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Re: What is the smallest positive integer x, such that 1,152x is a perfect [#permalink]

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30 Apr 2016, 02:32

Bunuel wrote:

What is the smallest positive integer x, such that 1,152x is a perfect cube?

A. 4 B. 6 C. 8 D. 12 E. 18

Take out the factors of 1152 that will come 2*2*2*2*2*2*2*3*3. for perfect cube you need every no. raise to the power 3. for 1,152x to be a perfect cube, you need two 2 and 1 3 that means 12. D is the answer.

What is the smallest positive integer x, such that 1,152x is a perfect [#permalink]

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25 Jun 2016, 04:00

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prime factorization 0f 1152 = 2*2*2*2*2*2*2*3*3 = \(2^{7}\) *\(3^2\) for a number to be perfect cube it should have powers of its prime factors in multiple of 3 (3,6 ,9 ...etc) we need \(2^2\) *3 = 12 to make the given number a perfect cube so x =12 Correct Answer - D

Re: What is the smallest positive integer x, such that 1,152x is a perfect [#permalink]

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20 Jul 2016, 09:03

Bunuel wrote:

What is the smallest positive integer x, such that 1,152x is a perfect cube?

A. 4 B. 6 C. 8 D. 12 E. 18

It seems the question is asking if x is an integer placed in front of the number making it either a 5 digit perfect cube or a six digit perfect cube. please post questions with clarity. The language here should have been : what should be the minimum value of integer x when multiplied by 1152 yields a perfect cube? Thanks!

What is the smallest positive integer x, such that 1,152x is a perfect cube?

A. 4 B. 6 C. 8 D. 12 E. 18

It seems the question is asking if x is an integer placed in front of the number making it either a 5 digit perfect cube or a six digit perfect cube. please post questions with clarity. The language here should have been : what should be the minimum value of integer x when multiplied by 1152 yields a perfect cube? Thanks!

Re: What is the smallest positive integer x, such that 1,152x is a perfect [#permalink]

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29 Nov 2016, 20:31

1152 = (2^7)(3^2)

To get a perfect cube, we need to get the exponents (listed above) so that they are multiples of 3 --> closest multiple for 2^7 is 2^9 --> we need 2^2 to do this (KEEP)

Let's do the same for 3^2: closest multiple is 3^3 --> we need 3 to do this (KEEP)

X should have both 2^2 & 3 in order to transform 1152 into a perfect cube.

What is the smallest positive integer x, such that 1,152x is a perfect cube?

A. 4 B. 6 C. 8 D. 12 E. 18

------ASIDE-------------------------------------- Let's examine a perfect CUBE.

1000 is a perfect cube since 1000 = 10^3 Now check out the prime factorization of 1000 1000 = (2)(2)(2)(5)(5)(5) = [(2)(5)][(2)(5)][(2)(5)] Notice that we can take the prime factorization of 1000 and divide the prime factors into THREE identical groups -----NOW ONTO THE QUESTION------------------------

1,152 = (2)(2)(2)(2)(2)(2)(2)(3)(3) So, 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(x) Let's test the answer choices....

A. x = 4 We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(4) = (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(2) In order for the above to be a perfect CUBE, we must be able to divide the prime factors into THREE identical groups. Since we cannot do that here, we can ELIMINATE A

B. x = 6 We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(6) = (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(3) Can we divide the above prime factors into THREE identical groups? NO! ELIMINATE B

C. x = 8 We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(8) = (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(2)(2) Can we divide the above prime factors into THREE identical groups? NO! ELIMINATE C

D. x = 12 We get: 1152x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(12) = (2)(2)(2)(2)(2)(2)(2)(3)(3)(2)(2)(3) Can we divide the above prime factors into THREE identical groups? YES! 1152x = [(2)(2)(2)(3)][(2)(2)(2)(3)][(2)(2)(2)(3)] So, if x = 12, then 1152x IS a perfect cube.

Answer: D

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Re: What is the smallest positive integer x, such that 1,152x is a perfect
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02 Dec 2017, 15:51