Trick for Numbers Divisible by 3 or 9

Date: 02/24/98 at 10:39:33
From: Leon Planken
Subject: Division by 3 and by 9

A lot of people know the trick that to see if a number is divisible by 
3 or by 9, the figures should be added, and if the result is divisible 
by 3 or by 9, the original number was too.

I'm looking for the proof.

Thanks in advance,
Leon Planken

Date: 02/24/98 at 10:54:07
From: Doctor Sam
Subject: Re: Division by 3 and by 9

Hi Leon,

The proof depends upon the observation that 9, 99, 999, 9999, ... etc. 
are all multiples of 3 and 9 and also 1 less than a power of 10.

Any number can be written as a sum of powers of tens.  I'll illustrate 
with a four digit number but the proof is easy to extend to an n-digit 
number.

   abcd = a(10^3) + b(10^2) + c(10) + d

Rewrite each power:

        = a(999+1) + b(99+1) + c(9+1) + d

Rearrange the terms:

        = a(999) + b(99) + c(9) + [a + b + c + d]

The first three terms are divisible by 3 and by 9, so

   abcd                                 a+b+c+d
   ---- =   A whole number quotient +   -------
    9                                      9

The division will be exact if the last division, the sum of the digits 
by 9, is exact. The proof actually shows even more. Not only is a 
number divisible by 3 or by 9 if the sum of its digits is, but the 
remainder of the division is the same as the remainder of the division 
of the sum of the digits.

Example:   
 
   9876543/9 has the same remainder as (9+8+7+6+5+4+3)/9, which is 6.

Okay?

-Doctor Sam, The Math Forum
Check out our web site http://mathforum.org/dr.math/