Let \(r\) and \(s\) be integers. Then \[\frac{6^{r+s}\times12^{r-s}}{8^r\times9^{r+2s}}\] is an integer if
\(r+s\le 0\),
\(s \le 0\),
\(r\le 0\),
\(r\ge s\).
We have \(6 = 2 \times 3, 12 = 2^2 \times 3, 8 = 2^3\), and \(9 = 3^2\).
If we separate the powers of \(2\) and \(3\), we have \[\frac{6^{r+s}\times12^{r-s}}{8^r\times9^{r+2s}}= \frac{(2\times3)^{r+s}\times(2^2\times3)^{r-s}}{(2^3)^r\times(3^2)^{r+2s}}\] \[= 2^{r+s+2r-2s-3r} \times3^{r+s+r-s-2r-4s}\] \[=2^{-s}\times 3^{-4s}.\]
This is an integer precisely when \(s\le 0\).
So the answer is (b).
