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

Question

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

skipjames · Accepted Answer

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.

If my comments were helpful to you, please give kudos.
Thanks,
Skip

vinay.kaipra · Answer

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.

-----------------------------------------------------
Please give kudos, if my comments were helpful.

marcusaurelius · Answer

Thanks, I didn't even think to use the prime factorization.

barron600 · Answer

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

kostyan5 · Answer

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.

manalq8 · Answer

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

diegocml · Answer

Prime factorize 288 then do your magic:

\left( { 2 }^{ x } ight) \left( 3^{ y } ight) =288\ \left( { 2 }^{ x } ight) \left( 3^{ y } ight) ={ 2 }^{ 5 }{ 3 }^{ 2 }\ x=5\ y=2\ \ \left( { 2 }^{ x-1 } ight) \left( { 3 }^{ y-2 } ight) =\left( { 2 }^{ 5-1 } ight) \left( { 3 }^{ 2-2 } ight) ={ 2 }^{ 4 }{ 3 }^{ 0 }=16

KSBGC · Answer

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

E is our answer.

CLIMBTHELADDER · Answer

Break down 288 into its prime factors. You find that X = 5 and Y = 2. The rest is...

vishalkazone · Answer

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
Answer A.

CrackverbalGMAT · Answer

If we break up the term as the following then  2^{x-1} * 3^{y-2} = \frac{2^x}{2} * \frac{3^y}{3^2}

Since  2^{x} * 3^{y} = 288 , substituting, we get  \frac{288}{2 * 9} = \frac{32}{2} = 16

Option A

Arun Kumar

BrentGMATPrepNow · Answer

APPROACH #1 : Use prime factorization 
 288 = (2)(2)(2)(2)(2)(3)(3) = (2^5)(3^2) 
So, the given equation becomes:  (2^x)(3^y) = (2^5)(3^2) , which means  x=5  and  y=2 
Thus  (2^{x-1})(3^{y-2}) = (2^{5-1})(3^{2-2}) = (2^{4})(3^{0}) = (16)(1) = 16 
 Answer: A

APPROACH #2 : Use algebraic manipulation 
Given:  (2^x)(3^y) = 288 
Divide both sides of the equation by  2^1  (aka  2 ) to get:  \frac{(2^x)(3^y)}{2^1} = \frac{288}{2^1} 
Simplify to get:  (2^{x-1})(3^y) = 144 
Divide both sides of the equation by  3^2  to get:  \frac{(2^{x-1})(3^y)}{3^2} = \frac{144}{3^2} 
Simplify:  (2^{x-1})(3^{y-2}) = 16 
 Answer: A

totaltestprepNick · Answer

A https://www.youtube.com/watch?v=Rxmn2eayJn4 Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience tutoring@totaltestprep.net

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)

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