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

It is currently 21 Dec 2014, 02:42

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

2^x)(3^y)=288 where x and y are positive integars, then

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 23 Jan 2006
Posts: 24
Followers: 0

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

2^x)(3^y)=288 where x and y are positive integars, then [#permalink] New post 07 Feb 2006, 20:14
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (02:07) correct 0% (00:00) wrong based on 2 sessions
(2^x)(3^y)=288 where x and y are positive integars, then (2^(x-1)))(3^(y-2))=

16
24
48
96
144

(Edited by Hong to show the correct question.)
VP
VP
avatar
Joined: 06 Jun 2004
Posts: 1061
Location: CA
Followers: 2

Kudos [?]: 36 [0], given: 0

Re: PS - Exponent [#permalink] New post 07 Feb 2006, 20:43
brutus wrote:
(2^x)(3^y)=288 where x and y are positive integars, then (2^x-1)(3^y-2)=

16
24
48
96
144


288 = 2^5*3^2
x = 5
y = 2

(2^x-1)(3^y-2) = 31*7 = 217

This is not any of the choice so I imagine there are supposed to be parenthesis around the exponents, in that case, A, 16
_________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

VP
VP
User avatar
Joined: 29 Dec 2005
Posts: 1349
Followers: 7

Kudos [?]: 29 [0], given: 0

Re: PS - Exponent [#permalink] New post 07 Feb 2006, 21:40
288 = 2^5*3^2

x = 5
y = 2

if (2^x-1)(3^y-2) = {(2^x) - 1} {(3^y) - 2}=31x7=217
if 2^(x-1) 3^(y-2) = 16x 1 = 16
Intern
Intern
avatar
Joined: 23 Jan 2006
Posts: 24
Followers: 0

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

 [#permalink] New post 07 Feb 2006, 21:46
Yes, you all are correct...there are supposed to be parenthesis around the exponent...

OA is 16

My trouble came from 3^(y-2) where y=2 does 3^0 equal 1 as opposed to 3 or 0?

Please explain if you have a moment...
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

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

 [#permalink] New post 07 Feb 2006, 23:58
16 is answer

Without parenthesis we can not solve, so up to the time 16 is one answer choice I am happy... :lol:
Senior Manager
Senior Manager
avatar
Joined: 19 Feb 2005
Posts: 488
Location: Milan Italy
Followers: 1

Kudos [?]: 10 [0], given: 0

 [#permalink] New post 08 Feb 2006, 02:33
brutus wrote:
Yes, you all are correct...there are supposed to be parenthesis around the exponent...

OA is 16

My trouble came from 3^(y-2) where y=2 does 3^0 equal 1 as opposed to 3 or 0?

Please explain if you have a moment...


hi brutus,
I guess there are some stickies about number properties and exponents.
Anyway, the basic convention of powers states that
x^0=1
x^1=x

to understand why x^0=1 see http://planetmath.org/?op=getobj&from=objects&id=3948

additional reference: http://en.wikipedia.org/wiki/Exponentiation

I'm almost sure the GMAC would never put 0^0 in a question...I'd rather think to be wrong on some of my calculations!
_________________

Please allow me to introduce myself: I'm a man of wealth and taste

Senior Manager
Senior Manager
avatar
Joined: 11 Nov 2005
Posts: 332
Location: London
Followers: 1

Kudos [?]: 8 [0], given: 0

 [#permalink] New post 08 Feb 2006, 04:29
16.

2^(x-1) = 2^x/2, 3^(y-2)= 3^y/(3*3)

so 2^x * 3^y /(2*3*3) =288/18 = 16
Intern
Intern
avatar
Joined: 23 Jan 2006
Posts: 24
Followers: 0

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

 [#permalink] New post 08 Feb 2006, 06:44
Thanks Thearch...I was not aware of that - made for a frustrating problem.
Manager
Manager
avatar
Joined: 23 Jan 2006
Posts: 192
Followers: 1

Kudos [?]: 4 [0], given: 0

 [#permalink] New post 08 Feb 2006, 09:13
Without assuming first that 2^5 x 3^2 = 288

Image
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2251
Followers: 13

Kudos [?]: 210 [0], given: 0

 [#permalink] New post 08 Feb 2006, 10:09
Yes, the simplest way to do this is to divide 288 by 2, and then by 9 (=3^2).
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5081
Location: Singapore
Followers: 19

Kudos [?]: 165 [0], given: 0

 [#permalink] New post 08 Feb 2006, 17:57
288 = 4*72 = 4*8*9 = 2^5*3^2

2^x* 3^y= 2^5*3^2

x = 5, y = 2

Then (2^(x-1)))(3^(y-2)) = 2^4 * 3^0 = 16
Senior Manager
Senior Manager
User avatar
Joined: 11 Jan 2006
Posts: 270
Location: Chennai,India
Followers: 1

Kudos [?]: 3 [0], given: 0

 [#permalink] New post 08 Feb 2006, 20:01
same approach 288 = 2^5 * 3^2 thus answer for the required = 16
_________________

vazlkaiye porkalam vazltuthan parkanum.... porkalam maralam porkalthan maruma

  [#permalink] 08 Feb 2006, 20:01
    Similar topics Author Replies Last post
Similar
Topics:
4 Experts publish their posts in the topic If (243)^x(463)^y = n, where x and y are positive integers, rohitgoel15 8 07 Apr 2010, 22:37
In the fraction x/y, where x and y are the positive what is Balvinder 1 12 Nov 2007, 20:11
If (2^x)(3^y)=288 where x and y are positive integers then Nayn02 5 07 Jan 2006, 14:05
in the fraction x/y, where x and y are positive integers, joemama142000 3 18 Dec 2005, 12:47
In the fraction x/y, where a and y are positive integers, rahulraao 4 14 Oct 2005, 01:59
Display posts from previous: Sort by

2^x)(3^y)=288 where x and y are positive integars, then

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

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®.