Sum of the Digits of Multiples of 9

Date: 04/08/2001 at 20:10:07
From: Jennifer Springer
Subject: The number nine

Why do the digits of nine times a number add up to 9? For example:

     9 x 4  = 36,   3 + 6 = 9

     9 x 10 = 90,   9 + 0 = 9

And why does this not work for 9 x 11?

     9 x 11 = 99,   9 + 9 = 18

Why not 9?

Date: 04/08/2001 at 23:08:46
From: Doctor Peterson
Subject: Re: The number nine

Hi, Jennifer.

First, let me point out that this observation can be extended to 
larger numbers in two ways: 

(1) As the rule for identifying a number divisible by 9. See our 
    Divisibility Rules FAQ at:

    http://mathforum.org/dr.math/faq/faq.divisibility.html   

(2) In the method of "casting out nines" to determine whether any 
    addition or multiplication is correct:

    Casting Out Nines to Check Arithmetic
    http://mathforum.org/library/drmath/view/55926.html   

In the simple case, we easily prove your rule algebraically. Suppose x 
is any single-digit number; then

     9x = (10-1)x

        = 10x - x

This almost looks like the expanded form of a decimal number, except 
that the digits would have to be x and -x, and we can't have negative 
digits. So we can "borrow," adding 10 to the units digit and taking 1 
from the tens digit:

     9x = 10x - x

        = 10(x-1) + (10-x)

This tells us that the tens digit is x-1, and the units digit is 10-x. 
In your example with x = 4, these are 4-1 = 3 and 10-x = 6 (making the 
number 36).

But what is the sum of the digits?

     (x-1) + (10-x) = 9

Now, why doesn't this work if x > 10? Because then x-1 and 10-x would 
not both be single digits, so there would be additional carries or 
borrows needed, which would mess up the trick a bit - but not 
completely. The sum of the digits will still be a MULTIPLE of 9 - and 
that's the trick for recognizing whether a number is divisible by 9.

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