Sep 19 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. One hour of live, online instruction. Sep 19 10:00 PM PDT  11:00 PM PDT Join a FREE 1day Data Sufficiency & Critical Reasoning workshop and learn the best strategies to tackle the two trickiest question types in the GMAT! Sep 20 08:00 AM PDT  09:00 AM PDT Feeling perplexed by the paradox? Despairing the disparity? In this webinar, Hailey Cusimano explains how we can use strategy and structure to address premises that don't seem to align. Sept 20, Friday, 8am PST. Sep 19 12:00 PM PDT  10:00 PM PDT On Demand $79, For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Sep 21 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Sep 22 08:00 PM PDT  09:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE Sep 23 08:00 AM PDT  09:00 AM PDT Join a free 1hour webinar and learn how to create the ultimate study plan, and be accepted to the upcoming Round 2 deadlines. Save your spot today! Monday, September 23rd at 8 AM PST
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 458
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
Updated on: 18 Nov 2013, 22:35
Question Stats:
68% (02:09) correct 32% (02:15) wrong based on 434 sessions
HideShow timer Statistics
A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? A. 33 B. 46 C. 49 D. 53 E. 86 OA is a and this is how I arrived at.
Let say n = 4x+1 and n = 5y+3 > From the question Stem
n=4x+1 n = 5y+3 5 8 9 13 13 18 17 23 25 33 29 38 33 43
I get these above values by putting the same values for x and y. Is my concept correct?
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730
Originally posted by enigma123 on 21 Jan 2012, 14:37.
Last edited by mau5 on 18 Nov 2013, 22:35, edited 3 times in total.
Edited the OA



Math Expert
Joined: 02 Sep 2009
Posts: 58117

Re: Group of Students
[#permalink]
Show Tags
21 Jan 2012, 14:59
enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86 Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one. A group of n students can be divided into equal groups of 4 with 1 student left over > n=4q+1 > n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4) A group of n students can be divided into equal groups of 5 with 3 students left over > n=5p+3 > n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5) Therefor two smallest possible values of n are 13 and 33 > 13+33=46. Answer: B. Else you can derive general formula based on n=4q+1 and n=5p+3. Divisor will be the least common multiple of above two divisors 4 and 5, hence 20. Remainder will be the first common integer in above two patterns, hence 13 > so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) > 13+33=46. Answer: B. For more about this concept see: manhattanremainderproblem93752.html#p721341whenpositiveintegernisdividedby5theremainderis90442.html#p722552whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope it helps.
_________________



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 458
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
21 Jan 2012, 15:34
Hi Bunuel  can the values of q and p be ZERO? I don't think they can be and therefore n cannot be 1 & 3. Am I wrong?
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Math Expert
Joined: 02 Sep 2009
Posts: 58117

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
21 Jan 2012, 15:44
enigma123 wrote: Hi Bunuel  can the values of q and p be ZERO? I don't think they can be and therefore n cannot be 1 & 3. Am I wrong? THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is nonnegative integer and always less than divisor).For example we are told that when positive integer n is divided by 25, the remainder is 13 > \(n=25q+13\). Now, the lowest value of \(q\) can be zero and in this case \(n=13\) > 13 divided by 25 yields the remainder of 13. Generally when divisor (25 in our case) is more than dividend (13 in our case) then the reminder equals to the dividend. For example: 3 divided by 24 yields a reminder of 3 > \(3=0*24+3\); or: 5 divided by 6 yields a reminder of 5 > \(5=0*6+5\). Also note that you shouldn't worry about negative numbers in divisibility questions, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers. OUR ORIGINAL QUESTION: We are told that "a group of n students can be divided into equal groups of 4 with 1 student left over" > n=4q+1. Here q also can be zero, which would mean that there is only 1 student and zero groups of 4. QUESTIONS TO PRACTICE: PS questions on remainders: search.php?search_id=tag&tag_id=199DS questions on remainders: search.php?search_id=tag&tag_id=198THEORY ON REMAINDERS: compilationoftipsandtrickstodealwithremainders86714.htmlHope it helps.
_________________



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 458
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
21 Jan 2012, 15:53
You are a true genius buddy.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Manager
Joined: 12 Nov 2011
Posts: 59

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
22 Jan 2012, 08:16
Stright and simple B just try numbers for 4*n+1=5*k+3 where k and n are integers, find 2 smallest and your solve it for 30 sec



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9643
Location: Pune, India

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
22 Jan 2012, 18:48
enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a)33 b)46 c)49 d)53 e) 86
OA is a and this is how I arrived at.
Let say n = 4x+1 and n = 5y+3 > From the question Stem
n=4x+1 n = 5y+3 5 8 9 13 13 18 17 23 25 33 29 38 33 43
I get these above values by putting the same values for x and y. Is my concept correct? I wrote a blog post discussing this concept in detail. I have discussed a couple of questions very similar to this one in the post. You can check it out if you like. http://www.veritasprep.com/blog/2011/05 ... spartii/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Status: DDay is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Joined: 12 Apr 2011
Posts: 162
Location: Kuwait
Schools: Columbia university

