Question

3. Suppose that a positive integer is written in decimal notation as n = akak-1… a2a1a0 where 0 ai 9. Prove that n is divisible by 9 if and only if the sum of its digits ak + ak–1 + … + a1 + a0 is divisible by 9.

Answer 1

Answer:

Therefore n is divisible by 9 if and only if  $a_k+a_{k-1}+.........+a_1+a_0$ is also divisible by 9.

Step-by-step explanation:

Given number is

$n= a_ka_{k-1}.....a_2a_1a_0$

This means

$n=a_k10^k +a_{k-1}10^{k-1}+.....+a_110^1+a_0$

Here we need to prove

$a_k+a_{k-1}+......+a_2+a_1+a_0$ is divisible by 9.

We know that

10  ≡ 1 mod 9

It means if 10 divides by 9 the remainder = 1.

$n=a_k10^k +a_{k-1}10^{k-1}+.....+a_110^1+a_0$

$\Rightarrow n \equiv a_k(1)^k+a_{k-1}(1)^{k-1}+.........+a_1(1)^1+a_0$ mod 9

$\Rightarrow n \equiv a_k+a_{k-1}+.........+a_1+a_0$ mod 9

Therefore n is divisible by 9 if $a_k+a_{k-1}+.........+a_1+a_0$ is also divisible by 9.

And conversely is also true.

Therefore n is divisible by 9 if and only if  $a_k+a_{k-1}+.........+a_1+a_0$ is also divisible by 9.