Divisibility by 9/Corollary

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Corollary to Divisibility by 9

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

From Divisibility by 9 we have that:

$N = \sqbrk {a_0 a_1 a_2 \ldots a_n}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $3^2$

if and only if:

$a_0 + a_1 + \ldots + a_n$ is divisible by $3^2$.


So:

\(\displaystyle \paren {a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n}\) \(\equiv\) \(\displaystyle \paren {a_0 + a_1 + a_2 + \cdots + a_n}\) \(\displaystyle \pmod {3^2}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n}\) \(\equiv\) \(\displaystyle \paren {a_0 + a_1 + a_2 + \cdots + a_n}\) \(\displaystyle \pmod 3\) Congruence by Divisor of Modulus

$\blacksquare$


Sources