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

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

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07 Jan 2006, 15:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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07 Jan 2006, 15:49
A

2^x * 3^y=288
=>2^x * 3^y = 2^5 * 3*2
so x=5 y=2
Put in the values

2^(5-1) * 3^(2-2)= 2^4*3^0= 16*1=16
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08 Jan 2006, 05:53
andy_gr8 wrote:
A

2^x * 3^y=288
=>2^x * 3^y = 2^5 * 3*2
so x=5 y=2
Put in the values

2^(5-1) * 3^(2-2)= 2^4*3^0= 16*1=16

May I suggest an easier way (May not be easier for you ) ?

2^ (x-1) = (2^x)/2^1 and 3 ^ (y-2) = (3^y)/(3^ 2)

So now its multiplication becomes
[(2^x) * (3^y)]/ 2 * 3^ 2 = 288/18 = 16
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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08 Jan 2006, 06:00
ps_dahiya wrote:
andy_gr8 wrote:
A

2^x * 3^y=288
=>2^x * 3^y = 2^5 * 3*2
so x=5 y=2
Put in the values

2^(5-1) * 3^(2-2)= 2^4*3^0= 16*1=16

May I suggest an easier way (May not be easier for you ) ?

2^ (x-1) = (2^x)/2^1 and 3 ^ (y-2) = (3^y)/(3^ 2)

So now its multiplication becomes
[(2^x) * (3^y)]/ 2 * 3^ 2 = 288/18 = 16

Thanks ps_dahiya. That made it look very easy.
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09 Jan 2006, 02:06
288 = 8 * 36 = 8 * 4 * 9 = 2^3 * 2^2 * 3^2 = 2^5 * 3^2

x = 5, y = 2

(2^x-1)(3^y-2)= (32-1)*(9-2) = 31*7 = 217 (which is not in the answer choice.

(2^(x-1))(3^(y-2)) = 2^4 * 3*0 = 16

Ans A
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10 Jan 2006, 10:15
How did you come up with =>2^x * 3^y = 2^5 * 3*2???

Thanks!
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