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

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

Case I:

If n = 5q

n is divisible by 5

Now, n = 5q

⇒ n + 4 = 5q + 4

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

Again, n = 5q

⇒ n + 8 = 5q + 8 = 5(q + 1) + 3

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

Again, n = 5q

⇒ n + 12 = 5q + 12 = 5(q + 2) + 2

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

Again, n = 5q

⇒ n + 16 = 5q + 16 = 5(q + 3) + 1

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

Case II:

When n = 5q + 1

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

Now, n = 5q + 1

⇒ n + 2 = 5q + 3

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

Again, n = 5q + 1

⇒ n + 4 = 5q + 5 = 5(q + 1)

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

Again, n = 5q + 1

⇒ n + 8 = 5q + 9 = 5(q + 1) + 4

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

Again, n = 5q + 1

⇒ n + 12 = 5q + 13 = 5(q + 2) + 3

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

Again, n = 5q + 1

⇒ n + 16 = 5q + 17 = 5(q + 3) + 2

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

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

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