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 350,000 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: Probability - Integer [#permalink]
24 Jul 2011, 20:07

1

This post received KUDOS

Expert's post

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For \(5N^3/8\) to be an integer, \(5N^3\) should be completely divisible by 8. This will happen only if \(N^3\) is divisible by 8 i.e. if N has 2 as a factor. So for \(5N^3/8\) to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that \(5N^3/8\) is an integer is 1/2. _________________

Re: Probability - Integer [#permalink]
25 Jul 2011, 00:51

VeritasPrepKarishma wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For \(5N^3/8\) to be an integer, \(5N^3\) should be completely divisible by 8. This will happen only if \(N^3\) is divisible by 8 i.e. if N has 2 as a factor. So for \(5N^3/8\) to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that \(5N^3/8\) is an integer is 1/2.

Re: Probability - Integer [#permalink]
25 Jul 2011, 22:54

If i can give my 2 cents tip. When you see the words non-negative you should immediately think about zero in your solution. Same way about non-positive of course. _________________

Re: Probability - Integer [#permalink]
27 Jul 2011, 09:11

VeritasPrepKarishma wrote:

abhi398 wrote:

As per my understanding zero is neither postive nor negitive integer.

why should zero be included in this question as the question is asking about the non-negative single-digit.

Since 0 is neither positive nor negative, 'non negative' means 'positive and 0'

Absolutely! I've fell to this GMAT's one of the favorite traps so many times that I had this pinned on my wall (the real one. not FB's :D _________________

Re: Probability - Integer [#permalink]
30 Jul 2011, 17:56

VeritasPrepKarishma wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For \(5N^3/8\) to be an integer, \(5N^3\) should be completely divisible by 8. This will happen only if \(N^3\) is divisible by 8 i.e. if N has 2 as a factor. So for \(5N^3/8\) to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that \(5N^3/8\) is an integer is 1/2.

The probability is not 1/2. You have assumed that the result would be an integer with n=0. _________________

ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED

Re: Probability - Integer [#permalink]
30 Jul 2011, 19:31

Sudhanshuacharya wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

is it (5N^3)/8 or 5N^(3/8)???

For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer

therefore all even number satisfies this condition i.e. 2,4,6,8. Number of terms = from 0 to 9. i.e. 10 numbers

Hence probability should be = 4/10 = 2/5

For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5

This issue would have occurred if the question had asked 5N^3^3^3/8 Now first calculate the exponents : 5N^3^3^3 from Right to Left the exponents and not Left to Right So 5N^3^3^3/8 = 5N^3^27/8 and not 5N^27^3/8....In other words 3^27 <> 27^3

Re: Probability - Integer [#permalink]
31 Jul 2011, 21:10

krishp84 wrote:

Sudhanshuacharya wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

is it (5N^3)/8 or 5N^(3/8)???

For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer

therefore all even number satisfies this condition i.e. 2,4,6,8. Number of terms = from 0 to 9. i.e. 10 numbers

Hence probability should be = 4/10 = 2/5

For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5

This issue would have occurred if the question had asked 5N^3^3^3/8 Now first calculate the exponents : 5N^3^3^3 from Right to Left the exponents and not Left to Right So 5N^3^3^3/8 = 5N^3^27/8 and not 5N^27^3/8....In other words 3^27 <> 27^3

Hope you got my point.

Hi..I didnt get thi spart..can u help elaborate pls..

Re: Probability - Integer [#permalink]
01 Aug 2011, 08:33

ok - you have a question asking - what is the value of 2^3^4 ? Is it 2^81 or 8^4 ?

It will be 2^81

This was easy...

Now let us say the question asked - what is the value of 2^3^4^2 ? Will it be 2^3^16 or 8^4^2 ?

It will be 2^3^16

Takeaway - Always evaluate exponents from Right to Left if they are of the form - x^y^z^.... DeeptiM - Hope you got the point. Kudos are welcome if the post is useful. _________________

Re: Probability - Integer [#permalink]
04 Aug 2011, 16:47

VeritasPrepKarishma wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For \(5N^3/8\) to be an integer, \(5N^3\) should be completely divisible by 8. This will happen only if \(N^3\) is divisible by 8 i.e. if N has 2 as a factor. So for \(5N^3/8\) to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that \(5N^3/8\) is an integer is 1/2.

how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8

Re: Probability - Integer [#permalink]
04 Aug 2011, 20:27

1

This post received KUDOS

Expert's post

pinchharmonic wrote:

how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8

Yes, 5 doesn't have any role to play except to tell you that it is not divisible by 8 and has no 2s in it. If instead of 5, we had 4, things would have been different (but let's not get into it right now).

