Why is this number a multiple of 9 Suppose you have two numbers, $n$ and $m$, such that $n$ and $m$ have the same digits, just in different orders. I have been told that $g:=n-m$ is always a multiple of $9$, but I can't seem to figure out why. If $n$ and $m$ are both multiples of $9$ then I know that their difference is, but this is not true in general.
 A: Since
$$ n\equiv a+b+c+\cdot+p \pmod{9}, \\m\equiv p+a+b+c+\cdot+o \pmod{9}  $$
you have
$$ n-m\equiv 0 \pmod{9}$$
or
$n-m$ is dividable by $9$.
A: You need to edit things in your question.
The answer to your question is simple
Any number when divided by 9 leaves the same remainder as it's sum of digits do
Since n and m leave same remainder on dividing by 9 hence their difference is divisible by 9
I had to edit my answer in response to asker's edit
A: You can work with $10^{m+r}-10^m=10^m(10^r-1)$ and $9=10-1|10^r-1$, or alternatively $10\equiv 1 \bmod 9 \implies 10^r\equiv 1\bmod 9$. However you work, the sum of the digits of each number modulo $9$ is equal to the remainder modulo $9$. So if the two numbers have the same digit sum, their difference is divisible by $9$.

A: The following numbers are not multiple of $9$
$abcd - bcda = 1000 a + 100 b + 10 c + d - (1000 b + 100 c + 10 d + a)=\\=999 a - 900 b - 90 c - 9 d = 9 (111 a - 100 b - 10 c - d)$
and
${abcde}-{edabc}=10000 a+1000 b+100 c+10 d+e-(10000 e+1000 d+100 a+10 b+c)=9900 a + 990 b + 99 c - 990 d - 9999 e=99 (100 a+10 b+c-10 d-101 e)$
Consider $m-n$ where $m$ and $n$ have the same digits only in different order, that is
$m=m_0+10 m_1+10^2 m_2+10^3 m_3+\ldots+10^{k-1} m_{k-1}+ 10^k m_k$
$n=m_0 10^{k_1}+m_1 10^{k_2}+m_2 10^{k_2}+m_3 10^{k_3}+\ldots+m_{k-1}10^{k_{k-1}}+m_k 10^{k_k}$
where $\{k_i\}$ is a permutation $\sigma$ of $\{k\}$
$m-n$ has decimal representation as
$m_0(1-10^{k_1})+m_1(10-10^{k_2})+\ldots+m_k(10^k-10^{k_k})$
any parenthesis is $0$ or a multiple of $9$ like $90,900,990,\ldots$ because they are  the differences of powers of $10$ and thus they have just $9$ and $0$ as digits.
This prove that the difference between a number and another number with the same digits after a permutation is always a multiple of $9$
Hope it helps
