Question

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

Answer 1

Answer:

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

We know that n is of the form 3q,3q+1 or, 3q+2 (As per Euclid Division Lemma),

So, we have the following

Case I When n=3q

In this case, n is divisible by 3 but n+1 and n+2 are not divisible by 3.

Case II When n=3q+1

In this case, n+2=3q+1+2=3(q+1) is divisible by 3 but n and n+1 are not divisible by 3.

Case III When n=3q+2

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.

Step-by-step explanation:

if it is helpful please mark me as brainlist....

Answer 2

$\huge\mathfrak\red{❥︎AnsweR:-}$

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

We know thɑt n is of the form 3q,3q+1 or, 3q+2 (ɑs per Euclid Division Lemmɑ),

So, we hɑve the following

Cɑse I When n=3q

In this cɑse, n is divisible by 3 but n+1 ɑnd n+2 ɑre not divisible by 3.

Cɑse II When n=3q+1

In this cɑse, n+2=3q+1+2=3(q+1) is divisible by 3 but n ɑnd n+1 ɑre not divisible by 3.

Cɑse III When n=3q+2

In this cɑse, n+1=3q+1+2=3(q+1) is divisible by 3 but n ɑnd n+2 ɑre not divisible by 3.

Hence one of n,n+1 ɑnd n+2 is divisible by 3.