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

To draw the perimeter of the 25×25 square we have no choice but to draw all 96 edge and corner squares. Divide the inner remaining 23×23 square into 264 ‘dominoes’ of size 2×1 and one 1×1 square. We need to draw one of the 1×1 squares in each ‘domino’ otherwise the line dividing the 2×1 square into two 1×1 squares will not be visible. Therefore in total we need to draw $96 + 264 = 360$ 1×1 squares.

From another perspective – 350 squares is sufficient. Consider a ‘chessboard’ shading of the 25×25 square, in which all of the corner 1×1 squares are black. Each line segment in the ‘grid’ we are trying to create is an edge of either a ‘border’ square – edge or corner squares – or a white square. Therefore it is sufficient to draw all of the border squares, of which there are 96, and all of the inside white squares, of which there are 264.

Answer

360 squares.

Hint

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Hint

Colour the blocks in black and white in an alternating order, like a chessboard. If the tourist passed two streets of one block, then he made a turn of 120$^{\circ}$.

Solution

The total area is equal to 15, therefore, the tourist passed through 14 streets connecting pairs of neighbouring squares. Colour the blocks in black and white in a chequered order. We end up with 10 black blocks, and each street is the boundary of exactly one black block. The tourist could not have passed through all three streets bordering the block. Hence, there are 14-10 = 4 black blocks, for which the tourist has walked on two streets. If the tourist has walked on two streets of one block, it is not difficult to see that he must have made a turn of 120$^{\circ}$ in one of the squares of this block. So, in no fewer than four black quarters, a turn of an angle of 120$^{\circ}$ is made.

Hint

Colour the blocks in black and white in an alternating order, like a chessboard. If the tourist passed two streets of one block, then he made a turn of 120$^{\circ}$.

Hint

Paint every over vertical of the board in black and the rest in white. Then the beetle will crawl from a black cell to a white, and vice versa.

Solution

Let’s paint every over vertical of the board in black and the rest in white. As a result, $5 \times 9$ = 45 cells $($5 verticals$)$ will be painted in black and only 36 cells in white. Note that from a black cell the beetle can only creep onto a white cell, and from a white cell only to a black one. Consequently, after the beetles crawled onto diagonally neighbouring cells, there were 36 beetles on 45 black cells. Hence, at least 9 black cells were not occupied.

Hint

Paint every over vertical of the board in black and the rest in white. Then the beetle will crawl from a black cell to a white, and vice versa.

Solution

The first way. Let our set contain an irrational number a. Then each of the remaining numbers has the form either p – a or p/a, where p is rational. We show that there are no more than 2 numbers of the form p – a. Let $b_1 = p_1 – a, b_2 = p_2 – a, b_3 = p_3 – a$, then the number $b_1 + b_2 = (p_1 + p_2) – 2a$ is not rational, then the number
$b_1b_2 = p_1p_2 – a (p_1 + p_2) + a^2$ is rational. Similarly, the numbers $b_1b_3, b_2b_3$ are rational. It follows that the subsequent numbers are rational: $A_3 = a^2 – a (p_1 + p_2), A_2 = a^2 – a (p_1 + p_3), A_1 = a^2 – a (p_2 + p_3)$ are rational. Hence, the number $A_3 – A_2 = a (p_3 – p_2)$ is rational, which is possible only for $p_3 = p_2$, that is, $b_3 = b_2$. This is a contradiction.
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Consequently, there are more than two numbers of the second kind, let be such numbers. The sum can be rational only for $q_2 = – q_1$. But $q_3 ≠ q_2$, which means that the sum is irrational. Then the number is rational, from where we get that $a^2$ is rational.
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The second way. Let’s consider six arbitrary numbers from our set. You can construct a graph yourself with six vertices corresponding to these numbers; vertices are connected by a blue edge if the sum of the corresponding numbers is rational, and a red edge if the product of the corresponding numbers is rational. In this graph, we can find a triangle of one colour $($see problem number 30815$)$. We can consider two cases.
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1) This triangle is blue, that is, there are three numbers x, y, z such that x + y, x + z, y + z are rational. Then the number $(x + y) + (x + z) – (y + z) = 2x$ is also rational. Similarly, the numbers 2y and 2z are rational. Consider any of the t remaining numbers. From the rationality of any of the numbers xt and x + t, the rationality of the number t follows $($all of the numbers are non-zero by hypothesis$)$. That is, all of the numbers in the set are rational.
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2) We found a red triangle, that is, there were three numbers x, y, z such that xy, xz, yz are rational. Then the number is rational. Similarly, the numbers $y^2$ and $z^2$ are rational. If at least one of the numbers x, y, z is rational, then, similarly to the previous case, we obtain the rationality of all of the numbers in the collection.
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Let where a is rational, m = ± 1. Since the number is rational, then , where c is rational. Consider any of the remaining numbers t. If xt or yt are rational, then analogous to the previous one, where d is rational, that is, $t^2$ is rational.
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If the numbers x + t and y + t are rational, then the number is rational. But this is not so. This is a contradiction.$\\$
That is, in any case, the squares of all of the numbers of the set are rational.$\\$
The statement of the problem is true for any number of numbers greater than 4.

Answer

See the solution above.

Hint