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2. [5] Baron Münchhausen came up with a theorem: if the polynomial $x^{n}-a x^{n-1}+b x^{n-2}+\ldots$ has $n$ natural roots, then on the plane there will be $a$ lines, which have exactly $b$ intersection points with each other. Is the baron mistaken?
(Fedor Ivlev) | Answer: it is not mistaken. Let the roots of the polynomial from the condition be the numbers $x_{1}, \ldots, x_{n}$. We choose $n$ different directions on the plane and take $x_{1}$ lines of the first direction, $x_{2}$ of the second, $\ldots$, $x_{n}$ of the $n$-th direction.
Then, by Vieta's formulas, the number of... | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,867 |
4. [7] At each of two round tables, there are $n$ gnomes sitting. Each gnome is friends only with their neighbors to the left and right at their table. A kind wizard wants to seat the gnomes at one round table so that every two neighboring gnomes are friends. He has the ability to make $2 n$ pairs of gnomes friends (th... | Answer: for all odd $n>1$.
Let $A_{1}, A_{2}, \ldots, A_{n}$ be the gnomes sitting at the first table, and $B_{1}, B_{2}, \ldots, B_{n}$ be the gnomes sitting at the second table.
Case of odd $n$. Let $n=2 k-1$. The good wizard's strategy: to befriend pairs of gnomes $\left(A_{i}, B_{i}\right)$ and $\left(A_{i}, B_{i... | foralloddn>1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,868 |
5. [7] Does there exist a rectangle that can be cut into 100 rectangles, all similar to it, but no two of which are the same?
(Mikhail Murashkin) | Answer: Yes, it exists.
To begin with, let's show that there exists a rectangle that can be cut into 4 similar rectangles of different sizes. The diagrams below show that for any $x$, there exists a rectangle that can be cut into 4 rectangles with the ratio of sides equal to $x$ (two methods):
1)
.
Proof of the lemma. We will prove this by induction on $k$. For $k=1$, the statement is obvious. Suppose the statement is true for $k=l-1$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,870 |
1. [3] For what largest natural number $m$ will the number $m! \cdot 2022!$ be a factorial of a natural number?
(B. Frenkin) | Answer: when $m=2022!-1$.
Solution. (2022! - 1)!$\cdot$2022! = (2022!)!. If $m \geq 2022!$, then $m!<m!\cdot 2022!<(m+1)!$. | 2022!-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,871 |
3. [5] On a line, 2022 points are marked such that every two adjacent points are at the same distance from each other. Half of the points are painted red, and the other half are painted blue. Can the sum of the lengths of all possible segments where the left endpoint is red and the right endpoint is blue equal the sum ... | Answer: It cannot. Solution. We can consider the marked points to be integers from 1 to 2022. It is sufficient to show that the sum $S$ of the lengths of all segments with endpoints of different colors is odd.
Method 1. Let the number of red-even points be $x$, then the number of red-odd and blue-even points is $y=101... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,873 |
4. [5] Given an acute scalene triangle. In one move, it is allowed to cut one of the existing triangles along a median into two triangles. Can all triangles become isosceles after several moves?
(E. Bakayev) | Answer: they cannot. Solution. We will prove that among the existing triangles, there will always be an isosceles non-right triangle. Initially, this is true.
Suppose that on a certain move, the existing non-isosceles non-right triangle $ABC$ was cut along the median $BM$. Since it is non-isosceles, the median $BM$ do... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,874 |
4. а) [3] Can a square be cut into 4 isosceles triangles, none of which are equal? б) [3] Can an equilateral triangle be cut into 4 isosceles triangles, none of which are equal?
Vladimir Rastorguev
Answers: yes in both parts. | Solution. See figures. On the left figure, we first draw the angle bisector $A K$ of angle $B A C$, and then reflect point $B$ over $A K$ to obtain point $E$.
 | yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,875 |
5. On a checkered board, dominoes are placed without touching even at the corners. Each domino occupies two adjacent (by side) cells of the board. The lower left and upper right cells of the board are free. Can you always walk from the lower left cell to the upper right cell, making moves only up and to the right to ad... | Solution. The figure in the upper right shows the placement of dominoes on a $6 \times 7$ board, which does not allow a path from the bottom-left cell to the top-right cell. Indeed, it is impossible to enter the (gray) area to the right of the lowest domino, as we must first move above the first domino, and then we are... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,876 |
1. [5] On the computer screen, a certain natural number divisible by 7 is printed, and a cursor marks a gap between some two of its adjacent digits. Prove that there exists such a digit that if it is typed into the marked gap any number of times, the resulting number will be divisible by 7.
A. Galochkin | Solution. Let the original number be of the form $\overline{A B}$, where $A$ gives a remainder $r$ when divided by 7. Take such a digit $a$ that $2 r+a$ is divisible by 7 (it obviously exists). We will divide a number of the form $\overline{A a \ldots . . a B}$ by 7 in a column. When we finish dividing $A$, the remaind... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,877 |
3. [7] Two non-intersecting wooden circles of not necessarily the same size - a gray one and a black one - are glued to a plane. An infinite wooden angle is given, one side of which is gray and the other is black. It is moved so that the circles are outside the angle, with the gray side touching the gray circle and the... | Solution. The desired ray is the geometric locus of points inside the angle, for which the ratio of distances to the gray and black sides is equal to the ratio $\frac{r_{1}}{r_{2}}$ of the radii of the gray and black circles. Further reasoning almost repeats the solution of problem 3 for younger classes. It is clear th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,878 |
5. The orthogonal projection of a tetrahedron onto the plane of one of its faces is a trapezoid with an area of 1. a) [4] Can the orthogonal projection of this tetrahedron onto the plane of another of its faces be a square with an area of 1? b) [4] And a square with an area of $1 / 2019$? | Answer: a) cannot; b) can. Solution. Let the unit square $ABCD$ be the projection of the tetrahedron $ABCD'$ onto the plane of the face $ABC$. Then this tetrahedron is "inscribed" in the rectangular parallelepiped $ABCDAB'C'D'$.
By symmetry with respect to the plane $DBD'$, it is clear that the projection of the tetra... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,879 |
6. [8] Petya and Vasya are playing a game. For each set of five different variables from the set $x_{1}, \ldots, x_{10}$, there is a unique card on which their product is written. Petya and Vasya take turns picking a card, with Petya starting first. According to the rules of the game, when all the cards are taken, Vasy... | Answer: Yes. Solution. We will show how Vasya can act to make the sum of the products on his cards greater than Petya's.
Suppose Petya did not take the card with $x_{6} x_{7} \chi_{8} x_{9} x_{10}$. Then Vasya can take this card and then take any cards. With $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0, x_{6}=x_{7}=x_{8}=x_{9}=x_... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,880 |
7. [12] Consider broken lines on a grid plane starting at the point $(0,0)$ and having vertices at integer points, such that each subsequent segment goes along the sides of the cells either upwards or to the right. Each such broken line corresponds to a worm - a figure consisting of cells of the plane that have at leas... | Solution. Different broken lines correspond to different worms: if two broken lines coincide up to point $A$, but diverge at it, then the corresponding worms contain different cells (Fig. 1): blue, if the move from $A$ is to the right, red - if upwards, none of them, if the broken line ends at $A$.
