Question

let n, n+1 and n+2 are three consecutive positive integers prove that exactly one of these is exactly divisible by 3

Answer 1

Start by a substitution method
Let n=1
The three numbers are 1,2,3
We have 3 in above sequence which is exactly divusible by 3
Now lets check by taking a bigger number
Let n=1001
The three numbers are 1001,1002,1003
From the above three numbers 1002 is exactly divisible by 3.
So wd can conclude that there is exactly one number from three consecutive numbers which is exactly divisible by 3
Hope this will help you ☺️

Answer 2

Hey

Here is your answer,

Let us three consecutive 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.
let 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 that n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.

Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So that n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.

But n (n + 1) (n + 2) is divisible by 2 and 3.

Hope it helps you!