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

 It is currently 21 Oct 2016, 04:29

### 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: 01 Nov 2007
Posts: 144
Followers: 1

Kudos [?]: 316 [1] , given: 0

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

### Show Tags

16 Apr 2008, 10:25
1
KUDOS
10
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

46% (02:59) correct 54% (01:58) wrong based on 232 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
Current Student
Joined: 27 Mar 2008
Posts: 416
Schools: Kellogg Class of 2011
Followers: 3

Kudos [?]: 42 [0], given: 1

### Show Tags

16 Apr 2008, 10:45

Suppose f(n) = n(n+1)(n+2)

We just need to calculate probability for a range of 8 numbers between 1 and 96 since every 8th number is divisible by 8 (8,16,24....)

Suppose we choose a range - 9, 10, 11, 12, 13, 14, 15, 16

Straight off, we can see that 16 is divisible. Also f(14) and f(15) are divisible becase (n+2) and (n+1) = 16 respectively.

Now f(9) and f(11) and f(13) are not divisible since they will comprise of 2 odd numbers and one even.

However, f(10) and f(12) are divisible because:

f(10) = 10*11*12 or 2*5*11*3*4 OR 8*3*5*11
Similarly for f(12).

Thus, out of 8 cases, 5 satisfy the condition of divisibility by 8.
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 273 [1] , given: 2

### Show Tags

16 Apr 2008, 11:56
1
KUDOS
i really didnt do any math on this...i realize as long as n=even its divisible by 8..

so right away i know that probability is greater than 1/2

now..3.4.5 is divisible too..soo really its down to D and E..

then i notice 9.10.11 is not divisible..

basically how many n or n+1 or n+2=2^3*M

5/8 makes sense..
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 273 [0], given: 2

### Show Tags

16 Apr 2008, 12:00
yes this is another neat way of looking at the problem..and you are absolutely correct that range will be divisible by 8 every 8 terms..
however, i dont think 5/8 is the exact right answer..it should be a little below 5/8

casue numbers > 8 yes every 5 numbers out of 8 will be divisible, however for number from 1-8 only 2 numbers exist 1.2.3 and 5.6.7 which are not divisble by 8..

yellowjacket wrote:

Suppose f(n) = n(n+1)(n+2)

We just need to calculate probability for a range of 8 numbers between 1 and 96 since every 8th number is divisible by 8 (8,16,24....)

Suppose we choose a range - 9, 10, 11, 12, 13, 14, 15, 16

Straight off, we can see that 16 is divisible. Also f(14) and f(15) are divisible becase (n+2) and (n+1) = 16 respectively.

Now f(9) and f(11) and f(13) are not divisible since they will comprise of 2 odd numbers and one even.

However, f(10) and f(12) are divisible because:

f(10) = 10*11*12 or 2*5*11*3*4 OR 8*3*5*11
Similarly for f(12).

Thus, out of 8 cases, 5 satisfy the condition of divisibility by 8.
Intern
Joined: 25 Jun 2008
Posts: 13
Followers: 0

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

### Show Tags

09 Jul 2008, 01:12
i think ans is 3/8.

1. If n is disvisible by 8, then n could be 8,16,24....96 ( 12 integers)
2. If n+1 is divisible by 8, then n could be 7,15,23,...95 ( 12 integers)
3. similarly for n+2

There probability = (12+12+12)/96
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Followers: 7

