AMC 12 Mock Exam 3 Problem 22

Power of $2$. If $n$ is odd, $3^n\equiv3\pmod4$, so $3^n-1$ is divisible by $2$ only once. If $n=2^km$ with $m$ odd and $k\ge1$, $$3^n-1=(3^m-1)(3^m+1)(3^{2m}+1)(3^{4m}+1)\cdots(3^{2^{k-1}m}+1).$$ Here $3^m-1$ contributes one factor of $2$; $3^m\equiv3\pmod8$ so $3^m+1$ contributes exactly two; and each $3^{2^jm}+1\equiv2\pmod4$ ($j\ge1$) contributes one. So $3^n-1$ is divisible by exactly $2^{k+2}$, and we need $k+2\ge20$, i.e., $2^{18}\mid n$.

Factor $7$. The powers of $3$ modulo $7$ are $3,2,6,4,5,1$, so $7\mid3^n-1$ iff $6\mid n$.

Both conditions hold iff $\operatorname{lcm}(2^{18},6)=3\cdot2^{18}$ divides $n$, so the answer is $\boxed{3\cdot2^{18}}$.