Divisibility by 8

Theorem

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 $8$.

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 $10^3 \equiv 0 \pmod 8$

$\blacksquare$