Question

If $n\in N$, then $n(n^2-1)$ is divisible by

• A. $6$
• B. $16$
• C. $26$
• D. $24$

Answer 1

The correct option is A $6$

$n{(n}^2-1)=n(n-1)(n+1)$

One of the $n$, $n+1$ and $n-1$ will be a multiple of $3$.

Since $n-1$, $n$ and $n+1$ are three consecutive integers, therefore at least one of them will be divisible by $2$.

Therefore $n{(n}^2-1)$ is divisible by $6$.

Option d can be rejected by putting $n=2$