3x^3y^-2/xy^3
We can bring the y^-2 down to the denominator and divide by the x to get 3x^2/y^5
:D
Using algebra, we can see after subtracting a few numbers, we notice there are 10 numbers, therefore there are 5 pairs and one number.
$1-2, 4-8, 16-32, 64-128, 256-512, 1024$
using normal subtraction
We get
683
LOL!