Processing math: 100%

Last visit was: 08 Apr 2025, 05:19

It is currently 08 Apr 2025, 05:19

Decision Tracker

New comers' posts

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

Request Expert Reply

Confirm Cancel

PREVIOUS TOPIC

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

Tags :

Find Similar Topics About Add a Tag

Show Tags

Hide Tags

Carcass

Verbal Expert

Joined: 18 Apr 2015

Posts: 31493

Own Kudos [?]: 37698 [1]

Given Kudos: 26234

Send PM

The sum of the positive integers from 1 through n can be cal [#permalink]

14 Apr 2018, 02:31

Expert Reply

1

Bookmarks

00:00

A

B

C

D

Question Stats:

84% (01:49) correct

15% (01:07) wrong

based on 86 sessions

Hide Show timer Statistics

This question is part of GREPrepClub - The Questions Vault Project

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

Official Answer

A

amorphous

Retired Moderator

Joined: 07 Jan 2018

Posts: 739

Own Kudos [?]: 1470 [3]

Given Kudos: 93

Send PM

Top Member of the Month

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

14 Apr 2018, 22:48

3

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

halam

Intern

Intern

Joined: 14 Apr 2018

Posts: 7

Own Kudos [?]: 16 [4]

Given Kudos: 0

Send PM

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

15 Apr 2018, 12:20

4

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

amorphous

Retired Moderator

Joined: 07 Jan 2018

Posts: 739

Own Kudos [?]: 1470 [0]

Given Kudos: 93

Send PM

Top Member of the Month

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

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?

halam

Intern

Joined: 14 Apr 2018

Posts: 7

Own Kudos [?]: 16 [0]

Given Kudos: 0

Send PM

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

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.

amorphous

Retired Moderator

Joined: 07 Jan 2018

Posts: 739

Own Kudos [?]: 1470 [0]

Given Kudos: 93

Send PM

Top Member of the Month

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

16 Apr 2018, 21:18

1

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

bumpbot

GRE Prep Club Legend

GRE Prep Club Legend

Joined: 07 Jan 2021

Posts: 5239

Own Kudos [?]: 77 [0]

Given Kudos: 0

Send PM

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

27 May 2024, 07:44

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.

gmatclubot

Re: The sum of the positive integers from 1 through n can be cal [#permalink]

27 May 2024, 07:44

Moderators:

GRE Instructor

108 posts

GRE Forum Moderator

37 posts

Moderator

1131 posts

GRE Instructor

234 posts

Hi Generic [Bot],

Here are updates for you:

ANNOUNCEMENTS

GRE Practice Quiz - Solve 20 Challenging Questions in Timed Conditions

Subscribe to our YouTube Channel to Get Quality GRE Content

GRE Quant & Verbal ADVANCED Daily Challenge 2025 NE

ETS Explains GRE - Test Format, Sections, Scoring Algorithm, Free Prep Resources

Subscribe to our YouTube Channel to Get Quality GRE Content

Average GRE Scores - Top 50 MBA Programs (2025)

GRE PREMIUM Quant Question Banks - Topic-Wise 2700 Questions

GRE Percentiles 2024-2025

GRE Math Handout

GMAT Club and Prodigy Finance scholarships

Prep Club for GRE REWARDS

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

Main navigation

Partners

GRE Resources

Copyright © 2025 Prep Club for GRE

GRE® is a registered trademark of the Education Testing Services® (ETS ®). Prep Club for GRE website has not been reviewed or endorsed by ETS

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