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Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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?

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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?

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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.

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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

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.....

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
25 Oct 2005, 04:43

1

This post received KUDOS

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.

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
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.

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. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
19 Nov 2014, 22:55

Almost everyone chose answer (A), even GMAT review says (A) is the answer.

I disagree, question is not complete.

In statement (1), it tells, K is an integer in the question. It never say it is a positive integer, hence K may have range of values for -ve to +ve including zero.

When you substitute -1, or 0 in place of K, you can not say it is divisible by 4.

People discussed on this problem, simply took for granted that, K is integer means, it will have values from 1,2...... which is I disagree.

In question stem, it talks about n as a positive integer, where as in statement (1), K is not specified as positive integer, hence pretty much it can have any integer values, ...-1,0,1...

Asking anyone answer if my understanding is correct or not...

Re: If n is a positive integer, is n^3 – n divisible by 4 ? [#permalink]
20 Nov 2014, 06:59

Expert's post

1

This post was BOOKMARKED

rjkbecse wrote:

Almost everyone chose answer (A), even GMAT review says (A) is the answer.

I disagree, question is not complete.

In statement (1), it tells, K is an integer in the question. It never say it is a positive integer, hence K may have range of values for -ve to +ve including zero.

When you substitute -1, or 0 in place of K, you can not say it is divisible by 4.

People discussed on this problem, simply took for granted that, K is integer means, it will have values from 1,2...... which is I disagree.

In question stem, it talks about n as a positive integer, where as in statement (1), K is not specified as positive integer, hence pretty much it can have any integer values, ...-1,0,1...

Asking anyone answer if my understanding is correct or not...

No, you are wrong. Yes, k could be any integer, positive, negative, or zero but this does not change the answer. For example, if k is 0 or -1, then n^3 - n = 0, which IS divisible by 4: 0 is divisible by every integer except 0 itself.

If n is a positive integer, is n^3 – n divisible by 4 ?

n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible bu 4.

(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.

(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.

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