1 1 vote If $n$ is any odd number greater than $1$, then $n(n^2 – 1)$ is divisible by $96$ always divisible by $48$ always divisible by $24$ always None of these Quantitative Aptitude cat2016 quantitative-aptitude number-systems + – go_editor 14.2k points 1.9k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote Let, $n = 2m+1; n>1;m \in \mathbb{Z},$ and $m>0$ Also let, $x = n(n^{2}-1)$ $\Rightarrow x = (2m+1)[ \,(2m+1)^{2}-1]$ $\Rightarrow x = (2m+1)(4m^{2}+1+4m-1)$ $\Rightarrow x = (2m+1)(4m^{2}+4m)$ $\Rightarrow \boxed{x = (2m+1)4m(m+1)}$ Now, we can substitute the various value of $m$ $m = 1 \Rightarrow x = 3 \times 4 \times 2 = 24$ $m = 2 \Rightarrow x = 5 \times 8 \times 3 = 120$ $m = 3 \Rightarrow x = 7 \times 12 \times 4 = 336$ $\vdots \quad \vdots \quad \vdots$ $\therefore x = n(n^{2}-1)$ is always divisible by $24.$ Correct Answer $:\text{C}$ Anjana5051 answered Jan 13, 2022 • edited Jan 20, 2022 by Lakshman Bhaiya Anjana5051 12.0k points comment Share Follow 0 reply Please log in or register to add a comment.