BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Divisibility & Prime Number

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 73
Joined: Tue Sep 16, 2008 4:33 pm
Thanked: 2 times
by relaxin99 » Wed Mar 11, 2009 8:16 pm
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?
86

53

49

46

33
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members
Thanked: 712 times
GMAT Score:770

by DanaJ » Wed Mar 11, 2009 9:31 pm
I'm thinking: go for the multiples of 4, add 1 and then check to see if that number fits the second condition:
4 + 1 = 5 - NO
8 + 1 = 9 - NO
12 + 1 = 13 - YES - first one
16 + 1 = 17 - NO
20 + 1 = 21 - NO
24 + 1 = 25 - NO
28 + 1 = 29 - NO
32 + 1 = 33 - YES - second one

This makes the sum, IMHO, 46.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Wed Mar 11, 2009 9:56 pm
Location: Whitler, BC, Canada

by franciskyle » Thu Mar 12, 2009 1:33 am
For this one, I would start with deciding which answers gave a remainder of 3 when divided by 5 (the units digit would have to be either 3 or 8 because to be divisible by the units would have to be 0 or 5... add 3 to get 3 and 8 respectively). That leaves us with B & E.

A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Thu Mar 12, 2009 2:13 pm
Good solutions!



n = 4k+1 and n=5q+3

Find the smallest number that satisfies this i.e. 13

Therefore the family of numbers will be

a*LCM(4,5) + 13

20a+13

When a=0 the number is 13 which u obtained above
When a = 1 the number will be 33

So 33+13 = 46

Hope this helps!

Regards,
CR
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

by Ian Stewart » Thu Mar 12, 2009 2:58 pm
franciskyle wrote:For this one, I would start with deciding which answers gave a remainder of 3 when divided by 5 (the units digit would have to be either 3 or 8 because to be divisible by the units would have to be 0 or 5... add 3 to get 3 and 8 respectively). That leaves us with B & E.

A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.
Ah, note that the correct answer shouldn't give the same remainders that are mentioned in the question. The question asks for the sum of two different numbers with certain remainders. We know each of those two numbers will have a remainder of 3 when divided by 5; when we add those two numbers we should now have a remainder of 1 when we divide by 5 (because 3+3 has a remainder of 1). We also want a number with a remainder of 2 when divided by 4, since the two numbers we're adding give a remainder of 1 when divided by 4.

Still, two answer choices remain, unfortunately, so some other method would be needed to choose between 46 and 86.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 32
Joined: Wed Mar 11, 2009 9:56 pm
Location: Whitler, BC, Canada

by franciskyle » Thu Mar 12, 2009 3:03 pm
Thanks... that is my Achilles heel - jumping the gun and making stupid mistakes.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1