Divisibility by 9

Theorem

A number:

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

is divisible by $9$ iff the sum:

$a_0 + a_1 + \ldots + a_n$

of its digits is divisible by $9$.

Or, commonly stated, a number is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

Direct Proof

If $N$ is divisible by 9, then

	\(\displaystyle \)	\(\displaystyle N\)	\(\equiv\)	\(\displaystyle 0\,\bmod\, 9\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n\)	\(\equiv\)	\(\displaystyle 0 \bmod 9\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle a_0 + a_1 1 + a_2 1^2 + \cdots + a_n 1^n\)	\(\equiv\)	\(\displaystyle 0 \bmod 9\)	\(\displaystyle \)	since $10 \equiv 1 \bmod 9$
	\(\displaystyle \iff\)	\(\displaystyle a_0 + a_1 + \cdots + a_n\)	\(\equiv\)	\(\displaystyle 0 \bmod 9\)	\(\displaystyle \)	as desired.

$\blacksquare$

Alternative Proof

It can be seen that this is a special case of Congruence of Sum of Digits to Base Less 1.

$\blacksquare$

Divisibility by 3

This same argument holds for divisibility by $3$. The proof is exactly the same as the proof above, replacing all instances of $9$ by $3$.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 2.6$: Example $41$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $2.8$