We will cover the w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,881 |
1. Given a triangular pyramid $S A B C$, the base of which is an equilateral triangle $A B C$, and all plane angles at vertex $S$ are equal to $\alpha$. For what least $\alpha$ can we assert that this pyramid is regular?
M. Malkin | Answer: $60^{\circ}$.
We will prove that when $\alpha=60^{\circ}$, the pyramid is regular. Let the sides of the triangle $ABC$ be equal to 1. Note that in any triangle with an angle of $60^{\circ}$, the side opposite this angle is the middle-length side (and if it is strictly less than one of the sides, it is strictly... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,882 |
2. Does there exist an integer $n>1$ satisfying the inequality
$$
[\sqrt{n-2}+2 \sqrt{n+2}]<[\sqrt{9 n+6}] ?
$$
(Here $[x]$ denotes the integer part of the number $x$, that is, the greatest integer not exceeding $x$.)
M. Malkin | Answer: No.
Suppose an integer $n>1$ satisfies this inequality. We have $[\sqrt{9 n+6}]^{2} \leqslant 9 n+6$, but the square of an integer cannot give a remainder of 6 or 5 when divided by 9, so $[\sqrt{9 n+6}]^{2} \leqslant 9 n+4$, which means $[\sqrt{9 n+6}] \leqslant[\sqrt{9 n+4}]$. Then the original inequality imp... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,883 |
5. Let's call the arrangement of $N$ grasshoppers on a line at various points of it $k$-successful if, after making the necessary number of moves according to the rules of leapfrog, the grasshoppers can achieve that the sum of pairwise distances between them decreases by at least $k$ times. For which $N \geqslant 2$ do... | Answer: for $N \geqslant 3$.
First solution. For $N=2$, no matter how the grasshoppers jump, the distance between them does not change.
Let $N \geqslant 3$. Place the 1st, 2nd, and 3rd grasshoppers on a line at points with coordinates $0,1, \sqrt{2}$, and call these grasshoppers $V, Q, R$. The others are seated arbit... | N\geqslant3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,884 |
6. A row of $N$ boxes, numbered consecutively $1,2, \ldots, N$, stands from left to right. Some of these boxes, standing in a row, will have a ball placed in them, leaving the others empty.
The instruction consists of a sequence of commands of the form "swap the contents of box № $i$ and box № $j$", where $i$ and $j$ ... | Answer: Yes.
Let's assume all the balls are blue. We will put a red ball in the empty boxes. Now there are no empty boxes. We will show an even stronger statement: that for any $N$ there is an instruction not longer than $3 N$ with the following property.
Suppose in $N$ boxes, arranged in a row, there are red and blu... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,885 |
1. Let's define the complexity of an integer $n>1$ as the number of prime factors in its prime factorization. For which $n$ do all numbers between $n$ and $2n$ have a complexity
a) [2] no greater than that of $n$; b) [2] less than that of $n$?
(Boris Frenkin) | Answer. a) For $n=2^{k}$; b) there are no such numbers. Solution. a) Obviously, $2^{k}$ is the smallest number of complexity $k$. Therefore, all numbers between $2^{k}$ and $2^{k+1}$ have a complexity no greater than $k$. Let $n$ be not a power of two. Then there is a power of two between $n$ and $2n$ (we can take the ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,886 |
2. [7] Two acute triangles $A B C$ and $A_{1} B_{1} C_{1}$ are such that points $B_{1}$ and $C_{1}$ lie on side $B C$, and point $A_{1}$ lies inside triangle $A B C$. Let $S$ and $S_{1}$ be the areas of these triangles, respectively. Prove that $\frac{S}{A B + A C} > \frac{S_{1}}{A_{1} B_{1} + A_{1} C_{1}}$.
(Nairi Se... | Solution. Let points $D$ and $D_{1}$ be symmetric to points $A$ and $A_{1}$ with respect to $B C$. Draw the angle bisectors $A K$ and $A_{1} K_{1}$ of our triangles. Note that $K$ and $K_{1}$ are the centers of the circles inscribed in quadrilaterals $A B D C$ and $A_{1} B_{1} D_{1} C_{1}$, and the required inequality ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,887 |
3. [7] There are 100 indistinguishable coins of three types: gold, silver, and copper (each type appears at least once). It is known that gold coins weigh 3 g each, silver coins weigh 2 g each, and copper coins weigh 1 g each. How can you determine the type of all the coins using a balance scale without weights in no m... | Solution 1. (A. Shapovalov) Let's call a situation a winning one if no more than $k+1$ weighings have been conducted and the weights of $k$ coins have been determined, with at least one being silver or two being copper. In a winning situation, by comparing an unknown coin with a weight of 2 g, we can determine its weig... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,888 |
4. [7] From the center $O$ of the circumscribed circle of triangle $ABC$, perpendiculars $OP$ and $OQ$ are dropped to the bisectors of the internal and external angles at vertex $B$. Prove that the line $PQ$ bisects the segment connecting the midpoints of sides $CB$ and $AB$. (Artemiy Sokolov) | Solution. Perform a homothety with center $B$ and coefficient 2. Point $O$ will map to point $D$, diametrically opposite vertex $B$ on the circumcircle $\Omega$, point $P$ - to point $R$ where the angle bisector of $\angle B$ intersects $\Omega$, point $Q$ - to the point $S$ diametrically opposite $R$, "the segment con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,889 |
5. [8] Let's call a pair ( $m, n$ ) of distinct natural numbers $m$ and $n$ good if $mn$ and $(m+1)(n+1)$ are perfect squares. Prove that for each natural number $m$ there exists at least one $n > m$ such that the pair $(m, n)$ is good.
^{2}\right)$ is good. Indeed,
$(m+1)\left(m(4 m+3)^{2}+1\right)=(m+1)\left(16 m^{3}+24 m^{2}+9 m+1\right)=(m+1)^{2}\left(16 m^{2}+8 m+1\right)=((m+1)(4 m+1))^{2}$.
Path to the solution. It is natural to try to find such an $n$ that it is a square multiplied by $m$, and at the same... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,890 |
6. [9] Petya had several hundred-ruble bills, and no other money. Petya started buying books (each book costs an integer number of rubles) and receiving change in small denominations (1-ruble coins). When buying an expensive book (not cheaper than 100 rubles), Petya paid only with hundred-ruble bills (the minimum neces... | Solution 1. Let's see how much small change Petya could have received.
Consider the very last cheap purchase that increased the amount of small change. Let the cost of this purchase be $x$, then before this, there was no more than $x-1$ rubles of small change, and thus, after this, it would become no more than $x-1+10... | 4950<5000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,891 |
7. [10] In a checkered wooden square, 102 cells are painted black. Petya, using the square as a stamp, applied it 100 times to a white sheet of paper, and each time these 102 cells (and only they) left a black mark on the paper. Could it be that in the end, a $101 \times 101$ square was formed on the sheet, all cells o... | Answer. Can. Solution. We will show that any square $(2 N+1) \times(2 N+1)$ without a corner cell can be obtained by applying a stamp of $2 N+2$ cells, $2 N$ times. For illustration, we will provide a diagram for $N=4$.
