Easier way to prove a really simple theorem While I was in class I noticed that if we take a number $n$ and we subtract from $n$ his digits we always get a multiple of $9$. More specifically:
Any number can be expressed as $10p + b$ (for example $42$ is $4\cdot 10 + 2$); the digits of $n$ will be $p$ and $b$ so:
$$10p + b-(p+b)=9p$$
Q.E.D
I'm asking if there is another way to prove this little theorem for example without using the $10p + b-(p+b)$ notation
 A: The remainder of a number after division by 9 is the same remainder as dividing the sum of its digits by 9, therefore the remainder of the difference is always zero.
Probably the shortest way to prove it is to use modular arithmetic. Suppose the number $X$ has decimal digit expansion $\overline{a_n a_{n-1} ... a_1 a_0}$, then
$ X = a_n \cdot {10}^n + a_{n-1} \cdot 10^{n-1} + ... + a_1 \cdot 10 + a_0 \equiv a_n + a_{n-1} + ... + a_1 + a_0 \text{ (mod 9)},$
since $10 \equiv 1$ and so $10^n \equiv 1^n = 1$ modulo $9$.
Therefore $X-(a_0+a_1+...+a_n)$ is zero modulo $9$, i.e. divisible by it.
A: Look-see proof, maybe?
X X X X X  <--- get rid of this row
* * * * *       (subtract left digit)     * * * * *
* * * * *                                 * * * * *     <--- you have a nine-
* * * * *                                 * * * * *          by-something
* * * * *                                 * * * * *          rectangle
* * * * *                                 * * * * *
* * * * *                                 * * * * *
* * * * * X   <--- and this column        * * * * *
* * * * * X        (subtract right        * * * * *
* * * * * X         digit)                * * * * *

53 = 5(10) + 3                          45 = 53 - 5 - 3 = 5(9)

A: Note that $$10≡1\mod 9.$$
Therefore for any $n \in N$ $$10^n≡1\mod 9.$$  Take any integer of the form $$a_n...a_2a_1a_0=a_n10^n+...+a_210^2+a_110+a_0.$$ Then $$a_n...a_2a_1a_0-(a_n+...+a_2+a_1+a_0)\\=a_n(10^n-1)+...+a_2(10^2-1)+a_1(10-1)$$ is divisible by $9.$  
For similar result:
(1) Take any integer with more than $1$ digit.
(2) Compute the sum of digits in even places and odd places separately.
(3) Obtain the difference of them.
(4) Subtract and add it to the initial number.
At least one of these number is divisible by $11.$
