Last visit was: 23 Apr 2026, 18:24 It is currently 23 Apr 2026, 18:24
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Hovkial
User avatar
Joined: 23 Apr 2019
Last visit: 24 Nov 2022
Posts: 802
Own Kudos:
2,600
 [33]
Given Kudos: 202
Status:PhD trained. Education research, management.
Posts: 802
Kudos: 2,600
 [33]
The sum of the first 'n' consecutive positive integers is a perfect 20 Sep 2019, 10:56
1
Kudos
Add Kudos
32
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

35% (02:38) correct 65% (02:31) wrong based on 157 sessions

History

Date
Time
Result
Not Attempted Yet
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6
ShowHide Answer
Official Answer
A
Most Helpful Reply
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 22 Apr 2026
Posts: 11,229
Own Kudos:
45,002
 [9]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,002
 [9]
The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 06:13
3
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6
Show more


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
49 and 50
Thus answer is 3, when n is 1, 8 and 49.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)
\(\frac{49*50}{2}=49*25=(7*5)^2\)

B
General Discussion
avatar
DonVito
avatar
Joined: 04 Jul 2019
Last visit: 11 Mar 2023
Posts: 25
Own Kudos:
14
 [1]
Given Kudos: 128
Location: India
Schools: Molson '22 (A) HEC Montreal '21 (A) Alberta "22 (A)
GMAT 1: 680 Q49 V33
Schools: Molson '22 (A) HEC Montreal '21 (A) Alberta "22 (A)
GMAT 1: 680 Q49 V33
Posts: 25
Kudos: 14
 [1]
The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 11:20
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
Thus answer is 2.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)

A
Show more

Just a little doubt here Sir, if n=49, sum will be 49*50/2=49*25 which is a perfect square(square of 35), shouldn't the answer be 3(total 3 values of n).

Please let me know where am I wrong?
User avatar
unraveled
User avatar
Joined: 07 Mar 2019
Last visit: 10 Apr 2025
Posts: 2,706
Own Kudos:
2,329
Given Kudos: 763
Location: India
WE:Sales (Energy)
Posts: 2,706
Kudos: 2,329
The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 21:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DonVito
chetan2u
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
Thus answer is 2.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)

A

Just a little doubt here Sir, if n=49, sum will be 49*50/2=49*25 which is a perfect square(square of 35), shouldn't the answer be 3(total 3 values of n).

Please let me know where am I wrong?
Show more

I was also in a fix what I am getting wrong about.
I marked A since I missed a little insight that its not Sum(n) < 100 but n < 100.
The catch is that after dividing n(n+1) by 2 both the digits(even one) should be a perfect square

However, after the expert solution by chetan2u i figured out that 49 and 50 do make up the solution but why 'A'. He corrected it now.
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 22 Apr 2026
Posts: 11,229
Own Kudos:
45,002
 [1]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,002
 [1]
The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 22:02
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DonVito
chetan2u
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
Thus answer is 2.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)

A

Just a little doubt here Sir, if n=49, sum will be 49*50/2=49*25 which is a perfect square(square of 35), shouldn't the answer be 3(total 3 values of n).

Please let me know where am I wrong?
Show more

Yes, you are correct.
Although my solution did include 49 and 50 as one of possibilities, i did not include that by mistake..
Thanks
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,964
Own Kudos:
1,117
Posts: 38,964
Kudos: 1,117
The sum of the first 'n' consecutive positive integers is a perfect 11 Jul 2024, 02:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109785 posts
Krunaal
Tuck School Moderator
853 posts