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If k and x are positive integers and x is divisible by 6

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If k and x are positive integers and x is divisible by 6 [#permalink] New post 29 Oct 2010, 16:10
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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.
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Re: Exponents & Roots MGMAT [#permalink] New post 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.

\sqrt{288kx}=\sqrt{2^5*3^2*k*6n}=\sqrt{2^6*3^3*kn}=2^3*3\sqrt{3kn}=24\sqrt{3kn}.

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

Answer: B.
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Re: Exponents & Roots MGMAT [#permalink] New post 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.

It looks so easy when you do it :-)

Thanks for that Bunuel.
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Re: Exponents & Roots MGMAT [#permalink] New post 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...
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Re: Exponents & Roots MGMAT [#permalink] New post 19 Feb 2011, 05:36
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?
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Re: Exponents & Roots MGMAT [#permalink] New post 19 Feb 2011, 07:01
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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.
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Re: Exponents & Roots MGMAT [#permalink] New post 28 Apr 2011, 12:08
katealpha wrote:
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.
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Re: If k and x are positive integers [#permalink] New post 06 Nov 2011, 03:26
k is a divisible by 6. So lets say k=6n

sqrt(288*k*x) = sqrt (12*12*2*2*3*n*x) = 12*2*sqrt(3*n*x)= 24sqrt(3*n*x).

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.
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Re: If k and x are positive integers [#permalink] New post 06 Nov 2011, 09:26
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X is given to be divisible by 6, so let X = 6n, where n is a positive integer.

Let's rearrange the given value, \sqrt{288kx}:

\sqrt{288kx}

= \sqrt{288k(6n)}

= \sqrt{1728kn}

= \sqrt{1728kn}

= \sqrt{576(3)kn}

= 24\sqrt{(3n)k}

Given this, the only option provided where n can't be set to a value which satisfies the given formula is B.
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Re: If k and x are positive integers [#permalink] New post 07 Nov 2011, 23:14
288= 4*4*3*3*2
thus (288)^1/2 = 12*(3)^1/2
x=6n,check for values n = 1,2,3,4 etc.
hence,
24*(k*6n)^1/2

for n=1, C is POE.
for n=2, D is POE.
for n=3, E is POE.
for n=6k, A is POE.
thus, B remains.
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Re: If k and x are positive integers and x is divisible by 6 [#permalink] New post 26 Sep 2013, 12:02
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Re: Exponents & Roots MGMAT [#permalink] New post 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.

\sqrt{288kx}=\sqrt{2^5*3^2*k*6n}=\sqrt{2^6*3^3*kn}=2^3*3\sqrt{3kn}=24\sqrt{3kn}.

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

Answer: B.



Can you explain how you know that? I got C for an answer...
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Re: Exponents & Roots MGMAT [#permalink] New post 06 Nov 2013, 02:16
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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.

\sqrt{288kx}=\sqrt{2^5*3^2*k*6n}=\sqrt{2^6*3^3*kn}=2^3*3\sqrt{3kn}=24\sqrt{3kn}.

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

Answer: B.



Can you explain how you know that? I got C for an answer...


That's because for any values of positive integers k and n, \sqrt{3kn}>\sqrt{k}
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Re: Exponents & Roots MGMAT [#permalink] New post 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.

\sqrt{288kx}=\sqrt{2^5*3^2*k*6n}=\sqrt{2^6*3^3*kn}=2^3*3\sqrt{3kn}=24\sqrt{3kn}.

Now, for any values of positive integers k and n 24\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!
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Re: Exponents & Roots MGMAT [#permalink] New post 12 May 2014, 02:26
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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.

\sqrt{288kx}=\sqrt{2^5*3^2*k*6n}=\sqrt{2^6*3^3*kn}=2^3*3\sqrt{3kn}=24\sqrt{3kn}.

Now, for any values of positive integers k and n 24\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.

Does this make sense?
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Re: Exponents & Roots MGMAT   [#permalink] 12 May 2014, 02:26
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