Question

Show that one and only one out of $n,n+1,n+2$ is divisible by $3$, where $n$ is any positive integer

Answer 1

Let $n$ be any positive integer and $b=3$
If $n$ is divided by $b$ then by division algorithm,
$n=3q+r$
where $q$ is the quotient and $r$ is the remainder such that
$0\le r<3$
So, the remainders may be $0,1$ and $2$
So, $n$ may be in the form of $3q,3q+1,3q+2$
Case: 1
If $n=3q$ then $n$ is divisible by $3$.
Adding $1$ on both sides,
$n+1=3q+1$
Thus, $n+1$ is not divisible by $3$.
Adding $2$ on both sides,
$n+2=3q+2$
Thus, $n+2$ is also not divisible by $3$.
Therefore, $n$ is only divisible by $3$ but $n+1,n+2$ is not divisible by $3$.

Case 2:
If $n=3q+1$ is not divisible by $3$
Adding $1$ on both sides,
$n+1=3q+1+1$
$n+1=3q+2$
Thus, $n+1$ is not divisible by $3$
Similarly add $2$ on both sides,
$n+2=3q+1+2$
$n+2=3q+3$
$n+2=3(q+1)$
Thus, $n+2$ is divisible by $3$.
Therefore, $n+2$ is only divisible by $3$ but $n,(n+1)$ is not divisible by $3$.
Similar way Case 3: if $n=3q+2$ is obviously not divisible by $3$.
$\Rightarrow n+1=3q+2+1$
$\Rightarrow n+1=3q+3$
$\Rightarrow n+1=3(q+1)$ is divisible by $3$.
But $n+2=3q+2+2\Rightarrow n+1=3q+4$ is not divisible by $3$.
Therefore, $n+1$ is only divisible by $3$ but $n,(n+2)$ is not divisible by $3$.
Hence, one and only one out of $n,n+1,n+2$ is divisible by $3$.