# I'm a Proper Divisor

Ann and Bob are playing a number game. Ann starts with the number 60. Then she subtracts a proper divisor of 60 from it. Bob then takes the number Ann made and subtracts one of its proper divisors from it. This keeps going until someone has no possible moves (aka when someone receives a 1, which has no proper divisor). The person who gets the 1 loses.

Questions:

1. What is the optimal strategy?

2. If Ann goes first and both Ann and Bob use the optimal strategy, who will win?

Bonus:

Ann and Bob get their friends Charles, David, and Edward to play the game with them. Every round, the loser gets eliminated. The last person in the game is the winner. Assuming Ann goes first, they go in the order Ann, Bob, Charles, David, Edward, and they all use the optimal strategy, who wins?

• What's the structure of the Bonus Game? Commented Jul 5, 2019 at 2:54
• What do you mean "structure"? Commented Jul 5, 2019 at 11:31
• Is it a series of single games containing every remaining player (5P-4P-3P-2P)? Matti's answer sounds like it would only work on some sort of tournament-style setting. Commented Jul 5, 2019 at 22:30
• So first its 5 player, and then the loser gets eliminated, and then they play again with the remaining 4 people, and so on (The person whos name is first in alphabetical order goes first) Commented Jul 5, 2019 at 23:01
• Who gets to select the number at the beginning? Commented Jul 6, 2019 at 5:30

My solution:

The one ($$A$$) who arrives at $$1$$ loses. So the one ($$B$$) who arrives at $$2$$ wins, because he subtracts one and his opponent arrives at $$1$$. Consequently, $$A$$ must be at $$3$$ on the series, because he only can subtract $$1$$, so $$B$$ arrives at $$2$$. If it were the other way $$A$$ would win, because he could follow the same way. Therefore $$B$$ must be on series at $$4$$, so he can bring $$B$$ to $$3$$, then $$B$$ subtract $$1$$ and brings $$A$$ to $$2$$, then $$A$$ subtract $$1$$ and $$B$$ loses. Thus $$A$$ must be on series at $$5$$, so he has to subtract $$1$$. And so on $$....$$
So $$A$$, the one who loses always must to be on series at odd numbers and $$B$$, the one who wins should be on series at divisible numbers. Ann starts with a divisble number, so if she plays right, she should win.

Proof:

Ann always puts Bob on an odd number $$(2 * n + 1)$$. So Bob only can subtract odd divisors $$(2 * k + 1)$$. Therefore Ann always arrives at a number divisible by $$2$$, because:
$$(2 * n + 1) - (2 * k + 1) = 2 * n - 2 * k = 2 * (n - k)$$
So it is impossible for Bob to reach $$2$$, if Ann plays good, he can't win.

1) The optimal strategy is to...

always leave the other with odd numbers.

2) If both are playing optimal...

Ann will win.

3)

If $$N$$ is odd Bob will win, else Ann will win.

Bonus:

Bob, Charles, David and Edward will connect against Ann and will try to kick Ann out, because all of them now, that if Ann will play with the strategy from point 2) they will lose in the final. In the next game Charles, David and Edward will connect against Bob, because of the same reasons as before. And so on$$...$$ Eduard know that he can't win the whole game, in every final he will only has the second turn and can't win against the strategy from point 2). So David will win the game. If the start number $$N$$ isn't $$60$$ rather any number, you can't say who will win.

• Nice, but what about the bonuses? Commented Jul 4, 2019 at 19:28
• I think it would be good to mention, that an even number can always be made an odd number by subtracting 1, as 1 is a divisor of any number in the sequence. Commented Jul 4, 2019 at 19:40
• I think this is correct. I'll check mark this tomorrow Commented Jul 4, 2019 at 22:18