hey guys,

I was wondering how to prove this

Problem:

Let m be a positive integer and let n be an integer obtained from m by rearranging the digits of m in some way. (for example 381762 reaaranging 378126) prove m-n is divisble 9

thanks :D

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- Aug 14th 2011, 12:35 AMthimellecxDivisible by 9
hey guys,

I was wondering how to prove this

Problem:

Let m be a positive integer and let n be an integer obtained from m by rearranging the digits of m in some way. (for example 381762 reaaranging 378126) prove m-n is divisble 9

thanks :D - Aug 14th 2011, 12:45 AMpickslidesRe: Divisible by 9
If the digits of any number sum to a multiple of nine then it is also divisible by nine. Rearranging them won't change this.

(Smirk) - Aug 14th 2011, 12:46 AMMooRe: Divisible by 9
Hello,

Well a simple way to do it is to remember that, S(.) being the sum of the digits of a number, $\displaystyle m\equiv S(m) \bmod 9$ (this can be easily proved with congruences, writing the number as powers of 10). And since $\displaystyle S(m)=S(n)$ (rearranging the digits won't change their sum !), then $\displaystyle m-n \equiv S(m)-S(n) \equiv 0 \bmod 9$ - Aug 14th 2011, 12:46 AMMooRe: Divisible by 9
- Aug 14th 2011, 12:48 AMthimellecxRe: Divisible by 9
what if the sum of its digits is not multiple to nine like 721 and 172

- Aug 14th 2011, 12:57 AMpickslidesRe: Divisible by 9
- Aug 14th 2011, 07:13 AMHallsofIvyRe: Divisible by 9
More generally, if the

**remainder**of the sum of digits of a number, divided by 9, is n, then the remainder of the number, divided by 9, is also n. Since rearranging the digits in a number does not change the sum of digits, it also does not change the remainder when divided by n. Let A be the original number, and let n be its remainder when divided by 9: A= 9k+ n for some integer k. If B is any rearrangement of the digits of A, then B= 9j+ n for some integer j. A-B= (9k+ n)- (9j+ n)= 9(k- j).

You can prove that statement about the remainders by looking at the numbers expansion in powers of 10:

$\displaystyle A= a_0+ 10a_1+ 100a_2+ \cdot\cdot\cdot+ 10^ma_m$

$\displaystyle = a_0+ (9+ 1)a_1+ (99+ 1)a_2+ \cdot\cdot\cdot+ (9(1+ 10+\cdot\cdot\cdot+ 10^{m-1})a_m$

$\displaystyle = (a_0+ a_1+ a_2+ \cdot\cdot\cdot+ a_m)+ 9(a_1+ 11a_2+ \cdot\cdot\cdot (1+ 10^{k-1})a_m)$ - Aug 14th 2011, 09:17 AMMooRe: Divisible by 9
Hm well, isn't that equivalent to what I wrote in #3 ? (Itwasntme) (I'm quite disturbed by your "more generally", not by the fact that you're giving another way to see through the answer)

The congruence gives the remainder, so...

And the OP seems to have understood the solution, given the private messages he's sent me... - Aug 14th 2011, 09:49 AMHallsofIvyRe: Divisible by 9
Yes, you are completely correct! I just did not read your post properly. The "more generally" was referring to PickSlides' post #2.

- Aug 14th 2011, 07:26 PMSammySRe: Divisible by 9
- Aug 15th 2011, 12:12 AMMooRe: Divisible by 9
- Aug 17th 2011, 02:48 AMlivingdreamRe: Divisible by 9
well sum adds up to 27 so the number is divisble by 9 and 3 and even if you rearrange it adds to same