+983 \(12^n=(3\cdot4)^n=3^n\cdot4^n\)
\(20!=1\cdot\boxed{2}\cdot\boxed{3}\cdot\boxed{4}\cdot5\cdot\boxed{6}\cdot7\cdot\boxed{8}\cdot\boxed{9}\cdot\boxed{10}\cdot11\cdot\boxed{12}\cdot13\cdot\boxed{14}\cdot\boxed{15}\cdot\boxed{16}\cdot17\cdot\boxed{18}\cdot19\cdot\boxed{20}\)
Boxed numbers are multiples of 2 and 3:
\(2\cdot3\cdot4\cdot6\cdot8\cdot9\cdot10\cdot12\cdot14\cdot15\cdot16\cdot18\cdot20\Rightarrow x\cdot3^8\cdot4^{9}\)
Therefore, the largest integer n is 8.
I recall there being a better way to do this using modular arithmetic, but I don't know how to.
