|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 23 Jan 2006
Posts: 25
Followers: 0
Kudos [?]:
0
[0], given: 0
|
2^x)(3^y)=288 where x and y are positive integars, then [#permalink]
 07 Feb 2006, 21:14
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 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: 1068
Location: CA
Followers: 2
Kudos [?]:
12
[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: 1356
Followers: 6
Kudos [?]:
17
[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: 25
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: 389
Followers: 1
Kudos [?]:
11
[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... 
|
|
|
|
|
|
Director
Joined: 19 Feb 2005
Posts: 508
Location: Milan Italy
Followers: 1
Kudos [?]:
2
[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: 339
Location: London
Followers: 1
Kudos [?]:
4
[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: 25
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: 199
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Without assuming first that 2^5 x 3^2 = 288
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[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: 5134
Location: Singapore
Followers: 9
Kudos [?]:
87
[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: 277
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 a and y are positive integers,
|
rahulraao |
4 |
14 Oct 2005, 02:59 |
|
|
|
|
in the fraction x/y, where x and y are positive integers,
|
joemama142000 |
3 |
18 Dec 2005, 13:47 |
|
|
|
|
If (2^x)(3^y)=288 where x and y are positive integers then
|
Nayn02 |
5 |
07 Jan 2006, 15:05 |
|
|
|
|
x and y are positive integers, where (x + y)/3 = 36. What is
|
consultinghokie |
2 |
06 May 2006, 11:43 |
|
|
|
|
In the fraction x/y, where x and y are positive integers,
|
mahesh004 |
6 |
05 Oct 2006, 01:23 |
|
|
|
|
|
|
close
