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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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Re: If y is an integer, is y^3 divisible by 9 ? [#permalink]

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29 Jan 2014, 03:07

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From S1:y=4k y^3=64*k^3.Insufficient as K could take any value and accordingly y^3 could or could not be div by 9. From S2:y=6k y^3=216k^3.Regardless of value of k,y^3 will be div by 9 because 216 is div by 9. Sufficient. Ans.B

IMO B takind A alone, if y is divisible by 4 then y= 4n therefore , y^3 = (4n)^3 = 64n^3 . thus we cant say anything about the divisibility by 9 since it will not depend totally on the value of n INSUFFICIENT

taking B alone, if y is divisible by 6 then y= 6n therefore , y^3 = (6n)^3 = (2*3*n)^3 = (8* 9*3) n^3. thus we can say that it is divisible by 9. SUFFICIENT.

Statement 1 : It says y is a multiple of 4 i.e. y=4k , but here k may or may not be a multiple of 3 and thus it is not sufficient to say if y^3 will be divisible by 9.

Statement 2 : It says y = 6k and thus y^3=216(k^3) ; 216 being divisible by 9, y^3 is divisible by 9 for sure irrespective of the value of k.It is sufficient.

Re: If y is an integer, is y^3 divisible by 9 ? [#permalink]

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07 May 2015, 08:54

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Re: If y is an integer, is y^3 divisible by 9 ? [#permalink]

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30 Nov 2016, 21:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If y is an integer, is y^3 divisible by 9 ? [#permalink]

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23 Jan 2017, 11:29

Is y = \(\frac{y^{3}}{9}\) 1, \(\frac{Y}{4}\) For 4(2x2 factors ) For 9 (3x3 factors ) so Insufficient 2, \(\frac{Y}{6}\) 6(2x3 factors) 6^3 (2x2x2x3x3x3) Divisible by 9(3x3 factors) B Bunuel is this a right approach? Pls help. thanks
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