Any positive integer will be of form 5*q*, 5*q* + 1, 5*q* + 2, 5*q *+ 3, or 5*q* + 4.

**Case I:**

If *n = *5*q*

*n* is divisible by 5

Now, *n = *5*q*

*⇒ n + *4 *= *5*q + *4

The number* *(*n + *4)* *will leave remainder 4 when divided by 5.

Again, *n* = 5*q*

*⇒ n + *8 = 5*q + *8 = 5(*q* + 1) + 3

The number* *(*n *+ 8) will leave remainder 3 when divided by 5.

Again, *n *= 5*q*

*⇒ n + *12 = 5*q + *12 = 5(*q + *2) + 2

The number (*n *+ 12) will leave remainder 2 when divided by 5.

Again,* n* = 5*q*

*⇒ n + *16 = 5*q + *16 = 5(*q + *3) + 1

The number (*n + *16) will leave remainder 1 when divided by 5.

**Case II:**

When *n = *5*q + *1

The number *n* will leave remainder 1 when divided by 5.

Now, *n* = 5*q* + 1

*⇒ n + *2 = 5*q* + 3

The number (*n* + 2) will leave remainder 3 when divided by 5.

Again, *n* = 5*q *+ 1

*⇒ n + *4* = *5*q + *5 *= *5(*q *+ 1)

The number (*n + *4)* *will be divisible by 5.

Again, *n* = 5*q* + 1

*⇒ n + *8 = 5*q *+ 9 = 5(*q* + 1) + 4

The number (*n *+ 8) will leave remainder 4 when divided by 5

Again, *n* = 5*q* + 1

*⇒ n* + 12 = 5*q *+ 13 = 5(*q* + 2) + 3

The number (*n *+ 12) will leave remainder 3 when divided by 5.

Again, *n *= 5*q *+ 1

*⇒ n + *16* = *5*q + *17 *= *5(*q* + 3) + 2

The number (*n *+ 16) will leave remainder 2 when divided by 5.

Similarly, we can check the result for 5*q* + 2, 5*q *+ 3 and 5*q *+ 4.

In each case only one out of *n, n *+ 2, *n *+ 4, *n *+ 8, *n *+ 16 will be divisible by 5.