Solution

We divide all the squares into 32 pairs so that the squares of one pair are connected by the course of the knight $($ Figure 1 $)$.
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Wherever the first player puts his knight somewhere for his turn, the second player moves it to the cell in its pair. Therefore, the second player will always have a move to make, and he can not lose.

Answer

His partner.

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

Hint

Try to ask not only about the road, but also about the Big Bad Wolf himself.

Solution

Prince Charming can, for example, ask: “What answer would you, Big Bad Wolf, give me yesterday to the question, which of the two roads leads to the magic kingdom?” And regardless of whether the Big Bad Wolf is telling the truth today or lying, the answer to this question will always be incorrect. It is interesting that if Prince Charming uses “the day before yesterday” instead of the word “yesterday”, ie he asks: “What would you, Big Bad Wolf, answer me the day before yesterday to the question of which of the two roads leads to magic kingdom? “, the answer will always be true, regardless of whether the Big Bad Wolf speaks the truth today, or if he spoke it yesterday.

Hint

Try to ask not only about the road, but also about the Big Bad Wolf himself.

Hint

Remember task numbers 140, 148, 149.

Solution

Joe may, for example, ask: “What answer would you give me yesterday to the question, which chair is defective?” And regardless of whether the guard is telling the truth today or lying, the answer to this question will always be wrong. It is interesting that if Joe uses “the day before yesterday” instead of “yesterday” in this question, that is, if he asks: “What answer would you give me the day before yesterday to the question, which chair is defective?”, the answer will always be true, regardless of whether the guard is telling the truth today or if he spoke it yesterday.

Answer

“What answer would you have given me yesterday to the question, which chair is faulty?”

Hint

Remember task numbers 140, 148, 149.

Solution

For the convenience of further reasoning, we replace all even numbers with vowels, and odd ones with consonants, bearing in mind that the same number can correspond to different letters. Therefore the rebus will take the following form:$\\$
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Note that the numbers AEB and YHK differ by the second digit, so C $>$ 1. For C $>$ 3 or A $>$ 2, the product $AEB \times C$ will be four-digit, therefore C = 3, A = 2. Hence, I is at most 2, that is it is equal to 2. It follows that D = 9 $($ for a smaller value of D, the expression $AEB \times D$ will be less than 2100, and IFUG $>$ 2100 $)$ Y = 8 $($ otherwise the entire product will not be a five-digit number $)$. E = 8 $($ if E $<$ 8, then YHK $<$ 810 $)$. If B varies from 1 to 9, then the number represented by IFUG will vary from 2529 to 2601, therefore F = 5, and B $<$ 9. H = 5 $($ with B changing from 1 to 7, YHK will vary from 843 to 861 $)$. K = 5 $($ otherwise the number 85K will not be divisible by 3 $)$.

Answer

285 × 39 = 11115.

Hint

Solution

Note that the number 111111 is divisible by 7, so 48 eights can be dropped from the front, and 48 nines from the end. Thus, the problem has been reduced to the same problem but for the number 88? 99. Subtracting an even multiple of 7 – 84049, we find that the number 4? 5 should be divisible by 7. Finally, subtracting 399, we obtain that ?6 is a multiple of 7. Hence ? = 5.

Answer

5.

Hint

Solution

To the number 4 you can add the number 2 many times, so that you can create all possible even numbers.
Let it be necessary to obtain an odd compound number mn, where m $>$ 2, n $>$ 2. First, we get the number 2m. Next we add to it the divisor m until we get the number mn.

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

Here is one possible example. At the ball there were 10 girls and 10 boys $($ we’ll give them numbers from 1, 2, …, 10 $)$. Larger numbers signify greater beauty of the girls. The girl with the number 10 was the least intelligent of all, and the rest had increasing intelligence from the first to the ninth. In the first dance, the pairs were made up of girls and boys with the same numbers. In the second dance, the young man # 1 danced with the girl # 2, the young man # 2 – with the girl # 3, …, the young man # 9 – with the girl # 10, the young man # 10 – with the girl # 1.
Thus, each young man from the first to the eighth danced the second dance with a girl who was at the same time more intelligent and more beautiful than the one with which he danced the first dance, the ninth young man got a girl more beautiful, and the tenth – a girl more intelligent than the one from the first dance.

Answer

Yes it could.

Hint

Solution

Obviously the numbers to the left of 1 stand in descending order, and on the right – in increasing order. Therefore, the string is uniquely determined by the choice of a subset of numbers that will stand to the left of the unit. There are known to be 29 such subsets in the set {2, …, 10} of 9 elements.

Answer

In 512 ways.

Hint

Solution

Let our square be ABCD. First we draw the diagonal AC. If it turns out that P lies on the same side from it as the vertex B, it will be sufficient to find out on which side of the lines AB and BC does the point P lie.
The case when the first person is asked to point to the half-plane containing the point D is similar.
If only one or two questions are given, then less than three lines are drawn. Whatever the answers $($ except for the case when the point belongs to the intersection of two straight lines $)$, we can only know whether the point belongs to those parts on which the plane is broken up by the drawn lines. But these parts are not limited, and simply belonging to them cannot be sufficient proof that the point lies in the square.

Answer

Three questions.

Hint

Solution

b$)$ The second player has a winning strategy. It consists of the fact that for each turn of the first it is necessary to place a zero in a cell symmetrical about 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 winning 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 exist such a strategy; b$)$ there does not exist such a strategy.

