Divisibility by 2

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Theorem

An integer $N$ expressed in decimal notation is divisible by $2$ if and only if the least significant digit of $N$ is divisible by $2$.


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 $2$

if and only if:

$a_0$ is divisible by $2$.


Proof

Let $N$ be divisible by $2$.

Then:

\(\displaystyle N\) \(\equiv\) \(\displaystyle 0 \pmod 2\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \sum_{k \mathop = 0}^n a_k 10^k\) \(\equiv\) \(\displaystyle 0 \pmod 2\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}\) \(\equiv\) \(\displaystyle 0 \pmod 2\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle a_0\) \(\equiv\) \(\displaystyle 0 \pmod 2\) as $10 \equiv 0 \pmod 2$

$\blacksquare$


Sources