Question

. Q3. Prove that exactly one out of every three consecutive integers is divisible by 3.

Answer 1

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.

Answer 2

Step-by-step explanation:

Let 3 consecutive positive integers be n, n + 1 and n + 2 .

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

n = 3p or 3p+1 or 3p+2, where p is some integer

If n = 3p = 3(p) , then n is divisible by 3

If n = 3p + 1, then n + 2 = 3p +1 + 2 = 3 p + 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

Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3

Hence it is solved.

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