Why the sum of the digits of $9k+a$ (where $k,a\in \mathbb{N}$, and $0<a<8$) is equal to $9l+a$ (where $\mathbb{N}\ni l< k$)?

Question

The sum of the digits of integers that are multiples of $9$ (let's name them $9k$) always add up to another multiple of $9$ (let's call it $9l$). However, I have observed that the sum of the digits of such multiples plus another integer (let's call it $a$, smaller than $9$ in order not to reach the next multiple) add up to another smaller multiple of $9$, $9l$, plus that added integer $a$.

I see this is obvious when the addition only changes one digit (e.g. $5463+6=546\mathbf{9}$, where the sums of the digits of each number equals the sum of the digits of the whole sum; in this case the sum operation has changed the count from $5+4+6+3=18=9\cdot 2=9l$ to $5+4+6+\mathbf{9}=18+6=9\cdot 2 + 6=9l+a$, however I am not seeing it as clearly when more digits are changed (e.g. $5463+7=54\mathbf{70}$; in this case the sum operation has changed the count from $5+4+6+3=18=9\cdot 2=9l_1$ to $5+4+\mathbf{7}+\mathbf{0}=9\cdot 1 +7=9l_0+a$, where it happens $l_0=1<l_1=2$).

Answer 1

Just to do it backward:

Let the digits of the number be, in order from right to left, $a_0, a_1, .... a_n$.

So $a_n + a_{n-1} + ..... + a_1 + a_0 = 9l + a$.

$a_n + a_{n-1} + ....... + 9a_1 + a_1 + a_0 = 9l + 9a_1 + a$

$a_n + a_{n-1} + ...... + 10a_1 + a_0 = 9(l + a_1) + a$

$a_n + a_{n-1} + .......99a_2 + a_2 + 10a_1 + a_0 = 9(l+a_1) +99a_2 + a$

$a_n + a_{n-1} + ..... 100a_2 +10a_1 + a_0 = 9(l+a_1 + 11a_2) + a$

.... and so one

$10^na_n + 10^{n-1} + ..... 100a_2 + 10a_1 + a_0 = 9(l+a_1 + 11a_2 + .....+1111...1a_{n-1} + 1111...11a_n) + a$.

But $10^na_n + 10^{n-1} + ..... 100a_2 + 10a_1 + a_0$ is our original number!

So $9k +a = 9l + a$ and $k = (l+a_1 + 11a_2 + .....+1111...1a_{n-1} + 1111...11a_n)$

Answer 2

Fact: $ 10^ i \equiv 1 \pmod{9} $. (Do you see why?)

So, if $ n = \overline{n_k n_{k-1} \ldots n_2 n_1 n_0}$, then
$$\sum_{i=0}^k n_i \times 10^i \equiv \sum_{i=0}^k n_i \times 1 \pmod{9}. $$

Hence, if $ n = 9k + a$, then $ \sum n_i = 9l + a $.