# Divisibility by 9

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## Contents

## 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 $9$.

### Corollary

A number expressed in decimal notation is divisible by $3$ if and only if the sum of its digits is divisible by $3$.

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 $3$

- $a_0 + a_1 + \ldots + a_n$ is divisible by $3$.

## Proof 1

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 $10 \equiv 1 \pmod 9$ | |||||||||

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

$\blacksquare$

## Proof 2

This is a special case of Congruence of Sum of Digits to Base Less 1.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.8$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 2.3$: Congruences: Exercise $9$ - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $9$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $9$