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Two players play with a pile of 2024 chips placed on a table. Each in turn, a player removes a certain number of chips from the gaming table: at least one chip, but no more than half of the chips still on the table. The person who leaves only one chip on the table is declared the loser. Who will win the game, the player who starts or the other one?

I have the feeling that the player who starts can win every time but I can't find the winning strategy. Can you help me to find which player is winning and what is their strategy? Thanks in advance.

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    $\begingroup$ With problems like these, start by looking at the end games, e.g. what happens if there are 2 chips left? 3? 4? Also, the fact that 2024 is a power of 2, combined with the fact that you can remove half the chips is probably a huge giveaway. So also try 8, 16 .... you'll probably see a pattern quickly enough! $\endgroup$
    – Bram28
    Mar 27 at 14:08
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    $\begingroup$ thank you for the fast reply ! since when is $2024$ a power of 2 ? $\endgroup$ Mar 27 at 14:11
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    $\begingroup$ For reference, these types of games are known as Nim and are well studied. $\endgroup$
    – JMoravitz
    Mar 27 at 14:12
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    $\begingroup$ @mattandmaths Ha yes, 2024 is not a power of 2 :P I confused 1024 and 2048, oops! Anyway, comment still stands ...powers of 2 are probably crucial numbers to look at $\endgroup$
    – Bram28
    Mar 27 at 14:13

1 Answer 1

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if it's your turn and there is

1 chip : victory

2 chips : loss

3 : victory

4 : victory

5 : loss

6 - 10 : victory (make it 5)

11: loss

23: loss

you can realize the loss pattern is 2n + 1, and pretty easy to prove. you can calculate, or think about the fact that 2n + 1 is always odd, so if there is an even amount(2024), the first player wins!

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