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# if y is an integer, is y^3 divisible by 9? 1) y is divisible

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Director
Joined: 23 May 2008
Posts: 739
if y is an integer, is y^3 divisible by 9? 1) y is divisible  [#permalink]

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29 Sep 2008, 22:11
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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Senior Manager
Joined: 29 Mar 2008
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29 Sep 2008, 22:24
if y is an integer, is y^3 divisible by 9?

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

ST1) Y can be 4 or 24. When Y is 4, Y^3 is not divisible by 9. When Y is 24, Y^3 is divisible by 9. Hence, Insuff.
ST2) Since 6 is a multiple of 3, Y^3 should be divisible by 9 (or 3^2). Eg, when y= 6, 12, 18..., Y^3 is divisible by 9.Hence Suff.

Ans: (B)
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Director
Joined: 23 May 2008
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29 Sep 2008, 22:28
yes, B

i figured out why though,

1) you can get 12^3 which is divisible by both 9 and 4
becase 12^3 = 4x3x4x3x4x3 or you can get 4^3 =4x4x4, not divisible by 9 so insufficient

2) if divisible by 6 then y^3 has to be at least 3x2x3x2x3x2, which is always divisible by 9 since there are three 3's
VP
Joined: 17 Jun 2008
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29 Sep 2008, 22:29
B.

From stmt1: y = 4a for a = 0,1,2,3,.....

for any values of a other than multiple of 3, y^3 will not be divisible by 9 whereas for values of a that are multiple of 3, y^3 will be divisible by 9. Hence, insufficient.

From stmt2: y = 6a = 3*2*a for a = 0,1,2,3..... and for any values of a y^3 will be divisible by 9. Hence, sufficient.

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Re: from veritas prep &nbs [#permalink] 29 Sep 2008, 22:29
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