# 2007_Question1a

Looking at the prime factors, we see that the numerator is 2^{r+s}\cdot3^{r+s}\cdot2^{2(r-s)}\cdot3^{r-s} = 2^{3r-s}\cdot3^{2r}. Similarly, the denominator is 2^{3r}\cdot3^{2(r+2s)} = 2^{3r}\cdot3^{2r+4s}. Dividing one by the other then gives 2^{-s}\cdot3^{-4s}, which is an integer if s\leqslant0, and so the answer is (b).

Correct!!! Well done