Last visit was: 19 Nov 2025, 21:27 It is currently 19 Nov 2025, 21:27
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,392
 [31]
Kudos
Add Kudos
31
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Dumsy_1711
Joined: 23 May 2023
Last visit: 24 Jul 2024
Posts: 57
Own Kudos:
44
 [8]
Given Kudos: 305
Location: India
Concentration: Real Estate, Sustainability
GPA: 3.7
WE:Other (Other)
Posts: 57
Kudos: 44
 [8]
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 19 Nov 2025
Posts: 8,422
Own Kudos:
4,982
 [4]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,422
Kudos: 4,982
 [4]
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,392
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
There are N students in a class. When the students are distributed into groups that contain 4A number of students each, 3 students are left without a group. When the students are distributed into groups that contain A/3 number of students each, no students are left without a group. Which of the following statements is correct?

I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2
II. If the students are distributed into groups that contain 3 students each, no students are left without a group
III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group


A. I only
B. II only
C. III only
D. I and II only
E. II and III only
 



This question is a part of Are You Up For the Challenge: 700 Level Questions collection.­
User avatar
RenB
Joined: 13 Jul 2022
Last visit: 18 Nov 2025
Posts: 391
Own Kudos:
Given Kudos: 303
Location: India
Concentration: Finance, Nonprofit
GMAT Focus 1: 715 Q90 V84 DI82
GPA: 3.74
WE:Corporate Finance (Consulting)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Archit3110
Let value of A be 3
such that 4A ; 12 and 3 students are left behind means that total 9 students are in groups
later A/3 ; 3/3 ; 1 group such that no student is left behind

I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2
A+1 ; 4 this is not sufficient as 4*4 ; A/3 ; 1 student is left behind ; incorrect

II. If the students are distributed into groups that contain 3 students each, no students are left without a group
A has to be factor of 3 A/3 will always leave no student without a group correct

III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group

This cannot be correct as 12 *x ; or 12*x/3 will always leave no students behind ; incorrect
option II is only correct
option B

Bunuel
There are N students in a class. When the students are distributed into groups that contain 4A number of students each, 3 students are left without a group. When the students are distributed into groups that contain A/3 number of students each, no students are left without a group. Which of the following statements is correct?

I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2
II. If the students are distributed into groups that contain 3 students each, no students are left without a group
III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group


A. I only
B. II only
C. III only
D. I and II only
E. II and III only

 
Hi,

Is this qs representative of a hard GMAT quant qs?
Also, can you please help me understand the logic behind taking A=3?

TIA!­
User avatar
RAldertree
Joined: 05 Jul 2025
Last visit: 09 Nov 2025
Posts: 2
Posts: 2
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
There are N students in a class. When the students are distributed into groups that contain 4A number of students each, 3 students are left without a group. When the students are distributed into groups that contain A/3 number of students each, no students are left without a group. Which of the following statements is correct?

I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2
II. If the students are distributed into groups that contain 3 students each, no students are left without a group
III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group


A. I only
B. II only
C. III only
D. I and II only
E. II and III only
This problem lacks any clear solutions. Here is mine:

For some constants k and c:
We know N = 4Ak + 3 and N = A/3 * c.
A/3 must be an integer.

I. Is it possible for N = (A+1)k + 2?

First let's find possible values for A:
4Ak + 3 = A/3 * c.
12Ak + 9 = Ac
k is an arbitrary constant, so we can say 12Ak = Ak.
Ak + 9 = Ac
9 = A(c-k)
Again, c and k are arbitrary constants, so we can say c-k = k
9 = Ak
A must be a factor of 9. Given that A/3 is an integer, A can be 3 or 9.