A square without the top right corner can be represented as a $2 N \times 2 N$ square with two stri... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,892 |
1. [4] The magician lays out a deck of 52 cards in a row and announces that 51 of them will be thrown off the table, leaving only the three of clubs. On each step, the spectator says which card to discard by its position from the edge, and the magician chooses whether to count from the left or the right edge and discar... | Answer: at the extreme positions. Solution. The three of clubs will have to be discarded only if it ends up in the center of the row at some point; otherwise, another card can be discarded. Since the row always contains more than one card, an end card can be kept until the end.
Let the three of clubs $T$ initially not... | attheextremepositions | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,894 |
2. [4] Given a circle $\omega$ with center $O$ and two distinct points $A$ and $C$ on it. For any other point $P$ on $\omega$, mark the midpoints $X$ and $Y$ of segments $AP$ and $CP$, and construct the point $H$ as the orthocenter of triangle $OXY$. Prove that the position of point $H$ does not depend on the choice of... | Solution. Since $Y H \perp O X \perp A P$, then $Y H \| A P$, and the line $Y H$ contains the midline of triangle APC. Similarly, the line $X H$ contains the midline of this triangle. These midlines intersect at point $H$ - the midpoint of side $A C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,895 |
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles required to rearrange the chips in reverse order?
(Egor Bakaev) | Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells i... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,896 |
4. [5] Given integers $a_{1}, \ldots, a_{1000}$. Their squares $a_{1}^{2}, \ldots, a_{1000}^{2}$ are written in a circle. The sum of any 41 consecutive squares on the circle is divisible by $41^{2}$. Is it true that each of the numbers $a_{1}, \ldots, a_{1000}$ is divisible by 41?
(Boris Frenkin) | Answer: Correct. Solution. From the condition, it follows that $a_{k+41}^{2} \equiv a_{k}^{2}\left(\bmod 41^{2}\right)$ (indices are considered cyclic, that is, 1 follows 1000). Therefore, $a_{k+41 n}^{2} \equiv a_{k}^{2}\left(\bmod 41^{2}\right)$ for any $n$. Since the numbers 41 and 1000 are coprime, the squares of a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,897 |
5. [5] Vasya has an unlimited supply of $1 \times 1 \times 3$ blocks and L-shaped corners made of three $1 \times 1 \times 1$ cubes. Vasya completely filled a box $m \times n \times k$, where $m, n$, and $k$ are integers greater than 1. Prove that it was possible to use only the L-shaped corners. | Solution. Since $m \times k$ is divisible by 3, one of the factors is divisible by 3; let this be the height $k$. It is sufficient to fill the box $m \times n \times 3$. From two corners, we can form a brick $1 \times 2 \times 3$. If $n$ is even, the base of the box can be divided into dominoes $2 \times 1$ and place a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,898 |
1. [5] The king summoned two sages and announced a task to them: the first sage thinks of seven different natural numbers that sum up to 100, secretly informs them to the king, and tells the second sage only the fourth largest of these numbers. Then, the second sage must guess the numbers. The sages have no opportunity... | Answer: they can. Solution. Let the first sage think of the numbers $1,2,3,22,23,24,25$ and name the number 22. Then the second can uniquely determine all the numbers, as the sum of 100 can be achieved in only one way in this case - by taking the smallest possible numbers: $1,2,3,22,23,24$ and 25.
Remark: if the first... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,899 |
2. [7] On a line, there are 2019 grasshoppers sitting. In one move, one of the grasshoppers jumps over another so that it ends up at the same distance from the other. By jumping only to the right, the grasshoppers can achieve that some two of them are exactly 1 mm apart. Prove that the grasshoppers can achieve the same... | Solution. Let's call the leftmost grasshopper Richard. Then let all the grasshoppers, except for Richard, jump over Richard. It is clear that now the grasshoppers are in a position that is symmetrical to the initial one. Then they can, using moves that are symmetrical to those they would make when jumping to the right,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,900 |
3. [7] Two non-intersecting wooden circles of the same size, one gray and one black, are glued to a plane. A wooden triangle is given, one side of which is gray and the other is black. It is moved so that the circles are outside the triangle, with the gray side touching the gray circle and the black side touching the b... | Solution. The points on the bisector of angle $A$ between the gray and black sides of the wooden triangle are equidistant from these sides (the gray color is represented by blue). Draw lines through the centers $O_{1}$ and $O_{2}$ of the gray and black circles, parallel to these sides. Let these lines intersect at poin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,901 |
4. [8] Each segment with endpoints at the vertices of a regular 100-gon was painted - red if there is an even number of vertices between its endpoints, and blue otherwise (in particular, all sides of the 100-gon are red). Numbers were placed at the vertices, the sum of whose squares is 1, and on the segments - the prod... | Answer: $1 / 2$. Solution: Let the numbers $x_{1}, \ldots, x_{100}$ be arranged in the vertices in a circle, and let $k$ be the sum of the "red" numbers, and $s$ be the "blue" ones. Then the sum of the red numbers is the sum of all monomials of the form $x_{i} x_{j}$, where $i$ and $j$ have different parity and $i<j$, ... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,902 |
6. [9] Inside an isosceles triangle $A B C$, a point $K$ is marked such that $C K=A B=B C$ and $\angle K A C=30^{\circ}$. Find the angle $A K B$.
E. Bakayev | Answer: $150^{\circ}$. Solution 1. Construct an equilateral triangle $B C L$ (see the figure; points $A$ and $L$ are on the same side of the line $B C$). Points $A, C$, and $L$ lie on a circle with radius $B A$ and center at point $B$. Since $K$ lies inside triangle $A B C$, angle $A B C$ is greater than $60^{\circ}$, ... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,904 |
7. [12] There are 100 piles, each with 400 stones. In one move, Petya selects two piles, removes one stone from each, and earns as many points as the absolute difference in the number of stones in these two piles. Petya must remove all the stones. What is the maximum total number of points he can earn?
M. Diden | Answer: 3920000. Solution. Estimation. We will assume that the stones in the piles are stacked on top of each other, and Petya takes the top (at the moment) stones from the selected piles. We will number the stones in each pile from bottom to top with numbers from 1 to 400. Then the number of points Petya gets on each ... | 3920000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,905 |
1. [4] The distance from a certain point inside a regular hexagon to three of its consecutive vertices is 1, 1, and 2, respectively. What is the side length of this hexagon?
M. Evdokimov | Answer: $\sqrt{3}$.
Let $A, B, C$ be consecutive vertices of a hexagon, $O$ be a point inside it, and let $O A=O B=1, O C=2$.
Solution 1. Consider the other neighboring vertex $F$ of $A$. Then $F A B C$ is an isosceles trapezoid (see figure). The point $O$ lies on the common perpendicular bisector of its bases $F C$ ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,906 |
2. [4] Natural numbers $a$ and $b$ are such that $a^{n+1}+b^{n+1}$ is divisible by $a^{n}+b^{n}$ for an infinite set of different natural $n$. Is it necessarily true that $a=b$?
