BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Probability Problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 458
Joined: Sun Aug 31, 2008 10:44 amFollowed by: 1 member
Thanked: 3 times
by gmat009 » Tue Oct 07, 2008 12:54 pm
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

Oa is D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 316
Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member
Thanked: 36 times

Re: Probability Problem

by Morgoth » Tue Oct 07, 2008 1:22 pm
gmat009 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

Oa is D
n(n + 1)(n + 2) is a product of 3 consecutive integers.

if n is even, n(n + 1)(n + 2), will be divisible by 8.

even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48

also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.

multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12


total probability = 48+12 = 60/96 = 10/16 = 5/8.

Hope this helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 458
Joined: Sun Aug 31, 2008 10:44 amFollowed by: 1 member
Thanked: 3 times

Re: Probability Problem

by gmat009 » Tue Oct 07, 2008 2:38 pm
Morgoth wrote:
gmat009 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

Oa is D
n(n + 1)(n + 2) is a product of 3 consecutive integers.

if n is even, n(n + 1)(n + 2), will be divisible by 8.

even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48

also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.

multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12


total probability = 48+12 = 60/96 = 10/16 = 5/8.

Hope this helps.
Thanks for explanation
I am not clear with just 1 thing.
You calculated " even integers from 1-96 divisible by 2=48"
and "multiples of 8 from 1-96 =12"
Total 48+12 but they will not be having some common numbers which are even and divisible by 8 and we are counting them twice.
Don't we need to subtract those numbers.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 316
Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member
Thanked: 36 times

Re: Probability Problem

by Morgoth » Tue Oct 07, 2008 4:55 pm
gmat009 wrote: Thanks for explanation
I am not clear with just 1 thing.
You calculated " even integers from 1-96 divisible by 2=48"
and "multiples of 8 from 1-96 =12"
Total 48+12 but they will not be having some common numbers which are even and divisible by 8 and we are counting them twice.
Don't we need to subtract those numbers.
No. what I did was find the odd integers which lead to one 8 in n(n+1)(n+2)

lets start with odd numbers that will lead to one 8 in n(n+1)(n+2)

n=7, 7*8*9
n=15, 15*16*17
n=23, 23*24*25
so on and so forth

We basically need to find the how many odd numbers for n will satisfy the above condition.

All numbers 1 less than the multiple of 8 will satisfy the above condition. Therefore, if we find the multiples of 8, we have found the exact number of odd integers.

Hope this helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Tue Mar 17, 2009 11:03 am
Thanked: 1 times

Re: Probability Problem

by bullshark » Wed Jul 08, 2009 5:14 pm
Morgoth wrote:
gmat009 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

Oa is D
n(n + 1)(n + 2) is a product of 3 consecutive integers.

if n is even, n(n + 1)(n + 2), will be divisible by 8.

even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48

also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.

multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12


total probability = 48+12 = 60/96 = 10/16 = 5/8.

Hope this helps.
can you explain why you do (96-2)/2? then you add one to the result?
why did you not simply do 96/2 = 48 and 96/8 = 12 ?

Thanks
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 316
Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member
Thanked: 36 times

Re: Probability Problem

by Morgoth » Thu Jul 09, 2009 4:45 am
bullshark wrote: can you explain why you do (96-2)/2? then you add one to the result?
why did you not simply do 96/2 = 48 and 96/8 = 12 ?

Thanks
I did that only to be safe. Here I could have simply done the division and it still would have worked because the starting number was 1. Had it not been 1 or some other whole number starting from 30s or 40s, the simply division method would have missed 2 important values and would have lead to the wrong answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Sat May 02, 2009 7:24 pm

Still confused..

by uw490 » Thu Jul 09, 2009 7:51 am
Perhaps someone could help clear this up still...

I understand where the 12 comes from in the calculations (multiples of 8), but still am confused where we get 48. Unless I've misread the problem, and the subsequent explanations 48 represents all the other even numbers? But obviously, numbers such as 20, 22, 36, etc are not divisible by 8. This is where I lose the explanation of the problem.

Any clarification on this portion of the problem would be great.

thanks

J
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Fri Aug 22, 2008 7:26 am
Location: Accra, Ghana
Thanked: 1 times
GMAT Score:720

n(n+1)(n+2)

by bbaah » Thu Jul 09, 2009 8:37 am
We are interested in numbers (n) such that n(n+1)(n+2) is divisible by 8.

if n=22, then 22(22+1)(22+2)= 22(23)(24): This product is divisible by 8 because it has a factor of 2 in 22= 11*2 and a factor of 4 in 24 = 4*6.

