Question

prove that one of every three consecutive positive integers is divisible by 3​

Answer 1

Answer:

the positive integers be 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

then one of every three is the multiple of three

that is 6 12 18 24 30

Answer 2

Proof :

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

Also, we know that a positive integer n is of the form 3q, 3q + 1 or 3q + 2.

Here, the following three cases arise:

CASE I: When n = 3q

∴ n is divisible by 3 but (n + 1) and (n + 2) are not divisible by 3.

[ ∵ When (n + 1) is divided by 3, remainder = 1 and when (n + 2) is divided by 3, remainder = 2]

CASE II: When n = 3q + 1.

∴ n + 2 = (3q + 1) + 2 = 3q + 3

⟹ n + 2 = 3 (q + 1)

∴. (n + 2) is divisible by 3 but n and (n + 1) are not divisible by 3.

CASE III : When n = 3q + 2

∴ n + 1 = (3q + 2) + 1 = 3q + 3

⟹ n + 1 = 3 (q + 1).

∴ (n + 1) is divisible by 3 but n and (n + 2) are not divisible by 3.

Hence, one of every three consecutive positive integers is divisible by 3.

Proved