B. Frenkin | Answer: necessarily.
Solution 1. Let, for example, $a>b$. The fraction $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ is less than $a$ for all natural $n$ (this is obvious after multiplying by the denominator), but it tends to $a$ as $n \rightarrow \infty$ (indeed, dividing the numerator and the denominator by $a^{n}$ and noti... | b | Number Theory | proof | Yes | Yes | olympiads | false | 8,907 |
4. [5] A magician and an assistant perform a trick. In a row, there are 13 closed empty boxes. The magician leaves, and a spectator hides a coin in any two boxes of their choice, in full view of the assistant. Then the magician returns. The assistant opens one box that does not contain a coin. Next, the magician points... | Solution. Let's mentally arrange the boxes in a circle, representing them as points dividing the circle into 13 equal arcs of length 1, and express distances between the boxes in terms of these arcs. Between any two boxes on one side of the circle, there are no more than 6 arcs of length 1. Therefore, it is sufficient ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,909 |
5. [5] Several natural numbers are written in a row with a sum of 2019. No number and no sum of several consecutive numbers equals 40. What is the maximum number of numbers that could have been written?
A. Shapovalov | Answer: 1019 numbers.
Lemma. The sum of any forty consecutive numbers is not less than 80.
Proof. Let the numbers $a_{1}, \ldots, a_{40}$ be written consecutively. Among the numbers $b_{0}=0, b_{1}=a_{1}$, $b_{2}=a_{1}+a_{2}, \ldots, b_{40}=a_{1}+a_{2}+\ldots+a_{40}$, there will be two $b_{i}$ and $b_{j}(i1019$ numbe... | 1019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,910 |
2. [4] Find all natural numbers n that satisfy the following condition: the numbers 1, 2, 3, ..., 2n can be paired in such a way that if the numbers in each pair are added and the results are multiplied, the product is a square of a natural number.
(Folklore) | Answer: all $n>1$. Solution. $1+2$ is not a square. Let $n>1$.
1st method. We will divide these numbers into groups of four consecutive numbers, and, if necessary, a group of six initial numbers. From the groups of four, we form $(a+(a+3))((a+1)+(a+2))=(2 a+3)^{2}$, and from the group of six - $(1+5)(2+4)(3+6)=18^{2}$... | n>1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,911 |
4. [5] Nastya has five identical-looking coins, among which three are genuine and weigh the same, and two are counterfeit: one is heavier than a genuine coin, and the other is equally lighter than a genuine coin. At Nastya's request, an expert will perform three weighings on a two-pan balance without weights, as she in... | Answer. Yes. The first method. Let's denote the coins by the letters $a, b, c, d, e$. Nastya will ask to conduct the weighings: $a ? b, c ? d, a b ? c d$. With accuracy up to symmetry, there are four possible outcomes.
1) $a>b, c>d, a b>c d$. Then $a$ is the heavy one, $d$ is the light one.
2) $a=b, c>d, a b=c d$. The... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,912 |
1. On the first day, $2^{n}$ schoolchildren played ping-pong in a "knockout" format: first, two played, then the winner played with the third, the winner of this pair played with the fourth, and so on until the last schoolchild (there are no draws in ping-pong). On the second day, the same schoolchildren played for the... | Answer: 3.
Solution. Let's construct the following graph: vertices are players, edges are played matches. According to the condition, for both tournaments, this graph is the same. Consider the first tournament and select the matches where the winners had not won before (for example, the first match). Then the correspo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,913 |
2. A sphere touches 99 edges of a certain convex 50-sided pyramid. Does it necessarily touch the 100th edge of this pyramid as well?
M. Evdokimov | Answer: No.
Solution. We will show that this is false in either of the two cases: (a) the hundredth edge is an edge of the base, and (b) the hundredth edge is a lateral edge.
(a) Consider a regular 51-gon $A_{1} A_{2} \ldots A_{51}$ with center $C$. Let $\omega$ and $\Omega$ be its inscribed and circumscribed circles... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,914 |
3. For positive numbers $x_{1}, \ldots, x_{n}$, prove the inequality:
$$
\left(\frac{x_{1}}{x_{2}}\right)^{4}+\left(\frac{x_{2}}{x_{3}}\right)^{4}+\ldots+\left(\frac{x_{n}}{x_{1}}\right)^{4} \geq \frac{x_{1}}{x_{5}}+\frac{x_{2}}{x_{6}}+\ldots+\frac{x_{n-3}}{x_{1}}+\frac{x_{n-2}}{x_{2}}+\frac{x_{n-1}}{x_{3}}+\frac{x_{n... | Solution. Let $x_{n+k}=x_{k}$. Then for each $k=1,2, \ldots, n$ by the Cauchy mean inequality we have
$$
\left(\frac{x_{k}}{x_{k+1}}\right)^{4}+\left(\frac{x_{k+1}}{x_{k+2}}\right)^{4}+\left(\frac{x_{k+2}}{x_{k+3}}\right)^{4}+\left(\frac{x_{k+3}}{x_{k+4}}\right)^{4} \geq 4 \cdot \frac{x_{k}}{x_{k+1}} \cdot \frac{x_{k+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,915 |
4. The cells of a $100 \times 100$ board are painted in black and white in a checkerboard pattern. Is it possible to repaint exactly 2018 different cells of this board to the opposite color so that in each row and each column there is the same number of black cells?
Y. Chekanov | Answer: No.
Solution 1. It is clear that the rows of the board can be rearranged in any way, and the columns can be rearranged as well. Let's move all the odd columns to the left and all the odd rows to the bottom. As a result, from the initial board, we get a board divided into 4 identical squares - two of them are w... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,916 |
5. Given triangle $XBC$. Different points $A_{H}, A_{I}, A_{M}$ are such that $X$ is the orthocenter of triangle $A_{H}BC$, the incenter of triangle $A_{I}BC$, and the centroid of triangle $A_{M}BC$. Prove that if $A_{H}A_{M}$ and $BC$ are parallel, then $A_{I}$ is the midpoint of $A_{H}A_{M}$. | Solution. Let $M$ be the midpoint of $BC$, and $Y$ and $Z$ be the points symmetric to $X$ with respect to the line $BC$ and point $M$, respectively. Then $\angle B Z C = \angle B Y C = \angle B X C = 180^{\circ} - \angle B A_{H} C$, so the points $A_{H}, B, C, Y$, and $Z$ lie on the same circle $\Omega$ with center $O$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,917 |
6. For which natural numbers $n$ is the following statement true: for any polynomial $P$ of degree $n$ with integer coefficients, there exist such different natural numbers $a$ and $b$ for which $P(a) + P(b)$ is divisible by $a + b$?
G. Zhukov | Answer. For all even $n$.
Solution. Odd $n$ do not work. Indeed, consider the polynomial $P(x)=x^{n}+1$ and different natural numbers $a, b$. Since $n$ is odd, $a^{n}+b^{n}$ is divisible by $a+b$, and then $P(a)+P(b)=$ $\left(a^{n}+b^{n}\right)+2$ is not divisible, since $a+b>2$.
