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If x and y are positive integers and 4500x = y^3, what is the minimum

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If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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New post 15 Dec 2016, 04:30
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A
B
C
D
E

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Re: If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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New post 15 Dec 2016, 05:34
minimum value for x should be 2*3 hence ans is B
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Re: If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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New post 15 Dec 2016, 08:33
Bunuel wrote:
If x and y are positive integers and 4500x = y^3, what is the minimum possible value of x?

A. 5
B. 6
C. 12
D. 30
E. 90


\(4500x = y^3\)

Or, 2^2 * 3^2 * 5^3 * x = y^3

So, The minimum value of x must be 2*3 = 6

Answer will be (B) 6..

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Re: If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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New post 16 Dec 2016, 10:29
Bunuel wrote:
If x and y are positive integers and 4500x = y^3, what is the minimum possible value of x?

A. 5
B. 6
C. 12
D. 30
E. 90


4500= (2^2)*(3^2)*(5^3)
4500x= y^3 ==> x=2*3=6
B
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Re: If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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New post 19 Dec 2016, 10:59
Bunuel wrote:
If x and y are positive integers and 4500x = y^3, what is the minimum possible value of x?

A. 5
B. 6
C. 12
D. 30
E. 90


We are given that 4500x = y^3, which means that 4500x is a perfect cube. We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So, let’s break down 4500 into its prime factors to determine the minimum value of x.

4500 = 45 x 100 = 5 x 9 x 10 x 10 = 5 x 3 x 3 x 2 x 5 x 2 x 5 = 2^2 x 3^2 x 5^3

In order to make 4500x a perfect cube, the smallest value of x is 2^1 x 3^1, so that 4500x = (2^2 x 3^2 x 5^3) x (2^1 x 3^1) = 2^3 x 3^3 x 5^3, which is a perfect cube.

Thus, x = 2^1 x 3^1 = 6.

Answer: B
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Re: If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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New post 20 Dec 2016, 01:15
To answer the question correctly it is needed to understand that if y^3=4500x, than root extraction is possible from 4500x.the
First, we need to do prime factorization of 4500, to discover prime factors and its powers.
So, prime factorization of 4500 = 2^2 * 3^2 * 5^3.
5^3 can be easily extracted from the root, but we still have 2^2 and 3^2, it is obvioud that in order to make it possible to extract from root^3 we need to add 1 power to each of factors 3 & 2, after addition of 3^1*2^1, root^3 extraction has become possible. 3*2 = 6, thus correct answer is B
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Re: If x and y are positive integers and 4500x = y^3, what is the minimum  [#permalink]

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Re: If x and y are positive integers and 4500x = y^3, what is the minimum   [#permalink] 21 Oct 2018, 22:49
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