Question

Prove that a positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. [See Exercise 28]

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 ${\mathit{a}}_{\mathbf{n}}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{a}}_{\mathbf{1}}{\mathit{a}}_{\mathbf{0}}$.

The number will be given by:

role="math" localid="1659420751609" $\mathbf{\sum }_{\mathbf{k}\mathbf{-}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{k}}{\mathbf{10}}^{\mathbf{k}}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{-}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{k}}\mathbf{(}{\mathbf{10}}^{\mathbf{k}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{+}\mathbf{\sum }_{\mathbf{k}\mathbf{-}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{k}}$

Proof using Definition:

The general form of number is given by:

$\sum _{\mathrm{k}-0}^{\mathrm{n}}{a}_{\mathrm{k}}{10}^{\mathrm{k}}=\sum _{\mathrm{k}-0}^{\mathrm{n}}{a}_{\mathrm{k}}({10}^{\mathrm{k}}-1)+\sum _{\mathrm{k}-0}^{\mathrm{n}}{a}_{\mathrm{k}}$

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.

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.