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If y is an integer, is y^3 divisible by 9 ?
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29 Jan 2014, 01:18
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Re: If y is an integer, is y^3 divisible by 9 ?
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29 Jan 2014, 01:18



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Re: If y is an integer, is y^3 divisible by 9 ?
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29 Jan 2014, 02:07
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



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Re: If y is an integer, is y^3 divisible by 9 ?
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29 Jan 2014, 02:09
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf y is an integer, is y^3 divisible by 9 ? (1) y is divisible by 4. (2) y is divisible by 6. 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.



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Re: If y is an integer, is y^3 divisible by 9 ?
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30 Jan 2014, 02:21
[quote="Bunuel"] The Official Guide For GMAT® Quantitative Review, 2ND EditionIf y is an integer, is y^3 divisible by 9 ? (1) y is divisible by 4. (2) y is divisible by 6. 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. B it is.



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Re: If y is an integer, is y^3 divisible by 9 ?
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02 Jan 2017, 15:01
Nice Official Question. Here is what i did in this one > Using test cases =>
Statement 1> y=4 NO y=12 YES
Hence not sufficient
Statement 2=> y=> divisible by 12 => y has a 3 Thus y^3 will have a 27 Hence sufficient
Hence B
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Re: If y is an integer, is y^3 divisible by 9 ?
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23 Jan 2017, 10:29
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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Re: If y is an integer, is y^3 divisible by 9 ?
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30 Sep 2018, 14:02
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf y is an integer, is y^3 divisible by 9 ? (1) y is divisible by 4. (2) y is divisible by 6. Data Sufficiency Question: 64 Category: Arithmetic Properties of numbers Page: 157 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! Basically the question is asking for \(\frac{y^3}{3^2}\) statement 1) says y/4 so y could be 4,8,12 etc.. if y = 4 then it is not divisible by 3^2 however if y = 12 then y^3 is divisible by 9. Insufficient. Statement 2) y is divisible by 6. This means that y is 6,12,18 etc.. since any multiple of 6 will contain a 3 in it so it will be divisible. B sufficient.



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Re: If y is an integer is y^3 divisible by 9
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03 Dec 2018, 05:08
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