by: Kevi
4 mins
Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 1312:30 AM EDT
-01:30 AM EDT
Struggling to find the right strategies to score a 99 %ile on GMAT Focus? Riya (GMAT 715) boosted her score by 100-points in just 15 days! Discover how the right mentorship, tailored strategies, and an unwavering mindset can transform your GMAT prep.
Sep 1301:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
ChandlerBong
Joined: 16 Jan 2022
Last visit: 19 Jan 2025
Posts: 223
Location: India
GRE 1: Q165 V165
GPA: 4
WE:Analyst (Computer Software)
GRE 1: Q165 V165
Posts: 223
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
72% (02:47) correct
28%
(02:58)
wrong
based on 1453
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?
A. 20
B. 21
C. 24
D. 26
E. 72
GMAT-Club-Forum-r79fsrvn.png [ 41.63 KiB | Viewed 12026 times ]
_________________
A. 20
B. 21
C. 24
D. 26
E. 72
Attachment:
GMAT-Club-Forum-r79fsrvn.png [ 41.63 KiB | Viewed 12026 times ]
Pro tip:
Combine official questions for familiarity with GMAT Club Tests for analytics and tougher Focus-level practice so the real exam feels manageable.
Explore →
_________________
Sarcasm | GMAT | Sleep
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
ChandlerBong
Let's assume that the number is \(n\)
...a remainder of 1 when divided by 7...
\(n = 7q_1 + 1\)
\(q_1\) → quotient when n is divided by \(7\)
...a remainder of 2 when divided by 3...
\(n = 3q_2 + 2\)
\(q_2\) → quotient when n is divided by \(3\)
Both the equations can be merged into a single equations
\(n = \text{LCM}(7,3)q + \text{first common term}\)
\(q\) → quotient when n is divided by \(21\)
To find the first common term, let's write a few of the terms of each of the sequences-
\(n = 7q_1 + 1\) ⇒ \(1, 8, 15, 22, 29, 36, .... \)
\(n = 3q_2 + 2\) ⇒ \(2, 5, 8, 11, .... \)
Hence, the first common term = 8
\(n = 21q + 8\)
This equation represents an arithmetic progression.
The first term of the equation is \(8\), i.e. when \(q = 0\)
Last term =
\(500 = 21q + 8\)
\(21q = 492\)
\(q = 23.XX\)
Number of terms = \(23 - 0 + 1 = 24\)
Option C
_________________
Want to discuss quant questions and strategies : Join the quant chat group today
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
AnkurGMAT20
6. Remainders
- Theory
REMAINDERS
Remainders: Tips and hints
Divisibility and Remainders on the GMAT
Remainders on the GMAT
A Remainders Shortcut for the GMAT
All About Negative Remainders on the GMAT
A Tricky Question on Negative Remainders
Compilation of tips and tricks to deal with remainders
Cyclicity and the remainders on the GMAT
GMAT Question Patterns: Remainders
For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
Hope it helps.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
General Discussion
Dumsy_1711
Joined: 23 May 2023
Last visit: 24 Aug 2026
Posts: 56
Location: India
Concentration: Real Estate, Sustainability
GPA: 3.7
WE:Other (Other)
Posts: 56
Updated on: 27 Jan 2024, 02:09
Originally posted by Dumsy_1711 on 08 Nov 2023, 13:03.
Last edited by Dumsy_1711 on 27 Jan 2024, 02:09, edited 3 times in total.
Last edited by Dumsy_1711 on 27 Jan 2024, 02:09, edited 3 times in total.
\(\frac{500}{7} =71.\) There are 71 integers less than 500 that are divisible by 7.
Quick check for the last integer: 7(71)+1 = 497+1 = 498. 498 is less than 500.
So, there are 71 integers less than 500 that have a remainder of 1 when divided by 7.
Integers that have a remainder of 1 when divided by 7 = 7k+1.
