Divisibility Test for 9: How and Why is it so?
Oct. 31, 2017

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.