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

Question

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

aditi1903 · Accepted Answer

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.

Bunuel · Answer

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.

AKG1593 · Answer

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

joe26219 · Answer

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.

stonecold · Answer

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

kanusha · Answer

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

Salsanousi · Answer

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.

RAHUL_GMAT · Answer

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

ricardorr04 · Answer

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.

Bunuel · Answer

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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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 ?

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