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|---|---|---|---|---|---|---|---|---|---|
[ Division with remainder ]
[ Evaluation + example ]
Author: Polish A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 222 nuts into two small boxes. After looking at
the distribution, Chichikov names any integer \( N \) from 1 to 222. Then Nozdryov must, if necessary, move one or several nuts to an emp... | Upper bound. By placing 74 and 148 nuts in the boxes, Nozdryov can, for any $N$, move no more than 37. Indeed, $N$ can be written in the form $74 k+r$, where $k=0,1,2,3$, and $-37 \leq r0$, and the numbers 74, 148, and 222 can be collected with one or two boxes, without moving anything. If $r0,74 k$ $=0,74$ or 148. By ... | 37 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,674 |
Bakayev E.V.
In some cells of an $11 \times 11$ square, there are plus signs, and the total number of plus signs is even. In each $2 \times 2$ sub-square, the number of plus signs is also even.
Prove that the number of plus signs in the 11 cells of the main diagonal of the square is even. | The stepped figure in the upper left corner (see fig.) consists of $2 \times 2$ squares, so it contains an even number of plus signs. The same is true for the figure in the lower right corner. Adding these numbers, we get an even number. At the same time, the plus in each cell of the square outside the diagonal is coun... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,675 |
Podolsky A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 1001 nuts among three small boxes. After looking at the distribution, Chichikov names any integer \( N \) from 1 to 1001. Then Nozdryov must, if necessary, move one or several nuts to an empty fourth box and show Chichikov one or several boxes... | Upper bound. By placing 143,286 = 2 * 143 and 572 = 4 * 143 nuts in the boxes, Nozdryov can, for any N, move no more than 71. Indeed, N can be represented as 143k + r, where 0 ≤ k ≤ 7, and -71 ≤ r < 71, and the number 143k can be collected with one or several boxes without moving anything. If r < 0, then 1001 - N = 143... | 71 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,676 |
Folklore
In a football championship, 18 teams are participating. As of today, 8 rounds have been played (in each round, all teams are divided into pairs, and the teams in each pair play against each other, with no pairs repeating). Is it true that there will be three teams that have not played a single match against e... | Consider one of the teams, denoted as A. Over 8 rounds, it played against eight teams and did not play against nine teams. If among these nine teams, there are two teams $B$ and $C$ that did not play against each other, then $A, B$, and $C$ form the desired triplet.
Otherwise, these 9 teams played a full round-robin t... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,677 |
[ Tasks with inequalities. Case analysis ] [Examples and counterexamples. Constructions]
## Authors: Khayaturyan A.V., Raskina I.V.
Thirty-three bogatyrs (Russian knights) were hired to guard Lukomorye for 240 coins. The cunning Chernomor can divide the bogatyrs into squads of any size (or record all in one squad), a... | From a detachment of $N$ bogatyrs, Chernomor will receive at most $N-1$ coins in the best case, since the remainder is less than the divisor. Therefore, he will receive no more than
$33-K$ coins, where $K-$ is the number of detachments. But if there is only one detachment, then since $240=33 \cdot 7+9$, Chernomor will... | )31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,678 |
Shaovalov A.v.
Fox Alice and Cat Basilio have grown 20 fake banknotes on a tree and are now filling in seven-digit numbers on them. Each banknote has 7 empty cells for digits. Basilio calls out one digit at a time, either "1" or "2" (he doesn't know any other digits), and Alice writes the called digit in any free cell... | Basilio can always get two banknotes: he knows where the last digit should be written and names it so that it differs from the digit in the same position on another banknote. Then the numbers on these two banknotes will be different, and the cat can take them.
Let's show how Alice can ensure that there are no more tha... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,679 |
Blinkov A.D:
The teams held a football tournament in a round-robin format (each team played one match against every other team, with 3 points for a win, 1 point for a draw, and 0 points for a loss). It turned out that the sole winner scored less than $50 \%$ of the maximum possible points for one participant. What is ... | Let's prove that there could not have been fewer than six teams. If, for example, there were five teams in the tournament, then they played $5 \cdot 4: 2=10$ matches and scored a total of at least 20 points. Therefore, the sole winner scored more than $20: 5=4$ points. However, according to the condition, he scored no ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,680 |
Folklore
Do there exist 2013 such distinct natural numbers that the sum of any two of them is divisible by their difference? | Let's first construct an auxiliary sequence $\left\{x_{n}\right\}$, which contains 2013 numbers. Starting this construction from the end, we set:
$x_{2013}=1, x_{2012}=2, x_{i}=\left(x_{i+1}+x_{i+2}+\ldots+x_{2013}\right)$ ! for $i$ from 1 to 2011. Then we set $a_{n}=x_{1}+x_{2}+\ldots+x_{n}$. We will prove that $\lef... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,681 |
Aвmo: : conskonpEach student in the class attends no more than two clubs, and for any pair of students, there is a club where they attend together. Prove that there will be a club where at least $2 / 3$ of the entire class attends. | If the entire class attends some club, then everything is fine. Further, we assume that such a club does not exist. Let the most numerous club be the mathematics club; we will call its members mathematicians. There is a student Vasya who does not attend it. Consider him and one of the mathematicians. They attend anothe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,682 |
[ Coloring $]$ [ Pigeonhole Principle (other) $]$
Each of the edges of a complete graph with 17 vertices is colored in one of three colors.
Prove that there are three vertices, all edges between which are of the same color.
# | From an arbitrary vertex, at least 6 edges of the same color (let's say red) emerge. Consider a complete graph on 6 vertices, to which these edges lead. If at least one of the edges of this graph is red, then there is a red "triangle". Otherwise, this graph is bicolored, and according to problem $\underline{30815}$, it... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,686 |
7,8,9 |
In a deck of 16 cards, numbered from top to bottom. It is allowed to take a portion of the deck from the top, after which the removed and remaining parts of the deck, without flipping, are "interleaved" with each other. Can it happen that after several such operations, the cards end up numbered from bottom to ... | Let's consider a method that allows achieving the required order in four operations. Each time, we will take exactly half of the deck - 8 cards from the top and "interleave" the removed part into the remaining part "one by one". The transformation of the deck during such operations is shown in the diagram:
| Top | | ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,690 |
Poyannsii A.
