January 22, 2019 January 22, 2019 10:00 PM PST 11:00 PM PST In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one. January 26, 2019 January 26, 2019 07:00 AM PST 09:00 AM PST Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 10 Feb 2006
Posts: 631

A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
03 Nov 2007, 04:32
Question Stats:
69% (01:59) correct 31% (02:09) wrong based on 491 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 I got this so far
n = 4q + 1 n = 5q + 3
4q+1 + 5q+3 = 9q+4
plugging in value for q
q=1 q=2 q=3 q=4 q=5 = 45+4 = 49 ? not sure please help
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
GMAT the final frontie!!!.




Math Expert
Joined: 02 Sep 2009
Posts: 52390

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
13 Sep 2012, 07:15
siddharthasingh wrote: Isn't there any arithmetic solution to this question. I mean, just Hit n Trial method. Indeed there must be an arithmetic way out. Using this hit and trial method sometimes takes much longer time, henceforth I needed to go with a systematic approach. 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 Given: \(n=4q+1\), so \(n\) could be: 1, 5, 9, 13, ... \(n=5p+3\), so \(n\) could be: 3, 8, 13, ... General formula for \(n\) based on above two statements will be: \(n=20m+13\) (the divisor should be the least common multiple of above two divisors 4 and 5, so 20 and the remainder should be the first common integer in above two patterns, hence 13). For more about this concept see: manhattanremainderproblem93752.html#p721341, whenpositiveintegernisdividedby5theremainderis90442.html#p722552, whenthepositiveintegeraisdividedby5and125591.html#p1028654From, \(n=20m+13\) we have that the two smallest possible values of \(n\) are 13 (for \(m=0\)) and 33 (for \(m=1\)). 13+33=46. Answer: B. Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Director
Joined: 30 Nov 2006
Posts: 573
Location: Kuwait

n = 4q + 1
n = 5q + 3
I'll start with the first equation: n = 5+k4 where k = 0,1,2,3, ... etc
also, n = 8+m5 where m = 0,1,2,3,.. etc
for first equation: 5,9,13,17,21,25,29,33,37,41,45
for second equation: 8,13,18,23,28,33,38,43,48,53
The sum of minimum n's = 13 + 33 = 46
B



VP
Joined: 28 Mar 2006
Posts: 1277

my eq is 4x+1 = 5y+3
so 4x = 5y + 2
if y=2 x=3
ify=6 x=8
is the smallest group 8*4 + 1 =33



Retired Moderator
Joined: 05 Jul 2006
Posts: 1722

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?
33
46
49
53
86
4x+1 = 5y+3...........ie: 4x5y = 2
x,y must be >1 and y is even ie ( 2,4,6,..etc)
if y = 2 thus x = 3 and thus n = 13
if y = 4 thus x is a fraction ( not possible)
if y = 6 thus x = 8 and n= 33
13+33 = 46..... B



SVP
Joined: 29 Mar 2007
Posts: 2421

Re: Remainder
[#permalink]
Show Tags
03 Nov 2007, 10:11
alimad 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?
33 46 49 53 86
I got this so far
n = 4q + 1 n = 5q + 3
4q+1 + 5q+3 = 9q+4
plugging in value for q
q=1 q=2 q=3 q=4 q=5 = 45+4 = 49 ? not sure please help
Man ughhhh haha, I couldnt figure this question out forever. Was wondering why everyone was getting 46. I was like comon its 33.
question is really asking what is the SUM of the two possible values of n.
so ya 13+33=46.



VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1109
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
13 Sep 2012, 06:58
Isn't there any arithmetic solution to this question. I mean, just Hit n Trial method. Indeed there must be an arithmetic way out. Using this hit and trial method sometimes takes much longer time, henceforth I needed to go with a systematic approach.
_________________
Prepositional Phrases ClarifiedElimination of BEING Absolute Phrases Clarified Rules For Posting www.UnivScholarships.com



Manager
Status: Work hard in silence, let success make the noise
Joined: 25 Nov 2013
Posts: 132
Location: India
Concentration: Finance, General Management
GMAT 1: 540 Q50 V15 GMAT 2: 640 Q50 V27
GPA: 3.11
WE: Consulting (Computer Software)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
11 Dec 2013, 04:52
4x + 1 = n (1) 5y + 3 = n (2) Equating (1) and (2) 4x + 1 = 5y + 3 4x = 5y + 2 Put y=1,2,3,4,etc. Since (5y + 2) need to be a multiple of 4 to satisfy the equation on the left side. The 2 minimum values of y are 2 and 6. So, n = 5y + 3 n = 5(2) + 3 = 13 and n = 5(6) + 3 = 33 Adding the 2 minimum values of n 13 + 33 = 46 So, the correct answer is B.
_________________
Sahil Chaudhary If you find this post helpful, please take a moment to click on the "+1 KUDOS" icon. My IELTS 7.5 Experience From 540 to 640...Done with GMAT!!! http://www.sahilchaudhary007.blogspot.com



Manager
Joined: 18 Oct 2013
Posts: 71
Location: India
Concentration: Technology, Finance
Schools: Duke '16, Johnson '16, Kelley '16, Tepper '16, Marshall '16, McDonough '16, Insead '14, HKUST '16, HSG '15, Schulich '15, Erasmus '16, IE April'15, Neeley '15
GMAT 1: 580 Q48 V21 GMAT 2: 530 Q49 V13 GMAT 3: 590 Q49 V21
WE: Information Technology (Computer Software)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
11 Dec 2013, 11:28
From question we get N= => 4K+1=5P+3 K=P+(P+2)/4
So for P=2 & 6 we get K an integer i.e. K=13 & 33
Sum=13+33=46. B is correct.



Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 374

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
16 Oct 2016, 10:07
B is correct. Here's why: Given the information in the question we can create the following two equations: n = 4x+1 (5,9, 13,17,21,25,29, 33) n = 5y+3 (8, 13,18,23,28, 33) The parentheses following the equations represent potential value of n given various values of x (i.e. 1+). We can see from these calculations that 13 and 33 line up with both equations. Therefore, we can add 13 and 33 and we are done



Manager
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 106

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
03 Nov 2016, 02:03
n=4q+1 n= 5k+3
The smallest numbers that satisfies the equations are 13 and 33...so Answer B
_________________
Visit my youtube channel : ACE THE GMAT



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13371
Location: United States (CA)

Re: A group of n students can be divided into equal groups of 4
[#permalink]
Show Tags
26 Jan 2018, 13:09
Hi All, You would likely find it easiest to 'brute force' this question (simply write down enough of the possibilities until you either spot the pattern involved or have the exact answer on your pad). Equal groups of 4 with 1 left over COULD be... 5, 9, 13, 17, 21, 25, 29, 33..... Equal groups of 5 with 3 left over COULD be... 8, 13, 18, 23, 28, 33.... The two SMALLEST values that fit BOTH groups are 13 and 33. We're asked for the sum of those values... Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****




Re: A group of n students can be divided into equal groups of 4 &nbs
[#permalink]
26 Jan 2018, 13:09






