, 2023

Solution:

In this solution, all logs are in base $e$. Let $p_{1}, p_{2}, \ldots$ be the primes in sorted order. Let $q_{i} = p_{i} \bmod 2027$. Since the residues of primes modulo $2027$ should be uniformly distributed, we can make the probabilistic approximation that the $q_{i}$ are random variables uniformly distributed among $1, \ldots, 2026$. This becomes the famous "coupon collector" problem: the random variables $q_{i}$ are coupons with $2026$ different types, and we keep collecting coupons until we have encountered one of each type. In other words, we seek to find the smallest $k$ such that $\{q_{1}, \ldots, q_{k}\} = \{1, \ldots, 2026\}$, and then the answer to the problem is $p_{k}$.

It is known that the expected value of $k$ is $2026\left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2026}\right) \approx 2026 \log 2026$. This is because we must draw an expected $\frac{2026}{2026}$ coupons until we get our first distinct coupon type, then an expected $\frac{2026}{2025}$ coupons until we get our second new coupon type, and so on. The standard deviation of $k$ is a small fraction of its expectation, so we can safely assume that $k$ is approximately $2026 \log 2026$. Since the $n$-th prime is approximately $n \log n$, our estimate is
$$\begin{aligned} E & \approx 2026 \log 2026 \log (2026 \log 2026) \\ & \approx 2026 \log^{2} 2026 \\ & \approx 117448 \end{aligned}$$
This achieves $A / E \approx 0.969$, which scores 19 out of 25 points.