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

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

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07 Feb 2006, 20:14
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(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.)
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Re: PS - Exponent [#permalink]

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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
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Re: PS - Exponent [#permalink]

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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
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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...
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07 Feb 2006, 23:58

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

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!
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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
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08 Feb 2006, 06:44
Thanks Thearch...I was not aware of that - made for a frustrating problem.
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08 Feb 2006, 09:13
Without assuming first that 2^5 x 3^2 = 288

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08 Feb 2006, 10:09
Yes, the simplest way to do this is to divide 288 by 2, and then by 9 (=3^2).
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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
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08 Feb 2006, 20:01
same approach 288 = 2^5 * 3^2 thus answer for the required = 16
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08 Feb 2006, 20:01
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# 2^x)(3^y)=288 where x and y are positive integars, then

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