jmc

Prime factorizing gives $n=2^{23}\cdot 3^{12}$. Since any positive factor of $n$ must be of the form $2^a\cdot 3^b$ where $0\le a\le 23$ and $0\le b\le 12$, there are $(23+1)(12+1)=24\cdot 13=\boxed{312}$.