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Manager  Joined: 28 Oct 2009
Posts: 78
If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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3
14 00:00

Difficulty:   15% (low)

Question Stats: 78% (01:35) correct 22% (01:57) wrong based on 683 sessions

### HideShow timer Statistics If $$(2^x)(3^y) = 288$$, where x and y are positive integers, then $$(2^{x-1})(3^{y-2})$$ equals:

A. 16
B. 24
C. 48
D. 96
E. 144

Originally posted by marcusaurelius on 21 May 2010, 11:33.
Last edited by Bunuel on 18 Jun 2018, 04:31, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
Manager  Joined: 08 May 2010
Posts: 126
Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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6
1
If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)(3^y-2) equals:

16
24
48
96
144

So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=5, y=2. now quickly 2^4(3^0)=16. Than answer should be number 1.

Thanks,
Skip
##### General Discussion
Manager  Joined: 28 Oct 2009
Posts: 78
Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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Thanks, I didn't even think to use the prime factorization.
Manager  Joined: 07 Oct 2006
Posts: 58
Location: India
Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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2
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.

-----------------------------------------------------
Intern  Joined: 12 Nov 2011
Posts: 1
Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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

16
24
48
96
144

So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=5, y=2. now quickly 2^4(3^0)=16. Than answer should be number 1.

Thanks,
Skip

Hi Skip, could you please explain prime factorization and how you count the three's. I think I see how you count the two's by taking 2, 8, and 8 seperately and see how many two's making up each number but I don't quite understand how you get the three's.

Thanks, Steven
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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Here's how I do prime factorization.

In one row you start with your number to be factored and in the other column you write the smallest possible prime that number is divisible by. Then you do the same for the (number/prime) and so on until you get to 1.

For example:

number:______288______144______72______36______18______9______3______1 (done)
prime factor:_______2_________2______ 2_______2_______2______3______3

In this case you it's easy to see that 288 = 2*2*2*2*2*3*3 = 2^5 * 3^2.
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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I approached it by using prime factorization. you need to know how many two's and three's are there in 288. using prime factorization, i got 4 two's and 2 three's.
you just need to plug in the values for x and y and solve the given equation.
(2^5-1)(3^2-2)= (2^4)(3^0)= (16)(1)= 16

hope that helps
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GMAT 1: 410 Q18 V27 GMAT 2: 490 Q35 V23 Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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Prime factorize 288 then do your magic:

$$\left( { 2 }^{ x } \right) \left( 3^{ y } \right) =288\\ \left( { 2 }^{ x } \right) \left( 3^{ y } \right) ={ 2 }^{ 5 }{ 3 }^{ 2 }\\ x=5\\ y=2\\ \\ \left( { 2 }^{ x-1 } \right) \left( { 3 }^{ y-2 } \right) =\left( { 2 }^{ 5-1 } \right) \left( { 3 }^{ 2-2 } \right) ={ 2 }^{ 4 }{ 3 }^{ 0 }=16$$
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Re: If 2^x*3^y = 288, where x and y are positive integers then 2^(x-1)*3  [#permalink]

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Bunuel wrote:
If $$2^x*3^y = 288$$, where x and y are positive integers then $$2^{x-1}*3^{y-2}$$ is equal to?

A. 2
B. 4
C. 6
D. 8
E. 16

$$288 = 3 *96 = 3 * 3*32 = 3*3* 4*8 = 3*3 * 2*2 * 2*2*2 = 3^22^5$$

$$2^x*3^y = 288$$

$$2^x3^y = 2^53^2$$

x = 5 and y = 2.

$$2^{x-1}*3^{y-2}$$

=$$2^{5-1} *3^{2-2}$$

=$$2^4 * 3^0$$

=16

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

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Break down 288 into its prime factors. You find that X = 5 and Y = 2. The rest is...
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Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  [#permalink]

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288= 2*2*2*2*2*3*3 = 2^5 * 3^2

(2^x-1) * (3^y-2) = 2^4 * 3^2-2
= 16 *1= 16
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Kudos if it deserve....one... Re: If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)   [#permalink] 22 Dec 2018, 00:19
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# If (2^x)(3^y) = 288, where x and y are positive integers, then (2^x-1)  