<s>[INST] How do I write out this proof? The following works with any number greater than the number 9. 
How do I write the following examples out in a proof?
Take the two digits in your age for example & add then add those two numbers. 
So let's say you're 28.  2+8=10 Now subtract that total from your age: 
28-10=18   now add those two digits.
1+8= 9
Alright let's try 35
3+5=8
35-8=27
2+7= 9
Let's just try the number 42
4+2=6   So now subtract that number from 42
32-6=36  Now add those two. 
3+6= 9
I just decided to try 1,2345
1+2+3+4+5=15
Subtract that from the original number:
12345-15=12330  add that last number up:
1+2+3+3+0= 9
Okay, let's do a harder one.
456789
4+5+6+7+8+9=39
456789-39=456750
now do it again with that last number:
4+5+6+7+5+0=27
and again:
2+7= 9
Now I'm going to type a random number in a calculator.
1563248
1+5+6+3+2+4+8=29  subtract that number from the original.
1563248-29=1563219  Now add up those numbers
1+5+6+3+2+1+9=27
2+7=9
So how do I write this out in a proof?
 [/INST] the main point here is that the remainder when a number is divided by $9$ is the same as the remainder when the sum of its digits is divided by $9$. in particular this means that if $\sigma(N)$ is the sum of digits (decimal) of $N$ then $N-\sigma(N)$ is divisible by $9$.
the number may be written as:
$$
N = \sum_{k=0}^n d_k10^k
$$ 
and 
$$
\sigma(N)= \sum_{k=0}^n d_k
$$
so
$$
N-\sigma(N) = \sum_{k=1}^n d_k(10^k-1)
$$
now you need the fact that $10^k-1$ is always divisible by $9$. can you prove this?
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