Question

prove that one out of every three consecutive 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

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 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.

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Answer 2

Answer:

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