Question

If $n$ is a positive integer and $P\left(n\right)=n^2(n^2-1)$, then $P\left(n\right)$ is always divisible by

• A. $12$
• B. $24$
• C. $36$
• D. $12-n$

Answer 1

The correct option is A $12$

$P\left(n\right)=n^2(n^2-1)=(n-1).n.n.(n+1)$

All the numbers $n-1,n,n+1$ are consecutive numbers so it must be divisible by $3$ and $2$.

Since it has square of the middle number, in every case, it is also divisible by $4$.

So in overall, it is divisible by $12$ but definitely not $24$ always.

Examples:

$n=14$

So, $p\left(14\right)=13\times 14\times 14\times 15$ is divisible by $12$.

$n=33$

So, $p\left(33\right)=33\times 34\times 34\times 35$ is divisible by $12$.