertion that if 10a+b is divisible
LP>by 7, then so is a-2b.  Which is equivalent to saying for integer
LP>variables, 10a+b=7c, a-2b=7d+e, prove e is zero.  adding twice the
LP>first equation to the second equations shows 21a=7*(2a+d)+e,
21a=7*(2c+d)+e
LP>so e must be divisible by 7
Dividing by 7 yields: 3a=(2c+d)+e/7
Since a, c, and d are integer, then for left and right side
to be equal, e/7 must also be integer...
LP>but is in the range 0-6 by construction,
Since we factored out a 7 from the right side originally,
"e" is the remainder and must be 6 or less....
LP>so e is 0 and a-2b=7d is divisible by 7.  QED.
Agreed...
LP>(Not ALL Americans learn math poorly)
I'm impressed... While I follow the logic (after some
thought), I would have found it difficult to come up with
the approach independently.....
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 * MR/2 2.26 * OS/2 WARP: The choice of the next generation.
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