Solution

For n = 3, the first clearly wins, and for n = 4, the second player wins. We show how the first player wins for n $>$ 4. After the first move of the second player, two adjacent sides are painted. The first one can paint the side “one further away” from them, forming an unpainted “hole” out of one side. This is his reserve, which the second can not paint. The reserve is the first players to use only if it has no other possible moves. The next move of the second player is to paint an edge adjacent to already painted, then the first player on their next move should paint an edge “one further away” from the last painted edge, creating another reserve until possible. When it is not possible to make a move creating reserve, it means that unpainted part consists of pairs of adjacent sides. Then the first player can always follow a move of the second player, painting a side of the same pair.

Answer

Only for n = 4.

Hint

Solution

It is clear that the result is influenced only by the last digits of the numbers written on the cards. Therefore, we can assume that there are 202 cards with the numbers 3, …, 9, 0 and 203 cards with the numbers 1 and 2. The first player can first take a card with the number 2, and then repeat the moves of the second player for as long as possible. At the moment when he can not repeat the move of the second $($ after the second takes the last card with the number 1 $)$, the first takes any of the remaining cards, and then again repeats the moves of the second. And so on. As a result, at the end both cards of each type will be equally divided, except for the cards with the numbers 1 and 2: the first will have one extra card with a deuce and one less card with a one.
Since the sum $100 \times $(0 + 1 + … + 9)$ ends in a zero, the sum of the first players ends in a two, and the sum of the second ends in a one.

Answer

The first player.

Hint

Solution

The second, for example, can act like this. Up to the 1010th digit, he goes arbitrarily. When the first writes the 1011th digit, the second compares it with the 1009th $($ symmetrical relative to the 1000th digit $)$. If these numbers are the same, the second switches them around. If the 1001st and 1009th numbers are different, then one of them does not coincide with the 1010th digit. Then the second player switches this number with the 1010th. In any case, after a move of the second player, the 1011th digit coincides with the 1009th. Similarly, the second acts further, achieving the coincidence of the 1012nd and 1008th, …, 2021th and first digits.

Answer

Yes it is always possible.

Hint

Solution

Solution 1
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One of the possible strategies of the first player: with his first move he places a cross at the center of the board; then for every move of the second player in any cell, it responds with a move in a cell symmetric with respect to the center. This is possible, since symmetry is violated each time by the second player.
In the final position in each pair of cells, symmetrical with respect to the center, there is a cross and a nought. Therefore, in the central row and in the central column, there are more crosses than zeros; and each row $($ column $)$ that does not pass through the center corresponds to a symmetrical row $($ column $)$, with one of them having more zeros than crosses and one of them having fewer zeros than crosses. Thus, with this strategy, the first player’s win is equal to 2.
Now we indicate the strategy of the second player. If he has the opportunity to occupy a cell that is a symmetrical $($ relative to the center $)$ cell, which has just been occupied by the first player then it does so. Otherwise, he makes any move. With such a strategy, after the turn of the second player, the set of occupied cells is either symmetrical $($ until the first has occupied the center $)$, or differs from the symmetrical one by exactly one cell. In this case, in each pair of occupied symmetrical cells there will be one cross and one zero. With his last move, the first player will have to create the position described in the previous paragraph, and his winnings are again 2.
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Solution 2
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We cover the board with tiles as shown in the figure. The strategy of the first: take the first lower left corner, and then go to the same domino tile on the board, where the second one went before it.
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The strategy of the second: if possible, take a cell in the same part of the board, where the first player went first; if it is impossible – any move.
If at least one of the players adheres to the strategy indicated for him, then in the end in the lower left corner there will be a cross, and in each of the other parts of the board there will be an equal number of noughts and crosses $($ for reasons similar to those stated in the first solution: after each move of the second player in all filled dominoes there will stand a cross and a nought, and, perhaps, in one there is just a nought $)$.
Consider the two upper rows. In total, they contain 9 crosses and 9 noughts $($ since they are divided into 9 “dominoes” $)$. Therefore, in one of them there are more crosses, in the other – more zeroes. A “Draw” is also achieved in the third and fourth, …, seventh and eighth rows from the top, and in the bottom rows the first “wins” $($ it has 5 crosses and 4 noughts $)$. The situation is similar in the columns $($ pairs of neighboring columns, starting counting from the right, also consist of 9 “dominos” $)$.

Answer

B = 2.

Hint

Solution

b$)$ The second player has a non-losing symmetric strategy. Namely for each turn of the first player the second places a nought in a cell symmetrical with respect to the center of the board to the cell into which the cross was just placed. Then, by the end of the game, the number of consecutively standing crosses will be exactly equal to the number of fives of consecutive noughts.
a$)$ Here is the first player’s non-losing strategy. The first move is arbitrary. Next, if the second player moves symmetrically $($ puts a zero into the cell, symmetric to the cell into which the cross has just been placed$)$, the first player again places the cross in an arbitrary cell. If the second player does not move symmetrically $($ we designate the cell that is symmetric to the last cross placed through A $)$, the first player intercepts the “symmetric” strategy and places the crosses in cells symmetrical to those in which the second places the nought, until the second player places the zero in cell A.

