Simple Proof

**DiscreteW** · November 3rd 2008, 03:24 PM

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).

**chiph588@** · November 3rd 2008, 04:51 PM

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.

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