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:

If \(x^3\) – x = n and x is a positive integer greater than 1, is n divisible by 8?

(1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer.

I will really appreciate if you can tell me whether I am right or wrong. I have gone for D as an answer.

Let's simply the question stem a bit:

\(x^3\) - x = n which will give x (\(x^2\)-1) = n which will give (x-1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.

Now moving on to the statements

Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.

X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.

Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.

Therefore for me both statements alone are sufficient to answer this question.

Yes, your reasoning is correct.

If x^3 - x = n and x is a positive integer greater than 1, is n divisible by 8?

x^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)x(x+1) will be divisible by 8*3=24).

(1) When 3x is divided by 2, there is a remainder --> 3x=odd --> x=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.

(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.

Quick question: I do not understand why, when considering the product (x-1)x(x+1), you conclude that one of the factor is divisible by 4.

Let me explain my line of reasoning here: we know x is odd, so both (x-1) and (x+1) are even, hence my conclusion is that (x-1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x-1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?

As you could see from my post, I did check the previous posts and quoted them but I cannot figure out what is wrong with my reasoning.

(x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 _________________

Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

Show Tags

03 Apr 2013, 10:08

2

This post was BOOKMARKED

Just list out all the even numbers - 0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0

so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.

Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

Show Tags

12 Oct 2013, 06:48

1

This post was BOOKMARKED

nt2010 wrote:

Just list out all the even numbers - 0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0

so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.

Hope this helps..

I believe the last statement would be correct only if the sequence starts with even integer 2. For instance, if there are two consecutive even numbers, 0 and 2, then 4 is not a factor. However, if the two consecutive numbers are 2 and 4, then 4 is a factor of 2*4.

Just list out all the even numbers - 0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0

so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.

Hope this helps..

I believe the last statement would be correct only if the sequence starts with even integer 2. For instance, if there are two consecutive even numbers, 0 and 2, then 4 is not a factor. However, if the two consecutive numbers are 2 and 4, then 4 is a factor of 2*4.

That's not true.

Zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, zero is a multiple of every integer except zero itself).

Thus if (x-1)x(x+1)=0*1*2=0, then the product is still divisible by 8.

\(x^3\) - x = n which will give x (\(x^2\)-1) = n which will give (x-1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.

Now moving on to the statements

Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.

X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.

Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.

Therefore for me both statements alone are sufficient to answer this question.

Statement 1: 3x/2 gives remainder . This means x is odd. odd^3 = odd and odd- odd = even. x >1 and an odd integer . Lets take x = 3 n = x^3 - x = 27-3 n= 24. 24/8 ->Remainder = 0

Lets take x = 5 n = x^3 - x = 125-5 n= 120. 120/8 ->Remainder = 0

Sufficient

Statement 2: x = 4y+1 . y->integer Put x in equation n = x^3 - x n = (4y+1)^3 - (4y+1) apply formula(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 (4y+1)^3 = 64y^3 + 3*16x^2 + 3*4y*1 + 1 = 64y^3 + 48x^2 + 12y + 1

n = 64y^3 + 48x^2 + 12y + 1 - 4y - 1 n = 64y^3 + 48x^2 + 8y n = 8(8y^3 + 6x^2+1) Hence n is divisible by 8. -> Sufficient Hence D

(x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8
_________________

Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]

Show Tags

14 Jan 2017, 05:43

Great Question. Here is what i did in this one=> As n=x^3-x=> (x-1)*x*(x+1)=> If x is odd => both x-1 and x+1 will be consecutive even integers.So one of them would be multiple of 4 and other would be a multiple of 2 for sure.Hence n would be a multiple of 8.

Statement 1-> 3x is odd => x is odd => Sufficient. Statement 2-> x=4y+1 => Even+odd=odd Hence Sufficient.

Hi, I don't understand why if x is odd then (x-1)x(x+1) will be divisible by 8*3=24. Can someone pls help?

x^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers, so one of (x-1), x or (x+1) is divisible by 3. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)x(x+1) will be divisible by 8*3=24).
_________________

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...