Question

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

Answer 1

We applied Euclid Division algorithm on $n$ and $3.$
$a=bq+r$ on putting $a=n$ and $b=3$
$n=3q+r$ , $0<r<3$

i.e

$n=3q$ -------- (1),

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

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

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

or $n+2=3q+1+2=3q+3$ also divisible by $3$

or$n+4=3q+2+4=3q+6$ is also divisible by $3$

Hence $n,n+2,n+4$ are divisible by $3.$