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

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Question Stats:

57% (02:17) correct
43% (01:26) wrong based on 358 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

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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).

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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.

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