Integers that have a remainder of 2 when divided by 3 = 3n+2
Possible values of
7k+1: 8 15 22 29 36 43 50 57 64 71.......498
3n+2: 8 16 23 29 38 44 50 56 65 71.......498
(It is easier and faster to check for the integers of the form 3n+2 that match with integers of the form 7k+1 than to make an entire list of all the integers of the form 3n+2.)
Every third term after the first term (8) is common between 7k+1 and 3n+2.
\(\frac{71}{3} = 23. \)
23+1 = 24. The number of common integers = 24.
Therefore, the answer is C : 24.
Quick check for the last integer: 7(71)+1 = 497+1 = 498. 498 is less than 500.
So, there are 71 integers less than 500 that have a remainder of 1 when divided by 7.
Integers that have a remainder of 1 when divided by 7 = 7k+1.
Integers that have a remainder of 2 when divided by 3 = 3n+2
Possible values of
7k+1: 8 15 22 29 36 43 50 57 64 71.......498
3n+2: 8 16 23 29 38 44 50 56 65 71.......498
(It is easier and faster to check for the integers of the form 3n+2 that match with integers of the form 7k+1 than to make an entire list of all the integers of the form 3n+2.)
Every third term after the first term (8) is common between 7k+1 and 3n+2.
\(\frac{71}{3} = 23. \)
23+1 = 24. The number of common integers = 24.
Therefore, the answer is C : 24.
SN9633
Joined: 01 Aug 2023
Last visit: 12 Mar 2024
Posts: 42
Location: India
Schools: IIMA PGPX'25 (WA) ISB '25 (WL)
GPA: 3.75
Schools: IIMA PGPX'25 (WA) ISB '25 (WL)
Posts: 42
Runjhun26
hope this helps: the LCM was taken and added with the lowest value. Then the given condition N<500 was substituted with the new arrived equation
Attachments
sol.jpg [ 53.82 KiB | Viewed 35654 times ]
Bunuel, do you have more such questions for practice?
How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?
A. 20
B. 21
C. 24
D. 26
E. 72
When 7 divides 500 it gives 71 integers as quotient 497 being the last divisible number.
Hence there are 71 integers that leave remainder 1 498(497+1) being the last one.
When 3 divides 500 it gives 166 integers as quotient 498 being the last divisible number.
Hence there are 166 integers that leave remainder 1 499(498+2) being the last one.
But this is not helpful since we are looking for a common integer that leaves 1 and 2 remainder when divided by 7 and 3 individually.
So, only 71 integers are there among which common ones would leave remainders 1 and 2. These common can be found by dividing 71 by 3 which gives us 23.xx
Since we are looking for integer decimal is not possible hence we round up it to 24. Answer C.
Estimation also helps but that depends on the choices available. Had 23 been there among the choices it would have been difficult. But that is unlikely in these tests.
Answer C.
_________________
A. 20
B. 21
C. 24
D. 26
E. 72
When 7 divides 500 it gives 71 integers as quotient 497 being the last divisible number.
Hence there are 71 integers that leave remainder 1 498(497+1) being the last one.
When 3 divides 500 it gives 166 integers as quotient 498 being the last divisible number.
Hence there are 166 integers that leave remainder 1 499(498+2) being the last one.
But this is not helpful since we are looking for a common integer that leaves 1 and 2 remainder when divided by 7 and 3 individually.
So, only 71 integers are there among which common ones would leave remainders 1 and 2. These common can be found by dividing 71 by 3 which gives us 23.xx
Since we are looking for integer decimal is not possible hence we round up it to 24. Answer C.
Estimation also helps but that depends on the choices available. Had 23 been there among the choices it would have been difficult. But that is unlikely in these tests.
Answer C.
_________________
Pain + Reflection = Progress | Ray Dalio
Good Books to read prior to MBA
Good Books to read prior to MBA
GMAT 1: 800 Q51 V51
Posts: 2,161
How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?
When faced with a GMAT Quant question you're not sure how to answer, keep in mind that GMAT Quant questions are often designed to be answerable in multiple ways. So, it can often work to just try anything that might get you to the answer.
