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If n is a positive integer, is n^3 – n divisible by 4? [#permalink]
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18 Aug 2009, 07:53
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If n is a positive integer, is n^3 – n divisible by 4 ? (1) n = 2k + 1, where k is an integer (2) n^2 + n is divisible by 6
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Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]
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05 Sep 2009, 18:35



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Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]
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06 Sep 2009, 03:14
[quote="bambazoon"]This question is from the 12th ed. of the official guide and I think there's something wrong with the solution. I'm sure others have done this question already, but I haven't been able to find a related topic. The question goes: If n is a positive integer, is n^3  n divisible by 4? 1) n = 2k + 1, where k is an integer 2) n^2 + n is divisible by 6 (n1)(n+1)n ........ 2 odd , even or 2 even,odd from 1 suff from 2 (n)(n+1) = 6n..........even , odd we dont know the 3rd..........insuff A



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Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]
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06 Sep 2009, 10:11
Is n(n+1)(n1) divisible by 4, n is positive stat1: n= 2K+1 => n is odd, in that case (n1) and (n+1) are even => the product should be divisible by 4 ....suff.
stat2:n(n+1) is divisible by 6 => one no. is even( multiple of 2) and other one is odd and odd no. is multiple of 3 => the product may or may not be divisible by 4 (divisibility will hold only for even no.s those are multiple of 4)...insuff. IMO A



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Re: OG, 11th edition, #147 [#permalink]
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03 Feb 2010, 19:57
luminousmocha wrote: If n is a positive integer, is n^3n divisible by 4?
1. n=2k+1, where k is an integer 2. n^2+n is divisible by 6
got stuck on this one for a while.. It's a divisibility question, and factoring (prime factorization or algebraic factorization, depending on the question) is pretty much always the key to divisibility questions. We have: n^3  n = n(n^2  1) = n(n1)(n+1) = (n1)(n)(n+1) so we can see that n^3n is just the product of the three consecutive integers, n1, n and n+1. Now S1 tells us that n is odd, so n1 and n+1 are both even (divisible by 2), and therefore (n1)(n)(n+1) is divisible by 4. S2 is less useful; it tells us that n(n+1) is divisible by 2*3, but it might be that n=2, in which case the answer to our question is 'no', or that n=3, in which case the answer to our question is 'yes'. So S1 is sufficient, S2 is not.
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Re: OG, 11th edition, #147 [#permalink]
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03 Feb 2010, 21:03
Agreed, should be A. I, incorrectly, thought that if n = 1 and so n(n1)(n+1) = 0 is not divisible by 4. but 0 is divisible by any number.



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Re: OG, 11th edition, #147 [#permalink]
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03 Feb 2010, 23:30
Answer is A, if we put n=2k+1 in n(n+1)(n1) then on simplifying it gives n(4k) which shows it is divisible by 4. Statement 2 is not useful to solve the problem.



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Re: OG, 11th edition, #147 [#permalink]
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15 Mar 2010, 03:46
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Answer is A stmt1: since the equation n^3  n = n (n^21) = n(n1)(n+1) and replace it with n = 2k+1 (2k+1)2k(2k+2) = 4(2k+1)(k+1) so it is clearly divisible by 4 stmt2 n^2+n = n(n+1) is divisible by 6 but still we have one more term n1 in our equation which we dont know if even so insuff
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Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]
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23 Nov 2011, 13:27
Hi there, I have a question: Can I not just test cases for statement 1? Say k= 1, 2 and 5, then n would be 3, 5 and 11, respectively. Then plug into n(cube)n/4: 3 and 5 would work, 11 wouldn't > insufficient.



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Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]
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14 Jan 2012, 05:53
If n is a positive integer, is n^3  n divisible by 4?
1) n = 2k + 1, where k is an integer
2) n^2 + n is divisible by 6
Statement 1: n = 2k + 1
(2k + 1)^3 =
(4k^2 + 4k + 1)(2k + 1) =
8k^3 + (4k^2 + 8k^2) + (4k + 2k + 1).
n^3  n is derived by subtracting 2k + 1 from the binomial.
8k^3 + (4k^2 + 8k^2) + (4k)
Each term of n^3  n is divisible by 4 where n is odd.
Statement 2: n^2 + n is divisible by 6
If n = 2, n^2 + n = 6, and (2)^3  2 = 6 is not divisible by 4. However, if n = 3, n^2 + n = 12, and (3)^3  3 = 24 is divisible by 4.
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Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]
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02 Jun 2014, 23:54
first statement says k is an integer....so if k=0 then n will be 1 and the product of n(n1)(n+1) will be zero....in that it is not divisible by 4
pls illustrate



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Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]
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Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]
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Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]
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13 Dec 2015, 02:54
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If n is a positive integer, is n^3 – n divisible by 4 ? (1) n = 2k + 1, where k is an integer (2) n^2 + n is divisible by 6 > When you modify the condition and problem, n^3n=4t?(t is a positive integer) > (n1)n(n+1)=4t? and n1=even?. Since n1=even and n+1=even, it becomes yes. In 1), n1=2k and it is sufficient. In 2), from n(n+1)=6m(m is a positve integer), n=3 yes n=2 no, which is not sufficient. Therefore, the answer is A. > For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]
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