# Tag Info

### Making a 2n-digit number divisible by 9

I believe that Bob can win if (and only if) n is
• 8,590

### Making a 2n-digit number divisible by 9

I'll add on to SQLnoob's answer and say Bob can't win if n isn't So she chooses $$m = \begin{cases} 7-k & k < 7 \\ 1 & k = 7, 8 \end{cases}$$ For example,
• 10.1k

### Moving a knight to a new square each turn: who wins?

Here is a solution without the 4x2 grids (although it is a generalization of that solution). Pair up the squares such that the two squares in each pair are separated by a knight's move. On each of Bob'...
• 3,674
Accepted

### Paper, pencil and a bunch of bars

This game is So in this case If we take "rational" to mean that when either player had a winning move they took it then Now So
• 121k
Accepted

### Double or Take game

Answer to the non-bonus part: Proof: Bonus part: From 12 Hence
• 121k
Accepted

### Best strategy for stick-taking game

Pretty much all Nim games can be solved by starting at the end and working backwards. Let's first solve the basic Nim, with only one pile, and simple actions (take 1-4). Let's enumerate the endgame ...
• 78.6k
Accepted

### Marbles on a Mancala Board

You should To prove, let us first do the following
• 138k
Accepted

### A pile of chips involving primes

Here's a little Python program to test it yourself: https://repl.it/repls/ScientificIdenticalPixels And here's C++ code written by user @im_so_meta_even_this_acronym https://ideone.com/SfMHqC
• 2,377

### Single-pile Nim with Three Players

Even though the question says I'm Bob, I'd like to start by stating my intention to be the Quetzalcoatlus, thanks very much. I believe that the key to this game is that Example game: However, ...
• 32.3k

### Single-pile Nim with Three Players

Partial strategy up to 11 candies. A is the player whose turn is next, B is the next player, and C is last. Don't think I can do it for 40 without turning this into a novel. But I have to say that, ...
• 79.7k

### The 15 Pebbles Game

As with pretty much all the nim variants, this one can be solved by starting from the end and working backwards. With the original total number of stones being an odd number (15, as given in the title)...
• 78.6k

### The 15 Pebbles Game

The strategy is to do a move that In particular, for $15$ pebbles, your first move would be The reason this works is more interesting than with other single-pile nim variants.
Accepted

### Another variation of the game of Nim

1. Proof that all games will end after a finite number of steps Proof by induction: Induction hypothesis H: (not known to be true yet) ...
• 78.6k
Accepted

### Single-pile Nim with Three Players

I will assume that Given several equally desirable options, the players will randomly choose one. Doing so, it is easy to work out what happens when there are $n$ candies; If $n=1$, then player 1 ...
• 32.6k

### Can you help me understanding the Stones game?

Here's a clearer statement of the rules. Each player has two options: Remove a stone on this turn. Remove a stone on the next turn. This is because not removing a stone causes one to be removed next ...
• 32.6k

### Single-pile Nim with Three Players

Using the same assumption as @Oray, Now
• 2,006

### Single-pile Nim with Three Players

With this information: To be perfectly rational and impartial, only wanting to maximise their own chance of winning. and it is assumed that If there were 5 candles If there were 6 candles, If ...
• 30.4k

### Double or Take game

Let’s see: I assume that all numbers must be greater than or equal to zero, otherwise the possibilities become too painful. If that was the intention, I’ll look harder. 1 2 3 4 5 6 7 8 9 10 ...
• 32.3k
Accepted

Reasoning:
• 6,159
Accepted

### A pile of chips involving powers of 2

Code to find the pattern: https://ideone.com/O5S4Qu (The third number printed out on each line is the winning move if the player to move is in a winning position)
• 938

### A pile of chips involving powers of 2

We have a winning strategy for: The first move is:
• 3,625
Accepted

### The 50 game between two players, selecting numbers between 1 and 10 inclusive + variations

Question 1: The status of a game can be encoded as a pair A-B, where B is the sum including the last number said, and A is the sum excluding it. Question 2:

### Marbles on a Mancala Board

User hexomino already figured out the puzzle, and managed to actually find the very complicated path that was exactly how I came up with the game. To recap: The game itself is a lot easier to play ...
• 78.6k

### Three-player Nim

If everyone is trying to maximize their points: If both of your opponents work together to minimize your chances to win the game itself:
• 6,879
Accepted

### The Box of Tic-Tacs

Partial (assuming I understood the requirements correctly): My assumptions: You (captive) start. You and the kidnapper always return on the even turn the same number of tic-tacs and you decide the ...
• 18.4k
Accepted

### A short nim game

If you have to take exactly 5: If you can take a number from 1 up to 5 (which is what I'm assuming) If you can take a number from 0 up to 5 For the alternate game (which actually isn't too much ...
• 9,924
Accepted

### Game Night at the Binomial Elks Club

Any permutation of the rows and columns is treated as an "equivalent" board. Part 1: Matt chooses to go second. (Case A) If Ben takes 2 or 3 tokens, then Matt can turn it into a 2x2 board and win. ...
• 146

### Removing green marbles from the table

As described above, the first player loses if $N$, the initial number of marbles, is a Fibonacci number, and wins for any other integer $N > 1$. Proof follows. First, a definition. The remainder ...
• 7,518
1 vote
Accepted

### Zigzag nim game

As you suspected, the correct approach is to start from the end. First, let's invent a notation for the game positions. Let's view each position from the POV of the player whose turn it is, so that &...
• 78.6k
1 vote

### Three-player Nim

First of all: Taking this: With more marbles: Following that logic: List of winners: Patterns: Applying to the problem: Minimum rounds:
• 2,436

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