It is currently 23 Oct 2017, 10:24

### 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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 22 Jul 2008
Posts: 96

Kudos [?]: 232 [0], given: 11

Location: Bangalore,Karnataka
If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

29 Dec 2009, 05:41
1
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

29% (01:46) correct 71% (01:38) wrong based on 23 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 20 Apr 2015, 08:16, edited 3 times in total.
Renamed the topic, edited the question and added the OA.

Kudos [?]: 232 [0], given: 11

Senior Manager
Joined: 30 Aug 2009
Posts: 283

Kudos [?]: 188 [0], given: 5

Location: India
Concentration: General Management
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

29 Dec 2009, 09:13
kirankp 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. 41
B. 83
C. 21
D. 85
E. 43

are you sure the options are correct?? ...as per my understanding all even numbers in this range 2 4 6 8 till 96 will be divisible by 8 for n(n+1)(n+2)....what am I missing here

Kudos [?]: 188 [0], given: 5

Manager
Joined: 22 Jul 2008
Posts: 96

Kudos [?]: 232 [0], given: 11

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

### Show Tags

29 Dec 2009, 09:28
i am sorry, the options were wrong, have edited it

Kudos [?]: 232 [0], given: 11

Senior Manager
Joined: 30 Aug 2009
Posts: 283

Kudos [?]: 188 [0], given: 5

Location: India
Concentration: General Management
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

29 Dec 2009, 09:31
kirankp wrote:
i am sorry, the options were wrong, have edited it

hmmm...will go with C-1/2

Kudos [?]: 188 [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129475 [0], given: 12201

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

### Show Tags

29 Dec 2009, 09:50
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. 41
B. 83
C. 21
D. 85
E. 43

$$n(n+1)(n+2)$$ 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)=48

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

_________________

Kudos [?]: 129475 [0], given: 12201

Senior Manager
Joined: 30 Aug 2009
Posts: 283

Kudos [?]: 188 [0], given: 5

Location: India
Concentration: General Management
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

29 Dec 2009, 09:52
Bunuel wrote:
n(n+1)(n+2) is divisible by 8 in two cases:

1. When n is even:
No of even numbers (between 1 and 96)=48

AND

2. When n+1 is divisible by 8. --> n+1=8p, n=8p-1 --> 8p-1<=96 --> p<=12.1 --> 12 such numbers

Also note that these two sets have no overlaps.

Total=48+12=60

Probability=60/96=5/8

oops

Kudos [?]: 188 [0], given: 5

Manager
Joined: 18 Nov 2009
Posts: 57

Kudos [?]: 9 [0], given: 7

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

### Show Tags

02 Jan 2010, 03:53
agree with Bunuel here. if n=even, then n(n+2) will always have at least three 2's as factors (2x2x2=8), therefore 8 will always be a factor, where n= even. there are 48.

if (n+1)= divisible by 8, then the whole equation will also be divisible by 8: n=7,15,23,31,39,47,55,63,71,79,87,95 = 12 possibilities.

48+12=60 -- number of times expression will be divisible by 8.
96= number of possibilities.

probability = 60/96 = 5/8

Kudos [?]: 9 [0], given: 7

Intern
Joined: 26 Jan 2010
Posts: 7

Kudos [?]: 2 [0], given: 0

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

### Show Tags

02 May 2010, 09:16
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:
D

Last edited by chaitu1315 on 02 May 2010, 09:24, edited 1 time in total.

Kudos [?]: 2 [0], given: 0

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2761

Kudos [?]: 1889 [0], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

02 May 2010, 09:22
chaitu1315 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

IMO A

This is only possible when n is multiple of 4.

now from 4 to 96 there are 24 such numbers, thus probability = 24/96 = 1/4
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1889 [0], given: 235

Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 151 [0], given: 6

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

### Show Tags

02 May 2010, 11:47
gurpreetsingh wrote:
chaitu1315 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

IMO A

This is only possible when n is multiple of 4.

now from 4 to 96 there are 24 such numbers, thus probability = 24/96 = 1/4

Gurpreet you did not consider the options mentioned below ..!

