Divisibility by 9/Corollary

(Redirected from Divisibility by 3)

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 $3$
$a_0 + a_1 + \ldots + a_n$ 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 $3^2$
$a_0 + a_1 + \ldots + a_n$ 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

$\blacksquare$