Question

Prove that one of every three consecutive positive integers is divisible by 3.

Answer 1

Let three consecutive positive integers be n, n+ 1 and n + 2.

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

Answer 2

let three consecutive nos. = (n-1) n (n+1)
If any no. is divided by 3, it is of form 3q, 3q+1 3q+2
Put n=3q
case 1
(3q-1) and (3q) and (3q+1)
Here only 3q is divisible by 3....
case 2
put n=3q+1
(3q+1-1) and (3q+1) and (3q+1+1)
3q , (3q+1) , (3q+2)
Here, only 3q is divisible by 3.

case 3
put n = 3q+2
(3q+2-1) and (3q+2) and (3q+2+1)
(3q+1) , (3q+2) , (3q+3)
Here only (3q+3) is divisible by 3 bcoz 3(q+1) i.e 3m.
Hence proved.
Hope it will help u....