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.

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 $9$.
