Stack of Coins 1B

You and your friend stumble on another pirate’s treasure: 99 copper coins and 1 gold coin. As before, these old coins are no longer legal tender, so the copper ones are worthless. Only the gold coin is worth something due to its material.

To determine who gets the gold coin, your friend comes up with a game of wits. He stacks the 100 coins in a tall stack with the gold coin at the bottom. You and him will take turns taking anywhere between 1 to 10 coins from the top of the stack, all the way until the final coin is taken. He lets you decide if you want to go first or second.

Should you go first or second? And what strategy do you use to ensure you’ll win the gold coin?

Hint

Try the previous puzzle first and see if you can extend the solution.

Solution

Choose to go first and take 1 coin. After that, when your friend makes a move, take $11 - x$ coins, where $x$ is the number of coins he took.

Recapping the solution for Stack of Coins 1A

The trick to Stack of Coins 1A is to notice that you always want to reduce the stack of coins by 3. 3 coins is the perfect number because:

Your friend cannot remove 3 coins at once on his turn
After your friend’s turn (taking 1 or 2 coins), you can always make a move (taking 2 or 1 coins) such that you both reduce the stack by 3 coins.

This ensures you can always keep the stack at a multiple of 3, until the number of coins reaches 0 (which is also a multiple of 3). But since you will be the last to move for every group of 3 coins, you will be the player to bring the number of coins to 0 (i.e. taking the gold coin).

After that, all that’s left to do is to ensure the stack of coins is a multiple of 3 whenever it’s your friend’s turn. Since 20 coins isn’t a multiple of 3, but 18 is, you should choose to go first and take 2 coins.

Extending the previous solution, we want to find this “magic number” for this version of the game. 3 will no longer cut it, because your friend can take more than 3 coins at once on his turn. Our new number is 11: one more than the maximum number of coins your friend can take on his turn. We can again check that it satisfies our earlier 2 conditions:

Your friend cannot remove 11 coins at once on his turn
After your friend’s turn (taking 1 to 10 coins), you can always make a move (taking 10 to 1 coins) such that you both reduce the stack by 11 coins.

Now, all you have to do is to remove the first coin, to bring the stack down to 99 coins: a multiple of 11. Whenever your friend takes 1 to 10 coins, you can always take $11 - x$ coins to bring the stack down to the next lower multiple of 11. This continues until the stack is 11 coins high. He can take between 1 to 10 coins, and you’ll take the remaining coins, including the gold coin.

Previous puzzle in this series

Stack of Coins 1A

Next puzzle in this series

Stack of Coins 1C

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