Show that $n-m$ is a multiple of 9 when $n$ and $m$ have same digits I have just proved the divisibility rule for 3 and 9.
Let $n\in\mathbb{N}$. Let $m$ be a number that appears when you shuffle the digits in $n$. Show that $n-m$ is a multiple of 9.
Can anyone offer some help?
 A: I don't want to get drowned in notation. So let's just look at three digit numbers and you can figure out the general case.
The most general three digit number would be
$n = (d_3 d_2 d_1)_{10} = 100d_3 + 10d_2 + d_1$.
Let's let $S(n) = d_1+d_2+d_3$ denote the digit sum of $N$.
Now
\begin{align}
   n - S(n)
   &= (100d_3 + 10d_2 + d_1)  - (d_1+d_2+d_3)\\
   &= (100d_3 - d_3) + (10d_2 - d_2) + (d_1 - d_1)\\
   &= 99d_3 + 9d_2\\
   &= 9(11d_3 + d_2)
\end{align}
So 9 divides n - S(n) 
Hence 9 divides (m - S(m)) - (n - S(n)) = (S(n) - S(m)) + (m - n)$.
If $m$ is just $n$ with the digits shuffled, then $S(m) - S(n) = 0$.
In which case 9 divides $n - m$.
A: The digit sum $s$ of $n$ and $m$ is the same. Since we have $n \equiv s\pmod 9$ and $m \equiv s\pmod 9$ we get
$$n-m \equiv s-s \equiv 0 \pmod 9,$$
so $9\mid n-m$ .
A: We have
$$10\equiv 1 \pmod9.$$
By induction we have
$$10^n\equiv 1 \pmod9$$
for any $n\in\mathbb N\cup\{0\}$.
Using this you should be able to show $$10^k\equiv10^l\pmod9$$
for any $k,l\in\mathbb N\cup\{0\}$.
Can you get 
$$a\cdot10^2+b\cdot10+c \equiv b\cdot10^2+c\cdot10+a \pmod 9$$
from this? Can you generalize this to an arbitrary permutation?
