If y is an integer, is y^3 divisible by 9? : GMAT Data Sufficiency (DS)
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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]

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08 Jul 2011, 01:04
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-y-is-an-integer-is-y-3-divisible-by-127796.html
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]

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08 Jul 2011, 01:45
Yes this can be done by plugging numbers.

(1) Take multiples of 4 and 9 and the answer is yes, but say for numbers that are only multiples of 4, the answer is No

(2) is sufficient because y is divisible by 6 means that y has 3 as factor, hence y^3 will have 3^3 as factor

=> y^3 is divisible by 9.

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

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08 Jul 2011, 02:09
siddhans wrote:
Can someone please explain in detail?
Can this be done using random/smart numbers?

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

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

I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem.
Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading

In this question the best strategy is to understand the logic;
to be divisible by 9 .. u need atleast two 3's

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

hence sufficient.
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]

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08 Jul 2011, 02:18
sudhir18n wrote:
siddhans wrote:
Can someone please explain in detail?
Can this be done using random/smart numbers?

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

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

I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem.
Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading

In this question the best strategy is to understand the logic;
to be divisible by 9 .. u need atleast two 3's

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

hence sufficient.

The question says divisible by 6... can you write divisible by 6 to be a multiple of 6?

Also, in 1 you have written:

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.

You said it gives you atleast one 3 ....I dont think so its always the case...only in some cases it gives you 3...like lets says 4*3 = 12 ...

In st2 you have written :

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

How can y^3 have atleast three 2's???
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]

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08 Jul 2011, 02:42
siddhans wrote:
sudhir18n wrote:
siddhans wrote:
Can someone please explain in detail?
Can this be done using random/smart numbers?

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

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

I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem.
Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading

In this question the best strategy is to understand the logic;
to be divisible by 9 .. u need atleast two 3's

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

hence sufficient.

The question says divisible by 6... can you write divisible by 6 to be a multiple of 6?

Also, in 1 you have written:

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.

You said it gives you atleast one 3 ....I dont think so its always the case...only in some cases it gives you 3...like lets says 4*3 = 12 ...

In st2 you have written :

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

How can y^3 have atleast three 2's???

What I meant in st:1 regarding 4, was some multiples of 4 will give you 3. not all ...

st2: since 6= 2*3 ... any multiple of 6 will have atleast one 2 and atleast one 3..test numbers here

6,12,24,36,42..( all will have atleast one 2 and one 3)
thus if you do a cube of these numbers .. you are bound to have three 2's and three 3's
test numbers again:
12*12*12 = 2^2*3 * 2^2*3 * 2^2*3 = 2^6*3^3 ... same is the case with any multple.

any number which is divisible by 6 .. is a multiple of 6.
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]

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08 Jul 2011, 07:18
siddhans wrote:
Can someone please explain in detail?
Can this be done using random/smart numbers?

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

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

Taking 1, y = 4x
y^3 = (4^3)(x^3) = 64(x^3) can't say whether divisible by 9 or not

Taking 2, y = 6x
y^3 = (6^3)(x^3) = (8*27)(x^3) so divisible by 9

Hence B.
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]

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08 Jul 2011, 19:43
1. Not sufficient

if y =4 then y^3 is not divisible by 9.
if y = 12 then y^3 is divisible by 9.

2. Sufficient

3 is a factor of 6.

so 3 must be there in all the multiples of 6.

so y^3 is divisible by 9 .

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

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16 Jul 2011, 00:34
[quote="sudhir18n"

I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem.
Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading

In this question the best strategy is to understand the logic;
to be divisible by 9 .. u need atleast two 3's

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

hence sufficient.[/quote]

The question says divisible by 6... can you write divisible by 6 to be a multiple of 6?

Also, in 1 you have written:

st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.

You said it gives you atleast one 3 ....I dont think so its always the case...only in some cases it gives you 3...like lets says 4*3 = 12 ...

In st2 you have written :

St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3.
thus Y^3 at minimum will have atleast three 2's and three 3's.

How can y^3 have atleast three 2's???[/quote]

What I meant in st:1 regarding 4, was some multiples of 4 will give you 3. not all ...

st2: since 6= 2*3 ... any multiple of 6 will have atleast one 2 and atleast one 3..test numbers here

6,12,24,36,42..( all will have atleast one 2 and one 3)
thus if you do a cube of these numbers .. you are bound to have three 2's and three 3's
test numbers again:
12*12*12 = 2^2*3 * 2^2*3 * 2^2*3 = 2^6*3^3 ... same is the case with any multple.

any number which is divisible by 6 .. is a multiple of 6.[/quote]

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

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20 Aug 2013, 05:26
Can someone clarify if my approach is correct for this problem.. ( initial part)

we need to know if$$Y^3 = 9 I$$

We can further solve this as $$Y^3 = 3^2 * I$$ Now given that Y is an integer there must be an additional 3 present for Y to be an integer

so $$Y^3 = 3^2 * 3 * I$$ when we further solve this we get $$\frac{Y}{3} = Integer$$

Statement 1 says $$\frac{Y}{4} = Integer$$ Y could be 4, Y could be 36 insufficient

Statement 2 says $$\frac{Y}{6} = Integer$$ So the values 6,12,18 etc all divisible by 3 sufficient!

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

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20 Aug 2013, 05:31
fozzzy wrote:
Can someone clarify if my approach is correct for this problem.. ( initial part)

we need to know if$$Y^3 = 9 I$$

We can further solve this as $$Y^3 = 3^2 * I$$ Now given that Y is an integer there must be an additional 3 present for Y to be an integer

so $$Y^3 = 3^2 * 3 * I$$ when we further solve this we get $$\frac{Y}{3} = Integer$$

Statement 1 says $$\frac{Y}{4} = Integer$$ Y could be 4, Y could be 36 insufficient

Statement 2 says $$\frac{Y}{6} = Integer$$ So the values 6,12,18 etc all divisible by 3 sufficient!

Yes, the question basically asks whether y is divisible by 3.

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 --> 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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-y-is-an-integer-is-y-3-divisible-by-127796.html
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Re: If y is an integer, is y^3 divisible by 9?   [#permalink] 20 Aug 2013, 05:31
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