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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Given a series of n consecutive positive integers, where n > [#permalink]

Show Tags

12 Jan 2012, 22:35

2

This post received KUDOS

12

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

51% (02:44) correct
49% (01:57) wrong based on 283 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.

Re: Series of n consecutive positive integers [#permalink]

Show Tags

13 Jan 2012, 03:19

1

This post received KUDOS

2

This post was BOOKMARKED

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.

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.

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.

Re: Series of n consecutive positive integers [#permalink]

Show Tags

13 Jan 2012, 03:45

GyanOne wrote:

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).

Here how did you take n=1, in the question he said n>1 right?...can you please explain??
_________________

Re: Series of n consecutive positive integers [#permalink]

Show Tags

13 Jan 2012, 22:40

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

Re: Given a series of n consecutive positive integers, where n > [#permalink]

Show Tags

24 Oct 2014, 07:58

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: Given a series of n consecutive positive integers, where n > [#permalink]

Show Tags

25 Oct 2014, 00:34

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.
_________________

Re: Given a series of n consecutive positive integers, where n > [#permalink]

Show Tags

22 May 2016, 03:35

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.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...