# Math Help - Simple Proof

1. ## Simple Proof

I feel like this proof should be easy.

Prove that for 3 consecutive natural numbers, 1 of the numbers will be divisible by 3.

Intuitively this makes sense (add 1, 2 or 3 and then it's divisible by 3, obviously).

2. let $n \in \mathbb{N}, A = \{n, n+1, n+2\}$

if $n \equiv 0 \mod{3}$, then $A \equiv \{0,1,2\} \mod{3}$

if $n \equiv 1 \mod{3}$, then $A \equiv \{1,2,0\} \mod{3}$

if $n \equiv 2 \mod{3}$, then $A \equiv \{2,0,1\} \mod{3}$

So in all three cases, $\exists \; a \in A$ s.t. $a \equiv 0 \mod{3} \Rightarrow a$ is divisible by three.