Question

0 prove that one of any three consecutive positive integers must be divisible by 3.

Answer 1

we can have a logical proof
if x-1 x x+1 are integers then if x can be divisible by 3
if is divisible it can be written as 3x. if not it is either 3k+1 or 3k+2 since 3k+3 is divisible by 3
if the no is 3k+1 the 2nd no is 3k+2 and 3rd is divisible
if it is 3k+2 the 2nd no. is divisible.
so any one is definetely divisible
we can understand in an another way the multiples of 3 repeats in a cycle of 3
so in 3 nos. there will be atleast one
ALL THE BEST............

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.