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 →

ds

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
Legendary Member
Posts: 876
Joined: Thu Apr 10, 2008 8:14 am
Thanked: 13 times

by ketkoag » Thu Jul 09, 2009 1:13 pm
i think b alone is not sufficient coz if m=1,
then the expression will be 34n + 2 +1 i.e.
34n + 3
now take n = 1 then 34n+3 = 37. hence remainder 7
take n = 2 , 34n+3 = 71. hence remainder 1.
hence insufficient..
IMO C.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 345
Joined: Wed Mar 18, 2009 6:53 pm
Location: Sao Paulo-Brazil
Thanked: 12 times
GMAT Score:660

by shibal » Thu Jul 09, 2009 1:53 pm
ketkoag i think so too, but oa in gmat prep was b
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Tue Jul 07, 2009 1:57 pm
GMAT Score:650

by matanga » Thu Jul 09, 2009 2:11 pm
shibal wrote:ketkoag i think so too, but oa in gmat prep was b
I know what is wrong with this. The question says [3^(4n+2)] + m, and here is why the answer is B:

A. INSUFF -> You don't know the value of m

B. SUFF. Check out the values of x = 3^(4n+2) below:

n= 1 x= 3^6 = 9^3 = 729
2 x= 3^10 = 9^5 = 729 * 81
3 x= 3^14 = 9^7 = 9 * (729)^2

See the pattern? Regardless of the value of n, the value of the last digit of x will always be 9 and, when you add 1, you are dividing by a number that is divisible by 10.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 345
Joined: Wed Mar 18, 2009 6:53 pm
Location: Sao Paulo-Brazil
Thanked: 12 times
GMAT Score:660

by shibal » Thu Jul 09, 2009 2:14 pm
hummmm u'r right... good job... a 25sec effort would be valid....
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 379
Joined: Wed Jun 03, 2009 3:05 amFollowed by: 1 member
Thanked: 19 times
GMAT Score:690

by sreak1089 » Fri Jul 10, 2009 5:32 am
If you look closely, 3 ^ 1 always ends in 3
3 ^ 2 always ends in 9
3 ^ 3 always ends in 7
3 ^ 4 always ends in 1
3 ^ 5 always ends in 3
3 ^ 6 always ends in 9
3 ^ 7 always ends in 7
3 ^ 8 always ends in 1
and so on this pattern repeats

3 ^ 2, 3 ^ 6, 3 ^ 10, and so on can be represented as 3 ^ (4n + 2)
which will always end in 9, since m = 1, which means 3 ^ (4n + 2) + m
ends in 0, hence the remainder will always be zero.

Also, 3 ^1, 3 ^ 5, 3 ^ 9, so on is represented as 3 ^ (4n + 1) which always ends in 3.

3 ^ 3, 3 ^ 7, 3 ^ 11, so on is represented as 3 ^ (4n + 3) which always end in 7.

3 ^ 4, 3 ^ 8, 3 ^ 12, so on is represented as 3 ^ (4n) always ends in 1.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 217
Joined: Sun Jan 10, 2010 1:39 pmFollowed by: 1 member
Thanked: 7 times

by zaarathelab » Mon Sep 05, 2011 12:41 am
It's imp to remember the cyclicity of different numbers

3 has a cyclicity of 4. This means that when 3 is raised to an exponential power, the last digit of the final value will repeat itself after every 4 powers.

Ex -

3^1 =3 (last digit 3)
3^2 = 9 (last digit 9)
3^3 = 27 (last digit 7)
3^4 = 81 (last digit 1)
3^5 = 243 (last digit 3 - - - repeat pattern)

Now here, Statement 1 is clearly insufficient as m can take on different value, correspondingly the remainder will change

Statement 2 -

Sufficient - cyclicity of 3 which will repeat itself after 4 powers. plug in different values of n and see
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1