, 2020

Solution:

Because we only care about when the ratio of $A$ to $B$ is an integer, the value of the first term in $S$ does not matter. Let the initial term in $S$ be $1$. Then, we can write $S$ as $1, r, r^{2}, \ldots, r^{2019}$. Because all terms are in terms of $r$, we can write $A = r^{a}$ and $B = r^{b}$. We will now solve for $a$ and $b$.

Observe that the geometric mean of two terms $r^{m}$ and $r^{n}$ is simply $r^{\frac{m+n}{2}}$, or $r$ raised to the arithmetic mean of $m$ and $n$. Thus, to solve for $a$, we can simply consider the sequence $0, 1, 2, \ldots, 2019$, which comes from the exponents of the terms in $S$, and repeatedly replace the smallest two terms with their arithmetic mean. Likewise, to solve for $b$, we can consider the same sequence $0, 1, 2, \ldots, 2019$ and repeatedly replace the largest two terms with their arithmetic mean.

We begin by computing $a$. If we start with the sequence $0, 1, \ldots, 2019$ and repeatedly take the arithmetic mean of the two smallest terms, the final value will be
$$a = \frac{\frac{\frac{0+1}{2} + 2}{2} + 3}{2} + \cdots + 2019 = \sum_{k=1}^{2019} \frac{k}{2} \frac{2^{2020-k}}{2}$$
Then, we can compute
$$\begin{aligned} 2a &= \sum_{k=1}^{2019} \frac{k}{2^{2019-k}} \\ \Longrightarrow a &= 2a - a = \sum_{k=1}^{2019} \frac{k}{2^{2019-k}} - \sum_{k=1}^{2019} \frac{k}{2^{2020-k}} \\ &= \sum_{k=1}^{2019} \frac{k}{2^{2019-k}} - \sum_{k=0}^{2018} \frac{k+1}{2^{2019-k}} \\ &= 2019 - \sum_{j=1}^{2019} \frac{1}{2^{j}} \\ &= 2019 - \left(1 - \frac{1}{2^{2019}}\right) = 2018 + \frac{1}{2^{2019}} \end{aligned}$$
Likewise, or by symmetry, we can find $b = 1 - \frac{1}{2^{2019}}$.

Since we want $\frac{A}{B} = \frac{r^{a}}{r^{b}} = r^{a-b}$ to be a positive integer, and $a-b = \left(2018 + \frac{1}{2^{2019}}\right) - \left(1 - \frac{1}{2^{2019}}\right) = 2017 + \frac{1}{2^{2018}}$, $r$ must be a perfect $\left(2^{2018}\right)^{\text{th}}$ power. Because $r > 1$, the minimal possible value is $r = 2^{2^{2018}}$. Thus, $d = 2^{2018} + 1$, and so $\log_{2} d$ is clearly closest to $2018$.