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?


2 Answers 2


Play tic-tac-toe: $$\begin{array}{c|c}70&77&357\\ \hline 65&858&221\\ \hline 285&209&646\end{array}$$

  • 2
    $\begingroup$ Can you give more explanation? $\endgroup$ Commented Mar 11, 2020 at 1:50
  • 8
    $\begingroup$ @Dmitry To win this game, a player must pick three numbers with a common factor. The grid shows that this is analogous to that other game which has a known optimal strategy. $\endgroup$ Commented Mar 11, 2020 at 2:02
  • $\begingroup$ Well done! I'd love it if you edited more explanation into your spoiler block, but up to you. $\endgroup$
    – histocrat
    Commented Mar 11, 2020 at 14:38
  • 1
    $\begingroup$ Ah dangit! I had basically everything laid out in front of me, and had managed to work out the correct strategy too, but I just couldn't figure out why the pattern felt familiar. Well done, would +2 if I had a premium account. $\endgroup$
    – Bass
    Commented Mar 11, 2020 at 21:20

My initial answer (without a proof):

I think that Rupert has the advantage. I wrote a program to have both players play all possible games and Rupert won 98% of the time.

Click this link to see the (bad) code.


The first version of the code did not take into account draws and it allowed games to continue after a player has won. After fixing these problems:

rupert: 36% | jerry: 30% | draws: 34%

These results come from players making thoughtless choices and are not representative of perfect games.

In order to efficiently win a game, a player needs to take one of these combinations (and optionally take the "$1$"):


Here is the code that generated that: https://jsfiddle.net/6qwm3bcs/1/

Like Daniel Mathias pointed out, these 8 combinations can be placed on a tic-tac-toe board and the game would be analogous. The only difference would be the extra "$1$" space. If it is taken at the beginning, it is a regular game of tic-tac-toe where the game will end in a tie if played perfectly. If the "$1$" is taken in the middle, the game will be lost since the opponent will get a chance to take a needed divisor. If the "$1$" is taken at the end, the game ends in a draw.

Therefore, if both players play perfectly, no player has an advantage and the game will end in a draw.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.