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You can prove that any multiple of 9's digits add up to a multiple of 9 (and thus if you keep doing it, you'll eventually get 9)
Proof:
The "digits" of any number 9 can be expressed as:
dn-1*10^n + dn-2 * 10^(n-1) + ... + d1*10^1 + d0 * 10^0
10^n -1 is 999...9 (has n 9s), so we can write this as
dn-1*(999..9 + 1) + dn-2 * (999...9 + 1) + ... + d1*(9 + 1) + d0 * (1)
multiply it out, and you get two groups of terms. The first group is when you multiply by the 999....9 term, and those are automatically divisible by 9, since 9999..9 is always divisible by 9 (proof of that: it's 111...1 * 9)
The remaining terms are simply the digits:
9dn-1 + dn-2 + ... + d1 + d0
This is the sum of the digits. Thus, if the sum of the digits is divisible by 9, then the entire term is divisible by 9, and since the term is a representation of n, then n is divisible by 9.
You may also be wondering about the validity of the sum of multiples of x are divisible by x. Here's my proof:
if each number added is divisible by x, then you can write it as follows:
x is what each is divisible by, and k1 through kn are integers
x*k1 + x*k2 + ... + x*kn = x(k1 + k2 + ... + kn). k1 + k2 + ... + kn is an integer, and so x*(k1 + k2 + ... + kn) is an integer that's divisible by x.
Here's an example of working through the proof:
83367
= 8*10^4 + 3 * 10^3 + 3 * 10^2 + 6 * 10^1 + 7 * 10^0
= 8*(10000) + 3*(1000) + 3*(100) + 6*(10) + 7*(1)
= 8*(9999+1) + 3*(999+1) + 3*(99+1) + 6*(9+1) + 7*1
= (8*9999 + 3*999 + 3*99 + 6*9) + (8 + 3 + 3 + 6 + 7)
--------TERM 1--------------- -----TERM 2------
Term 1 is divisible by 9, because each number in it is divisble by 9, so if term 2 is divisible by 9, then 83367 is divisible by 9.
8+3+3+6+7 = 27
27 = 9*3, so it's divisible by 9.
Hopefully by working through an example of my proof, you can see that a number being divisible by nine is soley contingent on its digits adding to 9. Note that this will only work in base 10!
There is a similar proof that if the digits add up to a multiple of 3, they are divisible by 3. You can try to work that out yourself, if you want (it's basically the same proof).
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