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If k and x are positive integers and x is divisible by 6 [#permalink]
 29 Oct 2010, 17:10
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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]
29 Oct 2010, 17:21
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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]
29 Oct 2010, 17:27
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]
19 Feb 2011, 04:22
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]
19 Feb 2011, 06:36
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]
19 Feb 2011, 08:01
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Re: Exponents & Roots MGMAT [#permalink]
28 Apr 2011, 13:08
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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If k and x are positive integers [#permalink]
06 Nov 2011, 02:52
Did not find this on searching hence posting
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)
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Re: If k and x are positive integers [#permalink]
06 Nov 2011, 04:26
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]
06 Nov 2011, 10:26
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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]
08 Nov 2011, 00:14
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 [#permalink]
12 Nov 2011, 18:00
got B. Used same strategy as others
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Re: If k and x are positive integers [#permalink]
04 Jan 2012, 23:26
B is the answer. Used a similar approach as did others. Nice question.
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Re: If k and x are positive integers
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04 Jan 2012, 23:26
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