# Divisibility by 9/Proof 1

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## 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 $9$

- $a_0 + a_1 + \ldots + a_n$ is divisible by $9$.

## Proof

Let $N$ be divisible by $9$.

Then:

\(\displaystyle N\) | \(\equiv\) | \(\displaystyle 0 \pmod 9\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n\) | \(\equiv\) | \(\displaystyle 0 \pmod 9\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a_0 + a_1 1 + a_2 1^2 + \cdots + a_n 1^n\) | \(\equiv\) | \(\displaystyle 0 \pmod 9\) | as $10 \equiv 1 \pmod 9$ | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a_0 + a_1 + \cdots + a_n\) | \(\equiv\) | \(\displaystyle 0 \pmod 9\) |

$\blacksquare$

## Sources

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