Last visit was: 21 Dec 2024, 23:00 It is currently 21 Dec 2024, 23:00
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

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

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [21]
Given Kudos: 26096
Send PM
If an integer n is to be chosen at random from the integers [#permalink] New post   31 Aug 2020, 10:31
3
Expert Reply
18
Bookmarks
00:00

Question Stats:

30% (02:15) correct 69% (02:45) wrong based on 112 sessions

Hide Show 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
ShowHide Answer
Official Answer
D
Most Helpful Community Reply
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12234 [6]
Given Kudos: 136
Send PM

Top Member of the Month
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 Aug 2020, 11:00
4
2
Bookmarks
Carcass 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

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D
Cheers,
Brent
General Discussion
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 30448
Own Kudos [?]: 36808 [0]
Given Kudos: 26096
Send PM
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 Aug 2020, 10:32
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!

For more theory and practice see our GRE - Math Book
avatar
ChandanPednekar
Intern
Intern
Joined: 06 Jan 2022
Posts: 6
Own Kudos [?]: 4 [0]
Given Kudos: 0
Send PM
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 May 2022, 05:47
GreenlightTestPrep wrote:
Carcass 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

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D
Cheers,
Brent



How did you know that we had to check every 8 numbers exactly?

If we check every 12 numbers (because 12 * 8 = 96 ) then every 12 numbers have 7 valid values.

thus 7 * 8 = 56 which would give incorrect answer.
User avatar
P
GreenlightTestPrep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12234 [1]
Given Kudos: 136
Send PM

Top Member of the Month
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   31 May 2022, 14:42
1
ChandanPednekar wrote:
GreenlightTestPrep wrote:
Carcass 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

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, 5/8 of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D
Cheers,
Brent



How did you know that we had to check every 8 numbers exactly?

If we check every 12 numbers (because 12 * 8 = 96 ) then every 12 numbers have 7 valid values.

thus 7 * 8 = 56 which would give incorrect answer.


The key here is that we're looking for products that are divisible by 8.
We start with n = 1 (which is 1 more than a multiple of 8)
Then n = 2 (which is 2 more than a multiple of 8)
.
.
.
Then n = 7 (which is 7 more than a multiple of 8)
Then n = 8 (which is 0 more than a multiple of 8)
Then n = 9 (which is 1 more than a multiple of 8) at this point we can see that the pattern begins repeating itself
.
.
.
etc

Does that help?
avatar
B
ighuma
Intern
Intern
Joined: 09 Jun 2022
Posts: 15
Own Kudos [?]: 7 [3]
Given Kudos: 114
Send PM
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   21 Jul 2022, 08:37
3
Here is how I approached the problem:

n(n+1)(n+2) divisible by 8 means n is divisible by 4 or n+1 is divisible by 8 or n+2 is divisible by 4. Worth noting that whenever n or n+2 is divisible by 4, the other is also an even number, so n(n+2) will be divisible by 8. Also worth noting that every consecutive 3 numbers have at most one multiple of 4 and, of course, alternate odd and even numbers.

Case 1: n divisible by 4: 96/4 = 24 numbers
Case 2: n+2 divisible by 4; similar to case 1: 24 numbers
Case 3: n+1 divisible by 8: 96/8 = 12 numbers

Total: 24+24+12=60

Probability: 60/96=5/8
avatar
B
vaijan
Intern
Intern
Joined: 28 Jan 2023
Posts: 4
Own Kudos [?]: 4 [1]
Given Kudos: 1
Send PM
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   13 Mar 2023, 02:50
1
we see a pattern that when the consecutive no's start with even, we have multiples of 8 but not when n is odd.
But we also have a case where (n+1) is a multiple of 8 i.e. 8,16,24,32 upto 96
So, in this case, the 'n' can be odd but be divisible by 8.
• Ex: 7,8,9
○ 15,16,17
• But this case is already included where (n+2) is multiple of 8
○ 14,15,16: here n is even and included by 1st case.

• Total number of even numbers between 1 to 96 = 48
• Total number of multiples of 9 between 1 to 96 = 12
○ Total = 60
• Probability = 60/96 = 5/8
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5088
Own Kudos [?]: 76 [0]
Given Kudos: 0
Send PM
Re: If an integer n is to be chosen at random from the integers [#permalink] New post   22 May 2024, 07:10
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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.
Prep Club for GRE Bot
gmatclubot
Re: If an integer n is to be chosen at random from the integers [#permalink]
22 May 2024, 07:10
Moderators:
adewale223
GRE Instructor
88 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1115 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact