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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] New post 19 Feb 2012, 08:43
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A
B
C
D
E

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89% (01:41) correct 11% (00:51) wrong based on 52 sessions
If y is an integer is y^3 divisible by 9

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

OG answer is B - reason if y is divisible by 3 then y is divisble by 9
y is divisible by 6 so y is divisible by 3 and y is dvisible by 9


How is this possible - statement 2 means that Y is divisible by 6. Now Y is an integer therefore to be evenly divisible by 6 - y must be a multiple of 6.
Not all multiples of 6 are divisible by 9.
eg if Y = 12 then y is divisible by 6 but not divisible by 9
if Y = 18 then y is divisible by 6 and also by 9
So My answer is E

Could please someone help explain the OG answer
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Feb 2012, 09:22, edited 1 time in total.
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Re: If y is an integer is y^3 divisible by 9 [#permalink] New post 19 Feb 2012, 09:27
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If y is an integer is y^3 divisible by 9?

Notice that we are asked whether y^3 is divisible by 9 not y itself.

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

(2) y is divisible by 6 --> since y is divisible by 6 then it's divisible by 3 too: y=3k. So, y^3 will definitely have more than two 3's and thus will be divisible by 9: y^3=(3k)^3=9*(3k^3). Sufficient.

Answer: B.

Hope it's clear.
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Re: If y is an integer is y^3 divisible by 9 [#permalink] New post 21 Feb 2012, 04:02
+1 B

For such questions I would suggest to break the divisor i.e 4 / 6 in this case into the prime factors.

Thus statement 1 : 2 x 2
Statement 2 : 3 x 2

Now you have to get Y x Y x Y to be divided by 9 => 3 x 3

So in short however you do it you need 3 3's which is only possible through statement 2

hence statement 2 is sufficient => B
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Re: If y is an integer is y^3 divisible by 9 [#permalink] New post 20 Aug 2013, 05:56
Add me to 'B' supporters.

6=3*2. If we multiply multiples of 6 three times, we have three '3' in that number. For a number to be divisible by 9, we need just two '3'. So statement 2 is sufficient

Happy days.
Re: If y is an integer is y^3 divisible by 9   [#permalink] 20 Aug 2013, 05:56
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