Let's call a point on the plane a node if both of its coordinates are integers. Given a triangle with vertices at nodes, inside which there are at least two nodes. Prove that among the nodes inside the triangle, one can choose two nodes such that the line passing through them contains one of the vertices ... | Let $A_{1}, B_{1}, C_{1}$ be the midpoints of the sides $B C, C A$, and $A B$ of triangle $A B C$, respectively. Take two arbitrary nodes $X$ and $Y$ inside the triangle. Suppose one of them lies outside the triangle $A_{1} B_{1} C_{1}$, for example, $X$ lies inside the triangle $A B_{1} C_{1}$. Construct the segment $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,691 |
In a circle, several (a finite number) different chords are drawn such that each of them passes through the midpoint of some other of the drawn chords. Prove that all these chords are diameters of the circle. | Consider the shortest chord.
## Solution
Note that the shorter the distance from the center O of the circle to the chord, the longer the length of the chord. Since there are a finite number of chords, there is a shortest one among them, say AB. By the condition, it passes through the midpoint K of some other chord, s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,693 |
On a bookshelf, 30 volumes of an encyclopedia stand in some order. In one operation, it is allowed to swap any two adjacent volumes. What is the minimum number of operations required to guarantee that all volumes can be arranged in the correct order (from the first to the thirtieth from left to right) regardless of the... | Let's call a disorder a pair of volumes where the volume with the larger number is to the left of the volume with the smaller number. In one operation, the number of disorders changes by no more than 1.
## Solution
Suppose we have some arrangement of volumes on a shelf. Consider all possible pairs of volumes (there a... | 435 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,694 |
9,10,11 |
Behind Chernomor, an infinite number of bogatyrs (warriors) of different heights are lined up. Prove that he can order some of them to step out of the line so that an infinite number of bogatyrs remain in the line and they are all standing in order of height (either increasing or decreasing). | Take the tallest hero, then the tallest among those standing behind him, then the tallest among those standing behind the second chosen one, and so on. If this is impossible - find an infinite
subsequence of heroes standing in decreasing order of height.
## Solution
Let's call the $B$-tail the entire sequence of her... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,698 |
Prove that a square can be inscribed in a regular pentagon such that its vertices will lie on four sides of the pentagon.
# | Let the perpendiculars erected to the line $A B$ at the vertices $A$ and $B$ of the pentagon $A B C D E$ intersect the sides $D E$ and $C D$ at points $P$ and $Q$. Each point on the segment $C Q$ is a vertex of a rectangle (with sides parallel to $A B$ and $A P$) inscribed in our pentagon, and as this point moves from ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,700 |
[ Regular polygons ]
[ Proof by contradiction ]
The vertices of a regular $2 n$-gon $A_{1} \ldots A_{2 n}$ are divided into $n$ pairs.
Prove that if $n=4 m+2$ or $n=4 m+3$, then two pairs of vertices are the endpoints of equal segments. | Suppose that all pairs of vertices define segments of different lengths. To the segment $A_{p} A_{q}$, we assign the smallest of the numbers $|p-q|$ and $2 n-|p-q|$. As a result, for the given $n$ pairs of vertices, we obtain the numbers $1,2, \ldots, n$. Notice that among these numbers, there are exactly $k=2 m+1$ eve... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,701 |
[ $\quad$ Trigonometric substitutions $\quad$ ]
Solve the system:
$$
\left\{\begin{array}{l}
y=x(4-x) \\
z=y(4-y) \\
x=z(4-z)
\end{array}\right.
$$ | Making the substitution $x=2-2 u, y=2-2 v, z=2-2 w$, we obtain the system
$$
\begin{aligned}
& v=2 u^{2}-1 \\
& w=2 v^{2}-1 \\
& u=2 w^{2}-1
\end{aligned}
$$
According to problem $\underline{61282}$, the solutions of this system are $(1,1,1),(-1 / 2,-1 / 2,-1 / 2),\left(\cos 2 \pi / 9, \cos 4 \pi / 9, \cos 8 \pi / 9\... | (0,0,0),(3,3,3),(4\sin^{2}\pi/9,4\sin^{2}2\pi/9,4\sin^{2}4\pi/9),(4\sin^{2}\pi/7,4\sin^{2}2\pi/7,4\sin^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,702 |
|
For which natural numbers $n$ in the expression
$$
\pm 1^{2} \pm 2^{2} \pm 3^{2} \pm \ldots \pm n^{2}
$$
can the signs + and - be arranged so that the result is 0?
# | If $n=4 k+1$ or $n=4 k+2$, then regardless of the arrangement of signs, an odd number will always result. Therefore, the problem will have no solution. Let's investigate the progressions $n=4 k+3$ and $n=4 k$. We will show that for numbers from the first progression, the problem has a solution starting from $n=7$, and ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,704 |
[ Principle of the Extreme ]
$[\underline{\text { Induction (etc.) }}] ~$
On an island, all countries are triangular in shape (borders are straight lines). If two countries border each other, they do so along an entire side. Prove that the countries can be colored with 3 colors such that adjacent countries along a sid... | Suppose that some islands satisfying the problem's condition cannot be colored with three colors such that each country is colored in its own color, and adjacent countries are colored in different colors. We will choose from them the island with the smallest number of countries. Consider any coastal country (that is, a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,705 |
$\underline{\text { Folklore }}$
On the plane, $n$ lines are drawn, none of which are parallel. No three of them intersect at the same point. Prove that there exists an $n$-segment non-self-intersecting broken line $A_{0} A_{1} A_{2} \ldots A_{n}$ such that each of the $n$ lines contains exactly one segment of this br... | Let's prove by induction a stronger fact: let $A_{0}$ be an arbitrary point on one of the given lines, through which no other lines pass; then there exists the required broken line starting from $A_{0}$.
Base case. For $n=1$, the broken line (consisting of one segment) is constructed trivially.
Inductive step. Let $l... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,706 |
Kozhevnikov P.A.
A 101-gon is inscribed in a circle. From each of its vertices, a perpendicular is dropped to the line containing the opposite side.
Prove that for at least one of the perpendiculars, its foot will fall on the side (and not on its extension). | Let's draw all the major diagonals $A_{1} A_{51}, A_{2} A_{52}, \ldots$ of the 101-gon $A_{1} A_{2} \ldots A_{101}$ (we assume that $A_{102}=A_{1}, A_{103}=A_{2}, \ldots$). This will form a star with 101 edges. We will color the diagonal $A_{k} A_{k+50}$ blue if the arc $A_{k} A_{k+1} A_{k+50}$ is less than half the ci... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,707 |
Mushroom C.