What about when n =6 or may be 15 then also the equation is divisible by 8 and hence we need to consider them as well..

so in my opinion the expression will be divisible by 8 whenever n is divisible by 4 or (n+ 2) is divisible by 4 or when (n+1) is divisible by 8

1st Option when n is divisible by = total numbers = 24 ( as calculated by Gurpreet above 96/4= 24)
2nd option when n +1 is divisible by 8 = 12 ( 96/8 )
3rd Option when n+2 is divisible by 4 = 24 ( 96/4)

in total = 24 + 24 + 12 = 60

the probability = 60/ 96 = 5/ 8

D

Kudos [?]: 151 [0], given: 6

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2761

Kudos [?]: 1889 [0], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

02 May 2010, 14:46
yes right, i missed many.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1889 [0], given: 235

Manager
Joined: 15 Mar 2010
Posts: 101

Kudos [?]: 101 [1], given: 30

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

### Show Tags

02 May 2010, 20:45
1
KUDOS
1
This post was
BOOKMARKED
kirankp 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. $$\frac{1}{4}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D.$$\frac{5}{8}$$
E. $$\frac{3}{4}$$

IMO 'D'
Consider the following possibilities:
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
.......
......
96 97 98

If you notice every alternate pair is divisible by 8 => 48 such pairs
If you notice the pair(marked in blue) it has '8' in it and hence divisible by '8'=>12 such pairs

=>prob=48+12 / 96 = 5/8
_________________

If you like my post, consider giving me a kudos. THANKS!

Kudos [?]: 101 [1], given: 30

Intern
Joined: 29 Mar 2010
Posts: 41

Kudos [?]: 14 [0], given: 5

Location: Leeds
Schools: SBS, JBS(ding w/o interview), HEC
WE 1: SCI-12 yrs
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

11 May 2010, 10:15
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

The problem can be solved in a crude way, but is there a better alternative?
[Reveal] Spoiler:
OA : D

Thanks

Kudos [?]: 14 [0], given: 5

Manager
Joined: 08 May 2010
Posts: 142

Kudos [?]: 67 [0], given: 39

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

### Show Tags

11 May 2010, 19:52
I do not know how to solve this. Where did you get this problem?

Kudos [?]: 67 [0], given: 39

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2761

Kudos [?]: 1889 [0], given: 235

Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: If an integer n is to be chosen at random from the integers [#permalink]

### Show Tags

11 May 2010, 20:09
ajitsah 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

The problem can be solved in a crude way, but is there a better alternative?
[Reveal] Spoiler:
OA : D

Thanks

Probability = required cases/ total cases.

required cases . Case 1. n(n + 1)(n + 2) will be divisible by 8 when n is even, in that case, n will be multiple of 2 and n+2 of 4.

so total even numbers between 1 and 96 = 48

Case 2. when n+1 = 8k form => n = 8k-1

such numbers from 1 to 96 are 12... it starts from 7 to 95 which diff of 8.This forms an AP series.

So required cases = 48+12 = 60

total cases = 96

probability = 60/96 = 5/8

Hence D
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Kudos [?]: 1889 [0], given: 235

Manager
Joined: 16 Feb 2010
Posts: 179

Kudos [?]: 34 [0], given: 17

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

### Show Tags

17 May 2010, 15:30
good problem great explanation.

Kudos [?]: 34 [0], given: 17

Manager
Joined: 24 Apr 2010
Posts: 60

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

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

### Show Tags

11 Aug 2010, 06:36
I am curious...
how should we approach this type of question...?
for eg this question says 8
if question says 11 or may be 12
do we test numbers?
what if there are certain exception in middle? (in case of 8 it seems it goes continously...)

do we remember for each numbers?

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

Manager
Joined: 06 Apr 2010
Posts: 141

Kudos [?]: 939 [0], given: 15

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

### Show Tags

08 Sep 2010, 11:18
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. 3/8

Last edited by udaymathapati on 09 Sep 2010, 05:47, edited 1 time in total.

Kudos [?]: 939 [0], given: 15

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16533

Kudos [?]: 274 [0], given: 0

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

### Show Tags

20 Apr 2015, 07:07
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.
_________________

Kudos [?]: 274 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129475 [0], given: 12201

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

### Show Tags

20 Apr 2015, 08:16
kirankp 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. $$\frac{1}{4}$$
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
_________________

Kudos [?]: 129475 [0], given: 12201

Re: If an integer n is to be chosen at random from the integers   [#permalink] 20 Apr 2015, 08:16
Display posts from previous: Sort by