Kudos [?]: 50 [6] , given: 0

### Show Tags

09 Jul 2008, 01:33
6
KUDOS
3
This post was
BOOKMARKED
Since 96 is divisible by 8 and since "divisibility by 8" repeats every 8 terms, you can just focus on the 8 first terms (from 1 to 8):

it works for n=2,4,6,7,8, that is 5 numbers out of the 8

==> Answer is (D) = 5/8
Intern
Joined: 09 Jul 2008
Posts: 1
Followers: 0

Kudos [?]: 4 [3] , given: 0

### Show Tags

09 Jul 2008, 03:49
3
KUDOS
1
This post was
BOOKMARKED
n(n+1)(n+2) will be divisible by 8 for all even numbers, i.e. total 48 numbers
also cases in which (n+1) is a multiple of 8 will be divisible by 8
for ex: n=7,15,23.. i.e. total 12 numbers

so total number of such cases is 48+12=60
probability = 60/96 = 5/8
Intern
Joined: 09 Jul 2008
Posts: 2
Followers: 0

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

### Show Tags

09 Jul 2008, 05:52
We want n(n+1)(n+2) to be divisible by 8, and that n has to be between 1 to 96. (inclusive)
So for any even from 2 to 96 when substituted in the expression is divisible by 8.
There are 48 even numbers till 96, inclusive.
So, the probablity is (48/96) = 1/2
Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10
Followers: 7

Kudos [?]: 50 [0], given: 0

### Show Tags

09 Jul 2008, 05:56
Prasanna1981 wrote:
We want n(n+1)(n+2) to be divisible by 8, and that n has to be between 1 to 96. (inclusive)
So for any even from 2 to 96 when substituted in the expression is divisible by 8.
There are 48 even numbers till 96, inclusive.
So, the probablity is (48/96) = 1/2

Nope.

n=7, 15, 23, etc... work as well even though they are not even. You missed those
Manager
Joined: 08 Jun 2008
Posts: 71
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

09 Jul 2008, 06:09
I solved it just like Prasanna did and got 1/2. But when I looked at pisces solution I realized I missed the n+1 part. I agree with 5/8 and I liked pisces way of solving. +1 to pisces.
Intern
Joined: 02 Aug 2009
Posts: 8
Followers: 0

Kudos [?]: 22 [0], given: 5

### Show Tags

22 Sep 2009, 20:39
JCLEONES 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

Take a range of 8 no.'s to find the no.s that will be divisible by 8 within those 8 no's which can be used as a representative for a broader range of numbers:

n(n+1)(n+2)

8x9x10 divisible by 8
9x10x11
10x11x12 divisible by 8
11x12x13
12x13x14 divisible by 8
13x14x15
14x15x16 divisible by 8
15x16x17 divisible by 8

Probability = 5/8
Manager
Joined: 04 Sep 2010
Posts: 51
Followers: 2

Kudos [?]: 1 [0], given: 1

### Show Tags

18 Oct 2010, 11:26
Guys,
Ans is 1/2 C..
See we have to find n so that n(n+1)(n+2)..will be div by 8..
if u select any even no as n..the above expression will be divided by 8..
so ans is 1/2...
Math Expert
Joined: 02 Sep 2009
Posts: 35238
Followers: 6618

Kudos [?]: 85286 [5] , given: 10236

### Show Tags

19 Oct 2010, 12:43
5
KUDOS
Expert's post
1
This post was
BOOKMARKED
sudhanshushankerjha wrote:
Guys,
Ans is 1/2 C..
See we have to find n so that n(n+1)(n+2)..will be div by 8..
if u select any even no as n..the above expression will be divided by 8..
so ans is 1/2...

Check the solution below. Answer is D.

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

_________________
Manager
Joined: 27 Jul 2010
Posts: 197
Location: Prague
Schools: University of Economics Prague
Followers: 1

Kudos [?]: 41 [0], given: 15

### Show Tags

02 Feb 2011, 07:23
Could you comment on my sollution please?

Write numbers:
1,2,3,4,5,6,7,8 = 8 numbers

now:
(n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8
-------
I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)
_________________

You want somethin', go get it. Period!

