Multiples of 9 Number Puzzle

Date: 03/23/2003 at 15:57:43
From: Brittney
Subject: Multiple of 9

If you take a two-digit number like 21, add the two digits (2+1=3), 
and subtract the answer from the original number (21-3=18), why is the 
final answer always a multiple of 9?

Date: 03/24/2003 at 03:21:21
From: Doctor Jeremiah
Subject: Re: Multiple of 9

Hi Brittney,

A two-digit number can be represented as multiples of 10.

   For example: 21 = 10x2 + 1  and 45 = 10x4 + 5.

   So a number with the left-hand digit of a and the right-hand
   digit of b can be written as 10a + b.

   If you sum the digits you get a + b.

   If you subtract that sum from the original number you get

      (10a + b) - (a + b)
       10a - a + b - b
        9a

   So the value is always 9 times the left-hand digit. And 
   anything times nine is a multiple of nine.

This also works with three-digit numbers.

   A three-digit number is 100a + 10b + c.

   The sum of the digits is a + b + c.

   If we subtract them we get:

      (100a + 10b + c) - (a + b + c)
       100a - a + 10b - b + c - c
        99a + 9b

   And that is also evenly divisible by nine.

In fact this works for all numbers of any length.

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