It remains to prove that all even $n$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,918 |
1. The hostess baked a square cake and cut several pieces from it. The first cut was made parallel to the side of the original square from edge to edge. The next cut was made in the remaining part from edge to edge perpendicular to the previous cut, and so on (several times). All the cut pieces have equal area. Can the... | Answer. No, it cannot. Solution. The part left after each cutting will be called the remainder. The length of the remainder will be the size of the side along which it is cut, and the width will be the size of the other side. The length of the cut rectangle (piece) will also be considered the size of the side along whi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,919 |
2. Let $X$ be a fixed point on side $A C$ of triangle $A B C$ ($X$ is different from $A$ and $C$). An arbitrary circle passing through $X$ and $B$ intersects segment $A C$ and the circumcircle of triangle $A B C$ at points $P$ and $Q$, different from $X$ and $B$. Prove that all possible lines $P Q$ pass through one poi... | Solution. Let the second intersection point of $P Q$ and the circumcircle $(A B C)$ be denoted as $S$.
Then $\angle(B X, X C)=\angle(B X, X P)=\angle(B Q, Q P)=\angle(B Q, Q S)=$ const (equality in directed angles). This means that the angle $(B Q, Q S)$, subtending the arc $B S$ of the circumcircle $(A B C)$, is cons... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,920 |
3. 16 cards with integers from 1 to 16 are laid face down in a $4 \times 4$ table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the cards are arra... | Answer. Eight cards
## Solution.
Evaluation. Let's number the cells as shown in Figure 1.
Notice that one of the cells numbered 1 must be open, otherwise the red and blue ways of filling the table in Figure 2 would be indistinguishable. One of the cells numbered 2 must also be open, otherwise the red and blue ways o... | Eight\cards | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,921 |
4. There is a 1001-digit natural number $A$. The 1001-digit number $Z$ is the same number $A$, written from end to beginning (for example, for four-digit numbers, these could be 7432 and 2347). It is known that $A > Z$. For which $A$ will the quotient $A / Z$ be the smallest (but strictly greater than 1)? | Answer. For $A$, whose notation (from left to right) is: 501 nines, an eight, and 499 nines.
Solution 1. Let $A=\overline{a_{1000} a_{999 \ldots a_{0}}}$. Since $A>Z$, among the digits $a_{0}, a_{1}, \ldots, a_{499}$ there is at least one non-nine. Therefore, $Z \leqslant Z_{0}=\underbrace{99 \ldots 9}_{499} 8 \underb... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,922 |
5. Is it possible to arrange five spheres in space so that for each sphere, a tangent plane can be drawn through its center to the other four spheres? The spheres may intersect and are not required to have the same radius. | Answer: Yes, it is possible.
Solution 1. Take in the horizontal plane $\alpha$ an equilateral triangle with height 2. Let $J-$ be the center of one of its exscribed circles, and $A, B, C$ be the midpoints of its sides.
Choose such spheres: three with centers at $A, B, C$ and radius 1; two with radius 2 and centers at... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,923 |
6. Given a natural number $n>1$. Which is greater: the number of ways to cut a $3 n \times 3 n$ grid square into $1 \times 3$ grid rectangles or the number of ways to cut a $2 n \times 2 n$ grid square into $1 \times 2$ grid rectangles? | Answer. More tilings with triominoes.
Solution. We will provide a construction of a mapping that associates each tiling of a $2n \times 2n$ board with dominoes to a tiling of a $3n \times 3n$ board with triominoes.
Suppose we have some tiling of a $2n \times 2n$ board with dominoes. Number all the vertical lines from... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,924 |
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev) | Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells i... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,925 |
1. a) [2] A convex pentagon is divided into three triangles by non-intersecting diagonals. Can the points of intersection of the medians of these triangles lie on one straight line?
б) [2] The same question for a non-convex pentagon.
Alexander Grybalko
Answers: a) no; b) yes. | Solution. It is clear that exactly two diagonals have been drawn, and they both emanate from one vertex (let it be $A$). Then the specified points of intersection of the medians are obtained by a homothety with center $A$ and coefficient $2 / 3$ from the midpoints of the sides $B C, C D$, and $D E$.
a) These midpoints... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,926 |
2. a) [2] Tanya has 4 visually identical weights with masses of 1000, 1002, 1004, and 1005 grams (it is unknown which is which), and a balance scale (indicating which of the two pans is heavier or if they are equal). Can Tanya guarantee to determine which weight is which in 4 weighings? (The next weighing is chosen bas... | Solution. No matter how Tanya places the weights on the scales, they will never show balance. Therefore, each weighing divides the set of suspicious permutations into no more than two parts. Initially, there were 24 suspicious permutations; after the first weighing, in the "unsuccessful" outcome, there will be no fewer... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,927 |
3. [5] For which natural numbers $n$ can one find $n$ consecutive natural numbers whose product is equal to the sum of (possibly different) $n$ consecutive natural numbers?
Boris Frenkin | Answer: for odd $n$.
Solution. The product of $n$ consecutive integers is divisible by $n$, so the sum of these integers is also divisible by $n$. Therefore, the arithmetic mean of these numbers is an integer. This means that $n$ is odd. Here is an example for $n=2 m+1:((2 m)!-m)+((2 m)!-m+1)+\ldots+((2 m)!+m)=(2 m+1)... | for\odd\n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,928 |
4. [5] As is known, a quadratic equation has no more than two roots. Can the equation $\left[x^{2}\right]+p x+q=0$ for $p \neq 0$ have more than 100 roots? ([ $\left.x^{2}\right]$ denotes the greatest integer not exceeding $x^{2}$.)
Alexei Tolpygo | Answer: can.
Solution. Consider, for example, the equation $\left[x^{2}\right]-100 x+2500=0$. It has 199 roots of the form $50+\frac{k}{100}$ (where $k=-99,-98, \ldots, 99$). Indeed,
$$
\left[\left(50+\frac{k}{100}\right)^{2}\right]=\left[2500+k+\left(\frac{k}{100}\right)^{2}\right]=2500+k=100 \cdot\left(50+\frac{k}{... | can | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,929 |
5. [6] Let $O$ be the center of the circumcircle of an acute-angled triangle $ABC$, and let point $M$ be the midpoint of side $AC$. Line $BO$ intersects the altitudes $AA_1$ and $CC_1$ at points $H_a$ and $H_c$ respectively. The circumcircles of triangles $BH_aA$ and $BH_cC$ intersect again at point $K$. Prove that $K$... | Solution 1. Let $B'$ be the point symmetric to point $B$ with respect to point $M$, and the circumcircle of triangle $ACB'$ intersects the median $BM$ at point $K$. Then the external angle $\angle AK'B$ of triangle $AKB$ equals $\angle ACB' = \angle A$ (see the left figure below). However, the external angle $\angle B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,930 |
1. [5] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 inhabitants were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The inhabitants answered in turn, and everyone could hear their answers. Knights answere... | Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadir) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values ... | 1009 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,931 |
2. [7] In an acute scalene triangle $ABC$ with the circumcenter $O$, altitudes $AH_{a}$ and $BH_{b}$ are drawn. Points $X$ and $Y$ are symmetric to points $H_{a}$ and $H_{b}$ with respect to the midpoints of sides $BC$ and $CA$, respectively. Prove that the line $CO$ bisects the segment $XY$.
(Folklore) | Solution. Note that $C X: C Y=B H_{a}: A H_{b}=\cos \angle B: \cos \angle A$.
Moreover, $\sin \angle O C X=\cos \angle A, \sin \angle O C Y=\cos \angle B$. Therefore, $Z X: Z Y=S_{C Z X}: S_{C Z Y}=C X \sin \angle O C X: C Y \sin \angle O C Y=1: 1$, Q.E.D.
