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:

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Check the solution below. Answer is D.

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

\(n(n+1)(n+2)\) 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 \(n(n+1)(n+2)\).

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.

answer is D n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers also cases in which (n+1) is a multiple of 8 will be divisible by 8 for ex: n=7,15,23.. i.e. total 12 numbers

so total number of such cases is 48+12=60 probability = 60/96 = 5/8

yes this is another neat way of looking at the problem..and you are absolutely correct that range will be divisible by 8 every 8 terms.. however, i dont think 5/8 is the exact right answer..it should be a little below 5/8

casue numbers > 8 yes every 5 numbers out of 8 will be divisible, however for number from 1-8 only 2 numbers exist 1.2.3 and 5.6.7 which are not divisble by 8..

yellowjacket wrote:

Answer is D

Suppose f(n) = n(n+1)(n+2)

We just need to calculate probability for a range of 8 numbers between 1 and 96 since every 8th number is divisible by 8 (8,16,24....)

Suppose we choose a range - 9, 10, 11, 12, 13, 14, 15, 16

Straight off, we can see that 16 is divisible. Also f(14) and f(15) are divisible becase (n+2) and (n+1) = 16 respectively.

Now f(9) and f(11) and f(13) are not divisible since they will comprise of 2 odd numbers and one even.

However, f(10) and f(12) are divisible because:

f(10) = 10*11*12 or 2*5*11*3*4 OR 8*3*5*11 Similarly for f(12).

Thus, out of 8 cases, 5 satisfy the condition of divisibility by 8.

1. If n is disvisible by 8, then n could be 8,16,24....96 ( 12 integers) 2. If n+1 is divisible by 8, then n could be 7,15,23,...95 ( 12 integers) 3. similarly for n+2

We want n(n+1)(n+2) to be divisible by 8, and that n has to be between 1 to 96. (inclusive) So for any even from 2 to 96 when substituted in the expression is divisible by 8. There are 48 even numbers till 96, inclusive. So, the probablity is (48/96) = 1/2

We want n(n+1)(n+2) to be divisible by 8, and that n has to be between 1 to 96. (inclusive) So for any even from 2 to 96 when substituted in the expression is divisible by 8. There are 48 even numbers till 96, inclusive. So, the probablity is (48/96) = 1/2

Nope.

n=7, 15, 23, etc... work as well even though they are not even. You missed those

I solved it just like Prasanna did and got 1/2. But when I looked at pisces solution I realized I missed the n+1 part. I agree with 5/8 and I liked pisces way of solving. +1 to pisces.

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

Take a range of 8 no.'s to find the no.s that will be divisible by 8 within those 8 no's which can be used as a representative for a broader range of numbers:

n(n+1)(n+2)

8x9x10 divisible by 8 9x10x11 10x11x12 divisible by 8 11x12x13 12x13x14 divisible by 8 13x14x15 14x15x16 divisible by 8 15x16x17 divisible by 8

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

now: (n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8 ------- I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)
_________________

now: (n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8 ------- I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)

As in EACH group of 8 numbers (1-8, 9-16, 17-24, ..., 89-96) there are EXACTLY 5 numbers for which n*(n+1)*(n+2) is divisible by 8 and 96 is a multiple of 8 (so there are integer # of such groups in the range from 1 to 96 inclusive) then the probability will be 5/8 overall, so yes your approach is correct.

now: (n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8 ------- I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)

As in EACH group of 8 numbers (1-8, 9-16, 17-24, ..., 89-96) there are EXACTLY 5 numbers for which n*(n+1)*(n+2) is divisible by 8 and 96 is a multiple of 8 (so there are integer # of such groups in the range from 1 to 96 inclusive) then the probability will be 5/8 overall, so yes your approach is correct.

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Check the solution below. Answer is D.

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

\(n(n+1)(n+2)\) 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 \(n(n+1)(n+2)\).

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.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Answer: D.

Bunuel, my question involves your second bullet. What is that n=8p-1 and then 8p-1 <= 96 piece? It looks like a remainder formula but have no idea how to use the remainder formula like this??

Guys, Ans is 1/2 C.. See we have to find n so that n(n+1)(n+2)..will be div by 8.. if u select any even no as n..the above expression will be divided by 8.. so ans is 1/2...

Check the solution below. Answer is D.

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

\(n(n+1)(n+2)\) 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 \(n(n+1)(n+2)\).

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.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Answer: D.

Bunuel, my question involves your second bullet. What is that n=8p-1 and then 8p-1 <= 96 piece? It looks like a remainder formula but have no idea how to use the remainder formula like this??

Not sure I understand your question completely, but n+1 is divisible by 8 means that n is 1 less than a multiple of 8. Can you please elaborate your question?
_________________

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

Show Tags

16 May 2014, 11:47

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 [#permalink]

Show Tags

20 Dec 2015, 05:41

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.
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

There is without a doubt a stereotype for recent MBA grads – folks who are ambitious, smart, hard-working, but oftentimes lack experience or domain knowledge. Looking around and at...