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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
asked in Quantitative Aptitude by (7.4k points) 44 134 360
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2 Answers

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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. 

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what if $n = 5$?
+2

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.

+1 vote

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.

 

answered by (2.5k points) 2 8 29

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