Question

Show that only one of the no $n,n+2$ and $n+4$ in divisible $3$.

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 $3$.