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 →

The integers k and n are such that 4 < k < n and k is

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 298
Joined: Sun Jun 03, 2012 6:42 amFollowed by: 7 members
Location: New delhi
Thanked: 10 times
GMAT Score:590
by varun289 » Sat Dec 15, 2012 3:45 am
258. The integers k and n are such that 4 < k < n and k is not a factor of n. If R is the remainder when n is divided by k, is R > 2?
1. Thre greatest common factor of k and n is 4.
2. The least common multiple of k and n is 84.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 131
Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members
Thanked: 39 times

by puneetkhurana2000 » Sun Dec 16, 2012 4:13 pm
Statement 1) The greatest common factor of k and n is 4.

Assuming some values we get
k = 8 and n = 12, remainder = 4
k = 8 and n = 16, remainder = 0

Not Sufficient!!!

Statement 2) The least common multiple of k and n is 84.

Assuming some values we get
k = 7 and n = 12, remainder = 5
k = 12 and n = 28, remainder = 4
k = 28 and n = 42, remainder = 14

Remainder is always > 2.

Sufficient!!!

Answer B.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 131
Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members
Thanked: 39 times

by puneetkhurana2000 » Sun Dec 16, 2012 4:21 pm
Can someone explain this in detail?

As both statements together we get, 4*a * 4*b = 84 * 4, solving we get a*b = 21 and only values that satisfy this keeping 4<k<n in mind are k = 4*3 = 12 and n = 4*7 = 28.

If we divide n by k we get a remainder as 4, which is greater than 2. Hence we are certain.

But can we answer the question with such certainty by taking into account only Statement 2(shown in my previous reply to this thread).

Thanks

Puneet
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Wed Oct 24, 2012 5:02 pm

by tranhang233 » Sun Dec 16, 2012 10:51 pm
I think the correct answer is D.
in your explanation:
+ Statement 1: The greatest common factor of k and n is 4.
Assuming some values we get
k = 8 and n = 12, remainder = 4
k = 8 and n = 16, remainder = 0 (8 is the factor of 16, but the stem is that k is not a factor of n)
--> statement 1 is sufficient
--> correct answer is D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 131
Joined: Wed Nov 14, 2012 2:01 pmFollowed by: 2 members
Thanked: 39 times

by puneetkhurana2000 » Sun Dec 16, 2012 11:32 pm
Thanks Tranhang, I missed that and Answer should be D.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pmFollowed by: 6 members
Location: India
Thanked: 47 times

by The Iceman » Mon Dec 17, 2012 9:22 am
Statement I:

If gcd(n,k)=4, let n=4a, k=4b

We have 4a = 4bx + R => R must be a multiple of 4. Hence, Statement 1 is sufficient.

Statement II:

LCM(n,k)= 84

If(n,k)= (12,7) => R=5

If(n,k)= (14,12) => R=2

So, statement II is not sufficient.

Hence, choice A is correct
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1