GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Oct 2019, 13:31

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

Given a series of n consecutive positive integers, where n >

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

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 442
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Given a series of n consecutive positive integers, where n >  [#permalink]

Show Tags

New post 12 Jan 2012, 23:35
2
45
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

50% (02:28) correct 50% (02:39) wrong based on 355 sessions

HideShow timer Statistics

Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.

_________________
Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: Series of n consecutive positive integers  [#permalink]

Show Tags

New post 16 Jan 2012, 18:05
6
8
enigma123 wrote:
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?
(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.


Responding to a pm.

Couple of things:
In any evenly spaced set (AP) the arithmetic mean (average) is equal to the median and can be calculated by the formula: (first term+last term)/2

Now, set of consecutive integers is an evenly spaced set (AP with common difference of 1) and in order mean=median to be an integer it has to have an odd number of terms. If there are an even number of terms mean=median will be integer/2.

For example:
{1, 2, 3} --> mean=median (middle term)=(3+1)/2=2;
{1, 2, 3, 4} --> mean=median=(1+4)/2=5/2=2.5.

Next, in AP if the first term is \(a\) and the common difference of successive members is \(d\), then the \(n_{th}\) term of the sequence is given by: \(a_ n=a+d(n-1)\). In case of consecutive integers (so when common difference=d=1) the formula becomes: \(a_ n=a+n-1\).

Check Number Theory for more on AP: math-number-theory-88376.html (specifically "Consecutive Integers" and "Evenly Spaced Set" chapters of it)

BACK TO THE ORIGINAL QUESTION.
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

Basically we are asked whether: \(average=\frac{first \ term+last \ term}{2}\) is divisible by 3 or whether \(average=\frac{a+(a+n-1)}{2}=a+\frac{n-1}{2}\) is divisible by 3 (where \(a\) is the first term and \((a+n-1)\) is the \(n_{th}\), so last term).

(1) n is odd --> as n=odd then the average is definitely an integer, though it may or may not be divisible by 3: {2, 3, 4} - YES, {1, 2, 3} - NO. Not sufficient.

(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3 --> we are directly given that \(a+\frac{n-1}{2}\) is divisible by 3. Sufficient.

Answer: B.

Important note: you should have spotted that there was something wrong with your solution as on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

So we can not have answer NO from statement (1) and answer YES from statement (2) (as you got in your solution), because in this case statements would contradict each other.

Hope it helps.
_________________
General Discussion
SVP
SVP
User avatar
P
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 1883
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740
Re: Series of n consecutive positive integers  [#permalink]

Show Tags

New post 13 Jan 2012, 04:19
3
4
The sum of n terms of a consecutive positive integer series = (n/2)[2a + n -1] where a is the first term

Average of the first n terms = (1/2) [2a+ n -1] = a + [(n-1)/2]

Using statement (1), if n is odd then the answer may or may not be divisible by 3. For example, take a=1 and n=5 to get average of first n terms as 3, which is divisible by 3. However, if a=1 and n=3 then the average of the first n terms is 2, which is not divisible by 3. Insufficient.

Using statement (2), a + [(n-1)/2] is divisible by 3. This means that a + [(n-1)/2] is divisible by 3. Therefore the average is divisible by 3. Sufficient.

Therefore the answer is (B).
_________________
GyanOne [www.gyanone.com]| Premium MBA and MiM Admissions Consulting

Awesome Work | Honest Advise | Outstanding Results

Reach Out, Lets chat!
Email: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services
Intern
Intern
avatar
Joined: 22 Aug 2011
Posts: 2
Re: Series of n consecutive positive integers  [#permalink]

Show Tags

New post 13 Jan 2012, 23:40
For Statement II:

lets assume n=11 and the consecutive positive integer series be 1,2,3,4,5,6,7,8,9,10,11

then 1+(11-1)/2 = 6 which is divisible by 3 and the average of (1+2+3+4+5+6+7+8+9+10+11) =33 which is divisible by 3
however, if we assume n=3 and take the consecutive positive integer series to be 2,3,4
the 2+(3-1)/2= 3 which is divisible by 3 but the average of (2+3+4) is not divisible by 3


I'm not sure if what im doing is right ;@
Senior Manager
Senior Manager
avatar
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 442
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Re: Series of n consecutive positive integers  [#permalink]

Show Tags

New post 19 Jan 2012, 16:54
You are a star Bunuel. Thanks for such a detailed explanation.
_________________
Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Manager
Manager
User avatar
S
Joined: 22 Jan 2014
Posts: 170
WE: Project Management (Computer Hardware)
Re: Given a series of n consecutive positive integers, where n >  [#permalink]

Show Tags

New post 25 Oct 2014, 01:34
enigma123 wrote:
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.


B.

(1) n is odd
1,2,3 ; Ans = No.
2,3,4 ; Ans = Yes.
So insufficient.

(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3
Let the series be a,a+1,a+2,...,a+(n-1)
acc to FS2 [a + (n-1)/2] mod 3 = 0
the avg of series is: na + ((n-1)(n)/2) = [n (a + (n-1)/2)]
since [a + (n-1)/2] mod 3 = 0
=> n*[a + (n-1)/2] is also div by 3.
hence, the avg is div by 3.
sufficient.
_________________
Illegitimi non carborundum.
Current Student
avatar
B
Joined: 22 Sep 2016
Posts: 159
Location: India
GMAT 1: 710 Q50 V35
GPA: 4
Given a series of n consecutive positive integers, where n >  [#permalink]

Show Tags

New post 27 Jul 2017, 09:26
enigma123 wrote:
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.


(1)
consider 2+3+4+5+6+7+8 =35
also, n= odd.

consider 2+3+4 = 9


(2)

according to second statement, 2+((7-1)/2) = 8


but for n=9, 2+ ((9-1)/2) = 6

So here you go, answer is E
_________________
Desperately need 'KUDOS' !!
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13243
Re: Given a series of n consecutive positive integers, where n >  [#permalink]

Show Tags

New post 30 Aug 2019, 05:20
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 Bot
Re: Given a series of n consecutive positive integers, where n >   [#permalink] 30 Aug 2019, 05:20
Display posts from previous: Sort by

Given a series of n consecutive positive integers, where n >

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





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