Hint

Solution

A single horizontal layer G and two vertical layers pass through the given cube K. The sum of all the numbers in 361 vertical bars that do not belong to the last two layers is 361. From the sum obtained, we subtract the sum of S numbers lying on the cubes at the intersection of these columns with G $($ there are 361 such cubes $)$. These cubes are completely covered by 19 columns lying in G. The sum of all the numbers in these columns $($ it is equal to 19 $)$ exceeds S by the sum of 19 numbers lying in the perpendicular column containing K. The last sum is obviously 1-10 = -9. Hence S = 19 – $(-9)$ = 28. Finally we have: 361 – 28 = 333.
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1. Those who know the inclusion-exception formula can easily find the answer with its help: 400 – $3 \times 20$ + $3 \times 1$ – 10 = 333.
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2. Compare with the task number 107636.

Answer

333.

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 divide the board into parts as shown in the figure. The strategy of the first: take the first lower left corner, and then go to the same part of 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

The second, for example, can act like this. Up to the 1000th digit, he goes arbitrarily. When the first writes the 1001th digit, the second compares it with the 999th $($ symmetrical relative to the 1000th digit $)$. If these numbers are the same, the second switches them around. If the 1001st and 999th numbers are different, then one of them does not coincide with the 1000th digit. Then the second player switches this number with the 1000th. In any case, after a move of the second player, the 1001th digit coincides with the 999th. Similarly, the second acts further, achieving the coincidence of the 1002nd and 998th, …, 1999th and first digits.

Answer

Yes it is always possible.

Hint

Solution

Let Conrad first weigh some two coins. We consider two cases.
1$)$ The scales are in balance. Conrad can compare the weights of two coins lying $($ in the first weighing $)$ on one cup. If their weights are equal, then the weights of all the coins are equal, that is, there are either 0 or 4 false coins. If the weights are not equal, then exactly one of them is fake. But then on the other cup exactly one is counterfeit, that is, there are exactly two counterfeit coins.
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2$)$ The weights of the cups are not equal. Then the counterfeit coins of which there are either one or two $($ in this case they lie on one cup $)$ or three. Conrad can take a coin from each cup and swap them. If there are two counterfeit coins, then the scales will be balanced, otherwise they will not be balanced.
In any case, Conrad’s verification will be successful.

Answer

Yes he will succeed.

Hint

Solution

Both of the thought of numbers were divisors of the number of 2002 $($ otherwise someone would understand that 2002 is the sum of the thought of numbers, and would be able to figure out the second number $)$. However, even knowing that Janine had thought of a divisor, Zahara can not rule out that 2002 is the sum. But the sum of divisors is equal to 2002 only in the case of 1001 + 1001 $($ other dividers are equal to 2002 or less than 1001 $)$. Hence, Masha thought of 1001 $($ and Janine – 2 or 1001, and therefore neither really could figure out each other’s numbers).

Answer

1001.

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 200 cards with the numbers 3, …, 9, 0 and 201 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

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 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. If such a moment occurs $($ the second may lose before that even happens $)$, then $($ after the “closure” of the reserve $)$ the unpainted part will consist 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

Jennifer has a problem indeed. Let’s first look at a relatively normal hexagon. Maybe convex, maybe slightly concave. In all the simple cases we find out that if we cut through the opposite vertices, it always divides the figure into two quadrilaterals:

This doesn’t seem to change if we just move points around randomly, the same thing happens with our hexagons from the Example 1, the same thing seems to happen every time. But maybe there is a way… The idea is to try to align the vertices. If the line drawn through the opposite vertices also goes through other vertices, the problem does not occur. Obviously we cannot align all the three pairs of opposite vertices like this. But if we look at the picture above, we see that such a line between $B$ and $E$ is partially outside the hexagon, and thus cuts a triangle off it and is not a problem. In total, we would like some of these lines to be already aligned with some of the sides of the hexagon, and the other ones to be outside of the hexagon. We can do something like this:

The line between $A$ and $D$ cuts the hexagon into two triangles, the line between $C$ and $F$ is the same line, so it does the same thing. The line between $B$ and $E$ is again partially outside the hexagon, so it looks like we are alright. $\\$
However, there is another way to divide our hexagon into two quadrilaterals! Extending the line $BC$ causes trouble – we cut of an “arrow” on the left and a convex quadrilateral on the right. This is still not good. We could fix it if we moved $C$ to the other side of the $BE$ line, but then the segment $BE$ is all contained in the hexagon and divides it into two quadraliterals just like in the first case. The solution is to put $C$ exactly on the line $BE$. Then $BC$ and $BE$ are the same line, which cuts the hexagon into a quadrilateral and a triangle. Let us draw another hexagon, starting from the lines:

But is it actually impossible to divide this into two quadrilaterals? Maybe there is some other weird way to do that? We can go point by point and check that. First, any lines that go through two of the vertices of the hexagon do not do that, because it was designed that way. Then, we have to look at the lines that go through one vertex and then cut a side of the hexagon somewhere. Now, the line that crosses through $G$ and the side $JL$ looks dangerous, but it actually makes a pentagon on the right. Lines going through $I, K$ are not dangerous, because they always cut at least one triangle. A line through $L$ that cuts $GI$ gets a pentagon – and if it cuts through the $HK$ it gets a triangle. We can worry about the lines going through $H$ and $J$, so let’s have a look:

We can see that however these lines might cut a quadrilateral, the other figure will always be something else, a pentagon or a triangle. $\\$
Lines that don’t go through any vertices are a separate issue, but the ones that are in a direction similar to $MH$ and $NJ$ on the above picture and on the right or left side of the hexagon cut a quadraliteral and a hexagon, and the ones going through the centregive us two pentagons. The ones roughly perpendicular to these directions give us two pentagons if they cross through the centre and cut some triangles if they don’t. After carefully checking all the cases (vertex-vertex, vertex-side, side-side), we can check that this hexagon cannot be divided into two quadrilaterals with a single line.

Hint