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If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)

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If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1) [#permalink] New post   16 Jan 2023, 01:38
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Question Stats:

83% (01:17) correct 16% (03:30) wrong based on 6 sessions

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If (2x)(3y)=288, where x and y are positive integers, then (2x−1)(3y−2) equals:

A. 16
B. 24
C. 48
D. 96
E. 144
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A
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GeminiHeat
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1) [#permalink] New post   18 Jan 2023, 06:47
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Algebraically we can approach the problem in another way as well.

If (2^x)(3^y) = 288
then (2^x-1)(3^y-2) = 288/(2[*]3[*]3)
The reason being the exponent of 2 is reduced by 1 and exponent of 3 is reduced by 2.
Hence, the answer would be 288/18 = 16.

Answer: A
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1) [#permalink] New post   23 Oct 2024, 11:28
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2 ^x* 3^y =288

288= 2*144
288=2*12^2
288=2*(2*6)^2
288=2*(2^2*3)^2
288=2*2^4*3^2
288=2^5 * 3^2

2 ^x* 3^y =2^5 * 3^2

x=5, y=2

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

Answer is : A

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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1) [#permalink]
23 Oct 2024, 11:28
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