So \(N^3/8\) should be an integer. This means \(N^3\) should be divisible by 8. But we have values for N, not \(N^3\) so how do we figure what it means for N? Let's say, N has 2 as a factor. Then \(N = 2a\) \(N^3 = (2a)^3 = 8a^3\) This number is divisible by 8 since it has 8 as a factor. Since N gets cubed, if it has a 2 in it, that becomes 8 and \(N^3\) is divisible by 8.

If we consider that \(N^3\) has 2 as a factor and is greater than 8 (as you suggested above), is it essential that \(N^3\) will be divisible by 8? Consider \(N^3 = 10\) (has 2 as a factor and is >8) This is not divisible by 8 and for that matter, N is a not an integer in this case.

Consider now, what would have happened if instead of 5, we had 4. We would need \(4N^3/8\) to be an integer i.e. \(N^3/2\) to be an integer. Yet again, we need N^3 to be even, so N should be even too i.e. it must have a 2 in it. _________________

Re: Probability - Integer [#permalink]
05 Aug 2011, 09:59

VeritasPrepKarishma wrote:

pinchharmonic wrote:

how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8

Yes, 5 doesn't have any role to play except to tell you that it is not divisible by 8 and has no 2s in it. If instead of 5, we had 4, things would have been different (but let's not get into it right now).

So \(N^3/8\) should be an integer. This means \(N^3\) should be divisible by 8. But we have values for N, not \(N^3\) so how do we figure what it means for N? Let's say, N has 2 as a factor. Then \(N = 2a\) \(N^3 = (2a)^3 = 8a^3\) This number is divisible by 8 since it has 8 as a factor. Since N gets cubed, if it has a 2 in it, that becomes 8 and \(N^3\) is divisible by 8.

If we consider that \(N^3\) has 2 as a factor and is greater than 8 (as you suggested above), is it essential that \(N^3\) will be divisible by 8? Consider \(N^3 = 10\) (has 2 as a factor and is >8) This is not divisible by 8 and for that matter, N is a not an integer in this case.

Consider now, what would have happened if instead of 5, we had 4. We would need \(4N^3/8\) to be an integer i.e. \(N^3/2\) to be an integer. Yet again, we need N^3 to be even, so N should be even too i.e. it must have a 2 in it.

karishma, thanks.

So is there a way to make this factoring process generic? It seems like your thought process is to find the lowest factor of the denominator, which is 2, ie not 4 or 8. Then make sure you have that same factor in the numerator. Then the product of all factors in the numerator must be >= than the denominator. Meaning if in numerator there are other products that are not factors you just ignore them. like 5 below

(5x2^3)/8

and if you happeneted to choose 3

(5x3^3)/8 = (5x3x3x3)/8

since 5 and 3 are both not factors you still dont have the factor portion of the numerator >=8

Re: Probability - Integer [#permalink]
10 Jul 2013, 14:10

VeritasPrepKarishma wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For \(5N^3/8\) to be an integer, \(5N^3\) should be completely divisible by 8. This will happen only if \(N^3\) is divisible by 8 i.e. if N has 2 as a factor. So for \(5N^3/8\) to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that \(5N^3/8\) is an integer is 1/2.

what if N=8 or 4 rather than 2 ? i think 8 should be a factor of n^3 means either N=8, or NxN=8 or NxNxN=8 hence we have 3 fav conditions ..... now i got stuck here....

Re: Probability - Integer [#permalink]
10 Jul 2013, 14:30

Expert's post

jimmy0220 wrote:

VeritasPrepKarishma wrote:

DeeptiM wrote:

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For \(5N^3/8\) to be an integer, \(5N^3\) should be completely divisible by 8. This will happen only if \(N^3\) is divisible by 8 i.e. if N has 2 as a factor. So for \(5N^3/8\) to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd. Probability that \(5N^3/8\) is an integer is 1/2.

what if N=8 or 4 rather than 2 ? i think 8 should be a factor of n^3 means either N=8, or NxN=8 or NxNxN=8 hence we have 3 fav conditions ..... now i got stuck here....

can not decide total cond.

m i going wrong way ? pls explain

If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an integer?

Set of non-negative single-digit integers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 10 elements.

Now, \(\frac{5N^3}{8}=\frac{5N^3}{2^3}\) to be an integer N should be any even number. There are 5 even numbers in the set (0, 2, 4, 6, and 8), so the probability is 5/10=1/2.

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...