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

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JeffTargetTestPrep

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21 Mar 2017, 06:21

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Burnkeal

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)\(72\sqrt{k}\)

Since x is divisible by 6, we can write x = 6n for some positive integer n. So:

√(288kx) = √(288k * 6n) = √(144 * 2 * k * 2 * 3n) = √(12^2 * 2^2 * k * 3n) = 12 * 2 * √(3kn) = 24√(3kn).

Let’s go through the choices:

A) 24k√3

If n = k, then 24√(3kn) = 24√(3k^2) = 24k√3.

B) 24√k

In order for 24√(3kn) to be equal to 24√k, the value of n would have to be ⅓, as shown here:

24√(3kn) = 24√(3k(⅓)) = 24√k

However, we know that n is a positive integer, and so it cannot be equal to ⅓.

So √(288kx) CANNOT be 24√k.

Answer: B

Anurag06

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23 Feb 2020, 06:45

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Bunuel

Burnkeal

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.

Hi Bunuel

I encountered the same query in MGMAT 1, but couldn't understand the reason of this LHS being greater than RHS.

Because then it doesn't mention about other options due to lack of information.

Please suggest.

valedipalo

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26 Mar 2020, 04:18

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First of all we have rewrite it:
288=25*32
x=6n
So, substituting these values in the initial formula, we obtain 12*sqrt(18*n*k) that is equal to 24*sqrt(3*n*k), in which we know that n has to be an integer to make x multiple of six and because x is positive also n has to be positive.

  A)n could be k
B)n can’t be 1/3
C)n can be 1
D)n can be 2
E)n can be 3

[Answer B]

mansi1997

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22 May 2020, 22:24

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can someone explain how we assumed that k must be equal to x for option A to be true?

ablatt4

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25 Aug 2025, 14:46

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If x is a multiple of 6 we can first factor out sqrt 288xk into sqrt 2^5*3^3*(2*3*x/6)k

If we assume the least possible quantity for x we get 2^6*3^3*k

we can get an even number of 2s or 3s for certain but k will always be to the first power, we can factor anything out of the root except k

we can rule out answer choice A

we have a minimum of 2^3*3^2 "escaping the root" leaving us with 24 sqrt k and maybe a 2 or 3 or both

8*9=72, unless we leave more under the root by making x = 0 then we can't have anything less.

Their is no possible way to get 24 sqrt k

B

Burnkeal

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√3
(B) 24√k
(C) 24√(3k)
(D) 24√(6k)
(E) 72√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.