Question

Prove that one of every three consecutive positive integer is divisible by $3$.

Answer 1

Let $n,n+1,n+2$ be three consecutive positive integers.

We know that $n$ is of the form $3q,3q+1$ or, $3q+2$ (As per Euclid Division Lemma),
So, we have the following

Case $I$ When $n=3q$
In this case, $n$ is divisible by $3$ but $n+1$ and $n+2$ are not divisible by $3$.

Case $II$ When $n=3q+1$
In this case, $n+2=3q+1+2=3(q+1)$ is divisible by $3$ but $n$ and $n+1$ are not divisible by $3$.

Case $III$ When $n=3q+2$
In this case, $n+1=3q+1+2=3(q+1)$ is divisible by $3$ but $n$ and $n+2$ are not divisible by $3$.
Hence one of $n,n+1$ and $n+2$ is divisible by $3$.