Taiwan IMO Selection Camp, 2022

All $r = (b+1)/b$ with $b = 1, \dots, 1010$.

Denote the red and blue beads by $\mathcal{R}$ and $\mathcal{B}$, respectively. Introduce coordinates on the line and identify the points with their coordinates so that $R = 0$ and $B = 1$. Then, during the game, the coordinate $\mathcal{R}$ is always smaller than the coordinate of $\mathcal{B}$. Moreover, the distance between the beads always has the form $r^\ell$ with $\ell \in \mathbb{Z}$, since it only multiplies by the numbers of this form. Denote the value of the distance after the $m^{\text{th}}$ move by $d_m = r^{\alpha_m}$, $m = 0, 1, 2, \dots$ (after the $0^{\text{th}}$ move we have just the initial position, so $\alpha_0 = 0$).

If some bead is moved in two consecutive moves, then Alice could instead perform a single move (and change the distance from $d_i$ directly to $d_{i+2}$) which has the same effect as these two moves. So, if Alice can achieve her goal, then she may as well achieve it in fewer (or the same) number of moves by alternating the moves of $\mathcal{B}$ and $\mathcal{R}$. In the sequel, we assume that Alice alternates the moves, and that $\mathcal{R}$ is shifted altogether $t$ times.

If $\mathcal{R}$ is shifted in the $m^{\text{th}}$ move, then its coordinate increases by $d_m - d_{m-1}$. Therefore, the total increment of $\mathcal{R}$'s coordinate, which should be 1, equals
$$\text{either}\quad (d_0 - d_1) + (d_2 - d_3) + \dots + (d_{2t-2} - d_{2t-1}) = 1 + \sum_{i=1}^{t-1} r^{\alpha_{2i}} - \sum_{i=1}^{t} r^{\alpha_{2i-1}},$$
$$\text{or}\quad (d_1 - d_2) + (d_3 - d_4) + \dots + (d_{2t-1} - d_{2t}) = \sum_{i=1}^{t} r^{\alpha_{2i-1}} - \sum_{i=1}^{t} r^{\alpha_{2i}},$$

$$\sum_{i=1}^{n} r^{\beta_i} = \sum_{i=1}^{n-1} r^{\gamma_i}, \quad \beta_i, \gamma_i \in \mathbb{Z}, \qquad (1)$$
for some $n \le 1011$. Thus, if Alice can reach her goal, then this equation has a solution for $n = 1011$ (we can add equal terms to both sums in order to increase $n$).

Conversely, if (1) has a solution for $n = 1011$, then Alice can compose a corresponding sequence of distances $d_0, d_1, \dots, d_{2021}$ and then realize it by a sequence of moves. So the problem reduces to the solvability of (1) for $n = 1011$.

Assume that, for some rational $r$, there is a solution of (1). Write $r$ in lowest terms as $r = a/b$. Substitute this into (1), multiply by the common denominator, and collect all terms on the left hand side to get
$$\sum_{i=1}^{2n-1} (-1)^i a^{\mu_i} b^{N-\mu_i} = 0, \quad \mu_i \in \{0, 1, \dots, N\}, \qquad (2)$$
for some $N \ge 0$. We assume that there exist indices $j_-$ and $j_+$ such that $\mu_{j-} = 0$ and $\mu_{j+} = N$.

Reducing (2) modulo $a-b$ (so that $a \equiv b$), we get
$$0 = \sum_{i=1}^{2n-1} (-1)^i a^{\mu_i} b^{N-\mu_i} \equiv \sum_{i=1}^{2n-1} (-1)^i b^{\mu_i} b^{N-\mu_i} = -b^N \pmod{a-b}.$$
Since $\text{gcd}(a-b, b) = 1$, this is possible only if $a-b=1$.

Reducing (2) modulo $a+b$ (so that $a \equiv -b$), we get
$$0 = \sum_{i=1}^{2n-1} (-1)^i a^{\mu_i} b^{N-\mu_i} \equiv \sum_{i=1}^{2n-1} (-1)^i (-1)^{\mu_i} b^{\mu_i} b^{N-\mu_i} = S b^N \pmod{a+b}$$
for some odd (thus nonzero) $S$ with $|S| \le 2n-1$. Since $\text{gcd}(a+b, b) = 1$, this is possible only if $a+b \mid S$. So $a+b \le 2n-1$, and hence $b = a-1 \le n-1 = 1010$.

her goal. For this purpose, in (1) we put $n = a$, $\beta_1 = \beta_2 = \cdots = \beta_a = 0$, and
$\gamma_1 = \gamma_2 = \cdots = \gamma_b = 1$.