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

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If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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03 Jan 2004, 22:38
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If y is an integer, is y^3 divisible by 9?

1) y is divisible by 4

2) y is divisible by 6

>>>>>>>>>>>> Answer is B ...why?? >>>>>>>>.
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Joined: 13 Nov 2003
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03 Jan 2004, 23:26
factorize 6: 2 and 3
Y^3 will be (2*3)^3 = 2^3 * 3^3
3^3 will always be divisible by 9

thus, B
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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02 Dec 2011, 21:21
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KUDOS
I'm a bit confused here.
The question does not suggest that y is not 0.
y = 0 satisfies both, y is divisible by 4 and y is divisible by 6.
In that case we can say that y^3 is divisible by 9 and that D is the answer.
Is it not correct?
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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02 Dec 2011, 22:59
shm0401 wrote:
I'm a bit confused here.
The question does not suggest that y is not 0.
y = 0 satisfies both, y is divisible by 4 and y is divisible by 6.
In that case we can say that y^3 is divisible by 9 and that D is the answer.
Is it not correct?

You are right. The question should mention that Y cannot be 0 or that Y is a positive integer.
If I read that question in the real test, I would assume that Y can be 0 too.

Kudos to you!
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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03 Dec 2011, 17:41
metallicafan wrote:
shm0401 wrote:
I'm a bit confused here.
The question does not suggest that y is not 0.
y = 0 satisfies both, y is divisible by 4 and y is divisible by 6.
In that case we can say that y^3 is divisible by 9 and that D is the answer.
Is it not correct?

You are right. The question should mention that Y cannot be 0 or that Y is a positive integer.
If I read that question in the real test, I would assume that Y can be 0 too.

Kudos to you!

While on the real GMAT, divisibility questions are always restricted to positive integers only, it doesn't actually make any difference to the answer here if you allow y to be zero. Statement 1 is still not sufficient (y might be 0 or 12, in which case y^3 is divisible by 9, or y might be 4 or 8, in which case y^3 is not divisible by 9), and Statement 2 is still sufficient, since y^3 must be divisible by 9 even if y=0 (since 0 is divisible by every positive integer).
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Joined: 11 May 2009
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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04 Dec 2011, 04:04
The technique that Manhattan GMAT guides use make D-S questions easier. The key lies in paraphrasing the question and then paraphrasing the statements. It would be like this:

Question: What integer is y if (y^3)/9 = integer?

1) y is 4 or divisible by 4. y^3 = 4*4*4 = 64; 64/9 = 7.111 thus y is not an integer

2) y is 6 or divisible by 6. 6^3 = 6*6*6 = 216; 216/9 = 24 thus y is an integer

There is also more accurate approach which could save you time if you know how to use it well. Since you know the value of y, you can use prime factorization to find the solution.

1) y is 4. prime factors are 2*2. y^3 = 2*2*2*2*2*2 the six twos are never divisible by 9, because prime factors of 9 are 3*3

2) y is 6. prime factors are 2*3. y^3 = 2*3*2*3*2*3 or 2^3*3^3. since y^3 has three 3's it must be divisible by two 3's

Hope it helps
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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04 Dec 2011, 10:17
It has helped me understand the way i have to look at the problem.
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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04 Dec 2011, 21:56
for 1) try 12 - yes divisible, try 4- not divisible
for 2) any number divisible by 6 is a multiple of 3, and 3^3 is 9 which is divisibe by 9.
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Manager
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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04 Dec 2011, 22:04
If y=0 for option 1 then y is divisible by 9....

But if y=8 then y is not divisible by 9...

So there are lots of possiblities there so we have to rule out the possibility of option A.
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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04 Dec 2011, 22:26
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KUDOS
Simple.. In option B, we have 6 which is multiple of 3. So obviously any number div by 3 - the number cube is obviously div by 3.
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Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible [#permalink]

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05 Dec 2011, 05:34
I went with B because 6 has 2 and 3 as factors. so Y has 3 as a factor. therefore Y cube has at least 3 3s as a factor. So 9 which has only 2 3s can divide it
Re: If y is an integer, is y^3 divisible by 9? 1) y is divisible   [#permalink] 05 Dec 2011, 05:34
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