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

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08 Jul 2011, 03:09

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.

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

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08 Jul 2011, 03:18

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.

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

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08 Jul 2011, 03:42

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.

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

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16 Jul 2011, 01:34

[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]

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!

Answer is B

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.

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