# Divisibility by 8

## Theorem

An integer $N$ expressed in decimal notation is divisible by $8$ if and only if the $3$ least significant digits of $N$ form a $3$-digit integer divisible by $8$.

That is:

$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $8$
$100 a_2 + 10 a_1 + a_0$ is divisible by $8$.

## Proof

Let $N$ be divisible by $8$.

Then:

 $\ds N$ $\equiv$ $\ds 0 \pmod 8$ $\ds \leadstoandfrom \ \$ $\ds \sum_{k \mathop = 0}^n a_k 10^k$ $\equiv$ $\ds 0 \pmod 8$ $\ds \leadstoandfrom \ \$ $\ds 100 a_2 + 10 a_1 + a_0 + 10^3 \sum_{k \mathop = 3}^n a_k 10^{k - 3}$ $\equiv$ $\ds 0 \pmod 8$ $\ds \leadstoandfrom \ \$ $\ds 100 a_2 + 10 a_1 + a_0$ $\equiv$ $\ds 0 \pmod 8$ as $10^3 \equiv 0 \pmod 8$

$\blacksquare$