Problem 1863

Given positive integer $n$ and $r$ pairwise distinct primes $p_1, p_2, \cdots, p_r$. Initially, there are $(n+1)^r$ numbers written on the blackboard: $p_1^{i_1} p_2^{i_2} \cdots p_r^{i_r}$ where $0 \le i_1, i_2, \cdots, i_r \le n$.

Alice and Bob play a game by making a move by turns, with Alice going first. In Alice's round, she erases two numbers $a, b$ (not necessarily different) and writes $\gcd(a, b)$. In Bob's round, he erases two numbers $a, b$ (not necessarily different) and writes $\mathrm{lcm}(a, b)$. The game ends when only one number remains on the blackboard.

To determine the minimal possible $M$ such that Alice could guarantee the remaining number no greater than $M$, regardless of Bob's move, we analyze the strategies for both players.

### Alice's Strategy for $n$ Odd
1. Alice toggles $1$ and $M^n$, setting $\alpha = 1$.
2. If Bob toggles $a$ and $\frac{M^n}{a}$, then Alice takes $\alpha$ and $[\alpha, \frac{M^n}{a}]$. Since $\alpha \mid M^{\frac{n+1}{2}}$, this pair of moves removes $a$ and $\frac{M^n}{a}$.
3. If Bob toggles $a$ and $b$ for $a, b \ne \alpha$ and $ab \neq M^n$, then Alice toggles $\frac{M^n}{a}$ and $\frac{M^n}{b}$. It can be shown that $[a, b] \cdot \gcd(\frac{M^n}{a}, \frac{M^n}{b}) = M^n$.
4. If Bob toggles $\alpha$ and $b$, then Alice toggles $t$ and $\frac{M^n}{t}$ and sets $\alpha$ as their gcd.

### Alice's Strategy for $n$ Even
1. Alice toggles $(1, M^n)$ as her first move, setting $\alpha = 1$ and $\beta = M^{\frac{n}{2}}$.
2. If Bob doesn't toggle $\beta$, Alice follows the same strategy as for $n$ odd.
3. If Bob toggles $\alpha$ and $\beta$, Alice sets $\alpha = \gcd(T, \frac{M^n}{T})$ and $\beta = \mathrm{lcm}(\alpha, \beta)$.
4. If Bob toggles $\beta$ and $x$ for $x \ne \alpha$, Alice toggles $[\beta, x]$ and $\frac{M^n}{x}$.

### Bob's Strategy for $n$ Even
1. Bob sets $\alpha = M^{\frac{n}{2}}$.
2. If Alice toggles $x$ and $y$, Bob toggles $\frac{M^n}{x}$ and $\frac{M^n}{y}$.
3. If Alice toggles $x$ and $\frac{M^n}{x}$ for $x \ne \alpha$, Bob toggles $\alpha$ and $\gcd(x, \frac{M^n}{x})$.

### Bob's Strategy for $n$ Odd
1. Bob sets $\alpha = M^{\frac{n-1}{2}}$ and $\beta = M^{\frac{n+1}{2}}$.
2. If Alice toggles $\beta$ and $\alpha$, Bob toggles $T$ and $\frac{M^n}{T}$, setting $\beta = \gcd(\beta, \alpha)$ and $\alpha = \mathrm{lcm}(T, \frac{M^n}{T})$.
3. If Alice toggles $\alpha$ and $x$, Bob toggles $\gcd(\alpha, x)$ and $\frac{M^n}{x}$.

The minimal possible $M$ such that Alice could guarantee the remaining number no greater than $M$ is $M^{\lfloor \frac{n}{2} \rfloor}$, where $M = \prod_{i=1}^r p_i$.

The answer is: $\boxed{M^{\lfloor \frac{n}{2} \rfloor}}$.