Question

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

Answer 1

By Euclid's division algorithm , $a=bq+r$ where $a,b,q,r$ are non-negative integers and $0\le r<b$.

On putting $b=$ we get

$a=3q+r$, where $r=0,1,2$

Thus any number is in the form of $3q,3q+1$ or $3q+2$.

Case I : if $n=3q$

$\Rightarrow n=3q=3\left(q\right)$ is divisible by $3.$

$\Rightarrow n+2=3q+2$ is not divisible by $3.$

$\Rightarrow n+4=3q+4=3(q+1)+1$ is not divisible by $3.$

Case II : if $n=3q+1$

$\Rightarrow n=3q+1$ is not divisible by $3.$

$⇒ n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1)$ is divisible by $3.$

$\Rightarrow n+4=3q+1+4=3q+5=3(q+1)+2$ is not divisible by $3.$

Case III : if $n=3q+2$

$\Rightarrow n=3q+2$ is not divisible by $3.$

$\Rightarrow n+2=3q+2+2=3q+4=3(q+1)+1$ is not divisible by $3.$

$\Rightarrow n+4=3q+2+4=3q+6=3(q+2)$ is divisible by $3.$

Thus one and only one out of $n,n+2,n+4$ is divisible by $3.$