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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? (1) y is [#permalink]

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24 May 2011, 07:07
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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.
[Reveal] Spoiler: OA

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24 May 2011, 08:16

Case 1- both 12 & 4 are multiples of 4 but cube of one is divisible and other is not.

Case 2-cube of any multiple of 6 will be divisible by 9.

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24 May 2011, 10:19
Stmt1: y is divisible by 4.
y/4 = n where n is an integer
y=4n
Is y^3/9= m an integer
or (4n)^3/9= m
or 64/9 * n^3 =m an integer
take n=1 m= 64/9 not an integer
take n=3 m 64*3 an integer. Two answers.
Not sufficient.

Stmt2: y is divisible by 6.
y/6 = n where n is an integer
y=6n
Is y^3/9= m an integer
or (6n)^3/9= m
or 216/9 * n^3 =m an integer
24*n^3 = m will be an integer. Sufficient.
OA B.
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24 May 2011, 10:47
If we follow MGMAT Number Properties guide and try to get prime factorization of 6 we can see that there is a 2 and a 3. Since 6 is a factor of a number that is multiplied by itself three times, we can see that our prime factorization of y^3 would give us more than two 2's and two three's, therefore, if we get at least two three's out of the prime factorization of (y^3), we know its divisible by 9 since (3 * 3 = 9).

Therefore, Statement B is sufficient since we will always get at least two three's out of (y^3) prime factorization.

However, if we take a look at statement (A), we find the prime factorization of 4 which is 2 and 2. There is no three. So there is already some suspicion. To double check, we plug in 8 as y so 8^3 = 512. 512 / 9 does not give us an integer. However, if we take a number, 12, and cube it, we get 1728, which is divisible by 9. This makes statement (A) insufficient.

How did I pick the test numbers for Statement (A)?
Well 8 and 12 have factors of 4 since statement A states y^3 is divisible by 4. Also, I chose 8 because 8 does not have a factor of 3. Why is it important to choose a number that does not have a factor of 3? because 9 has a factor of 3.
I chose 12 because it DOES have a factor of 3. Again because 9 has a factor of 3 and it makes the statement sufficient whereas an integer that does not have a factor of 3 makes statement insufficient.

Hope this makes sense!
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24 May 2011, 19:37
1. Not sufficient

we know y has 4 as factor. but we dont know whether y has 3 as factor or not.

if y has 3 as a factor , then y^3 is divisible by 9 or else not divisible by 9.

2. Sufficient

we know y has 2 and 3 as factors => y^3 has 8 and 27 as factors

=>y^3 is divisible by 9.

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If y is an integer is y^3 divisible by 9 [#permalink]

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19 Feb 2012, 09:43
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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

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

Last edited by Bunuel on 19 Feb 2012, 10:22, edited 1 time in total.
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Re: If y is an integer is y^3 divisible by 9 [#permalink]

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19 Feb 2012, 10:27
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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.

Hope it's clear.
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Re: If y is an integer is y^3 divisible by 9 [#permalink]

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21 Feb 2012, 05:02
+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]

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20 Aug 2013, 06:56

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

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Re: If y is an integer is y^3 divisible by 9 [#permalink]

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

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22 Nov 2016, 07:09
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Re: If y is an integer, is y^3 divisible by 9? (1) y is   [#permalink] 22 Nov 2016, 07:09
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