A magician has his eyes tied, and a spectator lays out a row of $N$ identical coins, choosing which ones to place heads up and which ones tails up. The magician's assistant asks the spectator to write down any integer from 1 to $N$ on a piece of paper and show it to everyone present. Upon seeing the number... | a) Mentally arranging the coins in the cells of a $2 \times k$ table, the magician writes O under each column of two cells if the coins there lie with the same side up, and R if they lie with different sides up. This combination tells him the number $n$ from 1 to $k$. If there is an even number of tails in the top row,... | N | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 53,708 |
Zvonkin d:
On a plane, two convex polygons $P$ and $Q$ are drawn. For each side of polygon $P$, polygon $Q$ can be clamped between two lines parallel to this side. Let $h$ be the distance between these lines, and $l$ be the length of the side, and compute the product $l \cdot h$. Summing these products over all sides ... | Let $\boldsymbol{a}_{i}, \boldsymbol{b}_{j}$ be the vectors of the sides of polygons $P$ and $Q$ respectively. The distance $h_{i}$, computed as indicated in the condition, for the side $\boldsymbol{a}_{i}$ of polygon $P$, is obviously the length of the projection of $Q$ onto the line perpendicular to $\boldsymbol{a}_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,709 |
99 children stand in a circle, each initially having a ball. Every minute, each child with a ball throws their ball to one of the two neighbors; if two balls land with the same child, one of these balls is lost irretrievably. What is the minimum time after which only one ball may remain with the children? | Number the children and balls clockwise from 1 to 99.
Example. Suppose children 1 and 2 toss the first ball to each other. The other balls with odd numbers are always thrown counterclockwise until they reach the second child, who discards them (this happens after an odd number of minutes, and at that moment he also re... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,710 |
Folklore
Construct a subset of a circle with an area equal to half the area of the circle, such that its image under reflection about any diameter intersects with it over an area equal to a quarter of the circle. | Let's construct a circle concentric with the given one, with half the area. Divide the inner circle in half with an arbitrary diameter, and the outer ring - with a perpendicular diameter. By combining half of the inner circle with half of the outer ring, we obtain the desired set.
Send a comment | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,712 |
$\underline{\text { Fon-der-Flaass }}$ D:
In the vertices of a cube, numbers $1^2, 2^2, \ldots, 8^2$ are placed (one number in each vertex). For each edge, the product of the numbers at its ends is calculated. Find the maximum possible sum of all these products. | We will color the vertices of a cube in two colors such that the ends of each edge are of different colors. Let the numbers $a_{1}, a_{2}, a_{3}, a_{4}$ be placed in the vertices of one color, and the numbers $b_{1}, b_{2}, b_{3}, b_{4}$ in the vertices of the other color, with numbers having the same indices placed in... | 9420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,713 |
Bogogonov I.I.
Five irreducible fractions with odd denominators greater than $10^{10}$ were written in red around a circle. Between each pair of adjacent red fractions, the irreducible sum of these fractions was written in blue. Could it happen that all the denominators of the blue fractions are less than 100? | Assume the opposite. Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be the original fractions in the order of their sequence around the circle. Note that
$2 a_{1}=\left(a_{1}+a_{2}\right)+\left(a_{3}+a_{4}\right)+\left(a_{5}+a_{1}\right)-\left(a_{2}+a_{3}\right)-\left(a_{4}+a_{5}\right)$. The last expression is an algebraic ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,714 |
A.K.
Does there exist a quadratic trinomial $f(x)$ such that for any natural $n$ the equation $f(f(\ldots f(x)))=0$ (n times "f") has exactly $2 n$ distinct real roots? | For example, $f(x)=2 x^{2}-1$. Let's restrict the domain of the function to the punctured interval $D=(-1,0) \cup(0, 1)$. By plotting the graph $y=2 x^{2}-1$, we can see that the corresponding range is the interval $(-1,1)$ and each value is taken exactly twice. Denote $f\left(f(\ldots f(x))\right.$ ( $n$ letters "f") ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,717 |
Kosukhin O.N.
Can a cube with an edge of 1 be cut into pieces by four planes so that for each of the pieces the distance between any two of its points is:
a) less than $4 / 5$;
b) less than $4 / 7$?
It is assumed that all planes are drawn simultaneously, and the cube and its parts do not move. | a) Let's choose three edges of the cube that have a common vertex. We will draw the first two planes perpendicular to the first edge so that it is divided into three equal parts by these planes. The third plane will be drawn perpendicular to the second edge through its midpoint. The fourth plane will be drawn perpendic... | )Yes;b)No | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,718 |
[ Linear Inequalities and Systems of Inequalities ] Evaluation + Example
Authors: Bogdanov I.I., Knop K.A.
King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, 2, ..., 11 kg. He also has a bag that will tear if more than 11 kg is placed in i... | Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then ingots weighing 1, 4, and 6 kg. In both cases, the bag does not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_{1}, \ldots, w_{6} \) kg instead of ingots ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,719 |
Berov S.l.
From a square of checkered paper $100 \times 100$, 1950 dominoes (two-cell rectangles) were cut out along the cell boundaries. Prove that from the remaining part, a four-cell figure of the form $\mathbf{T}$ can be cut out along the cell boundaries - possibly rotated. (If such a figure is already among the r... | Let's imagine that the dominoes (rectangles $1 \times 2$) are not yet cut out, and we will cut them out one by one. At each moment of the process, we will call the price of an uncut cell the number of its uncut side neighbors, decreased by 2 (for example, the price of a non-corner cell lying on the boundary of the squa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,720 |
In a certain convex $n$-gon ( $n>3$ ) all distances between vertices are distinct.
a) We will call a vertex uninteresting if the closest vertex to it is adjacent to it. What is the smallest possible number of uninteresting vertices (for a given $n$)?
b) We will call a vertex unusual if the farthest vertex from it is ... | a) Example. Let's take a segment $AB$ and a convex broken line $l$ close to it with the same ends and with links of the same length. The broken line $l$ and its symmetric counterpart relative to $AB$ form a convex polygon, in which the only uninteresting vertices will be $A$ and $B$. Thus, we can obtain a polygon with ... | )2;b)3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,722 |
Shapovalov A.V.