![](https://cdn.mathpix.com/cropped/2024_05_06_034089706a976f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,932 |
4. [8] Initially, a finite number of cells on a white checkered plane are colored black. On the plane lies a checkered paper polygon M, which contains more than one cell. It can be moved in any direction by any distance without rotating, but so that after the move it lies "by cells". If after a certain move exactly one... | Solution 1. Instead of cells, we will consider their centers, which we will call nodes. Draw a vertical line through the leftmost node of the paper figure. If this line does not contain other nodes of the figure, we will rotate the line clockwise until it hits another node of the figure. We have found such a line $l$, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,933 |
5. [8] The three medians of a triangle divide its angle into six angles, of which exactly $k$ are greater than $30^{\circ}$. What is the greatest possible value of $k$?
(N. Sedrakyan) | Answer. $k=3$.
Evaluation. First method. Let $h_{1} \leq h_{2} \leq h_{3}$ be the heights of the triangle, and $m_{1}, m_{2}, m_{3}$ be the medians from the corresponding vertices (in fact, $m_{1} \leq m_{2} \leq m_{3}$, but this will not be used). Then we have $m_{3} \geq h_{i}$ for $i=1,2$, 3. Dropping perpendicular... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,934 |
1. [3] Several natural numbers are written in a row with a sum of 20. No number and no sum of several consecutive numbers equals 3. Could there be more than 10 numbers written?
A. Shapovalov | Answer: could.
Solution. An example with 11 numbers: $1,1,4,1,1,4,1,1,4,1,1$. It can be proven that more than 11 numbers could not have been written down. See also the more general problem 5 for higher grades. | could | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,936 |
2. [4] There are $2 n+1$ coins lying in a circle, all heads up. Moving clockwise, they make $2 n+1$ flips: they flip some coin, skip one coin and flip the next, skip two coins and flip the next, skip three coins and flip the next, and so on, finally skip $2 n$ coins and flip the next. Prove that now exactly one coin li... | Solution. Let the $(n-1)$-th flipped coin be $-X$, and the $n$-th be $Y$. Then, between $X$ and $Y$ clockwise, there are $n-1$ coins, and since there are a total of $2n+1$ coins in the circle, there are $n$ coins between $Y$ and $X$ clockwise (see figure). This means that the $(n+1)$-th coin we will flip again is $X$.
... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,937 |
3. [4] The product of natural numbers $m$ and $n$ is divisible by their sum. Prove that $m+n \leq n^{2}$.
B. Frenkin | Solution 1. Since $n^{2}=n(m+n)-m n$, it follows from the condition that $n^{2}$ is divisible by $m+n$. Therefore, $n^{2} \geq m+n$.
Solution 2. Let $d=$ GCD $(m, n), m=a d, n=b d$. According to the condition, $a b d^{2}$ is divisible by $d(a+b)$, which means $a b d$ is divisible by $a+b$. However, since the numbers $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,938 |
4. [5] In a rectangle $A B C D$, isosceles triangles with a given angle $\alpha$ at the vertex opposite the base are inscribed such that this vertex lies on the segment $B C$, and the ends of the base lie on the segments $A B$ and $C D$. Prove that the midpoints of the bases of all such triangles coincide. | Solution. Let $K L M$ be one of such triangles, $O$ - the midpoint of its base $K M$ (see figure).
Then $L O$ is the median, and therefore, also the bisector and altitude of triangle $K L M$. Since angles $K B L$ and $L O K$ are right angles, points $B$ and $O$ lie on the circle with diameter $K L$, hence $\angle K B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,939 |
5. [5] A magician and an assistant perform a trick. In a row, there are 12 closed empty boxes. The magician leaves, and a spectator hides a coin in any two boxes of their choice, in full view of the assistant. Then the magician returns. The assistant opens one box that does not contain a coin. Next, the magician points... | Solution. Let's mentally arrange the boxes in a circle, representing them as points dividing the circumference into 12 equal arcs of length 1, and express distances between the boxes in terms of these arcs. Between any two boxes on one side of the circle, there are no more than 6 arcs of length 1. Therefore, it is suff... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,940 |
5. There are 101 coins lying in a circle, each weighing 10 g or 11 g. Prove that there is a coin for which the total weight of $k$ coins to its left is equal to the total weight of $k$ coins to its right, if
$3 \quad$ a) $k=50$
$3 \quad$ b) $k=49$.
Alexander Gribyalko | 5. There are 101 coins arranged in a circle, each weighing 10 g or 11 g. Prove that there exists a coin for which the total weight of $k$ coins to its left is equal to the total weight of $k$ coins to its right, if
a) $[3] k=50$;
b) $[3] k=49$.
Let's solve both parts at once. Let $k$ be any of the numbers 49 or 50. ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,941 |
1. [3] Is it possible to arrange numbers in the cells of a $6 \times 6$ table, all of which are distinct, such that in every $1 \times 5$ rectangle (both vertical and horizontal) the sum of the numbers is equal to 2022 or 2023?
(E. Bakayev) | Answer: no. Solution. Suppose it were possible. The numbers in adjacent corners differ by 1, since each of them completes four cells between them to form a $1 \times 5$ rectangle. Let $a$ be the smallest number among the corner numbers. Then in the adjacent corners, the numbers are $a+1$. Contradiction. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,942 |
2. [4] Does there exist a natural number that can be represented as the product of two palindromes in more than 100 ways? (A palindrome is a natural number that reads the same backward as forward.)
(E. Bakayev) | Answer: it exists. Consider the palindrome $1_{n}=\underbrace{1 \ldots 1}_{n}$.
Method 1. If $n$ is divisible by $k$, then $1_{n}$ is divisible by the palindrome $1_{k}$, and the quotient is also a palindrome, consisting of ones separated by groups of $k-1$ zeros. It remains to choose a number $n$ that has more than 1... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,943 |
3. [5] Pentagon $A B C D E$ is circumscribed around a circle. The angles at its vertices $A, C$ and $E$ are $100^{\circ}$. Find the angle $A C E$. | Answer: $40^{\circ}$. It is not hard to understand that such a pentagon exists.
Solution 1. The lines connecting the vertices with the center $O$ of the inscribed circle $\omega$ are the angle bisectors of the pentagon. Therefore, $\angle O A E=\angle O E A=50^{\circ}, \angle A O E=80^{\circ}$. Let $\omega$ touch the ... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,944 |
4. [5] Can all natural numbers greater than 1 be colored in three colors (each number in one color, all three colors must be used) so that the color of the product of any two numbers of different colors is different from the color of each of the factors?
(M. Evdokimov) | Answer: no. Solution. Suppose it was possible to color the numbers in blue, red, and green. We can assume that the number 2 is blue, and 4 is not red. Take a red number $k$. Then the number $2k$ is green, and $2 \cdot (2k)$ is red. On the other hand, $4 \cdot k$ is not red. Contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,945 |
5. [5] Petya has 8 coins, of which he knows only that 7 are genuine and weigh the same, while one is counterfeit and differs in weight from the genuine ones, but it is unknown in which direction. Vasya has a balance scale that shows which pan is heavier, but not by how much. For each weighing, Petya pays Vasya (before ... | Answer: can. Solution. Let the coins be denoted as $A, B, C, D, E, F, G, H$. Petya gives $H$ and asks Vasya to compare $A+B$ with $C+D$.
