Hint

Try to write these names.

Solution

Yes, it can be, for example: Natasha Jane Cooper, Natasha Mary Smith, Hannah Jane Smith, Hannah Mary Cooper.

Hint

Try to write these names.

Solution

a) Let’s say, for example, that only two people with scores of 3 and 5 live in country A, and in B – one person with a score of 1. After the movement of a person with a score of 3 from A to B in both countries, the rating will increase.

b) First, we note that if the entire population of the country is divided into two groups X and Y with ratings $q_X$ and $q_Y$, respectively, then the rating q of the whole country is between $q_X$ and $q_Y$ $($and the equality will be only in the case when $q_X = q_Y$$)$.

Denote by a and b the ratings of countries A and B before emigration from A to B, by $a_1$ and $b_1$ the ratings of these countries after this emigration, and by c, the rating of the group of emigrants. By the condition a $<$a_1. Hence, as shown above, it follows that c $<$a $<$a_1 $($before emigration, A is divided into a group of emigrants with a rating of c and a group of those remaining with a rating $a_1$$)$.

Similarly, $b $<$b_1 $<$c$. Thus, $b $<$a$ and $b_1 $<$a_1$. The first inequality shows that an increase in the ratings of both countries is possible only when emigrating from a country with a higher rating to a country with a lower rating. The second inequality shows that the rating of country A remained higher than that of country B. Thus, a simultaneous increase in ratings with emigration from B to A is impossible.

c) Suppose there are only two residents in country A with marks Q = 1 and 2, in country B, there are four inhabitants $($Q = 2, 2, 4, 10$)$, in country B, there is one inhabitant $($Q = 1$)$. After the first wave of emigration from A to B, one person with Q = 1 emigrated, and from B to C – two with Q = 2. For the second wave, from C in B, one person with Q = 1 moved, and from B to A – two with Q = 1 and 4. The ratings of the countries changed as follows: A – 1.5 → 2 → 2 1/3; B – 4.5 → 5 → 5.5; C – 1 → 1 2/3 → 2.

Answer

b) It is not possible; c) It can.

Hint

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Solution

a) In the initial position of the cogs there are 4 pairs of removed teeth present, where a pair consists of one tooth on the first cog and one tooth on the second. There are 16 such pairs possible in total, and each pair occurs at most once in all of the possible positions of the cog. There are 13 other possible positions of the cogs relative to one another, aside from the initial position. There are however only 12 pairs of removed teeth that can be present in these 13 positions as 4 have been used up from the initial position, therefore there will always be at least one position of the cogs where no pair of removed teeth is present and so the projection appears as a single complete cog.

b) Counter-example: Let us number the teeth in order and remove the teeth 1, 2, 5 and 7. It is easy to check that the differences of these number gives all 13 numbers ${0, 1, 2,…,12} mod 13$. Therefore after each rotation of one position there will be exactly one pair of removed teeth that coincide.

Answer

a) yes

b) no

Hint

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Solution

F $(x + 1)$ $(F (x) + 1)$ = -1, hence, F does not vanish anywhere. For each x, either
$\\$
a$)$ F $(x + 1)$ $>$ 0, F $(x)$ + 1 $<$ 0, or $\\$ b$)$ F $(x + 1)$ $<$ 0, F $(x)$ + 1 $>$ 0.
$\\$
If a$)$ holds for at least one x, the function is discontinuous: F $(x + 1)$ and F $(x)$ have different signs.
$\\$
If b$)$ holds on the whole axis, then at every point -1 $<$ F $(x)$ $<$0. Then | F $(x + 1)$ | $<$ 1, | F $(x)$ + 1 | $<$ 1, and therefore the product of these numbers is not equal to 1. Hence we have a contradiction.

Hint

Solution

An example of the coloration of the chequered plane into four colours, in which every two cells at a distance of 6 are coloured in different colours, is shown in the figure on the left.