Can a square with a side of 1 be cut into two parts and used to cover a circle with a diameter greater than 1? | Consider a circle of diameter 1 and the union of two squares circumscribed around it, differing by a $45^{\circ}$ rotation. This results in an eight-pointed star. Remove four gray triangles from it (left image). The circle lies within the remaining white figure and touches its boundary only at four points. Slightly mov... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,723 |
$\underline{\text { Keshukhin } 0 . H .}$
Three cyclists are riding in the same direction on a circular track 300 meters long. Each of them moves at their own constant speed, and all speeds are different. A photographer can take a successful photo of the cyclists if all of them end up on some segment of the track $d$ ... | Without loss of generality, we assume that all cyclists are riding counterclockwise on the track, with the first one being the fastest and the third one being the slowest. Consider the motion of the cyclists in a reference frame attached to the second cyclist. Then the second cyclist is always at some point $A$, while ... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,725 |
Bakayev E.V.
A grasshopper can jump along a strip of $n$ cells by 8, 9, and 10 cells in either direction. We will call a natural number $n$ jumpable if the grasshopper can, starting from some cell, visit the entire strip, visiting each cell exactly once. Find at least one $n>50$ that is not jumpable. | Suppose a grasshopper has jumped over a strip of 62 cells. We will paint the 8 leftmost cells of the strip white, the next 10 cells black, then 8 cells white again, and so on. In total, there will be 32 white cells and 30 black cells. Since the difference in the number of white and black cells is greater than 1, there ... | 62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,726 |
[ $\left.\quad \begin{array}{lc}{\left[\begin{array}{l}\text { Processes and operations }\end{array}\right]} \\ {[} & \text { Semivariants }\end{array}\right]$
Authors: Fadin M. Kovalenko K.
Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it,... | Let $N>1$, and $1=d_{1}<d_{2}<\ldots<d_{k}<d_{k+1}=N-$ be all the divisors of $N$.
Notice that $d_{i} d_{k+2-i}=N$. Therefore,
$d_{1}^{2}+d_{2}^{2}+\ldots+d_{k}^{2}=\frac{N^{2}}{d_{k+1}^{2}}+\frac{N^{2}}{d_{k}^{2}}+\ldots+\frac{N^{2}}{d_{2}^{2}} \leq N^{2}\left(\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{N^{2}}\r... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,729 |
Diding M.
In a virtual computer state, there are no fewer than two cities. Some pairs of cities are connected by a road, and from each city, it is possible to travel by road to any other city (transferring from one road to another is only allowed in cities). If, starting from some city and not passing twice along the ... | Consider a graph where the vertices are cities and the edges are roads.
a) The condition means that the graph is a tree. Petya chooses an arbitrary vertex. From each vertex, there is exactly one path to the chosen one. He orients all edges on this path towards the chosen vertex.
In the first move, Petya moves the tou... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,732 |
Bazyan A.I.
Given $n>1$ reduced quadratic trinomials $x^{2}-a_{1} x+b_{1}, \ldots, x^{2}-a_{n} x+b_{n}$, and all $2 n$ numbers $a_{1}, \ldots, a_{n}$, $b_{1}, \ldots, b_{n}$ are distinct.
Can it happen that each of the numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ is a root of one of these trinomials? | Suppose this is the case. Since all $2 n$ coefficients are distinct, they make up the entire set of roots of our trinomials, and each of them has two roots. Let $x_{i}, y_{i}$ be the roots of the trinomial $x^{2}-a_{i} x+b_{i}$. Then $a_{i}=x_{i}+y_{i}, b_{i}=x_{i} y_{i}$ and
$\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n}\left(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,733 |
Petrov $\Phi$.
At each vertex of a convex 100-gon, two different numbers are written. Prove that one can erase one number at each vertex so that the remaining numbers in any two adjacent vertices are different. | Let the same numbers $a$ and $b$ stand in each vertex; then it is enough to leave the number $a$ in the vertices with even numbers, and the number $b$ in the vertices with odd numbers.
Let this not be the case. We will number the vertices in order from 1 to 100 so that different pairs of numbers stand in vertices 1 an... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,736 |
4 [ Examples and counterexamples. Constructions
In some cells of a $10 \times 10$ board, $k$ rooks were placed, and then all cells that are attacked by at least one rook (a rook attacks the cell it stands on as well) were marked. For what largest $k$ can it happen that after removing any rook from the board, at least ... | Evaluation. Let's consider the placement of $k$ rooks satisfying the condition. There are two possible cases.
1) In each column, there is at least one rook. Then the entire board is under attack, and a rook can be removed from any column that has at least two rooks. Therefore, in this case, there is exactly one rook i... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,738 |
Golovanov A.S
In an infinite increasing sequence of natural numbers, each number is divisible by at least one of the numbers 1005 and 1006, but none are divisible by 97. Additionally, any two consecutive numbers differ by no more than \( k \). What is the smallest \( k \) for which this is possible? | Let's denote our sequence as $\left(a_{n}\right)$. It is clear that $a_{1}D$ (while $a_{n} \neq D$ by the condition). However, the largest numbers less than $D$ and divisible by 1005 and 1006 are $D-1005$ and $D-1006$, respectively; therefore, $a_{n} \leq D-1005$. Similarly, $a_{n+1} \geq D+1005$; hence, $a_{n+1}-a_{n}... | 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,739 |
## [ Rectangles and Squares. Properties and Characteristics Method of Coordinates
On the plane, a square $A B C D$ is given. Find the minimum of the ratio $\frac{O A+O C}{O B+O D}$, where $O-$ is an arbitrary point on the plane. | It is known that for any rectangle $ABCD$ and any point $O$ in the plane of this rectangle, $OA^2 + OC^2 = OB^2 + OD^2$.
We will prove that $\frac{OA + OC}{OB + OD} \geqslant \frac{1}{\sqrt{2}}$, or $\sqrt{2}(OA + OC) \geqslant OB + OD$. Indeed,
\[
\begin{gathered}
\sqrt{2}(OA + OC) \geqslant OB + OD \Leftrightarrow ... | \frac{1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,740 |
Shapovalov A.V.
A rectangular grid is divided into dominoes, each consisting of two cells. In each domino, one of the two diagonals is drawn. It turns out that no diagonals share endpoints. Prove that exactly two of the four corners of the rectangle are endpoints of diagonals. | It is sufficient to prove that in any two dominoes adjacent along a segment, the drawn diagonals both exit from the lower right corners or both from the lower left corners.
