Question

Prove that a positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.

Short answer

Hence, it is proved that a positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.

Step-by-step solution

Divisibility:

When any integer consists of the digits $a_n{a}_{n-1}....a_1{a}_0$.

The number will be given by:

$\sum _{K=0}^n{a}_K{10}^K=\sum _{K=0}^n{a}_K\left({10}^K-1\right)+\sum _{K=0}^n{a}_K$

Proof using Definition:

The general form of number is given by:

$\sum _{K=0}^n{a}_K{10}^K=\sum _{K=0}^n{a}_K\left({10}^K-1\right)+\sum _{K=0}^n{a}_K$

Now, the given number is of the form:${10}^K-1$.

The first term can be given as 999…….99.

This is divisible by 9.

So, we get the first term is divisible by 9.

Step-3

This implies that:

$\sum _{K=0}^n{a}_K{10}^K$is divisible by 9, if and only if $\sum _{K=0}^n{a}_K$is divisible by 9, which is the sum of the digits.

Hence, it is proved that a positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.