Question

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.

[Hint: Any positive integer can be written in the form 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4].

Answer 1

On dividing n by 5, let q be the quotient and r be the remainder.

Then n = 5q + r, where 0 ≤ r < 5

`\implies` n = 5q + r, where r = 0, 1, 2, 3, 4

`\implies` n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4

Case I: If n = 5q,

Then only n is divisible by 5.

Case II: If n = 5q + 1,

Then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) which is divisible by 5

So, in this case, only (n + 4) is divisible by 5.

Case III: If n = 5q + 2, 

Then n + 8 = 5q + 10 = 5(q + 2) which is divisible by 5

So, in this case, only (n + 8) is divisible by 5.

Case IV: If n = 5q + 3,

Then n + 12 = 5q + 3 + 12 = 5q + 15 = 5(q + 3) which is divisible by 5

So, in this case, only (n + 12) is divisible by 5.

Case V: If n = 5q + 4, 

Then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4) which is divisible by 5

So, in this case, only (n + 16) is divisible by 5.

Hence 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.