# CAT 1996 | Question: 128

1,398 views

If n is any odd number greater than $1$, then the largest number that divides $n(n^{2} - 1)$ is:

1. $48$
2. $24$
3. $6$
4. None of these

The smallest value of n will be 3.

And we can see that n(n2 - 1) = n(n - 1)(n + 1) which is the product of 3 consecutive numbers. For n = 3, This quantity will be 2.3.4 = 24. So the largest number that divides n(n2 - 1) will be 24.

As $n(n^{2}-1)$ = (n−1) n (n+1)

Since n is any odd number , n−1 and n+1 are consecutive even integers.

So, one is divisible by 2 and the other is divisible by 4.

Hence, (n−1) n (n+1) is divisible by 8.

Since n−1, n , n+1 are 3 consecutive integers, one of them will be divisible by 3.

Hence, (n−1) n (n+1) is divisible by 3.

Since (n−1) n (n+1) is divisible by both 8 and 3, it is divisible by lcm(8,3)=24.

then the largest number that divides  $n(n^{2}-1)$ is 24

Hence Option C is the Correct Answer.

## Related questions

1.8k
views
1,843 views
Find the smallest number $y$ such that $y \times 162$ $(y$ multiplied by $162)$ is a perfect cube.$A)24$ $B)27$ $C)32$ ...
871
views
871 views
What is the smallest number, which when increased by 5 is completely divisible by 8, 11 and 24?264 259269 None of these
762
views
762 views
For the product $n(n + 1)(2n + 1)$, $n \hat{I} N$, which one of the following is necessarily false?It is always evenDivisible by $3$.Always divisible by the sum of the sq...
937
views
$5^{4}-1$ is divisible by$13$ $31$ $5$ none of these