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 [#permalink]

Show Tags

29 Dec 2009, 09:13

kirankp wrote:

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. 41 B. 83 C. 21 D. 85 E. 43

are you sure the options are correct?? ...as per my understanding all even numbers in this range 2 4 6 8 till 96 will be divisible by 8 for n(n+1)(n+2)....what am I missing here

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. 41 B. 83 C. 21 D. 85 E. 43

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

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

# of even numbers (between 1 and 96)=48

AND

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note 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.

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

Show Tags

02 Jan 2010, 03:53

agree with Bunuel here. if n=even, then n(n+2) will always have at least three 2's as factors (2x2x2=8), therefore 8 will always be a factor, where n= even. there are 48.

if (n+1)= divisible by 8, then the whole equation will also be divisible by 8: n=7,15,23,31,39,47,55,63,71,79,87,95 = 12 possibilities.

48+12=60 -- number of times expression will be divisible by 8. 96= number of possibilities.

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

Show Tags

02 May 2010, 11:47

gurpreetsingh wrote:

chaitu1315 wrote:

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. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4

IMO A

This is only possible when n is multiple of 4.

now from 4 to 96 there are 24 such numbers, thus probability = 24/96 = 1/4

Gurpreet you did not consider the options mentioned below ..!

What about when n =6 or may be 15 then also the equation is divisible by 8 and hence we need to consider them as well..

so in my opinion the expression will be divisible by 8 whenever n is divisible by 4 or (n+ 2) is divisible by 4 or when (n+1) is divisible by 8

1st Option when n is divisible by = total numbers = 24 ( as calculated by Gurpreet above 96/4= 24) 2nd option when n +1 is divisible by 8 = 12 ( 96/8 ) 3rd Option when n+2 is divisible by 4 = 24 ( 96/4)

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

Show Tags

02 May 2010, 20:45

1

This post received KUDOS

1

This post was BOOKMARKED

kirankp wrote:

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. \(\frac{3}{4}\)

If you notice every alternate pair is divisible by 8 => 48 such pairs If you notice the pair(marked in blue) it has '8' in it and hence divisible by '8'=>12 such pairs

=>prob=48+12 / 96 = 5/8
_________________

If you like my post, consider giving me a kudos. THANKS!

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

Show Tags

11 May 2010, 10:15

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. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4

The problem can be solved in a crude way, but is there a better alternative?

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

Show Tags

11 May 2010, 20:09

ajitsah wrote:

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. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4

The problem can be solved in a crude way, but is there a better alternative?

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

Show Tags

11 Aug 2010, 06:36

I am curious... how should we approach this type of question...? for eg this question says 8 if question says 11 or may be 12 do we test numbers? what if there are certain exception in middle? (in case of 8 it seems it goes continously...)

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

Show Tags

08 Sep 2010, 11:18

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. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/8

Last edited by udaymathapati on 09 Sep 2010, 05:47, edited 1 time in total.

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

Show Tags

20 Apr 2015, 07:07

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.
_________________

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.

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...