Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

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.

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

Show Tags

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.

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

Show Tags

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.

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

Show Tags

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]

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.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...