Question

Prove that only 1 out of 3 consecutive positive int. Is divisible by 3

Answer 1

Step-by-step explanation:

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 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. So there is always exactly one multiple of 3 among them.

Answer 2

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.