If (2^x)(3^y) = 108, where x and y are positive integers, then (2^x-1)(3^y-2) equals:
Choose one:
A) 6  B) 24  C) 48  D) 96  E) 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=2, y=3. now quickly 2^1(3^1)=6. Than answer should be number 1.   6
<result>A</result>
If (2^x)(3^y) = 108, where x and y are positive integers, then (2^x-1)(3^y-2) equals: