Question

Show that exactly one of the numbers $n,n+2,n+4$ is divisible by $3$.

Answer 1

Case I:
If $n=3q$
$n+4=3q+4n+2=3q+2$
here $n$ is only divisible by $3$
Case II:
If $n=3q+1$
$n+4=3q+5n+2=3q+3=3(q+1)$
Here only $n+2$ is divisible by $3$
Case III:
If $n=3q+2$
$n+2=3q+4$

$n+4=3q+2+4$ $=3q+6$
Here only $n+4$ is divisible by $3$

Hence, only one of the numbers $n,n+2,n+4$ is divisible by $3$.