Assume the opposite and find a bad pair: two adjacent dominoes with diagonals of different directions. Clearly, the common segment cannot be an en... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,742 |
Shen A.H.
In the country, there are 100 cities and several roads. Each road connects two cities, and the roads do not intersect. From any city, it is possible to reach any other city by traveling along the roads. Prove that it is possible to declare some roads as main roads such that from each city, an odd number of m... | We will divide all cities into pairs and connect each pair with its own route. We will calculate the multiplicity for each road: the number of routes it is part of. The sum of multiplicities for roads leaving a city is odd: the "own" route contributes 1, while the others contribute 0 or 2. Therefore, an odd number of r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,743 |
Tomongeo A.K.
Given a natural number, it is allowed to insert plus signs between its digits in any way and calculate the sum (for example, from the number 123456789, one can obtain $12345+6+789=13140$). The same operation is allowed to be performed on the resulting number, and so on. Prove that from any number, one ca... | Four operations are enough. For numbers less than 1000, this is obvious. Otherwise, break the number into four-digit chunks (not starting with zero) plus possibly zeros, plus possibly one smaller number at the end (for example, $12300004500060=1230+0+0+0+4500+0+60$). If there are $k$ four-digit chunks, the sum is at le... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,744 |
[ Coloring
Given an $n \times n$ square. Initially, its cells are colored in white and black in a checkerboard pattern, with at least one of the corner cells being black. In one move, it is allowed to simultaneously recolor the four cells in some $2 \times 2$ square according to the following rule: each white cell is ... | Suppose we managed to repaint the cells as required by the problem. We will call cells of the first type those that were originally white, and cells of the second type those that were black. Note that if a cell is repainted three times, it does not change its color in the end. Therefore, for a cell of the first type to... | Forallndivisible3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,745 |
Shapoveaov A.B.
The plan of the palace is a $6 \times 6$ square, divided into rooms of size $1 \times 1$. There is a door in the middle of each wall between the rooms. The Shah told his architect: "Knock down some walls so that all rooms become $2 \times 1$, no new doors appear, and the path between any two rooms pass... | Consider an arbitrary route from the lower left corner of the palace to the upper right. Since one needs to "climb" 5 horizontal levels and "shift right" 5 vertical levels, one has to pass through at least 10 doors, visiting at least 11 rooms (including the starting and ending rooms).
11 rooms of size $1 \times 1$ cou... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,746 |
Kosukhin O.N.
The teacher wrote on the board in alphabetical order all possible $2^{n}$ words consisting of $n$ letters A or B. Then he replaced each word with a product of $n$ factors, changing each letter A to $x$, and each letter B to (1 $-x)$, and added together several of the first of these polynomials in $x$. Pr... | Induction on $n$. Base. When $n=1$, after the teacher's actions, $x$ and $(1-x)$ will remain written on the board. $x$ is an increasing function, and $x+(1-x)$ is a constant. We will also consider a constant function as increasing.
Inductive step. Suppose the teacher adds the first $k$ obtained expressions ($1 \leq k ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,747 |
Polansky $A$.
Let $p$ be a prime number. A set of $p+2$ natural numbers (not necessarily distinct) is called interesting if the sum of any $p$ of them is divisible by each of the two remaining numbers. Find all interesting sets. | Let $S$ be the sum of all numbers in an interesting set, $c$ be the largest number, and $a$ and $b$ be any two other numbers in the set. The sums $S-a-c$ and $S-b-c$ are divisible by $c$, so their difference $b-a$ is also divisible by $c$. Since this difference in absolute value is less than $c$, it must be zero. There... | (p,,\ldots,)(,,\ldots,) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,748 |
[ Angles between lines and planes ] [ Least or greatest distance (length) ]
Authors: Bogdanov I.I., Karasev R.
Inside a convex polyhedron, a point $P$ and several lines $l_{1}, \ldots, l_{n}$ passing through $P$ and not lying in the same plane are chosen. For each face of the polyhedron, we associate the line from $... | Consider the intersection points $A_{1}, A_{2}, \ldots$ of the planes of the faces with the corresponding lines. Choose the smallest segment among all segments $P A_{i}$ (if there are several, consider any one of them). Denote this segment by $PA$. Suppose that the point $A$ does not belong to its face. Then the line $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,749 |
Agakhanov N.K.
Positive real numbers $a_{1}, \ldots, a_{n}$ and $k$ are such that $a_{1}+\ldots+a_{n}=3 k, a_{1}^{2}+\ldots+a_{n}^{2}=3 k^{2}$ and
$a_{1}^{3}+\ldots+a_{n}^{3}>3 k^{3}+k$
Prove that some two of the numbers $a_{1}, \ldots, a_{n}$ differ by more than 1. | According to the condition
$\left(a_{1}^{3}+\ldots+a_{n}^{3}\right)\left(a_{1}+\ldots+a_{n}\right)-\left(a_{1}^{2}+\ldots+a_{n}^{2}\right)^{2}=$
$\left(a_{1}^{3} a_{2}-2 a_{1}^{2} a_{2}^{2}+a_{1} a_{2}^{3}\right)+\ldots+\left(a_{n-1}^{3} a_{n}-2 a_{n-1}^{2} a_{n}^{2}+a_{n-1} a_{n}^{3}\right)=$
$=a_{1} a_{2}\left(a_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,750 |
Dmitriev 0.
The mammoth figure moves like a bishop (diagonally), but only in three of the four possible directions (the missing direction can differ for different mammoths). What is the maximum number of non-attacking mammoths that can be placed on an $8 \times 8$ chessboard? | Evaluation. From each mammoth, we will release three arrows in the directions in which it can attack. We will match an arrow to a diagonal (not necessarily the main one) if the mammoth from which the arrow originates is on this diagonal, and the arrow travels along it. Then, no more than two arrows are matched to each ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,752 |
Given 51 different two-digit numbers (single-digit numbers are considered two-digit with the first digit 0). Prove that from them, you can select 6 such numbers that no 2 of them have the same digit in any place.
# | Let's arrange the given numbers in ascending order and divide them into groups based on the tens digit. The number $m$ of such groups satisfies the conditions $6 \leqslant m \leqslant 10$. Among the $m$ groups, there will be a group $A_{6}$, which contains no fewer than 6 numbers.
Similarly (by contradiction), the exi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,754 |
Yasinsky V.
