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If n is a positive integer, is n^3 - n divisible by 4? (1) n

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If n is a positive integer, is n^3 - n divisible by 4? (1) n [#permalink] New post 24 Oct 2005, 21:12
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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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 [#permalink] New post 24 Oct 2005, 21:31
I got A.

Q: (n^3 - n)/4 = int ?
or n(n^2-1)/4 = int?
or (n-1)n(n+1)/4 = int?
Since n is an int, (n-1), n, (n+1) indicate consecutive int's. Which means if n is ODD then it's SUFF

1) says n = 2K+1 that is 3,5,7... when K=1,2,3...
Since it's saying n is ODD, SUFF => AD

2) says (n^2 + n)/6 = int
or n(n+1)/6=int
This can be true when n=5 or n=6 (among other values of course)
So we don't know for sure if n is ODD
NOT SUFF => A
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Re: OG DS 232 [#permalink] New post 24 Oct 2005, 21:33
TeHCM wrote:
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


(1) n= 2k+1
n^3-n= n(n^2-1)= n (n-1)(n+1) = (2k+1)( 2k)(2k+2)= 4k ( 2k+1) (k+1)
since k is integer ==> n is divisible by 4 ---> suff
(2)n= 2---> n^2+n= 6
n^3-n= 6 is not divisible by 4----> insuff

Btw, since the product of three consecutive numbers is divisible by 6. The information in 2 is useless, it doesn't provide further information to conclude that n^3-n is divisible by 4
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 [#permalink] New post 24 Oct 2005, 21:36
mbaqst wrote:
I got A.

Q: (n^3 - n)/4 = int ?
or n(n^2-1)/4 = int?
or (n-1)n(n+1)/4 = int?
Since n is an int, (n-1), n, (n+1) indicate consecutive int's. Which means if n is ODD then it's SUFF

1) says n = 2K+1 that is 3,5,7... when K=1,2,3...
Since it's saying n is ODD, SUFF => AD

2) says (n^2 + n)/6 = int
or n(n+1)/6=int
This can be true when n=5 or n=6 (among other values of course)
So we don't know for sure if n is ODD
NOT SUFF => A


Great explanation. Could you explain why you say when n is odd, its divisible by 4?
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 [#permalink] New post 24 Oct 2005, 21:45
TeHCM wrote:
mbaqst wrote:
I got A.

Q: (n^3 - n)/4 = int ?
or n(n^2-1)/4 = int?
or (n-1)n(n+1)/4 = int?
Since n is an int, (n-1), n, (n+1) indicate consecutive int's. Which means if n is ODD then it's SUFF

1) says n = 2K+1 that is 3,5,7... when K=1,2,3...
Since it's saying n is ODD, SUFF => AD

2) says (n^2 + n)/6 = int
or n(n+1)/6=int
This can be true when n=5 or n=6 (among other values of course)
So we don't know for sure if n is ODD
NOT SUFF => A


Great explanation. Could you explain why you say when n is odd, its divisible by 4?


your can refer to my above explanation :wink:
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 [#permalink] New post 24 Oct 2005, 21:53
TeHCM wrote:
mbaqst wrote:
I got A.

Q: (n^3 - n)/4 = int ?
or n(n^2-1)/4 = int?
or (n-1)n(n+1)/4 = int?
Since n is an int, (n-1), n, (n+1) indicate consecutive int's. Which means if n is ODD then it's SUFF

1) says n = 2K+1 that is 3,5,7... when K=1,2,3...
Since it's saying n is ODD, SUFF => AD

2) says (n^2 + n)/6 = int
or n(n+1)/6=int
This can be true when n=5 or n=6 (among other values of course)
So we don't know for sure if n is ODD
NOT SUFF => A


Great explanation. Could you explain why you say when n is odd, its divisible by 4?


Because for any 3 consec positive int's if the middle number is odd then the the other two are even, which means the other two numbers are each divisible by 2.. their product effectively is then divisible by 2*2 or 4.
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 [#permalink] New post 24 Oct 2005, 21:54
laxieqv wrote:
TeHCM wrote:
mbaqst wrote:
I got A.

