in Quantitative Aptitude recategorized by
684 views
1 vote
1 vote

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
in Quantitative Aptitude recategorized by
7.9k points
684 views

2 Answers

1 vote
1 vote
Best answer

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. 

selected by
1.5k points

2 Comments

what if $n = 5$?
0
0

For n = 5, it will be 5.4.6 which is still divided by 24. For n = 7, it will be 7.6.8 which will be divided by 48. But the question is asking for any odd number greater than 3. So the apt choice should be 24.

2
2
2 votes
2 votes

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.

 

2.7k points

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true