Math Expert
Joined: 02 Sep 2009
Posts: 35238
Followers: 6618

Kudos [?]: 85286 [0], given: 10236

### Show Tags

02 Feb 2011, 07:48
craky wrote:
Could you comment on my sollution please?

Write numbers:
1,2,3,4,5,6,7,8 = 8 numbers

now:
(n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8
-------
I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)

As in EACH group of 8 numbers (1-8, 9-16, 17-24, ..., 89-96) there are EXACTLY 5 numbers for which n*(n+1)*(n+2) is divisible by 8 and 96 is a multiple of 8 (so there are integer # of such groups in the range from 1 to 96 inclusive) then the probability will be 5/8 overall, so yes your approach is correct.

Check this question for similar solution: beginner-s-forum-question-106168.html
_________________
Manager
Joined: 27 Jul 2010
Posts: 197
Location: Prague
Schools: University of Economics Prague
Followers: 1

Kudos [?]: 41 [0], given: 15

### Show Tags

02 Feb 2011, 08:02
Bunuel wrote:
craky wrote:
Could you comment on my sollution please?

Write numbers:
1,2,3,4,5,6,7,8 = 8 numbers

now:
(n)*(n+1)*(n+2) is divisible by 8 when n equals: 2,4,6,7,8= 5 numbers

the probability is 5/8
-------
I think this could be enough, as 96 is a multiple of 8. So you only come to solution 60/96 = (5*12) / (8*12)

As in EACH group of 8 numbers (1-8, 9-16, 17-24, ..., 89-96) there are EXACTLY 5 numbers for which n*(n+1)*(n+2) is divisible by 8 and 96 is a multiple of 8 (so there are integer # of such groups in the range from 1 to 96 inclusive) then the probability will be 5/8 overall, so yes your approach is correct.

Check this question for similar solution: beginner-s-forum-question-106168.html

Whoau, it looks I somehow discovered your 30 second approach. Thanks for clarification.
_________________

You want somethin', go get it. Period!

Manager
Joined: 21 Jul 2012
Posts: 69
Followers: 0

Kudos [?]: 8 [0], given: 32

### Show Tags

26 Mar 2013, 15:08
Bunuel wrote:
sudhanshushankerjha wrote:
Guys,
Ans is 1/2 C..
See we have to find n so that n(n+1)(n+2)..will be div by 8..
if u select any even no as n..the above expression will be divided by 8..
so ans is 1/2...

Check the solution below. Answer is D.

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

Bunuel, my question involves your second bullet. What is that n=8p-1 and then 8p-1 <= 96 piece? It looks like a remainder formula but have no idea how to use the remainder formula like this??
Math Expert
Joined: 02 Sep 2009
Posts: 35238
Followers: 6618

Kudos [?]: 85286 [0], given: 10236

### Show Tags

27 Mar 2013, 05:52
jmuduke08 wrote:
Bunuel wrote:
sudhanshushankerjha wrote:
Guys,
Ans is 1/2 C..
See we have to find n so that n(n+1)(n+2)..will be div by 8..
if u select any even no as n..the above expression will be divided by 8..
so ans is 1/2...

Check the solution below. Answer is D.

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

Bunuel, my question involves your second bullet. What is that n=8p-1 and then 8p-1 <= 96 piece? It looks like a remainder formula but have no idea how to use the remainder formula like this??

Not sure I understand your question completely, but n+1 is divisible by 8 means that n is 1 less than a multiple of 8. Can you please elaborate your question?
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538

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

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

### Show Tags

16 May 2014, 11:47
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 Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538

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

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

### Show Tags

20 Dec 2015, 05:41
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.
_________________
Re: If an integer n is to be chosen at random from the integers   [#permalink] 20 Dec 2015, 05:41
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, 20:59
198 If an integer n is to be chosen at random from the integers 28 28 Jan 2012, 04:03
6 If an integer n is to be chosen at random from the integers 8 27 Sep 2009, 03:01
3 If an integer n is to be chosen at random from the integers 19 29 Dec 2009, 05:41
27 If an integer n is to be chosen at random from the integers 9 11 Nov 2007, 21:09
Display posts from previous: Sort by