# Who would be the first to defeat an abundant number?

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

$$1,2,3,4,5,6$$

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

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

• Can you give more explanation? – Dmitry Kamenetsky Mar 11 '20 at 1:50
• @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. – Daniel Mathias Mar 11 '20 at 2:02
• Well done! I'd love it if you edited more explanation into your spoiler block, but up to you. – histocrat Mar 11 '20 at 14:38
• 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. – Bass Mar 11 '20 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.

### Edit:

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 "$$1$$"):

221,65,858
285,65,70
646,70,858
357,70,77
209,77,858
221,357,646
209,285,646
285,357,858

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.