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

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by metalhead » Fri Jan 29, 2010 4:05 pm
Express 288 in terms of it's prime factors 2 and 3. That gives you x and y. Then substitute and solve.
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by ace_gre » Fri Jan 29, 2010 5:14 pm
Proceeding the same way as metalhead,
288 = 2^5 * 3^2
x = 5
y = 2

2^x-1 = 2^4 = 16
3^y-2 = 3^0 = 1

IMO A
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by Stuart@KaplanGMAT » Fri Jan 29, 2010 6:04 pm
imane81 wrote:Please, could you help solving this one

If (2^x)(3^y)=288, where x and y are positive integers then (2^(x-1))(3^(y-2))=
A. 16
B. 24
C. 48
D. 96
E. 144
We can actually solve without breaking 288 down into primes.

We know that "x-1" is 1 lower than "x" and "y-2" is 2 lower than "y". So, the final expression will have 1 fewer factor of 2 and 2 fewer factors of 3 than 288.

So:

288/2*3*3 = 144/9 = 16... choose A.
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by shashank.ism » Sat Feb 20, 2010 7:54 am
imane81 wrote:Please, could you help solving this one

If (2^x)(3^y)=288, where x and y are positive integers then (2^(x-1))(3^(y-2))=
A. 16
B. 24
C. 48
D. 96
E. 144
288 = 2x2x2x2x2x3x3 = 2^5 x 3^2 --> x=5, y=2
so (2^(x-1))(3^(y-2)) = (2^(5-1))(3^(2-2)) = 2^4 x1 =16 Ans A
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by shashank.ism » Sat Feb 20, 2010 7:58 am
ok I just got a simpler solution in my mind.....it will take just 4 seconds.....
(2^x)(3^y)=288 --> (2^(x-1))(3^(y-2)) x2x3 =288 --> (2^(x-1))(3^(y-2)) = 288/2/3^2 =[spoiler]16 Ans A
[/spoiler]
why to make fuzz in factorizing 288...just divide it by 6 and get the solution ...(4seconds)
Last edited by shashank.ism on Sat Feb 20, 2010 8:33 am, edited 1 time in total.
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by thephoenix » Sat Feb 20, 2010 8:04 am
imane81 wrote:Please, could you help solving this one

If (2^x)(3^y)=288, where x and y are positive integers then (2^(x-1))(3^(y-2))=
A. 16
B. 24
C. 48
D. 96
E. 144
288=2^5 *3^2---->x=5 and y=2

putting values exp is 16

however stuart is just genius on the GMAT if i get his power than i will do more than half of problem solving without wasting ink

stuart pass us your grey matter
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