We have studied in our 4th or 5th grade that that if the sum of the digits of a number is divisible by 9 then the number is divisible by nine.
For example:
$72$ is divisible by $9$ because, $ 7 + 2 = 9 $ which is divisible by $9$.
$198$ is divisible by $9$ because, $1+9+8=18$ which is divisible by $9$.
Now, why is it so??
Any number can be written terms powers of $10$:
$ 1234 = 1 * 1000 + 2 * 100 + 3 * 10 + 4 * 1 $
$ 32 = 3 * 10 + 2 * 1 $
In similar fashion, any number can be written as,
$$n = a_1 * 1 + a_2 * 10 + a_3 * 100 +... $$
And,
$1 = 0 + 1,$
$10 = 9 + 1,$
$100 = 99 + 1$, and so on.
So,
$ n = a_1 * (0+1) + a_2 * (9+1) + a_3 * (99+1) ...$
Separating the sums (using associativity),
$or, n = (a_1 * 0 + a_2 * 9 + a_3 * 99 + ...) + (a_1 * 1 + a_2 * 1 + a_3 * 1+ ...)$
$or, n = 9*(a_1 * 0 + a_2 * 1 + a_3 * 11 + ...) + (a_1 + a_2 + a_3 + ...)$
$or, n = 9 * X + Sum of digits$
Now comes the reason.
If $n$ is divisible by $9$, the right hand side should also be divisible by $9$. It is obvious that $9 * X$ is divisible by $9$. And thus, $Sumofdigits$ is forced to be divisible by $9$.
That's where divisibility test for $9$ came from.