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 →

good one

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 371
Joined: Tue Apr 29, 2008 10:16 amFollowed by: 1 member
Thanked: 6 times
by vaivish » Sun Sep 21, 2008 10:36 am
If K is an integer greater than 1, is k equal to 2^r for some positive integer r.
(1)k is divisible by 2^6
(2)k is not divisible by any odd integer greater than 1.


I dont have the ans for this one...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
Junior | Next Rank: 30 Posts
Posts: 24
Joined: Thu Jun 26, 2008 7:21 am

by arvindm07 » Sun Sep 21, 2008 11:02 am
B.
from (1): k can be 2^6 * 5 and be divisible by 2^6 and cannot be expressed in terms of 2^r. hence, not sufficient.

from (2): k does not have any odd number as its factor, which means k is a product of only 2's and hence, can be expressed as 2^r.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 371
Joined: Tue Apr 29, 2008 10:16 amFollowed by: 1 member
Thanked: 6 times

by vaivish » Sun Sep 21, 2008 11:17 am
thanks arvind...

but why do u assume k is product of 2 only...6 is not an odd number but if its is factor of K, we can not express it as 2^r..any thought on this?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 683
Joined: Tue Jul 22, 2008 1:58 pmFollowed by: 2 members
Location: Dubai
Thanked: 73 times

by mals24 » Sun Sep 21, 2008 11:29 am
@ vaivish

You cannot take 6 or any even number that has an odd factor because the second st says that k is not divisible by any odd integer greater than 1. 6 is a multiple of 3 and hence any number divisible by 6 is also divisible by 3.

So we cannot choose numbers like 6, 18, 24, 36 etc which have odd factors greater than 1

Hope you get the logic
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 24
Joined: Thu Jun 26, 2008 7:21 am

by arvindm07 » Sun Sep 21, 2008 11:42 am
we need to look at all the prime factors of a number to see if the number is divisible by an odd number. If the number is not divisible by any odd number then, it shouldn't have any odd prime factors and must contain only even prime factor. we know that the only even prime factor is 2. Hence, it can only be multiples of 2.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 371
Joined: Tue Apr 29, 2008 10:16 amFollowed by: 1 member
Thanked: 6 times

by vaivish » Mon Sep 22, 2008 7:39 am
thanks guys
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Mon Sep 22, 2008 8:25 am

Re: good one

by Ankush Soni » Mon Sep 22, 2008 1:07 pm
[quote="vaivish"]If K is an integer greater than 1, is k equal to 2^r for some positive integer r.
(1)k is divisible by 2^6
(2)k is not divisible by any odd integer greater than 1.


I dont have the ans for this one...[/quote]

Hi Vaivish..This is my soln.

1. K = (2^6) P, where P is an integer
if P= 2, then K can be written as a power of 2
if P = 6, then K can't be written as a power of 2
Therefore, Insufficient

2. K is not divisble by 3, 5, 7...or any other prime no. (as all primes > 2
are odd) or multiples of primes no's. If K is divisible by an even no. 6
then k is also divisible by 3. Therefore. K is divisible by only powers of
2.

My answer (B) i.e. statement 2 is sufficient in itself
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1