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.

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 k and x are positive integers and x is divisible by 6 [#permalink]
29 Oct 2010, 16:10

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

54% (02:15) correct
46% (01:27) wrong based on 322 sessions

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT. I also thought it could be interesting as I have not found a previous thread with this question.

Re: Exponents & Roots MGMAT [#permalink]
29 Oct 2010, 16:21

5

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

Burnkeal wrote:

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> x=6n for some positive integer n.

Re: Exponents & Roots MGMAT [#permalink]
29 Oct 2010, 16:27

Thanks. It is kind of the same approach of the one in MGMAT so I guess I am gonna have to stick to this. I guess I don't have the initiative to say x=6n and then to compare.

Re: Exponents & Roots MGMAT [#permalink]
19 Feb 2011, 03:22

This is not an ez one... u can also put x=6, k=1 and you can see A,B,C goes out quickly bc u found a very ez to spot solution for them... _________________

Re: Exponents & Roots MGMAT [#permalink]
19 Feb 2011, 07:01

1

This post received KUDOS

Expert's post

beyondgmatscore wrote:

How can one be sure that the expression here is 288*k*x and not 28800 + 10*K + x where K and x are integers?

Only multiplication sign is usually omitted so 288kx means 288*k*x. If it were meant to be 5-digit integer 288kx, where k is a tens digit and x is a units digit then this would be explicitly mentioned in the stem. _________________

I still dont understand it. It is clear that 24\sqrt{3kn} > 24\sqrt{k}. But isn't it also greater than 24\sqrt{3k} ?

Not really. For, x=6, n=1; 24\sqrt{3kn}=24\sqrt{3k}

Likewise; For, n=k; 24\sqrt{3kn}=24\sqrt{3k^2}=24k\sqrt{3} For, x=12, n=2; 24\sqrt{3kn}=24\sqrt{6k} For, x=18, n=3; 24\sqrt{3kn}=24\sqrt{3^2k}=72\sqrt{k}

But to make 24\sqrt{3kn}=24\sqrt{k}, n should be 1/3 and we know that n is an integer. Not possible. _________________

Option a- Posible if n*x=k^2 Option b- Not possible as x is an integer, it cannot be a fraction. (6n=k=an integer, x is an integer). For this to be true, k=1 and x=2 not possible as k is a multiple of 6 or k=3 again not possible.

Re: If k and x are positive integers and x is divisible by 6 [#permalink]
26 Sep 2013, 12:02

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Exponents & Roots MGMAT [#permalink]
05 Nov 2013, 12:44

Bunuel wrote:

Burnkeal wrote:

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> x=6n for some positive integer n.

Re: Exponents & Roots MGMAT [#permalink]
06 Nov 2013, 02:16

1

This post received KUDOS

Expert's post

AccipiterQ wrote:

Bunuel wrote:

Burnkeal wrote:

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> x=6n for some positive integer n.

Re: Exponents & Roots MGMAT [#permalink]
11 May 2014, 14:08

Bunuel wrote:

Burnkeal wrote:

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> x=6n for some positive integer n.

Now, for any values of positive integers k and n24\sqrt{3kn} is always more than 24\sqrt{k} (B): 24\sqrt{3kn}>24\sqrt{k} --> \sqrt{3n}>1.

Answer: B.

Makes complete sense when the steps are listed out in such a way. Question though -- when I read the statement that x is divisible by 6, I started to work with the primes (2 & 3) and tried to correlate that to the prime of (288). Why is that approach incorrect in this problem?

I realize that there is no sure fire way to encompass every problem but it just seems odd that we solve a lot of problems by factoring down to their primes but in this case, we added an unknown variable to solve the problem?

Re: Exponents & Roots MGMAT [#permalink]
12 May 2014, 02:26

Expert's post

russ9 wrote:

Bunuel wrote:

Burnkeal wrote:

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> x=6n for some positive integer n.

Now, for any values of positive integers k and n24\sqrt{3kn} is always more than 24\sqrt{k} (B): 24\sqrt{3kn}>24\sqrt{k} --> \sqrt{3n}>1.

Answer: B.

Makes complete sense when the steps are listed out in such a way. Question though -- when I read the statement that x is divisible by 6, I started to work with the primes (2 & 3) and tried to correlate that to the prime of (288). Why is that approach incorrect in this problem?

I realize that there is no sure fire way to encompass every problem but it just seems odd that we solve a lot of problems by factoring down to their primes but in this case, we added an unknown variable to solve the problem?

Thanks!

We are told that a positive integer x is divisible by 6. It can be 6, 12, 18, 24, ... So, it can have some other primes than 2 and 3. That's why we need to represent it as 6n, where n is a positive integer and not just assume that it's 6.

As for factorization: as you can see that along the way we do factorize 6n into 2*3*n.