In this case, one approach that we can use is to look for a pattern starting with the smallest integer we can find such that it has a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3.
That integer is 8.
Then, we can continue with integers 1 greater than a multiple of 7 and look for a pattern to the ones that are 2 greater than a multiple of 3.
Doing so, we see the following:
8, 15, 22, 29, 36, 43, 50, 57, 64, 71
The integers in green fit our constraints. So, apparently, one in three integers that's 1 more than a multiple of 7 is two more than a multiple of 3.
There are 500 - 8 = 492 integers between 8 and 500.
492/7 = 70.X of those integers are one more than a multiple of 7.
70/3 = 23.X
23 + 1 (for 8) = 24 total integers that fit our constraints.
A. 20
B. 21
C. 24
D. 26
E. 72
Correct answer: C
_________________
When faced with a GMAT Quant question you're not sure how to answer, keep in mind that GMAT Quant questions are often designed to be answerable in multiple ways. So, it can often work to just try anything that might get you to the answer.
In this case, one approach that we can use is to look for a pattern starting with the smallest integer we can find such that it has a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3.
That integer is 8.
Then, we can continue with integers 1 greater than a multiple of 7 and look for a pattern to the ones that are 2 greater than a multiple of 3.
Doing so, we see the following:
8, 15, 22, 29, 36, 43, 50, 57, 64, 71
The integers in green fit our constraints. So, apparently, one in three integers that's 1 more than a multiple of 7 is two more than a multiple of 3.
There are 500 - 8 = 492 integers between 8 and 500.
492/7 = 70.X of those integers are one more than a multiple of 7.
70/3 = 23.X
23 + 1 (for 8) = 24 total integers that fit our constraints.
A. 20
B. 21
C. 24
D. 26
E. 72
Correct answer: C
_________________
Perfect scoring coach with a history of helping test-takers achieve their goals
https://martymurraycoaching.com/
Contact me for a free consultation.
https://martymurraycoaching.com/
Contact me for a free consultation.
gmatophobia
Is this the equation in the form of an=a0+ (n-1) d.
If yes, I am curious as to how you plugged in 500 for an?
Isn't it supposed to be integer values less than < 500 as per question prompt?
Is this the equation in the form of an=a0+ (n-1) d.
If yes, I am curious as to how you plugged in 500 for an?
Isn't it supposed to be integer values less than < 500 as per question prompt?
Yes2GMAT
21q + 8 < 500
21q < 492
q < 23.someting
Hence, q can take 24 values, from 0 to 23 inclusive. Consequently, 21q + 8 can also take the corresponding 24 values.
Hope it helps
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
ChandlerBong
n = 7a + 1
n = 3b + 2
Since there is no common remainder, the first such value is found by hit and trial.
If a = 1, n = 8. 8 is of the form 3*2 + 2 and hence 8 is the smallest such number.
Next such number will be 21 (LCM of 7 and 3) more than 8. Next will be another 21 more. and so on
So all numbers of this form will be 21k + 8 where k starts from 0 and takes all positive integer values. To get the greatest value less than 500,
21k + 8 < 500
21k < 492
k < 23.something
Hence k can take all values from 0 to 23. These will be 24 total values.
Answer (C)
Attachment:
GMAT-Club-Forum-7d1gqs83.png [ 41.63 KiB | Viewed 11884 times ]
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
GMAT 1: 800 Q51 V51
Posts: 2,161
Runjhun26
The first commmon term is a number such that the remainder when it's divided by 7 is 1 and the remainer when it's divided by 3 is 2.
Now, notice that any number that's greater than the first common term by a muliple of 7 will be such that, when it's divided by 7, the remainder is 1.
For example:
8 + 7 = 15 when divided by 7 has a remainder of 1.
15 + 7 = 22 when divided by 7 has a remainder of 1.
etc.
Similarly, any number that's greater than the first common term by a muliple of 3 will be such that, when it's divided by 3, the remainder is 2.
