Question
For any positive integer n, prove that $n^3 - n$ is divisible by $6.$
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Answer
$n^3-n=n\left(n^2-1\right)=n(n-1)(n+1)$
Whenever a number is divided by 3 , the remainder obtained is either 0 or 1 or 2 .
$\therefore n =3 p$ or $3 p +1$ or $3 p +2$, where p is some integer.
If $n=3 p$, then $n$ is divisible by 3 .
If $n=3 p+1$, then $n-1=3 p+1-1=3 p$ is divisible by 3 .
If $n=3 p+2$, then $n+1=3 p+2+1=3 p+3=3(p+1)$ 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 .
$\therefore n =2 q$ or $2 q +1$, where q is some integer.
If $n=2 q$, then $n$ is divisible by 2 .
If $n=2 q+1$, then $n-1=2 q+1-1=2 q$ is divisible by 2 and $n+1=2 q+1+1=2 q+2=2(q+1)$ is divisible by
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 .
$\therefore n(n-1)(n+1)=n^3-n$ is divisible by 6 . (If a number is divisible by both 2 and 3 , then it is divisible by 6 )
Whenever a number is divided by 3 , the remainder obtained is either 0 or 1 or 2 .
$\therefore n =3 p$ or $3 p +1$ or $3 p +2$, where p is some integer.
If $n=3 p$, then $n$ is divisible by 3 .
If $n=3 p+1$, then $n-1=3 p+1-1=3 p$ is divisible by 3 .
If $n=3 p+2$, then $n+1=3 p+2+1=3 p+3=3(p+1)$ 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 .
$\therefore n =2 q$ or $2 q +1$, where q is some integer.
If $n=2 q$, then $n$ is divisible by 2 .
If $n=2 q+1$, then $n-1=2 q+1-1=2 q$ is divisible by 2 and $n+1=2 q+1+1=2 q+2=2(q+1)$ is divisible by
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 .
$\therefore n(n-1)(n+1)=n^3-n$ is divisible by 6 . (If a number is divisible by both 2 and 3 , then it is divisible by 6 )
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