On the plane, there are $n(n>2)$ points, no three of which lie on the same line. In how many different ways can this set of points be divided into two non-empty subsets such that the convex hulls of these subsets do not intersect? | Since the convex hulls of two subsets do not intersect, they lie on opposite sides of some line. Thus, it is necessary to find out in how many ways the given set of points can be divided by a line into two subsets. Let's take a point $O$ in the plane, not lying on any of the lines connecting the given points, and consi... | \frac{1}{2}n(n-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,757 |
Bogdanov I.i.
Petya wants to list all possible sequences of 100 natural numbers, in each of which a triple appears at least once, and any two adjacent members differ by no more than 1. How many sequences will he have to list? | Let's find the answer for similar sequences of $p$ natural numbers.
The first method. We call a sequence of $n$ natural numbers, any two adjacent members of which differ by no more than 1, interesting. To each interesting sequence $a_{1}, a_{2}, \ldots, a_{n}$, we associate a difference sequence $b_{i}=a_{i+1}-a_{i}(i... | 3^{100}-2^{100} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,758 |
Bogonov I.I.
Petya wants to list all possible sequences of 100 natural numbers, in each of which the number 4 or 5 appears at least once, and any two adjacent members differ by no more than 2. How many sequences will he have to list? | Let's find the answer for similar sequences of $n$ natural numbers.
First method. We call a sequence of $n$ natural numbers, any two adjacent members of which differ by no more than 2, interesting. To each interesting sequence $a_{1}, a_{2}, \ldots, a_{n}$, we associate a difference sequence $b_{i}=a_{i+1}-a_{i}$ ( $i... | 5^{100}-3^{100} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,759 |
Berroov S.L.
In a volleyball tournament, 110 teams participated, and each team played exactly one game with each of the others (there are no ties in volleyball). It turned out that in any group of 55 teams, there is one team that lost to no more than four of the other 54 teams in this group. Prove that in the entire t... | Lemma. Let $k \geq 55$, and suppose that among any $k$ teams, there is one that has lost to no more than four of the remaining $k-1$ teams. Then among any $k+1$ teams, there is one that has lost to no more than four of the remaining $k$ teams.
Proof. Assuming the contrary, consider a set of $k+1$ teams $M=\left\{C_{1}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,761 |
$\underline{\text { Kleptsyn's } B . A}$.
An infinite checkered board is painted in a chessboard pattern, and a non-zero integer is written in each white cell. After that, for each black cell, the difference is calculated: the product of what is written in the horizontally adjacent cells, minus the product of what is ... | The simplest example is obtained by periodically repeating the arrangement shown in the figure.
If the neighbors of the black cell horizontally are equal to 1 in modulus, then their product... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,762 |
Berdnikov A.
On a spherical planet with an equator length of 1, they plan to lay $N$ circular roads, each of which will run along a circle of length 1. Then, several trains will be launched on each road. All trains will travel along the roads at the same positive constant speed, never stopping or colliding. What is th... | Evaluation. Let's take any two roads - great circles on a sphere. They intersect at some point-node. Mentally rotate one of these roads relative to the diameter containing the node to align the roads and the directions of movement on them. If in this experiment the arc-trains intersect, then after some time they will a... | )1.5;b)2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,765 |
Ivlev $\Phi$,
Given a tetrahedron in which a sphere can be inscribed, touching all its edges. Let the segments of the tangents from the vertices be $a, b, c$, and $d$. Is it always possible to form some triangle from these four segments? (Not all segments need to be used. It is allowed to form a side of the triangle f... | Let's take two circles $\beta$ and ү with radii 2 and 1, respectively, touching each other externally. We draw a common external tangent to both circles and construct a circle $\delta$ inscribed in the curvilinear triangle formed by both circles and the tangent. Clearly, the radius of this circle is less than 1, so a t... | Notalways | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,766 |
Protasov V.Yu.
If a criminal is located at point $X$, and three police officers, located at points $A, B$, and $C$, are blocking him, meaning that point $X$ lies inside triangle $A B C$. A new police officer replaces one of them as follows: he takes a position equidistant from all three police officers, after which on... | Obviously, on the first evening the police will be at the vertices of an isosceles triangle, and this condition will always be satisfied thereafter. Therefore, we can assume that at the beginning $A C = B C$. Let $O, R$ be the center and radius of the circumcircle of triangle $A B C$. Then, since $O C \perp A B$ and $X... | Cannot | Geometry | proof | Yes | Yes | olympiads | false | 53,769 |
In a set consisting of $n$ elements, $2^{n-1}$ subsets are chosen, any three of which have a common element.
Prove that all these subsets have a common element. | In the set $A$, consisting of $n$ elements, there exist $2^{n}$ different subsets, including the empty set and the set $A$ itself.
The complement of a subset $B$ will be denoted by $\bar{B}$. According to the condition, $2^{n-1}$ subsets are chosen, which is half of all possible subsets, and the intersection of any tw... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,771 |
For any $n$ real numbers $a_{1}, a_{2}, \ldots, a_{n}$, there exists a natural number $k \leq n$ such that each of the $k$ numbers $a_{k}, 1 / 2$ $\left(a_{k}+a_{k-1}\right)$
$1 / 3\left(a_{k}+a_{k-1}+a_{k-2}\right), \ldots, 1 / k\left(a_{k}+a_{k-1}+\ldots+a_{2}+a_{1}\right)$ does not exceed the arithmetic mean $c$ of... | Assume the opposite: there exist such real numbers $a_{1}, a_{2}, \ldots, a_{n}$, that for each $k$ ( $10$, then we will find $n_{3}$, and so on until some $n_{r+1}$ turns out to be zero, that is, the next arithmetic mean will be taken from $a_{1}$ to $a_{n_{r}}$. Since the arithmetic mean in each group is greater than... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,772 |
Given V. $\mathbf{L}$.
a) 12 liters of milk were poured into a bucket. Using only 5-liter and 7-liter containers, divide the milk into two equal parts.
b) Solve the general problem: for which $a$ and $b$ can you divide $a+b$ liters of milk in half, using only $a$-liter, $b$-liter, and $(a+b)$-liter containers?
In on... | a) See the table:
Table 1.
b) Let $a \geq b$. We will call a container with a capacity of $a+b$ liters a reservoir, a container with a capacity of $a$ liters - the first container, and a cont... | \frac{}{b}=\frac{2-1}{1-2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,773 |
The quadratic trinomial $a x^{2}+b x+c$ is a perfect fourth power for all integer $x$. Prove that then $a=b=0$.