Another example. We introduce a coordinate system on the cellular plane. It is enough to colour the cells with an even sum of coordinates $($the colouring of the remaining cells is obtained from this shift by 1 to the right$)$. The points for which the sum of the coordinates is a multiple of 4 will be coloured in two colours: if both coordinates are even – in blue, if both odd – in red. Points whose sum of coordinates is even, but is not a multiple of 4, we will likewise colour in the remaining two colours.

To prove that a smaller number of colours cannot be dispensed with, it is sufficient to consider the four cells shown in the figure on right. Any two of them are located at a distance of 6 from each other, and therefore, they must all be painted in different colours.

Answer

In 4 colours.

Hint

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Hint

Note that in order to fulfil the conditions of the problem, the distance between adjacent plants must be no greater than 40m

Solution

In each 40m segment of the path there must be one plant. The 400m path must contain exactly 10 such segments, therefore we need 10 plants with the plant planted at the centre of each segment. We would have the first plant 20m from the start of the path, and the last 20m before the end, with the rest being at 40m intervals.

Hint

Note that in order to fulfil the conditions of the problem, the distance between adjacent plants must be no greater than 40m

Hint

Think about who the peasant can leave unattended.

Solution

The peasant cannot leave the wolf with the goat or the goat with the cabbage, but he can leave the cabbage with the wolf. Let’s show, on the diagram, what the peasant should do next:

Thus, the peasant, with all his property, will be able to cross to the other side. Think about how the peasant should behave, if at the third crossing he takes with him not the wolf, but the cabbage?

Answer

The peasant travels to the other shore with the goat and then returns. He goes with the wolf and comes back with the goat. He goes with cabbage and then returns. Finally, he goes with the goat.

Hint

Think about who the peasant can leave unattended.

Solution

We propose h =$1969^2$/1968. Then $h \times 1969^{n-1}$ = $1969^{n + 1}$/1968= m + 1/1968 and $h \times 1969^n$ = 1969m + 1 + 1/1968 $($where m is a natural number$)$. For any positive integer n, the number m is greater than 1, so 1969m + 1 is not divisible by m.

Answer

Yes, there does exist.

Hint

Solution

We show first that from the given set, you can select the 655 numbers that fit the required condition. In fact, we take all numbers of the form 3k + 1; k = 0, 1, …, 654. The difference of any two such numbers is obviously divisible by 3, while the sum of any two of them gives the remainder 2 when divided by 3. It follows that the sum of any two chosen numbers is not divisible into their difference, that is, the correct ones were chosen. We now show that no more numbers can be chosen which satisfy conditions of the problem. Indeed, if we chose more than 655 numbers, we would find two numbers that differ from each other by less than 3 $($otherwise the difference between the largest and smallest numbers of the given set would be greater than $3\times654$ = 1962, which is impossible$)$. So, there must be two numbers that differ from each other by 1 or 2. In the first case, the difference between them is 1, and everything, including their sum, is divisible by 1. In the second case, the difference between the chosen numbers is 2, and so both of these numbers have the same parity, which means that their sum is an even number. And in this case the sum of the numbers taken is divisible by their difference. We showed that from this set it is impossible to select more than 655 numbers so that the conditions of the problem are fulfilled, and so we have constructed an example of how you can select 655 numbers. Therefore, the largest amount of numbers that can be selected from this set in accordance with the conditions of the task is 655.

Hint

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Solution

Let there be less than seven stars in the table. Then two cases are possible.

1. There is no star in any line $($for example, in the first line$)$, then by the Dirichlet principle there is a line $($let it be the second one$)$ in which there are no more than two stars. We remove from the table the third and fourth rows, and also the columns containing the stars. The resulting table does not contain any stars.

2. All the lines have stars. If there were two stars in three rows, the number of stars in the table would exceed 6. Consequently, there are two lines $($even the first and second$)$, each containing only one star. We cross out the third and fourth rows of the table and those columns in which the stars of the first and second rows are placed. The remaining table does not contain stars.

