elementary number theory - Observation about multiples of $999$

I think I have observed an interesting pattern in the multiples of $999$. I have observed that the sum of digits of first $30$ multiples of $999$ is $27$. I have read this

but it doesn't prove it. So, is it true that the sum of digits of all multiples of $999$ is a multiple of $27$? Please prove or disprove. What is the minimum multiple of $999$ that has a sum other than $27$?

Answer

Here are some ideas that might help answer the question:

What is the minimum multiple of 999 that has a sum other than 27?

For some integer $k$, $999 \cdot k = 1000k - k$. The digit sum of $999k$ should be greater than $27$ for at least one four-digit number $k$. If we take $k = 9999$, the digit sum of $1000k$ is already $36$, which is much greater than $27$ already, and in fact $9999 \cdot 999$ has a digit sum greater than $27$.

Glaringly obviously, with $k=1001$ we have $999 \ 999$ which has a sum of $54$. This proves that if there is any smaller $k$, it must be three digits or shorter.

But according to Szeto's answer, the sum of digits for any three-digit $k$ is $27$. For one-digit $k$, we can construct a similar argument: with $k = a$, $999k = \overline{a000} - a$. Then we have the following subtraction:

$$\begin{array}{r} &^1a \quad \quad \quad ^10\quad \ \ \ \ ^10 \quad \ \ \ \ ^10\\ -\!\!\!\!\!\!&a\\ \hline &a-1 \quad \quad \quad 9 \quad \quad \ \ 9 \ \ 10-a \end{array}$$

and the sum of digits is $a + 9 + 9 + (10-a) = 27$.

For two-digit $k$, we can construct a similar argument:

$$\begin{array}{r} &a \quad \quad \quad ^1b \quad \quad \quad ^10\quad \ \ \ \ ^10 \quad \ \ \ \ ^10\\ -\!\!\!\!\!\!&a \quad \quad b\\ \hline &a \quad \quad \quad b-1 \quad \quad \ \ 9 \quad \quad 9-a \quad 10-b \end{array}$$

and the sum of digits is $a + (b-1) + 9 + (9-a) + (10-b)= 27$.

Therefore $k = 1001$ should be the smallest $k$.

What still needs to be done is to prove this rigorously for all $k$, instead of just by guessing and checking.