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25 Sep 2010, 11:06
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
48% (01:57) correct
52%
(02:14)
wrong
based on 514
sessions
History
Date
Time
Result
Not Attempted Yet
Is positive integer n – 1 a multiple of 3?
(1) n^3 – n is a multiple of 3
(2) n^3 + 2n^2 + n is a multiple of 3
(1) n^3 – n is a multiple of 3
(2) n^3 + 2n^2 + n is a multiple of 3
For 1, for following values of n, \(n^3 - n\) becomes multiple of 3 : 2,3,4,5,6,7,8....
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
25 Sep 2010, 11:16
Kudos
Bookmarks
Orange08
Is positive integer n – 1 a multiple of 3?
(1) n^3 – n is a multiple of 3
(2) \(n^3 + 2n^2+ n\) is a multiple of 3
For 1, for following values of n, \(n^3 - n\) becomes multiple of 3 : 2,3,4,5,6,7,8....
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
(1) n^3 – n is a multiple of 3
(2) \(n^3 + 2n^2+ n\) is a multiple of 3
For 1, for following values of n, \(n^3 - n\) becomes multiple of 3 : 2,3,4,5,6,7,8....
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
Is positive integer n – 1 a multiple of 3?
(1) n^3 – n is a multiple of 3 --> \(n^3-n=n(n^2-1)=(n-1)n(n+1)=3q\). Now, \(n-1\), \(n\), and \(n+1\) are 3 consecutive integers and one of them must be multiple of 3, so no wonder that their product is a multiple of 3. However we don't know which one is a multiple of 3. Not sufficient.
(2) \(n^3 + 2n^2+ n\) is a multiple of 3 --> \(n^3 + 2n^2+ n=n(n^2+2n+1)=n(n+1)^2=3p\) --> so either \(n\) or \(n+1\) is a multiple of 3, as out of 3 consecutive integers \(n-1\), \(n\), and \(n+1\) only one is a multiple of 3 then knowing that it's either \(n\) or \(n+1\) tells us that \(n-1\) IS NOT multiple of 3. Sufficient.
Answer: B.
25 Sep 2010, 11:15
Kudos
Bookmarks
Orange08
Is positive integer n – 1 a multiple of 3?
(1) \(n^3 – n\) is a multiple of 3
(2) \(n^3 + 2n^2+ n\) is a multiple of 3
For 1, for following values of n, \(n^3 - n\) becomes multiple of 3 : 2,3,4,5,6,7,8....
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
(1) \(n^3 – n\) is a multiple of 3
(2) \(n^3 + 2n^2+ n\) is a multiple of 3
For 1, for following values of n, \(n^3 - n\) becomes multiple of 3 : 2,3,4,5,6,7,8....
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 :
2,3,5,6,7,8....
Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
(1) \(n^3-n = n(n^2-1) = n(n+1)(n-1)\)
This number will always be a multiple of three ... so not sufficient to answer our question
(2) \(n^3 + 2n^2+ n = n(n+1)^2\)
If this is a multiple of 3, then either n is a multiple of 3 or (n+1) is a multiple of 3. In either case, n-1 cannot be a multiple of 3.
Hence this is sufficient
Answer should be (b)
In your solution above, notice that for all the n's in case (2) n-1 is not a multiple of 3.