For example:
8 + 3 = 11 when divided by 3 has a remainder of 2.
11 + 3 = 14 when divided by 3 has a remainder of 2.
etc.
Thus, any number that is BOTH greater by a multiple of 7 than the common term and greater by a multiple of 3 than the common term will fit the criterion of having a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?
Of course, since 21 is the LCM of 3 and 7, any number that is both greater by a multiple of 7 than the common term and greater by a multiple of 3 than the common term will be greater than the common term by a multiple of 21.
So, since 8 is first common term, we'll find that the numbers we're looking for fit the following pattern:
8 + 21 = 29
29 + 21 = 50
50 + 21 = 71
etc.
_________________
Perfect scoring coach with a history of helping test-takers achieve their goals
https://martymurraycoaching.com/
Contact me for a free consultation.
https://martymurraycoaching.com/
Contact me for a free consultation.
can you please explain the concept after finding the first common term?
Please share concept. calculation part i understood
Please share concept. calculation part i understood
Posts: 1
gmatophobia
Posted from my mobile device
Why is the +1 added to 23 at the end of the solution?
ChandlerBong
Watch very simple two methods to this questions
Answer: Option C
Sample question for Method-2
if-a-line-which-has-the-equation-7x-17y-1000-is-plotted-on-x-y-pla-240113.html
Attachment:
GMAT-Club-Forum-n208a46u.png [ 41.63 KiB | Viewed 11861 times ]
_________________
GMATinsight (Tutor & Admission Consultant) | Contact for a FREE trial session
Our Recent scores: 735 | 715(3) | 705(2) | 695 (>5)
FREE Resources (Recommended BOOKMARKS):
8 FREE Focus Tests | Dynamic Focus Calculator | Data Insight (DI): Guide/Analysis/20+Video solution of Official Qns.
Our Recent scores: 735 | 715(3) | 705(2) | 695 (>5)
FREE Resources (Recommended BOOKMARKS):
8 FREE Focus Tests | Dynamic Focus Calculator | Data Insight (DI): Guide/Analysis/20+Video solution of Official Qns.
ANVEYSHIKA
There are 24 terms when you count 0 to 23, wherein 0 is the first term.
You might be missing the first term. Check Bunuel's explanation.
_________________
Pain + Reflection = Progress | Ray Dalio
Good Books to read prior to MBA
Good Books to read prior to MBA
given x%7=1 and x%3=2 ,where x is any integer and % sign reads as n divided by 7 will give a remainder of 1.
first such no. is 8
second is 29
and the last one is 491
the combination above will certainly form an AP
so very easily can be no. of terms satisfying this AP can be found as 8+(n-1)21=491 i.e. 24
first such no. is 8
second is 29
and the last one is 491
the combination above will certainly form an AP
so very easily can be no. of terms satisfying this AP can be found as 8+(n-1)21=491 i.e. 24
valeriugirlea
Joined: 01 Jan 2024
Last visit: 11 Nov 2025
Posts: 6
Location: Netherlands
Concentration: Economics, Economics
GMAT Focus 1: 585 Q78 V74 DI84
GPA: 3.97
Products:
GMAT Focus 1: 585 Q78 V74 DI84
Posts: 6
Why do we ad 1 to 23 in the end?
valeriugirlea because we calculate the number of terms between two integers by
(Last - First) + 1
For example: If first term is 2 and last term is 11, then
(11-2) + 1 = 10
Hence 10 numbers.
(Last - First) + 1
For example: If first term is 2 and last term is 11, then
(11-2) + 1 = 10
Hence 10 numbers.
Moderator:
Announcements
How to Tell If Your GMAT Prep Resources Are Actually Helping You
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Tuck at Dartmouth
AI at Dartmouth Tuck: How the School Is Preparing Leaders for the Age of Artificial Intelligence
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Top 5 GMAT Debrief Videos
Success stories and strategies from high-scoring candidates.