Answer

a$)$ There does not exist such a strategy; b$)$ there exists such a strategy.

Hint

Solution

The first player first breaks off a 1 × 6 strip and eats it. There remains a chocolate bar with an even number of longitudinal and transverse strips. Next further moves of the first player are repeats of the moves of the second: he breaks the strips in the same direction as the second. Therefore after the move of the second player in one direction there remains an odd number of strips, and after the turn of the first player – in both directions there is an even number of strips. Thus the first eats as many squares as the second. The game ends when the second player breaks off a strip of width 2 into two equal slices, which are taken by the players, so that they again eat an equal amount of chocolate. And so overall, the first player eats exactly six pieces more.

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See also the problem number 74569.

Hint

Solution

In the sequence 1987, 993, 496, 248, 124, 62, 31, 15, 7, 3, each consecutive number is an incomplete quotient left after dividing the previous number by 2 with a remainder. Let us prove that this is a sequence of winning numbers $($ that is, a player who creates one of these numbers has a winning strategy $)$. The number 1987 is a winning number by the condition. Let the number 2k or 2k + 1 $( k > 2)$ be a winning number. Therefore k is also a winning number. Indeed, if one player creates the number k, then the other can only name a number from the interval [k + 1, 2k – 1], after which the first one can name both 2k and 2k + 1. The starting player must name the number 3 and, following this strategy, will win.

Answer

The first player.

Hint

Solution

In the cases A, C it is sufficient for the mouse to walk to a node centrally symmetric to the node on which the cat is located each time.
$($ For case A, it’s still simpler. $)$ If we color the nodes of the labyrinth in a chess coloring format, we see that the cat, after its turn, is always in a node with an opposite color to that of the node of the mouse, so it can not eat the mouse with any sequence of moves.
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In the case B we also consider the chess coloring of the nodes $($ see Fig. $)$. First, the cat goes to node M. If the mouse during this time “passes” on the diagonal segment, the cat catches it with the next move.
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Otherwise, the cat itself goes diagonally and returns to M. If the mouse is not yet eaten, at that moment it is standing on a black node and must go to a white one. But there are only four white nodes, which have not yet been visited by the cat. Two of them are angular, if the mouse gets into one of them, the cat “clamps” it in this corner and eats it on the next move. Otherwise, the cat makes a move towards the mouse, and the mouse is forced to retreat into the black corner, where the cat will catch it in the next turn.

Answer

In the labyrinths A and B it can not, but in C it can.

Hint

Solution

In total there are 45 rocks. In the end we get 45 piles with 1 rock each. In order to divide the first pile into piles with 1 rock each, we need 9 moves. To do the same for the second pile we need 14 moves, and for the third pile we need 19 moves. (Note: the number of moves does not depend on whether we take one rock out of the pile per move, or a few rocks per move.) Thus the total number of moves is $9+14+19=42$ and this number does not depend on which moves the players make. The last, even, move is by the second player, and so the second player wins.

Hint

Solution

 Suppose the first guard (Peter) will run into the vertex B and the other guard (James) will run into the vertex C (pic. 1). Obviously, the monkey cannot be on segments AB and AC at the moment. If it ends up on the segment BC, the problem is solved.

   Suppose this did not happen. It will not limit the generality, if we assume that the monkey is in the right half of the zoo, i. e., on one of the sides of the CEF triangle. We will say that James controls the vertex B (C) if the distance from him to B (C) is no more than 1/3 of the distance (along the paths BD, DE, EC, EF, CF) from the monkey to B (C). In this case, he can not let the monkey pass through the corresponding vertex. Meanwhile at the moment James controls only C.

  Now Peter stands in the vertex B and James moves towards him. Let us assume BC = 1. Suppose at some moment the distance from James to C is equal to 1/3 of the distance from the monkey to C for the first time. Since the sum of the distances from the monkey to the vertices C and B is always at least 3, at this moment James controls both C and B. It is clear that he will be able to maintain this control in the future (i. e., the monkey will not be able to pass through these vertices). If this does not happen, then James will gain control over both vertices at the moment he reaches B.

   After this, Peter runs to the vertex E. If at that moment the monkey is in the left half of the zoo, then the guards will be able to catch it by moving towards D. If it remains in the right half, then James returns to C, and then the guards move either along CE, if the monkey is there, or to F. Either way, the monkey will be caught.

Answer

The guards will be able to catch the monkey.

Hint

Solution

In the first turn, the rook gets to an arbitrary cell of the middle line, which runs along the board. Then it moves in one direction, behind the bishop on the course of the knight.

Hint

–

Solution

Let the first player, starting with their second turn, make moves symmetric to the moves of the second player relative to the center of the field. Then, as soon as the second player takes one of the corner cells, the first one can occupy the opposite corner cell.