1) Vasya answers that the scales are in balance. Then the coins $A, B, C, D$ are genuine (regardless of which coin Vasya got). Petya gives $G$ for comparing $A$ with $E$. If Vasya an... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,946 |
1. [5] What is the greatest rational root that the equation of the form $a x^{2}+b x+c=0$ can have, where $a, b$ and $c$ are natural numbers not exceeding $100 ?$ | Answer: - $\frac{1}{\text {. }}$ Solution Note that $-\frac{1}{2}$ is a root of the equation $99 x^{2}+1$ Mikhail Evdokimov 99 . Solution. Note that - $\frac{1}{99}$ is a root of the equation $99 x^{2}+100 x+1=0$.
Let's prove the maximality. It is clear that the root $x$ of such an equation, as in the condition, is ne... | -\frac{1}{99} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,947 |
2. [5] Given two coprime numbers $p, q$, both greater than 1 and differing by more than 1. Prove that there exists a natural number $n$ such that the least common multiple (LCM) of $(p+n, q+n) < \operatorname{LCM}(p, q)$.
Mikhail Malkin | Solution. We can assume that the number $m=q-p$ is no less than 2. In this case, GCD( $p, m)=$ GCD $(p, q)=1$. The numbers $p$ and $q$ are congruent modulo $m$, but not divisible by $m$, so after increasing them by some natural number $n1 \cdot m \geq n+1$, that is, $p q>p+q+n$. Therefore, $p q m \geq p q(1+n)>p q+n(p+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,948 |
3. [6] Given two concentric circles $\Omega$ and $\omega$. A chord $AD$ of circle $\Omega$ is tangent to $\omega$. Inside the smaller segment $AD$ of the circle bounded by $\Omega$, an arbitrary point $P$ is taken. Tangents from $P$ to circle $\omega$ intersect the larger arc $AD$ of circle $\Omega$ at points $B$ and $... | Solution. Let $O$ be the center of the circles, $A D$ touches $\omega$ at point $Z$, and $P B$ and $P C$ touch $\omega$ at points $X$ and $Y$ and intersect $A D$ at points $K$ and $L$ respectively. Points $A$ and $D$ are symmetric with respect to $O Z$, so $Z$ is the midpoint of $A D$.
The tangents $P X$ and $P Y$ are... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,949 |
4. [7] In a square grid, there is a closed door between every two adjacent cells. A beetle starts from some cell and moves through the cells, passing through the doors. When it encounters a closed door, it opens it in the direction it is going and leaves the door open. The beetle can only pass through an open door in t... | Solution. If we can reach some closed door, we direct the bug to it and open it. We continue this process. The total number of closed doors decreases, so at some wonderful moment, there will be no closed doors that can be reached.
Suppose that at this moment there remains a non-empty set $N$ of unreachable cells. The ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 8,950 |
7. $N$ friends have a round pizza. It is allowed to make no more than 100 straight cuts, without rearranging the pieces until all cuts are completed, after which all the resulting pieces are to be distributed among all friends so that each receives the same total area of pizza. Will such cuts exist if
a) $[5] N=201$; ... | Answer: they will be found. Solution. a) Let the area of the pizza be 201. Then the area of the square inscribed in it is $\frac{402}{\pi}>121$. Draw a square with side 11 inside the pizza so that the centers of the square and the pizza coincide. By making 12 cuts parallel to the sides of the square, we get 121 pieces ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,952 |
1. [3] Can a segment be placed inside a regular pentagon such that it is seen from all vertices at the same angle?
(E. Bakayev, S. Dvoryaninov) | Answer: No. Solution: Suppose it was possible to place segment XY in such a way. On one side of the line $X Y$, there will be three vertices of the pentagon - let's say $A, B$, and $C$. Then points $A, B, C, X$, and $Y$ lie on the same circle. This circle will coincide with the circle circumscribed around the regular p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,953 |
2. [4] Find all natural n, satisfying the condition: the numbers 1, 2, 3, ..., 2n can be paired in such a way that if the numbers in each pair are added and the results are multiplied, the result is a square of a natural number.
(Folklore) | Answer: all $n>1$. Solution. $1+2$ is not a square. Let $n>1$.
1st method. We will divide these numbers into groups of four consecutive numbers, and, if necessary, a group of six initial numbers. From the groups of four, we form $(a+(a+3))((a+1)+(a+2))=(2 a+3)^{2}$, and from the group of six, $-(1+5)(2+4)(3+6)=18^{2}$... | n>1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,954 |
3. [5] In parallelogram $A B C D$, angle $A$ is acute. A point $N$ is marked on side $A B$ such that $C N=A B$. It turns out that the circumcircle of triangle $C B N$ is tangent to line $A D$. Prove that it is tangent to it at point $D$.
(M. Evdokimov) | Solution. Let it touch it at point $T$.
First method. Since $B C \| A D$, then $B T = C T$. From the equality of inscribed angles $N B T$ and $N C T$, we get the equality of triangles $A B T$ and $N C T$. Therefore, $\angle T A B = \angle T N C = \angle T B C = \angle T C B$. This means that $A B C T$ is a parallelogr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,955 |
3. [5] Petya places 500 kings on the cells of a $100 \times 50$ board so that they do not attack each other. And Vasya places 500 kings on the white cells (in chess coloring) of a $100 \times 100$ board so that they do not attack each other. Who has more ways to do this?
(E. Bakayev) | Answer. Vasya has more. Solution. To each of Petya's arrangements, we will correspond a certain arrangement of Vasya, and different arrangements will correspond to different ones. For each of these, there will be a Vasya's arrangement that cannot be obtained this way. For example, when the kings occupy all the white ce... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,956 |
5. [8] Let's call a pair ( $m, n$ ) of distinct natural numbers $m$ and $n$ good if $mn$ and $(m+1)(n+1)$ are perfect squares. Prove that for each natural number $m$ there exists at least one $n > m$ such that the pair $(m, n)$ is good.
^{2}\right)$ is good. Indeed,
$(m+1)\left(m(4 m+3)^{2}+1\right)=(m+1)\left(16 m^{3}+24 m^{2}+9 m+1\right)=(m+1)^{2}\left(16 m^{2}+8 m+1\right)=((m+1)(4 m+1))^{2}$.
Path to the solution. It is natural to try to find such an $n$ that it is a square multiplied by $m$, and at the same... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,957 |
5. [5] Petya has 8 coins, of which he knows only that 7 are genuine and weigh the same, while one is counterfeit and differs in weight from the genuine ones, but it is unknown in which direction. Vasya has a balance scale that shows which pan is heavier, but not by how much. For each weighing, Petya pays Vasya (before ... | Answer: can. Solution. Let the coins be denoted as $A, B, C, D, E, F, G, H$. Petya gives $H$ and asks Vasya to compare $A+B$ with $C+D$.
