Question

Prove the 'rule of 9': an integer is divisible by 9 if and only if the sum of its integers is divisible by 9.

Answer 1

Answer:

Rule of 9 : an integer is divisible by 9 if and only if the sum of its integers is divisible by 9.

Proof :

Let us consider a number x such that,

$a=a_n......a_3a_2a_2$

Where, $a_0, a_1, a_2......,a_n$ are the the digits of the number x,

So, we can write,

$x=a_0+a_1\times 10 + a_2\times 10^2 +a_3\times 10^3..........a_n\times 10^n$

Let,

$S=a_0+a_1+a_2+a_3+.......a_n$

$x-S=a_1(10-1)+a_2(10^2-1)+a_3(10^3-1)+.......a_n(10^n-1)$

Since, a number is in the form of $10^k-1$, where k is an positive integer, is always divisible by 9,

⇒ $a_1(10-1), a_2(10^2-1), a_3(10^3-1),.......a_n(10^n-1)$ are divisible by 9.

⇒ x-S is divisible by 9,

⇒ If S is divisible by 9 ⇒ x must divisible by 9,

Or if x is divisible by 9 ⇒ S must divisible by 9.

Hence, proved...