Let $$n=\frac{4^{31}-1}{3}$$. Find the remainder when $$2^{n-1}$$ is divided by $$n$$.

From the definition, $$2^{62} = 4^{31} = 3n + 1$$, so $$2^{62} \equiv 1 \pmod{n}$$. Also $$n = 1 + 4 + 4^2 + \cdots + 4^{30}$$ is odd, so $$n - 1$$ is even, and since $$4^5 = 1024 = 31 \times 33 + 1$$ we have $$4^{30} \equiv 1 \pmod{31}$$, so 31 divides $$3(n - 1) = 4(4^{30} - 1)$$ and hence divides $$n - 1$$. Therefore $$n - 1 = 62t$$ and $$2^{n-1} = (2^{62})^t \equiv 1 \pmod{n}$$, so the remainder is 1.