Hint

Solution

Here is one possible strategy. The first player first writes an arbitrary coefficient in front of z in the first equation. Next, if the second player writes some coefficient in front of x or in front of y, then the first writes the same coefficient in the same equation for y or at x. If the second writes some coefficient at z, then the first writes an arbitrary coefficient for z in the remaining equation. The resulting system has a solution $(1, -1, 0)$.

Hint

Solution

The one who starts wins in both variations of the game. This is also true for any chocolate with mn number of pieces (size m × n), where mn is even (except for the case of a 2 × n chocolate with odd n in variation b) – in this case the answer depends on mn.)

We are going to consider the general case at once. Interestingly, the winning strategies are almost the same in the “opposite” variations a) and b).

1. a) Here is the strategy that provides a win for the one who starts the game. At least one of the numbers m and n is even – let it be m (so m = 2k). In the first move, the one who starts breaks the chocolate into two identical halves (n × k slices each). Then he duplicates each move of the second player on the other half of the chocolate. Thus, after each move of the first player, both halves will be broken in exactly the same way. It is clear that in this case the first will not break off the 1 × 1 slice before the second one does.
2. b) In this case with m> 2 and n> 1, the one who starts the game can use the same “symmetric” strategy until the second one breaks off a strip which width is 1; the first player immediately breaks off a 1 × 1 piece of it and wins.

In variation b) of the game, the “symmetric” strategy for a 2 × n chocolate bar does not work – it is impossible to break off a strip which width is 1 (this immediately leads to a loss), so the chocolate bar can only be broken “horizontally”. This case creates a completely different problem which is much more complicated and a detailed study of which is left to the reader.

In the general case, the answer for odd mn is unknown to us neither for the variation a), nor for the variation b) of the game.

Answer

The one who starts wins in both variations of the game.

Hint

Solution

Let Hannah choose the coin “Scissors – paper”. If Emma chooses the coin “Stone-scissors”, then Hannah has a probability of ¼ of winning, and Emma has a probability of ½ $($with a probability of ¼ of a draw$)$.
Emma makes a similar choice when faced with the two other possibilities of coins that Hannah could pick.

Answer

There is.

Hint

–

Solution

Let N$>$ 1. We give the winning strategy of the second player. His first move is in the outer cell, and he then plays as he likes. After the k-th move of the first player, the crosses divide the strip into at least k parts, consisting of empty squares and noughts. But by this time only k-1 noughts have been placed, which means that there is no sign in one of the parts, and there the second player can make a move. Since the game will at some point end, the first player will lose.

Answer

For N = 1, the first player wins, while for N$>$ 1 the second player wins.

Hint

–

Hint

Each move the second player can return the rook to the diagonal a1-h8.

Solution

We describe the winning strategy of the second player. With his first move, the first player is obliged to remove the rook from the diagonal a1-h8. With his turn, the second player can return the rook to the diagonal. We arrive at a situation similar to the initial one. The only difference is that the rook was closer to the h8 square. Adhering to this strategy, the second player will first come to the h8 cell.

Answer

the second player wins.

Hint

Each move the second player can return the rook to the diagonal a1-h8.

Hint

Keep track of the parity (whether it is even or not) of the number of rows and columns.

Solution

a) A strategy for the second player: always leave an even number of rows not crossed out and an even number of columns not crossed out. Then the second player can always make a move.

b) A strategy for the first player: during their first turn, cross out a row in order to make the board be 8×10. Then proceed by the strategy from a).

c) A strategy for the second player: after the first player’s first turn, make the board be 8×10 (if the first player crosses out a row, then te second player crosses out a column and vice versa). Then proceed to play using the strategy from a).

Answer

a), c) There is a winning strategy for the second player. b) There is a winning strategy for the first player.

Hint

Keep track of the parity (whether it is even or not) of the number of rows and columns.

Hint

Sheep can walk on fairly distanced from one another parallel lines.

Solution

We introduce a coordinate system on the plane. Let the k-th sheep have the coordinates $($4k; 0$)$, k = 1,2, …, 100, and the initial position of the wolf is such that in one move it can not reach any sheep. We describe the strategy of the sheep. Sheep will only walk on straight lines where x = 4k. Let the wolf be at the point $($x; y$)$ on one of the vertical lines | x-4k | $<$ 2 $($before the sheep moves$)$ $($these bands do not intersect$)$. Then the sheep with the number k moves, and if this sheep is at the point $($4k; z$)$, then it moves to the point $($4k; z + 1$)$ if z $>$ y and to the point $($4k; z-1$)$ if z is not greater than y $($ie, the k-th sheep is moving away from the wolf by 1 along the axis Oy$)$. It is clear that after this the wolf within one turn again can not catch a single sheep. If the wolf before the movement of the sheep is outside the strips | x-4k | $<$ 2, then the sheep can make any move. Adhering to this strategy, the sheep will be able to achieve their goal of none of them being caught.

Answer

No.

Hint

Sheep can walk on fairly distanced from one another parallel lines.

Solution

Let the first player in their first turn draw a cross in the centre of the grid. The second player has two options: to put a nought in the corner or in the centre of a side of the grid. The it is not easy to check all possibilities to see that in each case the first player can prevent the second player from winning.

Hint