Find all School-related info fast with the new School-Specific MBA Forum

It is currently 30 Sep 2014, 15:57

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If an integer n is to be chosen at random from the integers

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 186
Followers: 1

Kudos [?]: 77 [0], given: 3

If an integer n is to be chosen at random from the integers [#permalink] New post 27 Sep 2009, 02:01
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

36% (02:21) correct 64% (01:37) wrong based on 77 sessions
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2012, 05:30, edited 1 time in total.
Edited the question.
4 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 22 Dec 2009
Posts: 365
Followers: 10

Kudos [?]: 209 [4] , given: 47

GMAT ToolKit User
Re: integers and prob [#permalink] New post 14 Feb 2010, 12:07
4
This post received
KUDOS
marcodonzelli 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


All even numbers would be divisible by 8... as we have two consecutive even numbers being multiplied... Hence 96-1/2 + 1 = 48

Also we have 8n-1 sequence numbers divisible by 8... i.e, if n =7,15,23,31....

Therefore 8n-1 = 96 to find the number of elements in this sequence ... gives n = 97/8 = 12

Therefore Probability = (48+12) / 96 = 60 / 96 = 5/8
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!! :beer

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice|
|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|


~~Better Burn Out... Than Fade Away~~

Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23064
Followers: 3538

Kudos [?]: 27236 [2] , given: 2725

Re: Probability [#permalink] New post 18 Sep 2010, 20:53
2
This post received
KUDOS
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. 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).

# of even numbers between 1 and 96, inclusive is \frac{96-2}{2}+1=48 (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

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.

Total=48+12=60

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

Answer: D.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Senior Manager
Senior Manager
avatar
Joined: 20 Jul 2010
Posts: 271
Followers: 2

Kudos [?]: 39 [0], given: 9

GMAT ToolKit User Reviews Badge
Re: Probability [#permalink] New post 19 Sep 2010, 15:31
This was difficult. I was thinking either N can be multiple of 8 or n+1 can be multiple of 8 pr n+2 a multiple 8. With this I just got 36 choices. I missed sequences like 2*3*4
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 487 [0], given: 25

GMAT ToolKit User Reviews Badge
Re: Probability [#permalink] New post 19 Sep 2010, 15:41
Notice what happens when you shift this product by 8 :

(n+8)(n+1+8)(n+2+8) = 8^3 + 8^2*(n+n+1+n+2) + 8*(n*(n+1)+(n+1)*(n+2)+n*(n+2)) + n*(n+1)*(n+2) = 8*K + n*(n+1)*(n+2)

Here K is a constant. Therefore, when we shift by 8, the remainder when divided by 8 remains the same !

So if we figure out what happens for the first 8 choices of n, for all the rest the pattern will repeat

n=1 1x2x3 No
n=2 2x3x4 Yes
n=3 3x4x5 No
n=4 4x5x6 Yes
n=5 5x6x7 No
n=6 6x7x8 Yes
n=7 7x8x9 Yes
n=8 8x9x10 Yes

So for every 8 numbers, exactly 5 will be divisible by 8.

Here there are exactly 96 numbers to consider starting with 1 ... So probability will be exactly 5/8
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Senior Manager
Senior Manager
User avatar
Joined: 25 Feb 2010
Posts: 459
Followers: 3

Kudos [?]: 35 [0], given: 5

Re: Probability [#permalink] New post 21 Sep 2010, 05:24
Bunuel 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

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, inclusive is \frac{96-2}{2}+1=48 (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

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.

Total=48+12=60

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

Answer: D.


Thanks for such a good explanation !!!

I was stuck in this for more then 3 minutes and couldn't find the solution.
_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Intern
Intern
avatar
Joined: 27 Jun 2010
Posts: 40
Followers: 0

Kudos [?]: 26 [0], given: 7

Re: Probability [#permalink] New post 09 Oct 2010, 21:29
Thanks Bunuel for a very nice explanation.

I was stuck with this question for a very long time and could not find the solution to it.

Even tried counting but that too went wrong.

Request you to advice for how to tackle such kind of questions and how you spotted there would be two cases to test for this question ?

Regards,
Sachin
CEO
CEO
User avatar
Joined: 09 Sep 2013
Posts: 2551
Followers: 200

Kudos [?]: 40 [0], given: 0

Premium Member
Re: If an integer n is to be chosen at random from the integers [#permalink] New post 01 Nov 2013, 04:45
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: If an integer n is to be chosen at random from the integers   [#permalink] 01 Nov 2013, 04:45
    Similar topics Author Replies Last post
Similar
Topics:
If an integer n is to be chosen at random from the integers tarek99 7 24 Jul 2008, 06:29
If an integer n is to be chosen at random from the integers kripalkavi 5 07 Jan 2007, 02:01
If an integer n is to be chosen at random from the integers topkool 1 13 Nov 2006, 07:20
If an integer n is to be chosen at random from the integers Dilshod 18 04 May 2006, 11:19
If an integer n is to be chosen at random from integers 1 to AJB77 7 05 Jul 2005, 14:30
Display posts from previous: Sort by

If an integer n is to be chosen at random from the integers

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.