Case 1, A = 3:
Can N = 4c + 2?
4Ak + 3 = 4c + 2
If k is arbitrary, then Ak = k
4k + 3 = 4c + 2
4k + 1 = 4c
This is never true for integer values of c and k
rejected

Case 2, A = 9:
Can N = 10c + 2?
From N = 4Ak + 3, N = 36k + 3
36k + 3 = 10c + 2
36k + 1 = 10c
2*(18k) + 1 = 2*(5c)
as k and c are arbitrary, k = 18k and c = 5c
2k + 1 = 2c
This is never true for integer values of c and k
rejected

Answer to I: No

II. Is N = 3k?
N = 4Ak + 3
A/3 is an integer
N = 4 * 3 * A/3 * k + 3
N = 3(4 * A/3 * k + 1)
k is arbitrary, so 4 * A/3 * k = k
N = 3(k+1)
k is arbitrary, so k+1 = k
N = 3k

Answer to II: Yes

III. Is N = 12k + 9?
N = 4Ak + 3
A/3 is an integer
N = 4 * 3 * A/3 * k + 3
N = 12 * A/3 * k + 3
k is arbitrary, so A/3 * k = k
N = 12k + 3
12k + 3 can never be 12k + 9, for some integer value of k

Answer to III: No

Final Answer: B) II only



Explanation using Modular Arithmetic:

N = 3 (mod 4A)
N = 0 (mod A/3)
A/3 is an integer

I. Can N = 2 (mod A+1)?
4Ak + 3 = 0 (mod A/3)
12Ak + 9 = 0 (mod A)
9 = 0 (mod A)
As A/3 is an integer, A can be 3 or 9.

Case 1, A = 3
N = 2 (mod 4)?
N = 3 (mod 4A) -> N = 3 (mod 4)
rejected

Case 2, A = 9
N = 2 (mod 10)?
36k + 3 = 2 (mod 10)
6k + 3 = 2 (mod 10)
6k = -1 (mod 10)
6k = 9 (mod 10)
2k = 3 (mod 10)
2 divides 10 but not 3, so k DNE
rejected

Answer to I: No

II. N = 0 (mod 3)?
N = 3 (mod 4A)
A/3 is an integer
N = 3 (mod 4*3*A/3)
N = 3 (mod 3)
3 = 0 (mod 3)
N = 0 (mod 3)

Answer to II: Yes

III. Is N = 9 (mod 12)?
N = 3 (mod 4A)
A/3 is an integer
N = 3 (mod 4*3*A/3)
N = 3 (mod 12 * A/3)
N = 3 (mod 12)

Answer to III: No

Final Answer: B) II only
User avatar
Aarya_San
Joined: 05 Feb 2024
Last visit: 17 Nov 2025
Posts: 15
Own Kudos:
Given Kudos: 23
Schools: ISB '26 (S)
Products:
Schools: ISB '26 (S)
Posts: 15
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Honestly I feel algebra might not be the way to do this question.

Given:

N/4A, remainder = 3 -> N = 4A + 3 -> (Even + Odd = Odd), So we can conclude N is an odd number.
Also: A is a multiple of 3, as A/3 is an integer. (No. of people cannot be a fraction or decimal).
Therefore N = 4 (Multiple of 3) + 3 -> N is also divisible by 3
N = (A/3)k {Now since N is odd, and both A and k are being divided by just 3 to give N, A and K should also both be odd} -> A is odd.


So what do we have: A is odd and multiple of 3. N is also odd and a multiple of 3


Possible values of A = {3,9,15,21, .... }
Possible values of N = 4A +3 = {15, 39, 63, 87, .....}

Now let's look at the statements:

I. Can N/(A+1) leave 2 as a remainder. Let's test it out for our values above: 15/4 -> rem = 3 ; 39/10 -> rem = 9 ; 63/16 -> rem is 15 and this will keep increasing. Don't see the value becoming 2. Not True.

II. Does N/3 leave 0. as a rem? Yes!! We know N is divisible by 3. Yes!
III. Does N/12 leave 9 as a rem? Let's test using our values -> 15/12, rem = 3; 39/12, rem = 3 ; 63/12, rem = 3; and this trend continues. So we never see remainder equalling 9. No!

Done!
Moderators:
Math Expert
105390 posts
Tuck School Moderator
805 posts