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### 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/
```
Associated Topics:
Elementary Division
Elementary Multiplication
Elementary Puzzles
Elementary Subtraction
Middle School Algebra
Middle School Division
Middle School Puzzles

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