, 2022

Solution:

Let $a_{k}$ denote Alice's number after $k$ seconds, and let $p_{k}$ be the smallest prime divisor of $a_{k}$. We are given that $a_{2022}$ is prime, and want to find $a_{0}$.

If $a_{0}$ is even, then $a_{n+1}=a_{n}-2$, since every $a_{n}$ is even. Then we need $a_{2022}=2$, so $a_{0}=4046$.

If $a_{0}$ is odd, then $a_{1}=a_{0}-p_{0}$ is even, so by similar logic to the even case, $a_{1}=4044$. Then since $p_{0} \mid a_{0}-p_{0}$ and $4044=4 \cdot 3 \cdot 337$, we must have $p_{0}=3$ or $337$. But if $p_{0}=337$, $a_{0}=12 \cdot 337+337=13 \cdot 337$, so $337$ is not the smallest prime divisor of $a_{0}$. Thus, we need $p_{0}=3$, so $a_{0}=4047$, which works.

Thus, the final answer is $4046+4047=8093$.