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 350,000 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 x is positive is x divisible by 2 [#permalink]
22 Mar 2010, 17:39

2

This post received KUDOS

Expert's post

anaik100 wrote:

If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4 (2) 5x+4 is divisible by 6

I don't think that this is a GMAT question. In it's current form it's way beyond the GMAT scope.

First of all note that we are not told that \(x\) is an integer.

Given: \(x>0\). Question: is \(x\) even (integer)?

(1) \(x^3+x=x(x^2+1)=4n\)

If x is an integer then either: A. \(x\) is a multiple of 4 and \(x^2+1\) is odd; OR B. \(x\) is odd and \(x^2+1\) is a multiple of 4. But this scenario is not possible. Why? Because in case \(x=2k+1\) (\(x\) is odd), \(x^2+1=4k^2+4k+2\), which is not a multiple of 4, (it's even never multiple of 4.) So if \(x\) is an integer it must be multiple of 4.

But \(x\) also can be a non-integer, for example equation \(x^3+x=4\) has one real root (\(x\approx{1.38}\)), which is not an integer OR equation \(x^3+x=8\) also has one real root (\(x\approx{1.83}\)), which is not an integer.

Not sufficient.

(2) \(5x+4=6k\)

If \(x\) is an integer, then \(x\) must be even as \(5x+4=even\) --> \(5x=even-4=even\) --> \(x=even\).

But \(x\) can also be a fraction, for example \(\frac{2}{5}\), or \(\frac{8}{5}\) - basically \(x\) can be a fraction of a form \(x=\frac{a}{5}\), where \(a\) is an integer.

Not sufficient.

(1)+(2) Now if \(x\) is an integer then it's even as we concluded.

But can x be reduced fraction of a form \(x=\frac{a}{5}\)? We'll have \(\frac{a^3}{125}+\frac{a}{5}=4n\) --> \(a^3+25a=a(a^2+25)=500n\) --> as \(a\) and \(n\) are integers, \(a\) must be multiple of 5 (at least) --> hence \(x=\frac{a}{5}=integer\), so \(x\) can not be reduced fraction --> and if \(x\) is an integer, then we know it must be even.

Answer: C.

If we were told that \(x\) is an integer, then the answer would be D. _________________

Re: If x is positive is x divisible by 2 [#permalink]
09 Apr 2011, 19:05

If we can assume that x is an integer :

Is x even ?

From (1)

x^3 + x = 4k

x(x^2+1) = 4k

if x = Odd, then X^2 + 1 = even and is divisible by 2, but x is not divisible by 4, and X^2 + 1 needs one more 2 in its factor, which is not possible it it's odd. So x is even.

(Sufficient)

From (2)

5x + 4 = 6k

=> 5x + 4 has 2 and 3 as factors

So, 5x + 4 is even

so 5x + even = even

=> 5x = even

=> x = even

(sufficient)

Answer - D _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Re: If x is positive is x divisible by 2 [#permalink]
07 Sep 2011, 01:59

Quote:

If x is positive is x divisible by 2 ?

(1) x^3+x is divisible by 4 (2) 5x+4 is divisible by 6

x>0. is x a multiple of 2 (i.e. is x an even number)?

I got B for answer. I am working based on the assumption that x is an integer, so that the question will be within the scope of GMAT..

Rules applied: even + even = even odd + even = odd odd + odd = even even * even = even even * odd = even Statement 1 x^3+x=4k ---> k being an integer x(x^2+1)=4k

if x is even, (x^2+1) will be odd, multiplying both results in 4k (even). if x is odd, (x^2+1) will be even, which also results in 4k (even) --> insufficient

Statement 2 5x+4=6w ---> w being an integer if x is even --> 5x will be even --> even + even = even --> results in even outcome i.e. 6w (true) if x is odd --> 5x will be odd --> odd + even = odd --> results in odd outcome i.e. cannot be 6w So x must be even. --> sufficient

Answer: B _________________

"The best day of your life is the one on which you decide your life is your own. No apologies or excuses. No one to lean on, rely on, or blame. The gift is yours - it is an amazing journey - and you alone are responsible for the quality of it. This is the day your life really begins." - Bob Moawab

gmatclubot

Re: If x is positive is x divisible by 2
[#permalink]
07 Sep 2011, 01:59

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...