Q: (n^3 - n)/4 = int ?
or n(n^2-1)/4 = int?
or (n-1)n(n+1)/4 = int?
Since n is an int, (n-1), n, (n+1) indicate consecutive int's. Which means if n is ODD then it's SUFF

1) says n = 2K+1 that is 3,5,7... when K=1,2,3...
Since it's saying n is ODD, SUFF => AD

2) says (n^2 + n)/6 = int
or n(n+1)/6=int
This can be true when n=5 or n=6 (among other values of course)
So we don't know for sure if n is ODD
NOT SUFF => A


Great explanation. Could you explain why you say when n is odd, its divisible by 4?


your can refer to my above explanation :wink:


:lol: Actually that's how I solved the problem too. But I wanted to see if there are other ways.

I still wanna know why when n is odd, its divisible by 4..... :wink:
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 [#permalink] New post 25 Oct 2005, 00:49
Quote:
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


In Statement 1, can't k=0? Doesn't that possibility make the statement insufficient?

Statement 2 combined with Statement 1 rules out the possibility of k being equal to 0, so at the least, the answer should be C.
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 [#permalink] New post 25 Oct 2005, 03:46
BumblebeeMan wrote:
Quote:
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


In Statement 1, can't k=0? Doesn't that possibility make the statement insufficient?

Statement 2 combined with Statement 1 rules out the possibility of k being equal to 0, so at the least, the answer should be C.


yes, k can be 0, then n=1, n^3-n = 1-1= 0 , 0 is still divisible by 4.
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 [#permalink] New post 25 Oct 2005, 03:51
Quote:
yes, k can be 0, then n=1, n^3-n = 1-1= 0 , 0 is still divisible by 4.


I was assuming that (n^3 - n) has to be a multiple of 4, not just divisible by 4. Every number is divisible by 4.
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Re: OG DS 232 [#permalink] New post 25 Oct 2005, 04:43
TeHCM wrote:
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


n^3 -n can be written as n(n^2-1) = n(n+1)(n-1), if n = (2K+1) then
n^3 -n = (2K+1)(2K+2)(2K)
If K = 1 then n^3-n = 3*4*1 divisible by 4
If k = 2 then n^3-n = 5*6*4 divisible by 4
if k =3 then n^3-n = 7*8*6 divisible by 4
if k =4 then n^3 -n is divisible by 4.
So A is sufficient.

From statement 2, we get n(n+1) = 0 mod 6 then if (n^3-n) is divisible by 4 then (n-1) must be divisible by 2. If n is odd then this is true, but if n is even then it is false. So statement 2 is not sufficient.

My answer is A.
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 [#permalink] New post 25 Oct 2005, 04:45
Krisrini,

What if k=0. Statement 1 is no longer sufficient.
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 [#permalink] New post 25 Oct 2005, 04:54
BumblebeeMan wrote:
Krisrini,

What if k=0. Statement 1 is no longer sufficient.


I don't think there're any difference between " be a multiple of" and " be divisible by"!
"be a multiple of" means that n can be written n= xk ( x is integer)
"be divisible by" means that there's a x ( an integer) that multiples with k yield n.
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 [#permalink] New post 25 Oct 2005, 05:01
I'm pretty sure that 0 is not a multiple. I thought that multiples had to be positive integers.

Anybody else care to chime in on this?
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 [#permalink] New post 25 Oct 2005, 08:47
http://www.gmatclub.com/phpbb/viewtopic ... 051#124051

These are the most basic principles. I suggest people who feel not clear on these principles download the file attached in the opening post of that thread and read through it. :)
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 [#permalink] New post 26 Oct 2005, 00:00
BumblebeeMan wrote:
Krisrini,

What if k=0. Statement 1 is no longer sufficient.


When we say a number is divisible by 4 it means the remainder is zero. or in other words it is 0 mod 4.

When k = 0 the expression would evaluate to zero.Zero when divided by 4 is zero. Hence statement1 still holds good.
  [#permalink] 26 Oct 2005, 00:00
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