It is currently 18 Dec 2018, 17:19
My Tests

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.

The sum of the positive integers from 1 through n can be cal

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Moderator
Moderator
User avatar
Joined: 18 Apr 2015
Posts: 5205
Followers: 77

Kudos [?]: 1051 [0], given: 4711

CAT Tests
The sum of the positive integers from 1 through n can be cal [#permalink] New post 14 Apr 2018, 02:31
Expert's post
00:00

Question Stats:

75% (01:29) correct 25% (00:06) wrong based on 16 sessions
The sum of the positive integers from 1 through n can be calculated by the formula \(\frac{n(n + 1)}{2}\)

Quantity A
Quantity B
The sum of the multiples of 6
between 0 and 100
The sum of the multiples of 8
between 0 and 100


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
[Reveal] Spoiler: OA

_________________

Get the 2 FREE GREPrepclub Tests

1 KUDOS received
Director
Director
User avatar
Joined: 07 Jan 2018
Posts: 560
Followers: 4

Kudos [?]: 485 [1] , given: 85

Re: The sum of the positive integers from 1 through n can be cal [#permalink] New post 14 Apr 2018, 22:48
1
This post received
KUDOS
Find the value of \(n\) for each quantity

For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))

sum \(= 17 * \frac{96}{2}= 816\)

For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))

sum = \(13 * 96/2 = 624\)

clearly \(A\) is bigger
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

1 KUDOS received
Intern
Intern
Joined: 14 Apr 2018
Posts: 7
Followers: 0

Kudos [?]: 3 [1] , given: 5

Re: The sum of the positive integers from 1 through n can be cal [#permalink] New post 15 Apr 2018, 12:20
1
This post received
KUDOS
amorphous wrote:
Find the value of \(n\) for each quantity

For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))

From formula, sum \(= 17 * \frac{18}{2} = 17*9\)

For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))

From formula, sum = \(13 * \frac{14}{2} = 13*7\)

clearly \(A\) is bigger



Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13

So A is bigger
Director
Director
User avatar
Joined: 07 Jan 2018
Posts: 560
Followers: 4

Kudos [?]: 485 [0], given: 85

Re: The sum of the positive integers from 1 through n can be cal [#permalink] New post 16 Apr 2018, 00:19
halam wrote:
amorphous wrote:
Find the value of \(n\) for each quantity

For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))

From formula, sum \(= 17 * \frac{18}{2} = 17*9\)

For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))

From formula, sum = \(13 * \frac{14}{2} = 13*7\)

clearly \(A\) is bigger




Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13

So A is bigger


can you clarify what is the mistake?
Intern
Intern
Joined: 14 Apr 2018
Posts: 7
Followers: 0

Kudos [?]: 3 [0], given: 5

Re: The sum of the positive integers from 1 through n can be cal [#permalink] New post 16 Apr 2018, 13:38
amorphous wrote:
halam wrote:
amorphous wrote:
Find the value of \(n\) for each quantity

For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))

From formula, sum \(= 17 * \frac{18}{2} = 17*9\)

For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)
value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))

From formula, sum = \(13 * \frac{14}{2} = 13*7\)

clearly \(A\) is bigger


It is the sum, the number of multiples of 6 is not continuous from


Hi, I found that the you made a mistake with the sum above.
I have a different solution for this question:
- The first multiple of 6 is: 0 = 6*0
- The second multiple of 6 is: 6 =6*1
- The third multiple of 6 is: 12=6*2
............
- The seventeenth multiple of 6 is: 96=6*16
--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17
Similarly,
--> the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13

So A is bigger


can you clarify what is the mistake?


The sum you come up with: sum of all multiples of 6 = (17 * 18)/2 = 17*9
Since multiples of 6 are not continuous integers from 1 to 17, you cannot calculate the sum as above.
Please let me know your thought.
1 KUDOS received
Director
Director
User avatar
Joined: 07 Jan 2018
Posts: 560
Followers: 4

Kudos [?]: 485 [1] , given: 85

Re: The sum of the positive integers from 1 through n can be cal [#permalink] New post 16 Apr 2018, 21:18
1
This post received
KUDOS
Yes you are correct the given formula is for continuous numbers.
We can use avg * number of terms or similar methods to reach the answer. I have modified the above solution
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Re: The sum of the positive integers from 1 through n can be cal   [#permalink] 16 Apr 2018, 21:18
Display posts from previous: Sort by

The sum of the positive integers from 1 through n can be cal

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.