Divisibility by 9

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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 $9$

if and only if:

$a_0 + a_1 + \cdots + a_n$ 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 $3$

if and only if:

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


Proof 1

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 $10 \equiv 1 \pmod 9$
\(\ds \leadstoandfrom \ \ \) \(\ds a_0 + a_1 + \cdots + a_n\) \(\equiv\) \(\ds 0 \pmod 9\)

$\blacksquare$


Proof 2

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

$\blacksquare$


Sources

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