Question

For any positive integer , $n^3-n$ is always divisible by ___?

• A. 6
• B. 12
• C. 15
• D. 18

Answer 1

The correct option is A

6

$n^3–n=n(n^2–1)=(n-1).n.(n+1)$

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
$\Rightarrow n=3por3p+1or3p+2$, where p is some integer.
If $n=3p$, then n is divisible by 3.
If$n=3p+1$, $thenn–1=3p+1–1=3p$ 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 + 1 is always divisible by 3.
$\Rightarrow n(n–1)(n+1)$ is divisible by 3.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
$\Rightarrow n=2qor2q+1$, where q is some integer.
If $n=2q$, then n is divisible by 2.
If $n=2q+1,\text{then}n–1=2q+1–1=2q$ is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
$\Rightarrow n(n–1)(n+1)$ is divisible by 2.
Since,$n(n–1)(n+1)$ is divisible by 2 and 3.
We know that if a number is divisible by both 2 and 3 , then it is divisible by 6.
$\Rightarrow n(n-1)(n+1)=n^3-n$ is divisible by 6.