1) Vasya answers that the scales are in balance. Then the coins $A, B, C, D$ are genuine (regardless of which coin Vasya got). Petya gives $G$ for comparing $A$ with $E$. If Vasya an... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,958 |
2. [7] On a line, there are 2019 grasshoppers sitting at points. In one move, one of the grasshoppers jumps over another so that it ends up at the same distance from the other. By jumping only to the right, the grasshoppers can achieve that some two of them are exactly 1 mm apart. Prove that the grasshoppers can achiev... | Solution. Let's call the leftmost grasshopper Richard. Then let all the grasshoppers, except for Richard, jump over Richard. It is clear that now the grasshoppers are in a position that is symmetrical to the initial one. Then they can, using moves that are symmetrical to those they would make when jumping to the right,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,959 |
2. $p_{1}=p_{3}-2, p_{2}=p_{3}+2$ (up to reordering). Since $p_{3} \neq 3$, one of the numbers $p_{3}-2$ and $p_{3}+2$ is divisible by 3. And due to the simplicity of the numbers $p_{1}$ and $p_{2}$, one of them equals 3. It is directly verified that $p_{2}$ cannot equal 3. Hence, $p_{1}=3, p_{3}=5, p_{2}=7$ | Answer: $p_{1}=3, p_{3}=5, p_{2}=7$ or $p_{1}=7, p_{3}=5, p_{2}=3$.
## Problem 3
To solve the problem, for each number, consider the number of neighbors - numbers with which it can be connected by segments.
| 3 | $=$ | $=$ | 4 | - | - | 3 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | $\mathbf{c}$ ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,960 |
6. Place point $C$ at the origin of the Cartesian coordinate system, and point $B$ on the x-axis, as shown in the figure. Then the points $C, B, A$ will have coordinates $C(0,0), B(3,0), A(-1 / 2, \sqrt{3} / 2)$. The geometric locus of points $K(x, y)$ such that $B K: C K=1: 3$ is a circle. Indeed,
$\frac{B K^{2}}{C K... | Answer: $\frac{93+9 \sqrt{409}}{40}$.
 | \frac{93+9\sqrt{409}}{40} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,962 |
6. Place point $C$ at the origin of the Cartesian coordinate system, and point $B$ on the x-axis, as shown in the figure. Then the points $C, B, A$ will have coordinates $C(0,0), B(2,0), A(1, \sqrt{3})$. The geometric locus of points $K(x, y)$ such that $B K: C K=1: 2$ is a circle. Indeed,
$\frac{B K^{2}}{C K^{2}}=\fr... | Answer: $16 / 3$.
 | \frac{16}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,964 |
6. Let $r_{257}(x)$ denote the remainder of the division of the number $x$ by 257. Since
| digit on 1st wheel digit on 3rd wheel | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{1}$ | 0 | 6 | 12 | 18 | 24 |
| $\mathbf{2}$ | 25 | 1... | Answer: $\mathbf{1 .}$
Let's replace pairs of digits with rotation values:
| 0 | | 4 | 5 | 14 | 6 | 5 | 12 | 10 | 19 | | 15 | 5 | 3 | 33 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
Assume that the first word is the preposition "у". Then we will find ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,965 |
Problem 6.
It is known that three numbers $a_{1}, a_{2}, a_{3}$ were obtained as follows. First, a natural number $A$ was chosen and the numbers $A_{1}=[A]_{16}, A_{2}=$ $[A / 2]_{16}, A_{3}=[A / 4]_{16}$ were found, where $[X]_{16}$ is the remainder of the integer part of the number $X$ divided by 16 (for example, $[... | # Answer: 16.
## Conditions and answers to the problems of the final stage of the 2012-13 academic year
# | 16 | Number Theory | MCQ | Yes | Yes | olympiads | false | 8,967 |
# Problem 4.
Functions $f_{0}(x), f_{1}(x), \ldots, f_{6}(x)$ with domain $\{0,1,2,3\}$ and range $\{0,1,2,3\}$ are defined by tables (Table 2). Table:
| $x$ | $f_{0}(x)$ | $f_{1}(x)$ | $f_{2}(x)$ | $f_{3}(x)$ | $f_{4}(x)$ | $f_{5}(x)$ | $f_{6}(x)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|... | Answer: (a) $a=3, b=5, c=6$.
## Problem 5
Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The initial layout of the digits on the keyboard changes depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is replaced by the value $a_{i}$, which is the last d... | =3,b=5,=6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,968 |
6. When entering a personal account on the terminal, a four-digit password consisting of 0 and $1 . \quad$ is required. For this, the terminal has 4 buttons and 4 windows. When a button is pressed, the current symbol in the corresponding window is replaced with the opposite (that is, if the digit 1 is currently lit in ... | Answer: 1.
## Problem 2
Notice that for \( k=0,1,2,3 \) the equality \( 2 \overline{y_{k}} - \overline{y_{k+1}} = 31 x_{k+1} \) holds. Additionally, \( 2 \overline{y_{4}} - \overline{y_{0}} = 31 x_{0} \). The numbers 31 and (-1) give the same remainder 31 when divided by 32, i.e., \( 31 = r_{32}(31) = r_{32}(-1) \). ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,969 |
# Problem 1.
At robot running competitions, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: $\mathbf{1}$ sec., $\mat... | Answer: $30,32,35,36$ or $30,31,34,36$. | 30,32,35,36or30,31,34,36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,970 |
Task 2.
In a table consisting of $n$ rows and $m$ columns, numbers (not necessarily integers) were written such that the sum of the elements in each row is 408, and the sum of the elements in each column is 340. After that, $k$ columns were added to the table, the sum of the elements in each of which is 476, and a col... | Answer: 4 when $n=65, m=78, k=18$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,971 |
Task 3.
Screws and Washers use the following encryption system. The original text, written without spaces, is divided sequentially into parts of 10 letters. In each part, the letters are numbered from left to right from 1 to 10 and then rearranged according to the rule specified by Table 1.
That is, the first letter ... | # Answer: I HAVE A MOWING MACHINE.
# | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,972 | |
1. Vintik and Shpuntik use the following encryption system. The original text, written without spaces, is divided sequentially into parts of 10 letters. In each part, the letters are numbered from left to right from 1 to 10 and then rearranged according to the rule specified by Table 1. That is, the first letter of eac... | 1. Vintik and Shpuntik use the following encryption system. The original text, written without spaces, is divided sequentially into parts of 10 letters. In each part, the letters are numbered from left to right from 1 to 10 and then rearranged according to the rule specified by Table 1. That is, the first letter of eac... | зоопарк | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,979 |
2. To store a password written in a $32$-letter alphabet ("ё" is identified with "е"), each letter is represented by its ordinal number - a pair of digits (i.e., А - 01, Б - 02, etc.). This results in a sequence of digits $y_{1}, y_{2}, \ldots$ At the same time, according to the rule $x_{i+1}=r_{10}\left(a x_{i}+b\righ... | 2. To store a password written in a $32$-letter alphabet ("ё" is identified with "е"), each letter is represented by its ordinal number - a pair of digits (i.e., А - 01, Б - 02, etc.). This results in a sequence of digits $y_{1}, y_{2}, \ldots$ Simultaneously, according to the rule $x_{i+1}=r_{10}\left(a x_{i}+b\right)... | yacht | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,980 |
5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $ik$. The user... | 5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The layout of the digits on the keyboard changes after entering the code based on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user ent... | 3212 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,983 |
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