Question

Prove that any one of the three consecutive positive integers should be divisible by 3

Answer 1

hey mate...

the answer is .....

let the numbers be 2 , 4 , 6

so the product is ...

2 × 4 × 6

= 48

48 is divisible by 3 as 4 + 8 = 12
sum of digits must be multiple of 3 .

hence proved.

hope it helps uu dear

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.