An example of the placement of seven stars:

Hint

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Solution

Let f $(x)$ = 1/2 $ax^2$ + bx + c.These values have different signs, so one of the roots of the trinomial is located between $x_1$ and $x_2$.

Hint

Solution

It follows from the condition that $C_{n + 1}C_{n + 2}$ is the bisector of the angle of the triangle $C_nC_{n + 1}O$ $($ see Fig. $)$, So $2γ_{n + 1}$ + $γ_n$ + α = π. $(1)$ Let us prove that $γ_n$ tends to the limit β = $(π-α)$ / 3. We rewrite $(1)$ thus: 2 $(γ_{n + 1} – β)$ = β – $γ_n$. It follows that | $γ_{n + 1}$ – β | = ½ | $γ_n$ – β |, that is, the difference between $γ_n$ and β in the transition from n to n + 1 is halved. Therefore, | $γ_n$ – β | = $2^{1-n}$ | $γ_1$ – β | tends to zero.

Answer

$(π-α)$ / 3.

Hint

Solution

a) A function satisfying the condition of the problem increases monotonically. In fact, if $x_2 \geq x_1$ and $x_2 \geq 1$, then $f (x_2) \geq f (x_1) + f (x_2-x_1)$ and $f (x_2-x_1) \geq 0$; consequently, $f (x_2) \geq f (x_1)$. Therefore, if $½ $<$x \geq 1$, since $f (1) = 1$, we have $f (x) \leq 1 \leq 2x$.

Now let $0 $<$x \leq ½$. We take a natural number n such that $½ $<$nx \leq 1$ $($it is clear that this can always be done$)$. It is easy to prove $($by induction$)$ that if $x \geq 0$ and $nx \leq 1$, then $f (nx) \geq nf (x)$. For $½ $<$nx \leq 1$, it is already proved that $f (nx) \leq 2 \times nx$.

Therefore, $nf (x) \leq f (nx) \leq 2 \times nx$, whence $f (x) \leq 2x$ and for $0 $<$x \leq ½$. Now, it only remains to find out what will happen at zero. We take $0 $<$x_1 \leq 1$; then $f (x_1) = f (0 + x_1) \geq f (0) + f (x_1)$.

But by definition $f (0) \geq 0$, therefore, $f (0) = 0$.

b) The answer: no, it is not true.

Let us give an example. We define the function as follows: f $($x$)$ =

If $1 \geq x_1 \geq x_2 \geq 0$ and $x_1 + x_2 \leq 1$, then $x_2 \leq ½$ and $f (x_2) = 0$. Consequently, $f (x_1 + x_2) \geq f (x_1) + f (x_2)$ and the function in question satisfies the requirements of the problem. At the same time, $f (\frac{51}{100}) = 1> 1.9 \times \frac{51}{100}$.

Answer

b) No, it is not true.

Hint

Solution

Suppose that the grasshopper jumped a strip of 62 cells. We will paint 8 of its cells starting from the left side white, the next 10 will be painted black, then again 8 will be painted white and so on. There will be 32 white cells and 30 black cells. Since the difference in the numbers of white and black cells is greater than 1, there was a jump between white cells. But such jumps are impossible.

Let us give an example of the movement of cyclists, in which they cannot be on any part of the track with a length of less than 75 m at any time. Let the first and third cyclists move with respect to point A with velocities equal in magnitude but different in direction, and in the initial moment they are at the point B, separated from A by a quarter of the circle $($Figure on the right$)$. Then they will meet alternately at the diametrically opposite points B and C $($a distance from A of 75 m$)$, and their positions at each moment in time will be symmetrical with respect to the straight line BC. Hence, at each moment in time the distance from one of them to point A will be no less than 75 m.

Answer

n=62

Hint

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Solution

a$)$ Example 1. Take the number 2 and 1024 numbers equal to ½. Then $a_n$ = $2^n$ + $1024 \times 2^{-n}$ = 32 $(2^{n-5} + 2^{5-n})$. The sum of two positive reciprocal inverses is smaller, the closer they are to each other. Therefore, the constructed sequence decreases until n = 5, and then increases.

