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: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

17 Apr 2009, 04:13

nightwing79 wrote:

so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get \(y\)'s value in integer. To do that, \(x\) has to be \(5*6*6*6\). Since \(y = 4sqrt(5x^3 / 6)\) has one \(5\), one \(6\) in denominator, and three \(x\), make the value of \((5x^3 / 6)\) a square of a square. To do that, \(x\) requires \(5^3\) to make \(5^4\) and \(6^9\) to make \(6^8\) after cancealing a 6. Hence 4th sqrt power of \((5x^3 / 6)\) is a minimum integer vale.
_________________

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

17 Apr 2009, 16:37

GMAT TIGER wrote:

nightwing79 wrote:

so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get \(y\)'s value in integer. To do that, \(x\) has to be \(5*6*6*6\). Since \(y = 4sqrt(5x^3 / 6)\) has one \(5\), one \(6\) in denominator, and three \(x\), make the value of \((5x^3 / 6)\) a squre of a square. To do that, \(x\) requires \(5^3\) to make \(5^4\) and \(6^9\) to make \(6^8\) after cancealing a 6. Hence 4th sqrt power of \((5x^3 / 6)\) is a minimum integer vale.

Good stuff!

However - why does it have to be 6^9 why not 6^5 to make 6^4 hence square root of would still be an integer value.

The key is root to the power of 4 - have to remember that and then looking to understand what a min integer value would look like - good learning.

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

18 Apr 2009, 15:32

1

This post received KUDOS

nightwing79 wrote:

GMAT TIGER wrote:

nightwing79 wrote:

so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get \(y\)'s value in integer. To do that, \(x\) has to be \(5*6*6*6\). Since \(y = 4sqrt(5x^3 / 6)\) has one \(5\), one \(6\) in denominator, and three \(x\), make the value of \((5x^3 / 6)\) a squre of a square. To do that, \(x\) requires \(5^3\) to make \(5^4\) and \(6^9\) to make \(6^8\) after cancealing a 6. Hence 4th sqrt power of \((5x^3 / 6)\) is a minimum integer vale.

Good stuff!

However - why does it have to be 6^9 why not 6^5 to make 6^4 hence square root of would still be an integer value.

The key is root to the power of 4 - have to remember that and then looking to understand what a min integer value would look like - good learning.

look forward to your correspondence.

che dg

How do you get that? If you can, then thats the answer. But again, how?:roll:
_________________

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

18 Apr 2009, 17:54

nightwing79 wrote:

GMAT TIGER wrote:

nightwing79 wrote:

so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get \(y\)'s value in integer. To do that, \(x\) has to be \(5*6*6*6\). Since \(y = 4sqrt(5x^3 / 6)\) has one \(5\), one \(6\) in denominator, and three \(x\), make the value of \((5x^3 / 6)\) a squre of a square. To do that, \(x\) requires \(5^3\) to make \(5^4\) and \(6^9\) to make \(6^8\) after cancealing a 6. Hence 4th sqrt power of \((5x^3 / 6)\) is a minimum integer vale.

How do you get that? If you can, then thats the answer. But again, how?:roll:

come to think of it .... i don't know either..... i think i went back to that special place....where the basic laws of math and physics don't apply....

the whole problem seems so much easier now...lol....

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

26 Apr 2009, 22:08

No..You are absolutely right nightwing..you can use 6^5 BUT with x having an exponent of 3, you can never get that. The least figure you can get is 6^9. Thus GMAT tiger is correct.

I think this is quite a likely question of the highest level of the actual GMAT..ie, if you keep getting the questions right.

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

06 May 2009, 15:45

Just a quick question: Why do you have "none of the above" in answer choice E in almost all question?

cicerone wrote:

Folks, here is the next question..

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A. 1/4 B. 7/32 C. 1/2 D. 1/56 E. None of these

840 = 2x2x2x3x5x7 = 2^3 x 3 x 5 x 7

No of factors = (3+1)(1+1)(1+1)(1+1) = 32 How many factors of 840 are divisible by 15? 16. How is 16? No of factors divisible by 15 are 15 and multiples of 15 = 15, 15x2, 15x2x2, 15x2x2x2....................840. No of factors divisible by 15 = 1 + (4c1+4c2+4c3+4c4) = 16

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

06 May 2009, 19:27

cicerone wrote:

sureshbala wrote:

Folks, sorry for the delay (I am on a vacation).

Anyway, here is the next one.....

