Lesson PROOF of divisibility by 3 rule

There is a simple rule how to determine if a number is divisible by three.

Add up the digits and see if the sum of those digits (a much smaller number) is divisible by three.

This rule simplifies finding out if a big number is divisible by three, by reducing it to a small number.
Why is this proposition true?
Consider, for simplicity, a 3 digit number 'abc", such as 321.
Suppose that a+b+c is divisible by 3.
Our number, written as 'abc' is actually
100a+10b+c.
For instance, 321 = 100*3+10*2+1.
Let's rewrite 100a+10b+c as 99a+a + 9b+b + c.
99a+a + 9b+b + c = (99a + 9b) + (a + b + c)
The first part is always divisible by 3 since numbers with all nines are always divisible by 3 (9 = 3*3, 99=33*3, 999=333*3 etc).
That the second part (a+b+c) is divisible by 3, is a given.
So, we have a sum of two parts, both of which are divisible by 3. This sum is, therefore, also divisible by three. Theorem proven.