Example 2. Take $a_n$ = $1.02^n + 0.5^n$. Note that $a_{n + 1} > a_n ⇔ 0.02 \times 1.02^n > 0.5^{n + 1} ⇔ 2.04^n > 25 ⇔ n ≥ 5$. That is, the sequence decreases until n = 5, and then increases.

b$)$ Suppose this happens. The first method. Then $a_n$ $<$ $a_5$ for any n. Among the initial numbers was the number x $>$ 1, otherwise the sequence would never increase. Notice that
$a_n$ $>$ $x^n$ for any n. But the sequence $x^n$ is not bounded. So we have a contradiction.

The second method. Then $a_4$ $<$ $a_5$ and $a_6$ $<$ $a_5$, that is, $a_4$ + $a_6$ $<$ $2a_5$. Hence, $x^4$ + $x^6$ $<$ $2x^5$ for at least one of the original numbers x. But this contradicts the Cauchy inequality.

Answer

a$)$ It could; b$)$ it could not.

Hint

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Solution

Without loss of generality, we assume that all cyclists travel the track counter-clockwise, the first of them being the fastest and the third being the slowest. Consider the movement of the cyclists with reference to the second cyclist. Then the second cyclist is constantly at some point A, and the first and third move counter clockwise and clockwise, respectively. They periodically meet with each other at regular intervals. Denote by $B_1, B_2, B_3,$ … the consecutive points of their meetings from the beginning of the observation of these athletes $($Figure left$)$. Every two adjacent points $B_n$ and $B_{n + 1}$ $($where n is an arbitrary natural number$)$ are different, since the first cyclist cannot make a full circle counter clockwise, without meeting the third one. We denote by $β_n$ the smaller of the two arcs of the track with endpoints $B_n, B_{n + 1}$. Its length does not exceed 150 m. Since the meetings occur periodically with a shift in one direction, all the arcs βn are equal to each other and the union of several of them covers the whole track. Hence, there is an arc $β_m$ containing the point A. The length of one of the arcs $B_m$A or A$B_{m + 1}$ does not exceed 75 m. Thus, at some point of the meeting of the first and third cyclists they will be no more than 75 m from the second cyclist. Therefore, the photographer will be able to make a successful shot, if d$\geq$75.

Let us give an example of the movement of cyclists, in which they cannot be on any part of the track with a length of less than 75 m at any time. Let the first and third cyclists move with respect to point A with velocities equal in magnitude but different in direction, and in the initial moment they are at the point B, separated from A by a quarter of the circle $($Figure on the right$)$. Then they will meet alternately at the diametrically opposite points B and C $($a distance from A of 75 m$)$, and their positions at each moment in time will be symmetrical with respect to the straight line BC. Hence, at each moment in time the distance from one of them to point A will be no less than 75 m.

Answer

75

Hint

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Solution

An example of the arrangement with 16 crosses according to the condition can be seen on the figure on the left.

We can divide the board into a central cell and 8 rectangles $3\times1$ in size $($see right$)$. In each rectangle, there should be at least one nought. Therefore, there are at least eight of them on the board. If there is a nought in the central cell, then there are no more than 16 crosses, and the problem is solved. It remains to consider the case when there is a cross in the central cell.

The first method:

Consider all cells of large diagonals $($see Fig.$)$.

Each pair of adjacent coloured cells, together with the central cell, forms a triple, in which at least one nought must stand. Hence, in eight shaded cells there are at least four noughts. Therefore, in each of the $3\times1$ rectangles located on the edge of the board, there must be at least one nought. And so, in total there are no fewer than 8 noughts. But if there are crosses in the other cells, then three crosses “in a row” form crosses that are adjacent to the central one horizontally and vertically, which contradicts the condition. So, there are at least nine of them.

