-=> Quoting Ron Mcdermott to William Lipp <=-
 RM> Ok... Now WHY does this work?
 
 WL>The method is equivalent to the assertion that if 10a+b is divisible
 WL>by 7, then so is a-2b.  Which is equivalent to saying for integer
 WL>variables, 10a+b=7c, a-2b=7d+e, prove e is zero.  adding twice the
 WL>first equation to the second equations shows
 RM> 21a=7*(2c+d)+e
 
 RM> Dividing by 7 yields: 3a=(2c+d)+e/7
 RM> Since a, c, and d are integer, then for left and right side
 RM> to be equal, e/7 must also be integer...
 
 RM> Since we factored out a 7 from the right side originally,
 RM> "e" is the remainder and must be 6 or less....
 
 WL>so e is 0 and a-2b=7d is divisible by 7.  QED.
 RM> I'm impressed...
Thank you.  And thanks for fixing the errors and filling in some
of the details.   I was pleased with the proof, but it isn't
much more complicated than the ancient Greeks' (Pythagoreans')
proof the square root of 2 is irrational. I learned that back in
high school - does that make this "on topic?"
I decided to see if the same argument leads to simple divisibility rules
for other numbers.  The first ones I didn't know simple rules for were
11 and 13.  The same proof outline shows that 11 can be tested by
subtracting the last digit from the rest of the number.  To test 2354,
  235-4 = 231;  23-1=22;  2-2=0; so 2354 is divisible by 11.
To test for divisibility by 13, add 4 times the last digit.  For 2782,
 278+4*2=286;  28+4*6=52; 5+4*2=13  (stuck here - 1+4*3=13)
So 2782 is divisible by 13.
--- Maximus/2 3.01
---------------