Question

Show that only one of the number $n,n+2$ and $n+4$ is divisible by $3$.

Answer 1

We know that any positive integer of the form $3q$ or, $3q+1$ or $3q+2$ for some integer $q$ and one and only one of these possibilities can occur.

So, we have following cases:

Case-I: When $n=3q$

In this case, we have

$n=3q$, which is divisible by $3$

Now, $n=3q$

$n+2=3q+2$

$n+2$ leaves remainder $2$ when divided by $3$

Again, $n=3q$

$n+4=3q+4=3(q+1)+1$

$n+4$ leaves remainder $1$ when divided by $3$

$n+4$ is not divisible by $3$.

Thus, $n$ is divisible by $3$ but $n+2$ and $n+4$ are not divisible by $3$.

Case-II: when $n=3q+1$

In this case, we have

$n=3q+1,$

$n$ leaves remainder $1$ when divided by $3$.

$n$ is divisible by $3$

Now, $n=3q+1$

$n+2=(3q+1)+2=3(q+1)$

$n+2$ is divisible by $3$.

Again, $n=3q+1$

$n+4=3q+1+4=3q+5=3(q+1)+2$

$n+4$ leaves remainder $2$ when divided by $3$

$n+4$ is not divisible by $3$.

Thus, $n+2$ is divisible by $3$ but $n$ and $n+4$ are not divisible by $3$.

Case-III: When $n=3q+2$

In this case, we have

$n=3q+2$

$n$ leaves remainder $2$ when divided by $3$.

$n$ is not divisible by $3$.

Now, $n=3q+2$

$n+2=3q+2+2=3(q+1)+1$

$n+2$ leaves remainder $1$ when divided by $3$

$n+2$ is not divisible by $3$.

Again, $n=3q+2$

$n+4=3q+2+4=3(q+2)$

$n+4$ is divisible by $3$.

Hence, $n+4$ is divisible by $3$ but $n$ and $n+2$ are not divisible by $3$.