Question

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

Answer 1

Answer:

$\mathbb{question}$

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

Step-by-step explanation:

$\mathbb{solution}$

Let n,n+1,n+2 be three consecutive positive integers. ... In this case, n+1=3q+1+2=3(q+1) is divisible by 3 but n and n+2 are not divisible by 3. Hence one of n,n+1 and n+2 is divisible by 3.

Answer 2

Answer :

Let the three positive integers be n, n + 1 and n + 3

i.e., n is in the form of 3q, 3q + 1 or 3q + 2

$\hrulefill$

Case I :

When n = 3q, n is divisible by 3

Now n + 1 = 3q + 1, which is not divisible by 3

Again n + 2 = 3q + 2, which is not divisible by 3

$\hrulefill$

Case II :

When n = 3q + 1, n is not divisible by 3

Now, n + 1 = 3q + 2, which is not divisible by 3

Again, n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3

$\hrulefill$

Case III :

When n = 3q + 2, n is not divisible by 3

Now, n + 1 = 3q + 3 = 3(q + 1), which is divisible by 3

Again n + 2 = 3q + 4, which is not divisible by 3

Hence one of n, n + 1 and n + 3 is divisible by 3.