Prove that one and only one out of every three consecutive positive integers is divisible by $3$.

(N/A) Let the three consecutive positive integers be $n$,$n+1$,and $n+2$.
We know that any positive integer $n$ can be expressed in the form $3m$,$3m+1$,or $3m+2$,where $m$ is a non-negative integer.
Case $1$: If $n = 3m$,then $n$ is divisible by $3$. Here,$n+1 = 3m+1$ and $n+2 = 3m+2$,which leave remainders $1$ and $2$ respectively when divided by $3$. Thus,only $n$ is divisible by $3$.
Case $2$: If $n = 3m+1$,then $n+1 = 3m+2$ and $n+2 = 3m+3 = 3(m+1)$. Here,only $n+2$ is divisible by $3$.
Case $3$: If $n = 3m+2$,then $n+1 = 3m+3 = 3(m+1)$ and $n+2 = 3m+4 = 3(m+1)+1$. Here,only $n+1$ is divisible by $3$.
Conclusion: In all possible cases,exactly one of the three consecutive integers $n$,$n+1$,and $n+2$ is divisible by $3$.