Divisibility by 2

Theorem

That is:

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

$a_0$ is divisible by $2$.

Let $N$ be divisible by $2$.

Then:

$\ds N$

$\equiv$

$\ds 0 \pmod 2$

$\ds \leadstoandfrom \ \ $

$\ds \sum_{k \mathop = 0}^n a_k 10^k$

$\equiv$

$\ds 0 \pmod 2$

$\ds \leadstoandfrom \ \ $

$\ds a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}$

$\equiv$

$\ds 0 \pmod 2$

$\ds \leadstoandfrom \ \ $

$\ds a_0$

$\equiv$

$\ds 0 \pmod 2$

as $10 \equiv 0 \pmod 2$

$\blacksquare$