Puzzles
18 December
The final round of game show starts with £1,000,000. You and your opponent take it in turn to take any value between £1 and £900.
At the end of the round, whoever takes the final pound gets to take the money they have collected home, while the other player leaves with nothing.
You get to take an amount first. How much money should you take to be certain that you will not go home with nothing?
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To win this game, you need to always end your turn on a multiple of 901, as whatever you opponent does, you can always get back to a multiple of 901.
Therefore, you should take £791 on your first turn to get to a multiple of 901.
Turning squares
Each square on a chessboard contains an arrow point up, down, left or right. You start in the bottom left square. Every second you move one square in the direction shown by the arrow in your square. Just after you move, the arrow on the square you moved from rotates 90° clockwise. If an arrow would take you off the edge of the board, you stay in that square (the arrow will still rotate).
You win the game if you reach the top right square of the chessboard. Can I design a starting arrangement of arrows that will prevent you from winning?
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No, I can't.
If I could, then my arrangement would cause you to follow an infinitely long pattern without visiting this pattern. As there are only a finite number of squares, within this pattern there must be a square that you visit infinitely often. The arrow on this square will point in each direction an infinite number of times, so you must also visit the squares next to this one infinitely often.
For the same reason, you must visit the squares next to them infinitely often, and the squares next to them, and so on. In this way, we see that you visit every square, including the all-important winning square, infinitely often.
Placing plates
Two players take turns placing identical plates on a square table. The player who is first to be unable to place a plate loses. Which player wins?
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The first player can always win by first placing a plate on the exact centre of the table. Then the first player can copy what the second player does, but rotated 180°, and hence can always place a plate if the second player could.
Extension
What if the two players play on a regular hexagonal table? Or a regular octagonal table? Or a regular pentagonal table? Or a regular \(n\)-gonal table?
More doubling cribbage
Brendan and Adam are playing lots more games of
high stakes cribbage: whoever
loses each game must double the other players money. For example, if Brendan has £3 and Adam has £4 then Brendan wins, they will have £6
and £1 respectively.
In each game, the player who has the least money wins.
Brendan and Adam notice that for some amounts of
starting money, the games end with one player having all the money; but for other amounts, the games continue forever.
For which
amounts of starting money will the games end with one player having all the money?
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If Adam has £\(a\) and Brendan has £\(b\), we will write this as £\(a\):£\(b\).
First, we can take
\(a\) and \(b\) to have no common factors, as dividing both by a common factor gives an equivalent starting point. For example,
£3:£6 and £6:£12 will have exactly the same behaviour (imagine £6 and 3 £2 coins).
£\(a\):£\(b\) is also clearly equivalent to £\(b\):£\(a\) (but with the two players swapping places).
A game starting £\(a\):£\(b\) will end with one player having the money if \(a+b\) is a power of two. This is because:
If \(a+b=2^n\), then if \(n=1\), either the game has already ended or it sits at £1:£1 and is about to end. If \(n>2\), then the starting
position can be written as £\(2^n-k\):£\(k\) with \(k<2^n-k\). After another game this will be £\(2^n-2k\):£\(2k\). This is equivalent to
£\(2^{n-1}-k\):£\(k\). Therefore by
induction, if \(a+b=2^n\) then the
game ends with one player having all the money.
It can also be shown by induction, that if a game ends then it must be £\(a\):£\(b\)
with \(a+b=2^n\).
Extension
What would happen if the losing player has to triple the winning player's money?
Doubling cribbage
Brendan and Adam are playing high stakes cribbage: whoever loses each game must double the other players money. For example, if Brendan has £3 and Adam has £4 then Brendan wins, they will have £6 and £1 respectively.
Adam wins the first game then loses the second game. They then notice that they each have £180. How much did each player start with?
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Working backwards: before the second game, Brendan must have had £90 and so Adam had £270.
Before the first game, Adam must have had £135, so Brendan had £225.
Extension
After the next game, one player will have all the money and no more games can be played. Hence £135 and £225 lead to a finite number of games being played.
If the player with the most money always loses, which starting values £\(A\) and £\(B\) will lead to finite and infinite numbers of games?
Twenty-one
Scott and Virgil are playing a game. In the game the first player says 1, 2 or 3, then the next player can add 1, 2 or 3 to the number and so on. The player who is forced to say 21 or above loses. The first game went like so:
Scott: 3
Virgil: 4
Scott: 5
Virgil: 6
Scott: 9
Virgil: 12
Scott: 15
Virgil 17
Scott: 20
Virgil: 21
Virgil loses.
To give him a better chance of winning, Scott lets Virgil choose whether to go first or second in the next game. What should Virgil do?
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Virgil should go second. Whatever Scott adds, Virgil should then add to make four. For example, if Scott says 3, Virgil should say 1.
Using this strategy, Virgil will say 4, 8, 12, 16 then 20, forcing Scott to go above 21.
Extension
(i) If instead of 21, 22 cannot be said/beaten, how should Virgil win? How about 23? Or 24? How about \(n\)?
(ii) If instead of adding 1 to 3, 1 to 4 can be added, how should Virgil win? How about 1 to 5? Or 2 to 5? How about \(m\) to \(l\)?
(iii) Alan wants to join the game. Can Virgil win if there are three people? Can he win if there are \(k\) people?
