Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

It is currently 26 May 2017, 00:42

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If 1=<n<=99, what is the probability that n(n + 1) is perfec

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

Hide Tags

1 KUDOS received
Intern
Intern
avatar
Joined: 31 Oct 2010
Posts: 31
Followers: 0

Kudos [?]: 69 [1] , given: 25

If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 11 Dec 2010, 23:52
1
This post received
KUDOS
10
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

70% (02:04) correct 30% (01:40) wrong based on 38 sessions

HideShow timer Statistics

If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

[Reveal] Spoiler:
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?
1 KUDOS received
Manager
Manager
avatar
Joined: 11 Jul 2010
Posts: 224
Followers: 3

Kudos [?]: 91 [1] , given: 20

GMAT ToolKit User
Re: Beginner's Forum Question [#permalink]

Show Tags

New post 12 Dec 2010, 00:08
1
This post received
KUDOS
this formula is in the MGMAT math book..

really useful to know... you add 1 as you generally underestimate by 1 at the extreme -- this evenly spaced sets formula is the basis for the simpler formula: (last-First) + 1 when you have to count out the number of elements in a consecutive sequence 1, 2, 3....
You are basically doing (last-first)/1 + 1 ---> you are basically dividing by 1 there as that is the common space b/w the elements...

Knowing that you can apply this formula for evenly spaced sets should be enough...
Intern
Intern
avatar
Joined: 31 Oct 2010
Posts: 31
Followers: 0

Kudos [?]: 69 [0], given: 25

Re: Beginner's Forum Question [#permalink]

Show Tags

New post 12 Dec 2010, 00:22
thanks for the quick reply! kudos for the help.
Expert Post
Math Expert
User avatar
P
Joined: 02 Sep 2009
Posts: 38894
Followers: 7737

Kudos [?]: 106173 [0], given: 11607

Re: Beginner's Forum Question [#permalink]

Show Tags

New post 12 Dec 2010, 01:17
Expert's post
12
This post was
BOOKMARKED
mmcooley33 wrote:
If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?


First about the multiple of x in the given range:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totally-basic-94862.html?hilit=multiple%20range).

So, there will be \(\frac{99-3}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99;

n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n;

Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3.

30 second approach:

Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3.

Answer: 2/3.

Hope it helps.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
User avatar
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 402
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
Followers: 7

Kudos [?]: 46 [0], given: 46

Re: Beginner's Forum Question [#permalink]

Show Tags

New post 13 Dec 2010, 08:41
Ah, that 30 sec approach looks good.
_________________

Go Blue!

GMAT Club Premium Membership - big benefits and savings

1 KUDOS received
Manager
Manager
avatar
Joined: 14 Nov 2011
Posts: 150
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
Followers: 0

Kudos [?]: 18 [1] , given: 103

GMAT ToolKit User
Re: Beginner's Forum Question [#permalink]

Show Tags

New post 07 Jul 2013, 04:12
1
This post received
KUDOS
1
This post was
BOOKMARKED
Bunuel wrote:
mmcooley33 wrote:
If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?


First about the multiple of x in the given range:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totally-basic-94862.html?hilit=multiple%20range).

So, there will be \(\frac{99-3}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99;

n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n;

Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3.

30 second approach:

Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3.

Answer: 2/3.

Hope it helps.



Bump for the 30 sec approach.
Expert Post
Math Expert
User avatar
P
Joined: 02 Sep 2009
Posts: 38894
Followers: 7737

Kudos [?]: 106173 [0], given: 11607

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 07 Jul 2013, 06:36
mmcooley33 wrote:
If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

[Reveal] Spoiler:
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?


Similar questions to practice:
if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
if-integer-c-is-randomly-selected-from-20-to-99-inclusive-121561.html
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 15456
Followers: 649

Kudos [?]: 209 [0], given: 0

Premium Member
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 09 Jul 2014, 07:29
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Current Student
User avatar
Joined: 04 Jul 2014
Posts: 290
Location: India
GMAT 1: 640 Q47 V31
GMAT 2: 640 Q44 V34
GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)
Followers: 19

Kudos [?]: 259 [0], given: 403

Reviews Badge
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 22 Nov 2014, 03:19
Wow! I did this one very fast and correctly! :D (After getting a series of other questions in this type wrong)

Here is a very simple solution. For n(n+1) to be divisible by 3, either
a) n must be a multiple of 3 --> Number of possibilities = 99/3 = 33 OR
b) (n -1) must be a multiple of 3 --> Number of possibilities = 99/3 = 33

Now, favorable possibilities = 66 and total possibilities = 100 (100 because 99 and 1 are both included).

