Prove that one and only one out of $n,n+2andn+4$ is divisible by $3$, where $n$ is any positive integer.

To Prove that one and only one out of $n,n+2andn+4$ is divisible by $3$

As it is given that, $n,n+2andn+4$ three integers where $n$ is any positive integer

So, One and only one out of $n,n+2andn+4$ is divisible by$3$, where $n$ is any positive integer.

For this Let the positive integer $=n$ And $b=3$

According to Euclid’s division lemma, we know that

$n=3q+r$, where $q$ is the quotient and $r$ is the remainder

$0<r<3$ implies remainders may be $0,1and2$

Therefore, $n$ may be in the form of $3q,3q+1,3q+2$

Case 1: When $n=3q$

$$\begin{gathered} \Rightarrow n+2=3q+2 \\ \Rightarrow n+4=3q+4 \end{gathered}$$

Here $n$ is only divisible by $3$

Case 2: When$\mathrm{n}=3\mathrm{q}+1$

$$\begin{gathered} \Rightarrow \mathrm{n}+2=3\mathrm{q}+3=3(\mathrm{q}+1) \\ \Rightarrow \mathrm{n}+4=3\mathrm{q}+5 \end{gathered}$$

Here only $n+2$ is divisible by $3$

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

$$\begin{gathered} \Rightarrow n+2=3q+4 \\ \Rightarrow n+4=3q+2+4=3q+6 \end{gathered}$$

Here only $n+4$ is divisible by $3$

So, we note that one and only one out of $n,n+2andn+4$ is divisible by $3$.

Hence Proved.