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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
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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: If k and x are positive integers and x is divisible by 6 [#permalink] New post 03 May 2015, 05:46
I am really bad with number properties... So what I did was estimating the squareroot of 288XX which I thought to be something around 170.
Then I went through the solutions estimating which of them could be around 170, and I could elimate B.
24 times squareroot of 9 ( the highest digit for k) would def be less than 170
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Re: If k and x are positive integers and x is divisible by 6 [#permalink] New post 03 May 2015, 07:34
I did an enormous mistake on this one. Might as well include my poor reasoning.

I mistook 288 for 2*8*8 and put 2^7*k*2*3*n under the root.

:)
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Re: If k and x are positive integers and x is divisible by 6 [#permalink] New post 04 May 2015, 02:50
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ZLukeZ wrote:
I am really bad with number properties... So what I did was estimating the squareroot of 288XX which I thought to be something around 170.
Then I went through the solutions estimating which of them could be around 170, and I could elimate B.
24 times squareroot of 9 ( the highest digit for k) would def be less than 170


Hi ZLukeZ,

This is definitely not the way to solve this question. Even though you got the right answer, let me tell you the gaps in your methodology:

1. Please note that k and x are not single unit digits at the tens and units place of the number 288kx. The number is 288 *k *x. Hence, the number is not around 288XX for which the square root will be around 170. x can be any multiple of 6 and k can be any other positive integer. So value of 288kx can be far greater than 288XX.

2. Since k and x may not be single unit digits, assuming maximum value of k to be 9 is not correct. k is a positive integer and can take any value which may be greater than 9.

Simplifying the number by prime factorizing and then evaluating the options would be the right way to go about this question.

Hope its clear! Let me know if you need help at any point of this question.

Regards
Harsh
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Re: If k and x are positive integers and x is divisible by 6   [#permalink] 04 May 2015, 02:50
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