arithmetic - Looking for a theorem talking about the remainder when a number is divided by 9

I have come across an interesting property of the number 9, which some people call it casting out nines.

This is the property: If any number is divisible by 9, then you can keep adding the digits until you get a 9. For example 9117 is divisible by 9, because 9+1+1+7=18, then from the 18 we can see that 1+8=9. But 9113 is not divisible by 9, because 9+1+1+3=14, and then from the 14 we deduce 1+4=5 which is the remainder when this number, 9113, is divided by 9.

My question is: Is there a theorem that summarises this concept? If the theorem exists, then what is its proof?

Thanks in anticipation.

Answer

It is essentially an extension of the rule that a number is divisible by 9 if and only if the digits in the base ten representation add up to a number divisible by 9. You can prove that the digits of a number have to add up to a multiple of 9 by using the fact that every power of 10 is congruent to 1 modulo 9, viz.\begin{equation} 10^n = 1+ 9\times \sum_{j=0}^{n-1} 10^j. \quad (1)\end{equation}
Hence you have for any natural number $d=\sum_{j=0}^n d_j 10^j$, we have that $$d \cong l \mod 9$$ if and only if $$\sum_{j=0}^n d_j 10^j \cong l\mod 9$$
if and only if $$\sum_{j=0}^n d_j 1^j =\sum_{j=0}^n d_j\cong l \mod 9,$$
where we used (1) to get to the last line.

Now use the fact that $\sum_{j=0}^n d_j 10^j > \sum_{j=0}^n d_j > 0$ and the result should be clear.