Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

24 Oct 2005, 22:31

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]

Show Tags

24 Oct 2005, 22:33

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]

Show Tags

24 Oct 2005, 22:36

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]

Show Tags

24 Oct 2005, 22:45

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]

Show Tags

24 Oct 2005, 22:53

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]

Show Tags

24 Oct 2005, 22:54

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]

Show Tags

25 Oct 2005, 05:43

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]

Show Tags

25 Oct 2005, 05:54

1

This post received KUDOS

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]

Show Tags

19 Nov 2014, 23:55

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...

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.

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...