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# If y is an integer, is y^3 divisible by 9 ?

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

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29 Jan 2014, 02:18
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25% (medium)

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68% (00:57) correct 32% (01:18) wrong based on 500 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If 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

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

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29 Jan 2014, 02:18
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SOLUTION

If y is an integer, is y^3 divisible by 9 ?

(1) y is divisible by 4. Clearly insufficient: y can be 4 or 9*4.

(2) y is divisible by 6 --> y is divisible by 3 --> y^3 is divisible by 9. Sufficient.

Answer: B.
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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

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

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29 Jan 2014, 03:09
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If 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 ? [#permalink]

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30 Jan 2014, 03:21
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[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

If 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 ? [#permalink]

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01 Feb 2014, 10:07
SOLUTION

If y is an integer, is y^3 divisible by 9 ?

(1) y is divisible by 4. Clearly insufficient: y can be 4 or 9*4.

(2) y is divisible by 6 --> y is divisible by 3 --> y^3 is divisible by 9. Sufficient.

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

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02 Jan 2017, 16:01
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 ? [#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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Re: If y is an integer, is y^3 divisible by 9 ?   [#permalink] 23 Jan 2017, 11:29
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# If y is an integer, is y^3 divisible by 9 ?

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