# | Clearly, $a \geq 0$ and $c \geq 0$. Consider the values of $x$ equal to $1, 2, \ldots, n$. If at least one of the numbers $a$ and $b$ is not zero, then the quadratic polynomial $a x^{2} + b x + c$ for such $x$ takes at least $n / 2$ different values. These values are between 0 and $a n^{2} + |b| n + c$. The number of d... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,774 |
On $n$ cards, numbers are written on both sides - on the 1st: 0 and 1; on the 2nd: 1 and $2 ; \ldots$; on the $n$-th: $n-1$ and $n$. One person takes several cards from the stack and shows one side of each to the second person. Then, he takes another card from the stack and also shows one side. Indicate all cases in wh... | Let $k$ be the last number shown. We will prove that the second player can determine the number written on the back of the last card if and only if one of the following possibilities is met: 1) All numbers from $k$ to $n$ inclusive have been shown. 2) All numbers from 0 to $k$ inclusive have been shown. 3) All numbers ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,775 |
Can a point in space be covered by four spheres?
# | Let $A B C D$ be a regular tetrahedron with its center at a given point $O$. Describe a sphere around it. Consider the sections of the sphere by the planes $A B C, A B D, A C D$, and $B C D$. We obtain four circles on the sphere. Place non-intersecting spheres in the cones with a common vertex $O$ and bases described b... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,777 |
Fomin D:
The hostess baked a pie for the guests. At the table, there can be either $p$ people or $q$ people ($p$ and $q$ are coprime). Into what minimum number of pieces (not necessarily equal) should the pie be cut in advance so that in any case it can be distributed equally? | Considering the pie as a long rectangle, we will divide it into $p$ equal pieces ($p-1$ cuts) and into $q$ equal pieces ($q-1$ cuts). Since the number of cuts is $p+q-2$, the number of pieces is $p+q-1$.
Consider an arbitrary partition of the pie with volume $p q$. Let the guests be the vertices of a graph (a total of... | p+q-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,780 |
Voronin C.M.
A deck contains $n$ different cards. It is allowed to move any number of adjacent cards (without changing their order and without flipping) to another place in the deck. It is required to reverse the order of all $n$ cards using several such operations.
a) Prove that for $n=9$ this can be done in 5 opera... | a) Let's denote the cards by numbers. Suppose they are initially arranged in descending order: 9, 8, 7, 6, 5, 4, 3, 2, 1. We will show the results of the moves, highlighting the moved cards with underscores:
| 5, | 4, | 3, | 2, | 9, | 8, | 7, | 6, | $1 ;$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,781 |
Avoor: Prrossov B.B.
Point $P$ lies on the circumcircle of triangle $ABC$. Construct triangle $A_{1} B_{1} C_{1}$, the sides of which are parallel to segments $PA$, $PB$, $PC$
$\left(B_{1} C_{1} \| PA, C_{1} A_{1} \| PB, A_{1} B_{1} \| PC\right)$. Through points $A_{1}$, $B_{1}$, $C_{1}$, lines are drawn parallel to ... | Let triangle $ABC$ with angles $\alpha, \beta, \gamma$ be positively oriented (i.e., the traversal $A \rightarrow B \rightarrow C$ is counterclockwise). Then, extending sides $AB$ and $AC$ beyond vertex $A$ and drawing a line through $A$ parallel to $BC$, we obtain 6 angles arranged in the order $\alpha, \beta, \gamma,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,782 |
Fomin D:
A circle is divided into $n$ sectors, and some of the sectors contain chips - there are a total of $n+1$ chips. Then the position is subjected to transformations. One step of the transformation consists of the following: any two chips that are in the same sector are taken and moved in opposite directions to t... | Since there are more chips than sectors, at any moment there will be at least two chips in some sector. This means the movement continues indefinitely.
We number all sectors, starting from the given one, with numbers $1, \ldots, n$ in the order of their traversal clockwise. For each chip, we compute the square of the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,783 |
Shaovalov A.V.
There is a board $1 \times 1000$, initially empty, and a pile of $n$ chips. Two players take turns. The first player, on their turn, "places" no more than 17 chips on the board, one on any free cell (they can take all 17 from the pile, or part of them - from the pile, and part - by moving them on the bo... | a) Let's outline the strategy of the first player. Initially, he builds 12 series of 8 chips each, such that adjacent series are separated by one space, sequentially restoring a removed series and adding another. Then, after restoring the configuration following the second player's move, he inserts two chips into the o... | 98 | Combinatorics | proof | Yes | Yes | olympiads | false | 53,784 |
Tokarev S.I.
A lottery card represents a $10 \times 10$ grid of cells. The player marks 10 cells and sends the card in an envelope. After that, a set of 10 losing cells is published in the newspaper. Prove that
a) it is possible to fill out 13 cards in such a way that among them there will definitely be a "winning" c... | a) 13 cards $K_{1}, K_{2}, \ldots, K_{13}$ can be filled as follows: in $K_{i}$ for $i=1,2, \ldots, 9$ mark all cells in the $i$-th row, in $K_{10}$ - the left halves of the first and last (10th) rows, in $K_{11}$ - the right half of the first and the left half of the last row, in $K_{12}$ - the left half of the second... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,785 |
Razin M.
There is a set of 20 weights with which any integer weight from 1 to 1997 g can be measured (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the minimum possible weight of the heaviest weight in such a set, if:
a) the weights in the set are all integers
b) the... | Let's order the weights of the weights (in grams) in ascending order: $p_{0} \leq p_{1} \leq \ldots \leq p_{19}$. Clearly, $p_{0} \leq 1$ (otherwise, it's impossible to weigh a load of 1 g). Similarly, $p_{1} \leq 2$ (to weigh 2 g). Since these two weights are insufficient to weigh 4 g, $p_{2} \leq 4=2^{2}$. Continuing... | 146 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,786 |
Shaovalov A.V. Inside a rectangular sheet of paper, $n$ rectangular holes with sides parallel to the edges of the sheet have been cut out. What is the smallest number of rectangular pieces the sheet can be guaranteed to be cut into? (The holes do not overlap and do not touch.) | Evaluation. Let's assume that one of the sides of the sheet is vertical. Along each vertical "side" of one of the holes, we will draw a vertical cut to the "limit" at the horizontal side of the adjacent hole or to the edge of the sheet. We will repeat this procedure for all holes (see the left figure). After this, the ... | 3n+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,787 |
Shapovalov A.V.