A similar explanation can be given for n = 24, 26, etc.

Hope this helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Wed Jun 17, 2009 6:15 am
Thanked: 4 times

by cameronwu » Thu Jul 09, 2009 8:56 am
48 comes from the fact that every single even number works in this equation, and here's why.

Take n=2. It'll end up being 2 * (3) * (4)... now if you factor this out, you'll see that you get 2^1 * 3 * 2^2 .... This leads to 2^3 = 8*3.

Take n=4.... 4*5*6.... same thing as 2^2 * 5 * (2*3)...another 2^3 in there.

If you take the pattern out further, you'll see you'll be to extract a 2^3 factor for EVERY SINGLE EVEN NUMBER.

There are 48 even numbers from 1-96 inclusive. n*(n+2) with even numbers will always give you a multiple of 2^3 in there.

The 12 comes from the (n+1) term. It's every number that's a multiple of 8, minus 1. Like 7, 15, 23, etc... all the way through 95. If you counted all those terms up, then it would come to 12 terms.

12 odd terms + 48 even terms = 60 total N's that work.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Sat May 02, 2009 7:24 pm

Makes sense...

by uw490 » Thu Jul 09, 2009 11:10 am
Ok, gotcha.

Makes sense, but... (I'm probably making this way more difficult than I need to)

How do we know to look for the "number that's a multiple of 8, minus 1. Like, 7, 15, 23, etc..." Still not clear why we know to look for this portion of the solution.

bear with me

thanks!

J
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Wed Jun 17, 2009 6:15 am
Thanked: 4 times

Re: Makes sense...

by cameronwu » Thu Jul 09, 2009 12:46 pm
uw490 wrote:
How do we know to look for the "number that's a multiple of 8, minus 1. Like, 7, 15, 23, etc..." Still not clear why we know to look for this portion of the solution.
J
I guess it just comes with experience with number properties. However, if you notice, the applicability of the 48 even numbers is derived completely from the N and (N+2) term. The (N+1) term isn't even a factor for this part.

Well, you might think to yourself, why is it even here? At this point, you just have to reason out that (a multiple of 8) * (N) * (N+2) will be divisible by 8. Thus, any number such that N+1 is divisible by 8 will work, and that's the 7, 15, 23 etc... part. I know it can be a little tricky to see by yourself, but in these kinds of comprehensive number property questions, it's best to try different combinations of even/odd, positive/negative, and any other way to dichotimize numbers that you can think of if you can't recognize the number properties.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 447
Joined: Sun Apr 19, 2009 9:08 pm
Location: Kolkata,India
Thanked: 7 times
GMAT Score:670

by uptowngirl92 » Sun Sep 27, 2009 7:16 pm
Amazing explanation but jus one querie..
when we are dealing with the second part..we are dealing only with ODD nos. rite?So when we are counting the nos. divisible by 8 shouldnt we be counting only 6?The even part has already been dealt with...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Mon Jan 05, 2009 4:59 pm
Thanked: 2 times

by Simon » Sat Nov 13, 2010 9:39 am
For even . odd. even scenario total number of even numbers = 96-2/2 +1 = 47+1 = 48

For odd. even . odd scenario in which the even number is a multiple of 8 are only 11.

for instance 7.8.9...... 15.16.17..........23.24.25..... up to 87.88.89 but there is no twelfth number (i.e) there is no 95.96.97 as 97 is not part of the numbers provided ( 1 to 96 inclusive)

so 48+11 = 49

probability = 49/96 approx = 5\8


Can someone correct me if i'm wrong????
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Mon Jan 05, 2009 4:59 pm
Thanked: 2 times

by Simon » Sat Nov 13, 2010 9:45 am
also we probably need to take out one more number from the even. odd. even

we can have 2.3.4 or 4.5.6 up to 94.95.96 but not 96.97.98

so we only use 47 numbers

so 47 + 11 = 58 / 96 approx 5/8
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 131
Joined: Fri Oct 15, 2010 2:39 am
Location: New Delhi, India
Thanked: 11 times

by Rezinka » Mon Nov 15, 2010 11:56 pm
Can we do it using elimination ?
This is how I did it :
Writing down first few sets I discovered every alternate set is divisible, so prob > 1/2 or prob = 1/2. Also wherever a multiple of 8 comes in, additional sequences are divisible.
So, prob is slightly greater than 1/2. Now A,B,C are easily eliminated. E is much greater so I chose D.
I am fine with calculations as well, so should I stick to calculations or I can use such eliminations wherever possible ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply