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Bunuel
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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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[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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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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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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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.

Data Sufficiency
Question: 64
Category: Arithmetic Properties of numbers
Page: 157
Difficulty: 600

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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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Whenever you see language like this "y is divisible by 4"

Always consider Y = 4 A

(4A)^3 = May or may not be divisible by 9

(6A)^3 = is surely divisible by 9
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Hi Bunuel - i have a problem with this question. in the stimulus, they only say that Y is an integer, so it could be zero. Isnt zero divisible by any integer but itself?

In case that were true, wouldn't the answer be E? otherwise then it would clearly be B.

Thank you and appreciate the help.
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ricardorr04
Hi Bunuel - i have a problem with this question. in the stimulus, they only say that Y is an integer, so it could be zero. Isnt zero divisible by any integer but itself?

In case that were true, wouldn't the answer be E? otherwise then it would clearly be B.

Thank you and appreciate the help.

Yes, 0 is an integer and it is divisible by every integer except 0 itself. But if y = 0, then still from (2) we have the same yes answer: y^3 = 0^3 = 0 is divisible by 9.
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Bunuel
If y is an integer, is y^3 divisible by 9 ?

(1) y is divisible by 4.
(2) y is divisible by 6.





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