The second method:

Consider the 8 pairs of cells identified in the figure on the left. In one of the cells of each pair there should be at least one nought, so there are no less than eight noughts.

Suppose that in each selected pair of cells there is exactly one nought, and there are no noughts outside of these pairs. Then the remaining cells all contain crosses $($see figure in the center$)$. Consider the four triplets of cells highlighted in this figure. In each of them there are already two crosses, so there are noughts between them. Again, consider four of the eight previously identified pairs of cells $($Figure on the right$)$. We already found out that in each pair exactly one cell contains a zero, and the other contains a cross. If crosses are placed as such, then in the centre of the board there will be three triplets, going in a row. Consequently, this arrangement also does not satisfy the condition.

Answer

16 crosses.

Hint

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Solution

a) See problem number 64440.

b) We number the fighters from the weakest by the number 1 to the strongest by the number 100. Let $I_k$ be the indicator of the event “the k-th wrestler won both matches”. The probability of defeating an opponent in each fight for the k-th wrestler is equal to the probability that the wrestler’s opponent is one of those k-1 wrestlers who is weaker. Hence, the probability of winning is $\frac{k-1}{99}$. The probability of winning both times is $(\frac{k-1}{99})^2$, so $EI_k$ = $(\frac{k-1}{99})^2$.

The total number of winners, X, is equal to the sum of all the indicators, that is

Answer

a) 1; b) ≈33.5.

Hint

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Solution

Evaluation. Suppose that among the rational numbers there is a 0 and it is written on the top row of the table. Suppose that x irrational and 50 – x rational numbers are written along the left side of the table. Then along the upper side there are written 50 – x irrational and x rational numbers. We note that the product of a nonzero rational number and an irrational number is irrational. Hence, in the table there are at least x $(x – 1)$ + $(50 – x)^2$ = $2x^2$ – 101x + $50^2$ irrational numbers. The parabola’s vertex y = $2x^2 – 101x + 50^2$ is at the point 25,25, therefore the minimum value of this function at an integer point is reached at x = 25 and is equal to $25 \times 24 + 25^2 = 1225$.
If 0 is replaced by a non-zero rational number, then the number of irrational numbers can only increase. Therefore in the table in any case there are no more than 2500 – 1225 = 1275 rational numbers.

Example, when the rational number is 1275. Along the left side, we put the numbers 1, 2, …, 24, 25, $\sqrt{2}, 2\sqrt{2}, …, 25 \sqrt{2}$, and along the upper side we place the numbers 0, 26, 27, …, 49, $26\sqrt{2}, 27\sqrt{2}, …, 50 \sqrt{2}$. Then only $25 \times 24$ + $25^2$ will be irrational = 1225 products of nonzero rational and irrational numbers.

Answer

1275 products.

Hint

Solution

2017 = $43 \times 46$ + 39. Suppose that among Fred’s numbers there are 39 twos, 46 numbers equal to 44, and the rest are ones.
To achieve the required goal in 43 moves, Fred chooses 39 boxes, in which he always places 2 stones – 43 turns will mean there are $43\times2$ = 86 stones placed in these boxes. The remaining boxes he splits into 43 groups of 46 boxes; on the i-th move, he places 44 stones in all the boxes of the i-th group and one stone in the boxes of the remaining groups. Then after 43 moves in each box of each group there will be 44 + $42 \times 1$ = 86 stones, that is, in all boxes there will be an equal number of stones.
Let Fred make k $<$ 43 moves. Then in any box A there were 44 stones per move, and in it there will be no fewer than $44 + (k-1) \times 1 $ = 43 + k stones. On the other hand, since 46k $<$ 2017, in some box B none of the moves will be of 44 stones, that is, there will not be more than 2k stones in it. Since k $<$ 43, then 2k $<$ k + 43, which means that in the box B there are fewer stones than in A. Thus, Fred has not yet achieved the required goal.

Answer

He could.

Hint

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