Author
Message
TAGS:
Intern

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 ]
07 Feb 2006, 20:14

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 1 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

Joined: 06 Jun 2004

Posts: 1062

Location: CA

Followers: 2

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

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

Joined: 29 Dec 2005

Posts: 1351

Followers: 6

Kudos [? ]:
28
[0 ] , given: 0

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

Joined: 23 Jan 2006

Posts: 24

Followers: 0

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

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

Joined: 05 Jan 2006

Posts: 383

Followers: 1

Kudos [? ]:
16
[0 ] , given: 0

16 is answer

Without parenthesis we can not solve, so up to the time 16 is one answer choice I am happy...

Senior Manager

Joined: 19 Feb 2005

Posts: 492

Location: Milan Italy

Followers: 1

Kudos [? ]:
9
[0 ] , given: 0

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

Joined: 11 Nov 2005

Posts: 336

Location: London

Followers: 1

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

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

Joined: 23 Jan 2006

Posts: 24

Followers: 0

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

Thanks Thearch...I was not aware of that - made for a frustrating problem.

Manager

Joined: 23 Jan 2006

Posts: 193

Followers: 1

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

Without assuming first that 2^5 x 3^2 = 288

SVP

Joined: 03 Jan 2005

Posts: 2255

Followers: 12

Kudos [? ]:
197
[0 ] , given: 0

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

Joined: 07 Jul 2004

Posts: 5097

Location: Singapore

Followers: 17

Kudos [? ]:
136
[0 ] , given: 0

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

Joined: 11 Jan 2006

Posts: 273

Location: Chennai,India

Followers: 1

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

same approach 288 = 2^5 * 3^2 thus answer for the required = 16

_________________

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

Similar topics
Author
Replies
Last post
Similar Topics:

In the fraction x/y, where x and y are positive integers,
ywilfred
4
27 Sep 2007, 08:16

In the fraction x/y, where x and y are positive integers,
jaynayak
1
29 Jul 2006, 10:43

In the fraction x/y, where x and y are positive integers,
bewakoof
8
01 Apr 2006, 15:40

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 a and y are positive integers,
rahulraao
4
14 Oct 2005, 01:59