Rolling a d10 until the lcm passes 2000

You roll a fair ten-sided die (faces $1$ to $10$) repeatedly and track the least common multiple of all the values rolled so far. Let the expected number of rolls until this lcm first exceeds $2000$ be $\tfrac{p}{q}$ in lowest terms. Compute $p+q$.

$119$. The lcm of $1,\dots,10$ is $2520$, and every proper divisor of it is at most $1260$, so you need the full $2520$: a $7$, an $8$, a $9$ and a multiple of $5$ ($5$ or $10$). A coupon-collector style calculation (or a small Markov chain over lcm values) gives $\tfrac{113}{6}$ expected rolls.