Divisibility by 9

Date: 03/22/2002 at 08:18:34
From: Tonya Wagner
Subject: Divisibility by 9

Explain why an integer that is rearranged and then subtracted from 
itself is always divisible by nine.

I can show numerous examples and have looked through the math 
resources that I have, but I can not seem to come up with a good 
explanation.

Date: 03/22/2002 at 08:41:16
From: Doctor Paul
Subject: Re: Divisibility by 9

A number is divisible by nine if and only if the sum of the digits is 
divisible by nine.

This follows from the fact that a number is divisible by three if and 
only if the sum of the digits is divisible by three and that a number 
is divisible by nine if and only if it is divisible by three twice.

More about that here:

   Why do Divisibility Rules work?
   http://mathforum.org/k12/mathtips/division.tips.html   

Now, a permutation of the digits will not change the sum of those 
digits (since addition of real numbers is a commutative operation).

Suppose that x = a - b, and that the number b is obtained by simply 
permuting the digits of a.  Then a is either going to be:

a multiple of nine (0 mod 9)
1 more than a multiple of nine (1 mod 9)
2 more than a multiple of nine (2 mod 9)
3 more than a multiple of nine (3 mod 9)
4 more than a multiple of nine (4 mod 9)
5 more than a multiple of nine (5 mod 9)
6 more than a multiple of nine (6 mod 9)
7 more than a multiple of nine (7 mod 9)
8 more than a multiple of nine (8 mod 9)

Suppose that a is y more than a multiple of nine (a = y mod 9), where 

   y = 0, 1, 2, ..., or 8.  

Then the sum of the digits of a will be y more than a multiple of 
nine.

Since b is just a permutation of the digits of a, the sum of the 
digits of b will also be y more than a multiple of nine.

Notice that in general, a - b will be divisible by nine if and only if 
they are congruent to the same number mod 9. I.e., if a is congruent 
to 7 mod 9, but b is congruent to 5 mod 9, then their difference will 
be congruent to (7 - 5) mod 9 = 2 mod 9 and hence will not be 
divisible by nine (a number that is two more than a multiple of nine 
is obviously not divisible by nine).

Notice that what we have above is that regardless of the choice of a, 
that the sum of the digits of a will be the same as the sum of the 
digits of b and hence they will always be congruent to the same number 
mod 9. Thus their difference will be congruent to 0 mod 9 and hence 
will be divisible by nine.

I hope this helps. Please write back if you'd like to talk about this 
some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/