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
23 Jan 2012, 10:01
n=4q+1 > 1,5,9,13,17,21,25,29,33 n=5q+3>3,8,13,18,23,28,33 the first two common numbers are 13 and 33, so add those numbers, you get 13+33=46 so, asnwer is B, 46 hope this helps
_________________



Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 458
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
24 Jan 2012, 16:09
Thanks Karishma. A very helpful post.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730



Manager
Joined: 28 Apr 2013
Posts: 120
Location: India
GPA: 4
WE: Medicine and Health (Health Care)

Re: Group of Students
[#permalink]
Show Tags
18 Nov 2013, 20:15
Again since the max student left over are 3 so any no with 3 can be common between the two groups ; 3, 13, 23, 33, 43, 53 etc. using in the formula pf n= 4q+1 = 5r+ 3 ; will get 13 and 33 ; addd then =46
_________________
Thanks for Posting
LEARN TO ANALYSE
+1 kudos if you like



Manager
Joined: 26 Sep 2013
Posts: 188
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

Re: Group of Students
[#permalink]
Show Tags
19 Nov 2013, 10:48
Bunuel wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86 Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one. A group of n students can be divided into equal groups of 4 with 1 student left over > n=4q+1 > n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4) A group of n students can be divided into equal groups of 5 with 3 students left over > n=5p+3 > n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5) Therefor two smallest possible values of n are 13 and 33 > 13+33=46. Answer: B. Else you can derive general formula based on n=4q+1 and n=5p+3. Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.
Remainder will be the first common integer in above two patterns, hence 13 > so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) > 13+33=46.
Answer: B. For more about this concept see: manhattanremainderproblem93752.html#p721341whenpositiveintegernisdividedby5theremainderis90442.html#p722552whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope it helps. Could you explain what you mean by 'divisor'? Where is there a divisor in those two equations....Is there work here that you did in your head, but left out? I'm trying to learn how to do these, so if you could show any work that was done mentally I would appreciate it! Thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 58117

Re: Group of Students
[#permalink]
Show Tags
19 Nov 2013, 15:19
AccipiterQ wrote: Bunuel wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86 Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one. A group of n students can be divided into equal groups of 4 with 1 student left over > n=4q+1 > n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4) A group of n students can be divided into equal groups of 5 with 3 students left over > n=5p+3 > n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5) Therefor two smallest possible values of n are 13 and 33 > 13+33=46. Answer: B. Else you can derive general formula based on n=4q+1 and n=5p+3. Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.
Remainder will be the first common integer in above two patterns, hence 13 > so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) > 13+33=46.
Answer: B. For more about this concept see: manhattanremainderproblem93752.html#p721341whenpositiveintegernisdividedby5theremainderis90442.html#p722552whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope it helps. Could you explain what you mean by 'divisor'? Where is there a divisor in those two equations....Is there work here that you did in your head, but left out? I'm trying to learn how to do these, so if you could show any work that was done mentally I would appreciate it! Thanks! 20 is the divisor in n=20m+13. Please follow the links in my post.
_________________



SVP
Joined: 12 Dec 2016
Posts: 1502
Location: United States
GPA: 3.64

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
06 Sep 2017, 12:21
start with 8, then 13,18,23,... try every combination



VP
Joined: 07 Dec 2014
Posts: 1233

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
06 Sep 2017, 14:31
enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
A. 33 B. 46 C. 49 D. 53 E. 86 assume difference of 1 between quotients (n1)/4(n3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B



SVP
Joined: 12 Dec 2016
Posts: 1502
Location: United States
GPA: 3.64

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
06 Sep 2017, 15:19
gracie wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
A. 33 B. 46 C. 49 D. 53 E. 86 assume difference of 1 between quotients (n1)/4(n3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B this can be applied to any problem of this kind?



SVP
Joined: 12 Dec 2016
Posts: 1502
Location: United States
GPA: 3.64

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
06 Sep 2017, 15:24
gracie wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
A. 33 B. 46 C. 49 D. 53 E. 86 assume difference of 1 between quotients (n1)/4(n3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B also, how can you assume that the difference is 1, it can be 2 or 3 or 4



Director
Joined: 19 Oct 2013
Posts: 524
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
05 Oct 2018, 16:41
\(\frac{n}{x}= 4 + \frac{1}{x}\)
\(n = 4x + 1\)
n = 5,9,13,17,21,25,29,33
\(\frac{n}{y}= 5 + \frac{3}{y}\)
\(n = 5y + 3\)
n = 8,13,18,23,28,33
13 + 33 = 46
Answer choice B



Intern
Joined: 09 Apr 2018
Posts: 31
GPA: 4

A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
14 Oct 2018, 23:16
Deriving the general formula n=kx+r, based on both statements, where k is the LCM of the 2 divisors and r is the first common integer: n=4q+1> n=1,5,9, 13... n=5p+3> n=3,8, 13... general formula: n=20x+13> n=13, 33... sum=13+33=46Please hit Kudos if you liked this approach




A group of n students can be divided into equal groups of 4
[#permalink]
14 Oct 2018, 23:16