There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there? A. 4

B. 6

C. 8

D. 10

E. Cannot be determined

Folks, this can be answered very quickly if you can conclude the following two results.

Result 1: Since the number of chairs occupied by the girls in both the cases is same there is no need to consider the number of girls.

Result 2: In the second case we are able to seat 2 boys per chair and no boy is left unseated, hence the number of boys must be even.

Let the number of boys be 2x.

Seats occupied in the first case = 2x-1

Seats occupied in the second case = x

Given that 2x-1-x = 3

So x = 4

Hence the number of boys 2x = 8.

thanks. my approach: Number of chair=B+G-1=B/2+G+3 solving it will give b=8

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

06 May 2009, 20:00

Nice answer. Specially the method to calculate the number of factors divisible by 15. +1

GMAT TIGER wrote:

Just a quick question: Why do you have "none of the above" in answer choice E in almost all question?

cicerone wrote:

Folks, here is the next question..

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A. 1/4 B. 7/32 C. 1/2 D. 1/56 E. None of these

840 = 2x2x2x3x5x7 = 2^3 x 3 x 5 x 7

No of factors = (3+1)(1+1)(1+1)(1+1) = 32 How many factors of 840 are divisible by 15? 16. How is 16? No of factors divisible by 15 are 15 and multiples of 15 = 15, 15x2, 15x2x2, 15x2x2x2....................840. No of factors divisible by 15 = 1 + (4c1+4c2+4c3+4c4) = 16

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

07 May 2009, 18:00

walker wrote:

cicerone wrote:

A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A factor could be not only a positive number but also negative one. That is why GMAC often writes "number of positive factors".

So, for 840 = 2^3*3*5*7: N = 2 * (1+3)*(1+1)*(1+1)*(1+1) = 64 N15 = 2 * (1+3)*1*1*(1+1) = 16

p= 16/64 = 1/4

Thats definitely a good point but in standerized tests such as GMAT "a factor is always a +ve one". In other word, the factors of a given number/integer is only natural numbers. I am further confirmed by the following statement that factors are only +ve numbers, at least in standarized tests.

HongHu wrote:

When we talk about multiples and factors, we are talking about natural numbers, I believe. Otherwise we will not be able to find the least common multiple for any two numbers. The LCM of two numbers is the smallest number (not zero) that is a multiple of both. If we could use negative number here, a negative number with a large absolute value is smaller than a negative number with a small absolute value. There won't be a limit to this. We won't be able to find the smallest number.

Thats definitely a good point but in standerized tests such as GMAT "a factor is always a +ve one". In other word, the factors of a given number/integer is only natural numbers. I am further confirmed by the following statement that factors are only +ve numbers, at least in standarized tests.

Hi, Tiger!

Honestly, I have never seen any GMAT question that test factor / positive factor concept. So, to be GMAT-like the question should say "A positive factor of 840..." rather than "A factor of 840 ...".

By the way, it doesn't change answer.
_________________

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

08 May 2009, 18:58

GMAT TIGER wrote:

Just a quick question: Why do you have "none of the above" in answer choice E in almost all question?

cicerone wrote:

Folks, here is the next question..A factor of 840 is chosen at random. What is the probability that it is divisible by 15?

A. 1/4 B. 7/32 C. 1/2 D. 1/56 E. None of these

840 = 2x2x2x3x5x7 = 2^3 x 3 x 5 x 7 No of factors = (3+1)(1+1)(1+1)(1+1) = 32 How many factors of 840 are divisible by 15? 16. How is 16? No of factors divisible by 15 are 15 and multiples of 15 = 15, 15x2, 15x2x2, 15x2x2x2....................840. No of factors divisible by 15 = 1 + (4c1+4c2+4c3+4c4) = 16 The prob = 16/32 = 1/2

I noticed that no of factors divisible by 15 = 8 not 16. It would be 16 if the factors were not repeated.

So prob = 8/32 = 1/4.

But walker how did you get 64 and 16? Why you multiplied by 2 in both cases?
_________________

But walker how did you get 64 and 16? Why you multiplied by 2 in both cases?

I remember that GMAT always says "number of positive factors" to avoid ambiguity. So, I used "2" to count both negative and positive set of factors. Anyway, it would be better to use word "positive" in the question.
_________________

Re: Problem Solving for 780+ Aspirants [#permalink]

Show Tags

10 May 2009, 21:36

Folks, the answer for this question remains the same even if you consider negative divisors. But as our friends pointed out, GMAT talks of only positive divisors and hence I have modified the question to reflect the same.

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...