State whether n (n² - 1) is divisible by 24

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State whether n (n² - 1) is divisible by 24 [#permalink]

29 Aug 2009, 00:03

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33% (00:00) wrong

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Question:
n is a natural number. State whether n (n² - 1) is divisible by 24.
(1) 3 divides 'n' completely without leaving any remainder.
(2) 'n' is odd.

My solution:
n (n² - 1) = n (n-1) (n+1) = 3 consecutive integers
Any three consecutive integers will be divisible by 24 if the middle integer is odd.

Examples:
if n=1, then (n+1)=2 and (n-1)=0 => n (n-1) (n+1) = 0, which is divisible by any number
if n=3, then (n+1)=4 and (n-1)=2 => n (n-1) (n+1) = 24, which is divisible by 24
if n=5, then (n+1)=6 and (n-1)=4 => n (n-1) (n+1) = 120, which is divisible by 24
=> n (n-1) (n+1) is divisible by 24 for all odd values of n

Statement 1 states that n is a multiple of 3
if n=3 sufficient
if n=6 not sufficient ----> not sufficient

Statement 2 states that n is odd -> sufficient

Answer is B
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Re: State whether n (n² - 1) is divisible by 24 [#permalink]

29 Aug 2009, 13:03

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Answer is B...
It is a standard question...
n(n^2-1 ) is always divisible by 24 for all odd numbers..
I have always known this as a rule from standard number theory books...

Re: State whether n (n² - 1) is divisible by 24 [#permalink] 29 Aug 2009, 13:03

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State whether n (n² - 1) is divisible by 24

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State whether n (n² - 1) is divisible by 24

State whether n (n² - 1) is divisible by 24

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