Question

prove that one of any 3 consecutive positive integer must be divisible by 3?!

Answer 1

Let the three consecutive numbers be n, n+1 and n+2

Number 3 being a divisor can always give 0, 1 or 2 as a remainder for any dividend.
Hence, we've three possibilities using formula Dividend = Divisor * Quotient + Remainder,
{n = 3p + 0}
{n = 3p + 1}
{n = 3p +2}
where p is the Quotient.

Case 1, n = 3p
Therefore, n is divisible by 3

Case 2, n = 3p + 1
Adding 2 on both sides,
n + 2 = 3p + 3
n + 2 = 3(p + 1)
As there is a common factor 3
n + 2 is divisible by 3

Case 3, n = 3p + 2
Adding 1 on both sides
n + 1 = 3p + 3
n + 1 = 3(p + 1)
As there is a common factor 3
n + 1 is divisible by 3

Therefore, there will always be a number divisible by 3 among 3 consecutive positive integers.

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.