Rupert and Jenny are playing a game. Each player's score starts out as the number $94083986096100$, which is the square of the product of the first 8 primes and therefore has a lot of divisors. We'll concern ourselves with ten of those divisors:
$1, 65, 70, 77, 209, 221, 285, 357, 646, 858$
Rupert goes first. He picks one of the above ten numbers and divides his score by that number. Then Jenny picks a different one of those numbers, and divides her score by that number (so that now their scores are different). Players continue taking turns in this way, each time picking a number that neither player has already picked. If the number divides their current score, they divide it. The first player to pick a number that their current score doesn't evenly divide wins the game. If the players run out of numbers to pick, the game is a tie.
To illustrate, let's analyze the same game with smaller numbers: a starting score of $60$, and a number pool of
In this case, Rupert has a winning strategy. He picks $6$, reducing his score to $60\div6 = 10$. Say Jenny picks $4$, reducing her score to $60\div4 = 15 $. Then Rupert can pick $3$. Since $10$ isn't divisible by $3$, Rupert wins. If Jenny had picked $3$, Rupert could've picked $4$ and won anyway.
For the game with larger numbers, if both players play perfectly, who wins? Or is it a tie?