Answer:
Therefore n is divisible by 9 if and only if [tex]a_k+a_{k-1}+.........+a_1+a_0[/tex] is also divisible by 9.
Step-by-step explanation:
Given number is
[tex]n= a_ka_{k-1}.....a_2a_1a_0[/tex]
This means
[tex]n=a_k10^k +a_{k-1}10^{k-1}+.....+a_110^1+a_0[/tex]
Here we need to prove
[tex]a_k+a_{k-1}+......+a_2+a_1+a_0[/tex] is divisible by 9.
We know that
10 ≡ 1 mod 9
It means if 10 divides by 9 the remainder = 1.
[tex]n=a_k10^k +a_{k-1}10^{k-1}+.....+a_110^1+a_0[/tex]
[tex]\Rightarrow n \equiv a_k(1)^k+a_{k-1}(1)^{k-1}+.........+a_1(1)^1+a_0[/tex] mod 9
[tex]\Rightarrow n \equiv a_k+a_{k-1}+.........+a_1+a_0[/tex] mod 9
Therefore n is divisible by 9 if [tex]a_k+a_{k-1}+.........+a_1+a_0[/tex] is also divisible by 9.
And conversely is also true.
Therefore n is divisible by 9 if and only if [tex]a_k+a_{k-1}+.........+a_1+a_0[/tex] is also divisible by 9.
