# 3 is Divisor of one of n, n+2, n+4

## Theorem

Let $n \in \Z$ be an integer such that $n > 3$.

Then at least one of $n$, $n + 2$, $n + 4$ is divisible by $3$.

## Proof

Let $n \in \Z_{>0}$.

Then $n$ can be represented as either:

 $\ds n$ $=$ $\ds 3 k$ $\ds n$ $=$ $\ds 3 k + 1$ $\ds n$ $=$ $\ds 3 k + 2$

Let $n = 3 k$.

Then $n$ is divisible by $3$.

Let $n = 3 k + 1$.

Then:

$n + 2 = 3 k + 3 = 3 \paren {k + 1}$

which is divisible by $3$.

Let $n = 3 k + 2$.

Then:

$n + 4 = 3 k + 6 = 3 \paren {k + 2}$

which is divisible by $3$.

Hence the result.

$\blacksquare$