If an integer n is to be chosen at random from the integers : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 06:10

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 144 [0], given: 3

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

Show Tags

27 Sep 2009, 02:01
00:00

Difficulty:

95% (hard)

Question Stats:

40% (02:32) correct 60% (01:33) wrong based on 117 sessions

HideShow timer Statistics

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.
Senior Manager
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 377 [4] , given: 47

Re: integers and prob [#permalink]

Show Tags

14 Feb 2010, 12:07
4
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!!

|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~~

Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7090

Kudos [?]: 93312 [2] , given: 10557

Show Tags

18 Sep 2010, 20:53
2
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: $$\frac{60}{96}=\frac{5}{8}$$

_________________
Senior Manager
Joined: 20 Jul 2010
Posts: 269
Followers: 2

Kudos [?]: 78 [0], given: 9

Show Tags

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
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 957 [0], given: 25

Show Tags

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
_________________
Senior Manager
Joined: 25 Feb 2010
Posts: 481
Followers: 4

Kudos [?]: 84 [0], given: 10

Show Tags

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

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
Joined: 27 Jun 2010
Posts: 40
Followers: 0

Kudos [?]: 129 [0], given: 7

Show Tags

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
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13475
Followers: 576

Kudos [?]: 163 [0], given: 0

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

Show Tags

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.
_________________
Manager
Joined: 23 May 2013
Posts: 167
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5
Followers: 2

Kudos [?]: 70 [0], given: 39

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

Show Tags

17 Oct 2014, 07:23
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

What are the cases when $$n(n+1)(n+2)$$ is divisible by 8?

This means that $$n(n+1)(n+2)$$ must have 2*2*2 in its prime factorization.

If n is even, we'll have 2 even numbers in $$n(n+1)(n+2)$$, one of which is divisible by 4. This means that $$n(n+1)(n+2)$$ is divisible by 8. Therefore, our probability is at least 1/2. We can eliminate choices A,B, and C immediately.

The other possibility is if we have a multiple of 8 in the middle. This means that n will have to be odd, so we're not considering overlapping cases. This happens exactly 1/8 of the time, or (96/8) = 12 times.

1/2 + 1/8 = 5/8.

Re: If an integer n is to be chosen at random from the integers   [#permalink] 17 Oct 2014, 07:23
Similar topics Replies Last post
Similar
Topics:
3 If an integer n is to be chosen at random from the integers 5 30 Sep 2013, 19:59
227 If an integer n is to be chosen at random from the integers 30 28 Jan 2012, 03:03
3 If an integer n is to be chosen at random from the integers 19 29 Dec 2009, 04:41
29 If an integer n is to be chosen at random from the integers 19 16 Apr 2008, 09:25
29 If an integer n is to be chosen at random from the integers 9 11 Nov 2007, 20:09
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.