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If y and n are +ve integers and 450y = n^3 which

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If y and n are +ve integers and 450y = n^3 which [#permalink] New post 29 Aug 2018, 07:05
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If y and n are +ve integers and \(450y = n^3\) which of the following must be an integer?

A)\(\frac{y}{3*2^2*5}\)

B)\(\frac{y}{3^2*2*5}\)

C)\(\frac{y}{3*2*5^2}\)

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[Reveal] Spoiler: OA
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Re: If y and n are +ve integers and 450y = n^3 which [#permalink] New post 29 Aug 2018, 12:02
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We first have to predict Y in it's least possible form. Y must be such that after multiplication with 450 it becomes a cube. If we find the prime divisible of 450 we see that
450=3*3*5*5*2

So, to turn the product into a perfect square we need at least two more 2(s), one 3 and one 5. So the least desired value of
Y=3*5*2^(2)=60.
Thus, Y should be a multiple of at least 60 for n^(3) to be a perfect cube.

Now, when we insert the value of Y in the given options only option A pans out. The second option has an extra 3 in its denominator and the third option has an extra 5. So the correct answer is
Option A.

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Scrat
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Re: If y and n are +ve integers and 450y = n^3 which [#permalink] New post 31 Aug 2018, 21:11
Nice Question.
Re: If y and n are +ve integers and 450y = n^3 which   [#permalink] 31 Aug 2018, 21:11
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If y and n are +ve integers and 450y = n^3 which

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