Show that one and only one out of $n, n+4, n+8, n+12$ and $n+16$ is divisible by $5,$ where $n$ is any positive integer.

(N/A) By Euclid's division lemma,any positive integer $n$ can be expressed in the form $5q, 5q+1, 5q+2, 5q+3,$ or $5q+4$,where $q$ is a non-negative integer.
We examine the divisibility of the set ${n, n+4, n+8, n+12, n+16}$ for each case:
$1$. If $n = 5q$,then $n$ is divisible by $5$.
$2$. If $n = 5q+1$,then $n+4 = 5q+1+4 = 5q+5 = 5(q+1)$,which is divisible by $5$.
$3$. If $n = 5q+2$,then $n+8 = 5q+2+8 = 5q+10 = 5(q+2)$,which is divisible by $5$.
$4$. If $n = 5q+3$,then $n+12 = 5q+3+12 = 5q+15 = 5(q+3)$,which is divisible by $5$.
$5$. If $n = 5q+4$,then $n+16 = 5q+4+16 = 5q+20 = 5(q+4)$,which is divisible by $5$.
In every case,exactly one of the five expressions is divisible by $5$.