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 an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

23 Jun 2006, 16:06

i dont get it, if one of the ways of an odd # is even why cant it be divisible by odd, i mean a number like 7 would be divisible by 8. N+1=8 hence thats at least one odd # tat can be divisible by 8, so i dont understand why its 48.....and cant be any odd #'s

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

23 Jun 2006, 16:10

You are right..

If I pick 7 then 7 x 8 x 9 will also be divisible by 8....

GMATCUBS21 wrote:

i dont get it, if one of the ways of an odd # is even why cant it be divisible by odd, i mean a number like 7 would be divisible by 8. N+1=8 hence thats at least one odd # tat can be divisible by 8, so i dont understand why its 48.....and cant be any odd #'s

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

23 Jun 2006, 16:16

well u said n is even, and i just told u that n can also be odd, so how do u get 48, please explain......im not as good at math as u...so how do u do this one....its not that u do all even's because i just showed u 7 works....so how do u do it

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

22 Jul 2006, 11:32

If I pick any no from the set 7, 8, 9, 14, 15, 16 ..............94,95,96. then I can satisfy the condition n(n+1)(n+2) divisibel by 8. Such 36 no I can pick so probability is 36/96 So the answer is 3/8.

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

22 Jul 2006, 12:07

I think I am right now. One need to pick up even integer. If 'n' is even and not divisible by 4 then n+2 will be divisible by 4 so n*(n+2) will be divisible by 8. There are total 48 even integers and ans should 1/2. I think I am right now. I need to be careful about trap.

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

22 Jul 2006, 12:16

pravinkawadkar wrote:

I think I am right now. One need to pick up even integer. If 'n' is even and not divisible by 4 then n+2 will be divisible by 4 so n*(n+2) will be divisible by 8. There are total 48 even integers and ans should 1/2. I think I am right now. I need to be careful about trap.

Pravin

When n is
n n+1
7 8
15 16
....
95 96.. total 12.

48+12 = 60 possible outcome. Thus 5/8 => D _________________

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

24 Apr 2015, 06:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

Show Tags

24 Apr 2015, 08:26

Expert's post

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D. \(\frac{5}{8}\) E. \(\frac{3}{4}\)

\(n(n + 1)(n + 2)\) is divisible by 8 in two cases:

A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8;

(Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...