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

Express 288 in terms of it's prime factors 2 and 3. That gives you x and y. Then substitute and solve.

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

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.

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

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)

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 awesome