Step-by-Step Solution
Key Concept: The binomial theorem combined with modular arithmetic allows us to determine remainders and fractional parts of exponential expressions.
We expand $2^n - 2^{-n} = 8(2^3)^{(n-1)/2} = 8(31 + 1)^n = 8(\binom{n}{0}3^0 + \binom{n}{1}3^1 + ... + \binom{n}{n}3^n)$. Simplifying using binomial coefficients: $2^n = 8(31k + 1) = 8 \cdot 31k + 8$ and $2^n - 8 = 31k$, so $2^n \equiv 8 \pmod{31}$. We verify: $2^5 = 32 = 31 + 1$, so $2^5 \equiv 1 \pmod{31}$. Therefore $\frac{2^n}{31} = 8k + \frac{1}{31}$, giving the answer as $\frac{1}{31}$.
Correct Answer: 1