If an integer n is to be chosen at random from the integers 1 to 96, : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 14:33

### 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 1 to 96,

Author Message
TAGS:

### Hide Tags

Intern
Joined: 11 May 2004
Posts: 46
Location: USA
Followers: 0

Kudos [?]: 4 [0], given: 0

If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jun 2006, 14:39
3
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

45% (03:31) correct 55% (02:00) wrong based on 92 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. $$\frac{1}{4}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{4}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Apr 2015, 07:25, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 80 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jun 2006, 14:57
Messed up this ... deleting.

Last edited by haas_mba07 on 23 Jun 2006, 15:29, edited 2 times in total.
Intern
Joined: 11 May 2004
Posts: 46
Location: USA
Followers: 0

Kudos [?]: 4 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jun 2006, 15:06
i dont get it, if one of the ways of an odd # is even why cant it be divisible by odd, i mean a number like 7 would be divisible by 8. N+1=8 hence thats at least one odd # tat can be divisible by 8, so i dont understand why its 48.....and cant be any odd #'s
VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 80 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jun 2006, 15:10
You are right..

If I pick 7 then 7 x 8 x 9 will also be divisible by 8....

GMATCUBS21 wrote:
i dont get it, if one of the ways of an odd # is even why cant it be divisible by odd, i mean a number like 7 would be divisible by 8. N+1=8 hence thats at least one odd # tat can be divisible by 8, so i dont understand why its 48.....and cant be any odd #'s
Intern
Joined: 11 May 2004
Posts: 46
Location: USA
Followers: 0

Kudos [?]: 4 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jun 2006, 15:16
well u said n is even, and i just told u that n can also be odd, so how do u get 48, please explain......im not as good at math as u...so how do u do this one....its not that u do all even's because i just showed u 7 works....so how do u do it
VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 80 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jun 2006, 15:41
I think I finally got the right one.. although my first answer 1/2 still remains the same.

Between 1-10 : you have 5 numbers you can pick to get nx (n+1) x (n+2) divisible by 8 : {2, 4, 6, 7, 8}

Continuing this from 10-20 : 5 more and so on.

In total from 1-90 you have 9x5 = 45 numbers.

Between 91-96 you can pick 3 more to get the satisfying condition: 92, 94, 96

Therefore in total you can pick : 48
Probability = 48/96 = 1/2
Senior Manager
Joined: 22 May 2006
Posts: 371
Location: Rancho Palos Verdes
Followers: 1

Kudos [?]: 110 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

22 Jul 2006, 09:52
Is N(N+1)(N+2) multiple of 8?

1 2 3
2 3 4
3 4 5
.....
7 8 9

When N is even number, "N(N+1)(N+2)" is always multiple of 8.
Thus probability = 48/96 = 1/2

When N is odd number and N+1 is multiple of 8, N(N+1)(N+2) is multiple of 8
Thus, probability = 12/96 = 1/8

1/2+1/8 = 5/8
or
(48+12)/96 = 5/8

Ans is D.
_________________

The only thing that matters is what you believe.

Intern
Joined: 14 May 2006
Posts: 32
Followers: 0

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

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

22 Jul 2006, 10:32
If I pick any no from the set 7, 8, 9, 14, 15, 16 ..............94,95,96. then I can satisfy the condition n(n+1)(n+2) divisibel by 8. Such 36 no I can pick so probability is 36/96 So the answer is 3/8.

Pravin
Intern
Joined: 14 May 2006
Posts: 32
Followers: 0

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

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

22 Jul 2006, 10:35
The set should be 6,7,8, 14,15,16.....94,95,96. Ans is same. I commit typing mistake.

Pravin
Intern
Joined: 14 May 2006
Posts: 32
Followers: 0

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

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

22 Jul 2006, 11:07
I think I am right now. One need to pick up even integer. If 'n' is even and not divisible by 4 then n+2 will be divisible by 4 so n*(n+2) will be divisible by 8. There are total 48 even integers and ans should 1/2. I think I am right now. I need to be careful about trap.

Pravin
Senior Manager
Joined: 22 May 2006
Posts: 371
Location: Rancho Palos Verdes
Followers: 1

Kudos [?]: 110 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

22 Jul 2006, 11:16
I think I am right now. One need to pick up even integer. If 'n' is even and not divisible by 4 then n+2 will be divisible by 4 so n*(n+2) will be divisible by 8. There are total 48 even integers and ans should 1/2. I think I am right now. I need to be careful about trap.

Pravin

When n is
n n+1
7 8
15 16
....
95 96.. total 12.

48+12 = 60 possible outcome. Thus 5/8 => D
_________________

The only thing that matters is what you believe.

Director
Joined: 28 Dec 2005
Posts: 755
Followers: 1

Kudos [?]: 14 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

22 Jul 2006, 16:09
GMATCUBS21 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

D IMO. If n>=2 then for any even n, n(n+1)(n+2) is divisible by 2 and 4, and therefore by 8.

Further, every 8th number will be divisible by 8
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 24

Kudos [?]: 273 [0], given: 0

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

23 Jul 2006, 00:52
5/8

For all N = even N(N+1)(N+2) will be divisible by 8: Total = 48
For all N+1 = multiple of 8, N(N+1)(N+2) will be divisible by 8: Total = 12

Prob = 60/96 = 5/8
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

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

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

26 Jul 2006, 02:34
For all even numbers N . n(n+1)(n+2) is divisible by 8. Even numbers = 48

For all odd numbers
From 1-10 - 1 odd number
2- 10 - 1 odd number

90-96 - 1

hence total odd numbers = 12

Total numbers = 60

Probabiltiy = 60/96 = 5/8
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13456
Followers: 575

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

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

24 Apr 2015, 05:51
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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36567
Followers: 7081

Kudos [?]: 93200 [0], given: 10553

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

### Show Tags

24 Apr 2015, 07:26
Expert's post
1
This post was
BOOKMARKED
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. $$\frac{1}{4}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{3}{4}$$

$$n(n + 1)(n + 2)$$ is divisible by 8 in two cases:

A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. $$n+1$$ is itself divisible by 8;

(Notice 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.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

Similar question: divisible-by-12-probability-121561.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
_________________
Re: If an integer n is to be chosen at random from the integers 1 to 96,   [#permalink] 24 Apr 2015, 07:26
Similar topics Replies Last post
Similar
Topics:
227 If an integer n is to be chosen at random from the integers 30 28 Jan 2012, 03:03
6 If an integer n is to be chosen at random from the integers 8 27 Sep 2009, 02:01
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