Find all School-related info fast with the new School-Specific MBA Forum

It is currently 27 Nov 2014, 15:11

Happy Thanksgiving:

Free Access to GMAT Club Tests until 11 AM PST on Fri 11/28


Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If k is an integer greater than 1, is k wqual to 2^r for

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 27 Aug 2005
Posts: 332
Location: Montreal, Canada
Followers: 2

Kudos [?]: 23 [0], given: 0

If k is an integer greater than 1, is k wqual to 2^r for [#permalink] New post 19 Sep 2005, 18:37
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

50% (00:00) correct 50% (01:09) wrong based on 6 sessions
If k is an integer greater than 1, is k wqual 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
VP
VP
avatar
Joined: 22 Aug 2005
Posts: 1126
Location: CA
Followers: 1

Kudos [?]: 33 [0], given: 0

 [#permalink] New post 19 Sep 2005, 19:06
Not sure what does "some positive integer means"??
if it mean for ANY integer r > 0, ans will be C.

if it mean for ALL integers r > 0, answer will be E.
Senior Manager
Senior Manager
avatar
Joined: 27 Aug 2005
Posts: 332
Location: Montreal, Canada
Followers: 2

Kudos [?]: 23 [0], given: 0

 [#permalink] New post 19 Sep 2005, 19:30
"Some positive integer" means "any positive integer".

OA is B. Unfortunately GMATPrep doesn't provide any OEs.
VP
VP
avatar
Joined: 22 Aug 2005
Posts: 1126
Location: CA
Followers: 1

Kudos [?]: 33 [0], given: 0

 [#permalink] New post 19 Sep 2005, 20:20
coffeeloverfreak wrote:
"Some positive integer" means "any positive integer".

OA is B. Unfortunately GMATPrep doesn't provide any OEs.


I could not understnad OA.

ii)

if k = 2, not divisible by x, x > 1 and x is odd

for some integer r = 1, k = 2 ^ r

ans is YES

now if k = 6

k <> 2^r

ans is NO

ii) therefore, is INSUFFICIENT.
Senior Manager
Senior Manager
avatar
Joined: 27 Aug 2005
Posts: 332
Location: Montreal, Canada
Followers: 2

Kudos [?]: 23 [0], given: 0

 [#permalink] New post 19 Sep 2005, 20:27
But 6 is divisible by an odd integer other than 1. It's divisible by 3.

The only positive integers greater than 1 that don't have any odd factors greater than 1 are powers of 2. So statement 2 is sufficient on its own.

... And once again I just answered my own question. (I feel dumb now). I'd initially answered D, thinking that statement 1 was also sufficient in itself. But of course, 192 is a multiple of 64, and 192 is not a power of 2. So D is wrong. B it is.
VP
VP
avatar
Joined: 22 Aug 2005
Posts: 1126
Location: CA
Followers: 1

Kudos [?]: 33 [0], given: 0

 [#permalink] New post 19 Sep 2005, 20:31
coffeeloverfreak wrote:
But 6 is divisible by an odd integer other than 1. It's divisible by 3.

The only positive integers greater than 1 that don't have any odd factors greater than 1 are powers of 2. So statement 2 is sufficient on its own.

... And once again I just answered my own question. (I feel dumb now). I'd initially answered D, thinking that statement 1 was also sufficient in itself. But of course, 192 is a multiple of 64, and 192 is not a power of 2. So D is wrong. B it is.


You right. This is really tricky if one doesnt know these number system properties (especially if she is not good in number picking like me).
Manager
Manager
avatar
Joined: 06 Aug 2005
Posts: 198
Followers: 3

Kudos [?]: 5 [0], given: 0

 [#permalink] New post 20 Sep 2005, 01:37
Under the unique factorisation theorem, any positive integer > 1, like k can be expressed as the product of prime powers.

So

k = 2^a * 3^b * 5^c * 7^d * .... *

where a,b,c,d >=0 integers.

We are asked if k = 2^r
this will be when r=a, b=c=d=0

(1) says that a>=6
but it does not give any information about b,c,d, ...
So not sufficient.
(examples a=6, b=c=d=0, k=64, and
a=6, b=1, c=d=0, k=192)

(2) says that b=c=d=0
but k>1, so k>=2, so a>=1
So sufficient

B

This abstract approach is not the quickest way,
but if you don't know this conceptual material,
you could struggle with a real question.
Intern
Intern
avatar
Joined: 16 Aug 2005
Posts: 25
Location: India
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 20 Sep 2005, 05:59
I think the easiest way to see that A is not sufficient is by considering that 2^6*3 can not be represented as 2^n.

And B is suffiecient for the reason explained by coffee.
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3399
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 173 [0], given: 2

 [#permalink] New post 05 Oct 2005, 16:15
B it is...

baiscally you want to know if K is multiple of 2 ONLY....

which means numbers like 6, 10, 12 are out cause they are multiples of other primes such as 3, 5 etc...

as long as K is a multiple of 2...we can express it as 2^r...where r is some number (positive)
Intern
Intern
avatar
Joined: 02 Oct 2005
Posts: 39
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 05 Oct 2005, 18:24
i think it is E.

II) states that it is not divisible by any odd number > 1, which implies k is even. however, what we need to say if K is a power of 2. For example, 10 is even , but is not a power of 2. I am assuming here that the notation, 2^ r , denotes, 2 raised to r and not 2 multiplied by r.
Intern
Intern
avatar
Joined: 02 Oct 2005
Posts: 39
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 05 Oct 2005, 18:26
sorry , i understood my mistake. B it is.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5092
Location: Singapore
Followers: 19

Kudos [?]: 162 [0], given: 0

 [#permalink] New post 05 Oct 2005, 19:04
We're given k > 1, r > 0. We're asked if k = 2^r

From statement 1, we're told k is a multiple of 2^6. This could be 3(2^6), 4(2^6). Since there are so many values of k, we can't tell.

From statement 2, we're told k is not divisbile by any odd integer greater than 1. This means k is even and a power of 2. Any other even number that has a prime factor other than 2 will be divisible by any odd integer > 1. For e.g., if k=18, then 18 = 3*3*2. It's divisible by 3. However, if k = 16, then 2^4 and this is equal to 2^r if r is 4.

The answer B is correct if we're asked if k can be represented in the form 2^r where r is a positive value.

However, we're asked here if k EQUALS 2^r and we do not have information for r.
  [#permalink] 05 Oct 2005, 19:04
    Similar topics Author Replies Last post
Similar
Topics:
16 Experts publish their posts in the topic If k is an integer greater than 1, is k equal to 2^r for arjunrampal 18 29 Dec 2009, 09:00
2 If k is an integer greater than 1, is k equal to 2^r for tarek99 13 27 Oct 2008, 09:49
2 If k is an integer greater than 1, is k equal to 2^r for sondenso 5 22 Jul 2008, 20:23
If K is an integer greater than 1, is K equal to 2^r for Hermione 11 29 Nov 2006, 09:57
If k is an integer greater than 1, is k equal to 2^r for mrmikec 5 16 Jun 2006, 17:18
Display posts from previous: Sort by

If k is an integer greater than 1, is k wqual to 2^r for

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.