Therefore, probability = 66/100 or 2/3 :D

There's a discussion going on here for this question: an-integer-n-between-1-and-99-inclusive-is-to-be-chosen-at-160998.html#p1445924
_________________

Cheers!!

JA
If you like my post, let me know. Give me a kudos! :)

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 15456
Followers: 649

Kudos [?]: 209 [0], given: 0

Premium Member
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 13 Dec 2015, 15:05
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 15456
Followers: 649

Kudos [?]: 209 [0], given: 0

Premium Member
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 11 Jan 2017, 13:40
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Expert Post
Target Test Prep Representative
User avatar
B
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 1027
Location: United States (CA)
Followers: 34

Kudos [?]: 539 [0], given: 2

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 16 Jan 2017, 18:00
mmcooley33 wrote:
If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?


We are given that an integer n is to be selected at random from 1 to 99 inclusive, and we must determine the probability that n(n+1) will be divisible by 3.

Since probability = favorable outcomes/total outcomes and we know that the total number of outcomes is 99, because there are 99 integers from 1 to 99 inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 3.

First, we can determine the number of values of n that are divisible by 3, that is, the number of multiples of 3 that are between 3 and 99 inclusive. To calculate this, we can use the formula:

(Largest multiple of 3 – smallest multiple of 3)/3 + 1

(99 - 3)/3 + 1

96/3 + 1

32 + 1 = 33

Thus, there are 33 multiples of 3 between 1 and 99 inclusive. That is, the value of n can be any one of these 33 multiples of 3 so that n(n + 1) will be divisible by 3.

Similarly, if n + 1 is a multiple of 3, n(n + 1) will be also be divisible by 3. Since we know that there are 33 values of n that are multiples of 3, there must be another 33 values of n such that n + 1 is a multiple of 3. Let’s expand on this idea:

When n = 2, n + 1 = 3, and thus n(n+1) is a multiple of 3.

When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 3.

When n = 98, n + 1 = 99, and thus n(n+1) is a multiple of 3.

We can see that there are 33 values of n that are multiples of 3, and 33 more values of n for n + 1 to be multiples of 3. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 3 is:

66/99 = 2/3
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Manager
User avatar
B
Joined: 20 Jan 2017
Posts: 65
Location: United States (NY)
GMAT 1: 750 Q48 V44
GMAT 2: 610 Q34 V41
Followers: 2

Kudos [?]: 5 [0], given: 15

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

Show Tags

New post 24 Jan 2017, 20:05
1) n(n+1) is divisible by 3 when either n is divisible by 3 or (n+1) is divisible by 3.
2) n is divisible by 3 in 99/3=33 cases
3) n+1 is divisible by 3 in 99/3=33 cases
4) 33+33=66
5) 66/99=2/3
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec   [#permalink] 24 Jan 2017, 20:05
    Similar topics Author Replies Last post
Similar
Topics:
8 Experts publish their posts in the topic n is an integer from 1 to 50, what is the probability that n(n+1) is n siddharthsinha123 4 27 Jan 2017, 10:31
11 Experts publish their posts in the topic ((n-1)!+n!+(n+1)!)/n! jahanafsana 11 11 May 2017, 02:49
1 Experts publish their posts in the topic If 1<=n<=99 and n is an integer, what is the probability that n(n + 1) vinyasgupta 2 08 Feb 2015, 19:21
8 If A(n)=1/(n(n+1)) for all positive integers n, what is the Dixon 4 01 Dec 2016, 00:27
6 Experts publish their posts in the topic If n is an integer from 1 to 96, what is the probability for GMATmission 7 29 Aug 2014, 04:33
Display posts from previous: Sort by

If 1=<n<=99, what is the probability that n(n + 1) is perfec

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


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.