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: Tough and tricky: Probability of integr being divisible by 8 [#permalink]
11 Oct 2009, 18:08

Sure. Here is what I think, and I might be wrong (In fact most of the times I am wrong!):

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8. Now, there are 96/8 = 12 integers from 1 to 96 which are divisible by 8. For n*(n+1)*(n+2) to be divisible by 8: n, (n+1) or (n+2) must be divisible by 8. There could be 12*3 = 36 such integers. So the probability for n*(n+1)*(n+2) being divisible by 8 = 36/96. = 3/8 = 37.5%

Re: Tough and tricky: Probability of integr being divisible by 8 [#permalink]
11 Feb 2011, 04:39

2

This post received KUDOS

this questions tests the concepts of consecutive integers.

three consecutive integers will .. 1. have an integer that is a multiple of 3 e.g. {2,3,4} or {3,4,5} 2. either have two odds or two even integers e.g. {2,3,4} or {3,4,5} 3. have two even integers if n (1st integer) is even. the product n*(n+1)*(n+2) must be a multiple of 8 because one even integer will be a multiple of 4. e.g. {2,4,5}, {4,5,6} or {14,15,16} 4. have two odd integers if n (1st integer) is odd. the product can be a multiple of 8 only if (n+1) is a multiple of 8 because the other two are odd. e.g. {7,8,9} or {23,24,25}. on the other hand, the product of {3,4,5} or {11,12,13} is not multiple of 8 because (n+1) is not multiple of 8.

we can use the above rules to calculate probability.

no. of cases where n is even = \(\frac{96}{2} = 48\) no. of cases where n+1 is multiple of 8 = \(\frac{96}{8} = 12\) total cases in which product is multiple of 8 = \(48+12 = 60\)

probability = \(\frac{60}{96}\) = 62.5%

Ans: C

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

http://blog.davidbbaker.com/wp-content/uploads/2015/11/12249800_10153820891439090_8007573611012789132_n.jpg When you think about an MBA program, usually the last thing you think of is professional collegiate sport. (Yes American’s I’m going...