Divisibility by 4

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Theorem

An integer $N$ expressed in decimal notation is divisible by $4$ if and only if the $2$ least significant digits of $N$ form a $2$-digit integer divisible by $4$.


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 $4$

if and only if:

$10 a_1 + a_0$ is divisible by $4$.


Proof

Let $N$ be divisible by $4$.

Then:

\(\ds N\) \(\equiv\) \(\ds 0 \pmod 4\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sum_{k \mathop = 0}^n a_k 10^k\) \(\equiv\) \(\ds 0 \pmod 4\)
\(\ds \leadstoandfrom \ \ \) \(\ds 10 a_1 + a_0 + 10^2 \sum_{k \mathop = 2}^n a_k 10^{k - 2}\) \(\equiv\) \(\ds 0 \pmod 4\)
\(\ds \leadstoandfrom \ \ \) \(\ds 10 a_1 + a_0\) \(\equiv\) \(\ds 0 \pmod 4\) as $10^2 \equiv 0 \pmod 4$

$\blacksquare$


Sources