Modular arithmetic (casting out 9's)

[playa007]

Homework Statement

Let m be a positive integer and m' be an integer obtained from m by rearranging its digits. Prove that m-m' is a multiple of 9

Homework Equations

Casting out 9's method

The Attempt at a Solution

So I found that by applying the casting out 9's method on m and m'; the values are the same. This means that when one divides m by 9 and m' by 9; it is the same remainder. I'm wondering how do I use this to prove that 9 divides m-m'

 

[tiny-tim]

Hi playa007!

So I found that by applying the casting out 9's method on m and m'; the values are the same. This means that when one divides m by 9 and m' by 9; it is the same remainder. I'm wondering how do I use this to prove that 9 divides m-m'

erm … m = 9a + r, m' = 9a' + r, so m - m' = … ?

 

[HallsofIvy]

Also (tiny-tim is exactly right), if a number is written, say m= "abcd"= 1000a+ 100b+ 10c+ d, then, rearranging, m'= "cadb"= 1000c+ 100a+ 10d+ b and m- m'= 1000a+ 100b+ 10c+ d- 1000c- 100a- 10d- b= (1000-100)a+ (100-1)b+ (10-1000)c+ (1- 10)d= 900a+ 99b- 990c- 9d= 9(100a+ 11b- 110c-d), a multiple of 9.

In our "base 10 numeration system", the coefficient of each digit is a power of 10, of course, so no matter how you rearrange the digits, subtracting, the coefficient of each digit is "a power of 10 minus another power of 10" which is always a multiple of 9.