Divisibility by 9/Proof 1

Theorem

A number expressed in decimal notation is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

That is:

$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $9$

$a_0 + a_1 + \cdots + a_n$ is divisible by $9$.

Let $N$ be divisible by $9$.

Then:

$\ds N$

$\equiv$

$\ds 0 \pmod 9$

$\ds \leadstoandfrom \ \ $

$\ds a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$

$\equiv$

$\ds 0 \pmod 9$

$\ds \leadstoandfrom \ \ $

$\ds a_0 + a_1 1 + a_2 1^2 + \cdots + a_n 1^n$

$\equiv$

$\ds 0 \pmod 9$

as $10 \equiv 1 \pmod 9$

$\ds \leadstoandfrom \ \ $

$\ds a_0 + a_1 + \cdots + a_n$

$\equiv$

$\ds 0 \pmod 9$

$\blacksquare$

1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 2.6$. Algebra of congruences: Example $41$