With a chain of domino stones laid out according to the usual rules, it is allowed to perform such an operation: a segment consisting of several consecutive dominoes with the same number of pips on the ends of the segment is selected, flipped as a whole, and inserted back into the same place. Prove tha... | Let both chains lie horizontally on the table. We will denote a domino tile by a pair of numbers $(a, b)$, where $a$ and $b$ are the number of points on the halves.
If chain $A$ can be transformed into chain $B$, then chain $B$ can also be transformed into chain $A$. Therefore, it is sufficient to prove that, by apply... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,788 |
## Cooperative algorithms $\quad]$ Dirichlet's Principle (etc.). $\quad]$ [Pairings and groupings; bijections]
Before the clairvoyant lies a deck of 36 cards face down (4 suits, 9 cards of each suit). He names the suit of the top card, after which the card is revealed to him. Then he names the suit of the next card, a... | a) The first two cards can "encode" the suit of the second card, the next two cards can encode the suit of the fourth card, and so on. When only two cards are left in the deck, it is sufficient to encode only their order, which can be done using the back of the 35th card. Thus, the clairvoyant will guess the suits of 1... | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,791 |
Kanel-Belov A.Ya.
Consider the powers of five: $1,5,25,125,625, \ldots$ Form a sequence of their first digits: $1,5,2,1,6, \ldots$ Prove that any segment of this sequence, written in reverse order, will appear in the sequence of the first digits of the powers of two $(1,2,4,8,1,3,6,1, \ldots)$. | It is sufficient to prove that any initial segment of the sequence of the first digits of the powers of five appears (in reverse order) in the sequence of the first digits of the powers of two.
Consider the numbers: $2-1,2^{-2}, \ldots, 2^{-n}$. The sequence of the first non-zero digits of their decimal representation... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,792 |
Spivak A.V.
Does there exist a finite word made of letters of the Russian alphabet, in which there are no two adjacent identical substrings, but such substrings appear when any letter of the Russian alphabet is appended (either to the right or to the left)?
Comment. A word is any sequence of letters of the Russian al... | Consider the sequence of words:
## A, ABA, ABAVABA, ABAVABAGABAVABA,
The next word is obtained from the previous one as follows: the previous word is written, then the first of the letters that are not in it, and then the same word again.
We will prove by complete induction the following statement: in the p-th word,... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,793 |
Sharygin I.F.
In trapezoid $ABCD$, $AB$ is the base, $AC=BC$, $H$ is the midpoint of $AB$. Let $l$ be a line passing through point $H$ and intersecting lines $AD$ and $BD$ at points $P$ and $Q$ respectively. Prove that either angles $ACP$ and $QCB$ are equal, or their sum is $180^{\circ}$. | Let $M$ and $N$ be the points of intersection of the lines $CP$ and $CQ$ with the line $AB$, and let $K$ be the point of intersection of the line $PQ$ with $CD$ (see figure).
Then $DC: AM =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,794 |
Kuikuikin B.N.
On the sides $AB$ and $BC$ of triangle $ABC$, points $M$ and $N$ are chosen, respectively. Segments $AN$ and $CM$ intersect at point $O$, and $AO = CO$. Is triangle $ABC$ necessarily isosceles if a) $AM = CN$; b) $BM = BN$? | a) Consider triangle $ABC$, where $\angle B=60^{\circ}, \angle A=45^{\circ}, \angle ACB=75^{\circ}$.
Mark a point $O$ on the perpendicular bisector of side $AC$ such that $\angle OAC=\angle OCA=30^{\circ}$ (see figure).
Let ray $AO$ intersect side $BC$ at point $N$, and ray $CO$ intersect side $AB$ at point $M$. Then... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,795 |
[Ribamko A.V. In a $3 \times 3$ square, numbers are arranged (see figure). It is known that the square is magic: the sum of the numbers in each column, each row, and each diagonal is the same. Prove that a) $2(a+c+g+i)=b+d+f+h+4 e$. b) $2\left(a^{3}+c^{3}+g^{3}+i^{3}\right)=b^{3}+d^{3}+f^{3}+h^{3}+4 e^{3}$.
| $a$ | $b... | a) Adding $b+d+f+h$ to both sides, we get the obvious equality
$$
(a+b+c)+(a+d+g)+(c+f+i)+(g+h+i)=2(b+e+h)+2(d+e+f)
$$
b) 1) Let $S$ be the sum of the numbers in a row. Then $a+i=c+g=b+h=d+f=S-e$. Substituting into the equality from part a), we get $4(S-e)=2(S-e)+4 e$, from which $2 S=6 e$, that is, $S=3 e$.
2) Firs... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,796 |
Karpov, D.V.
In the cells of a $2000 \times 2000$ table, numbers 1 and -1 are written. It is known that the sum of all numbers in the table is non-negative. Prove that there exist 1000 rows and 1000 columns of the table such that the sum of the numbers written in the cells at their intersections is at least 1000. | The sum of all numbers in the table is non-negative, so there exists a row containing at least 1000 ones. Rearrange the columns of the table so that the first 1000 cells of this row contain ones. Denote by A and B the rectangles $2000 \cdot 1000$, formed by the first 1000 and the last 1000 columns of the table, respect... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,797 |
Kanel-Belov A.Y.
Numbers from 1 to 1000000 are painted in two colors - black and white. In one move, it is allowed to choose any number from 1 to 1000000 and repaint it and all numbers that are not coprime with it to the opposite color. Initially, all numbers were black. Is it possible to achieve, in several moves, th... | Lemma. Let a set of prime numbers $p_{1}, \ldots, p_{n}$ be given. Then it is possible to achieve, through several recolorings, that only those numbers that are divisible by all the prime numbers in this set change color.
Proof. For each non-empty subset of our prime numbers, we recolor the numbers that are not coprim... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,798 |
Karpov D.V.
In the country, there are several cities, some pairs of which are connected by roads. Moreover, from each city, at least three roads lead out.
Prove that there exists a cyclic route whose length is not divisible by 3. | Suppose there exists a graph where the degree of all vertices is more than two, but the length of each cycle in this graph is divisible by 3. Consider such a graph $G$ with the smallest number of vertices.
Obviously, in this graph there exists a cycle $Z$, let this cycle sequentially pass through the vertices $A_{1}, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,800 |
Podlissky o.k.
In a country with $n$ cities, there is either a road or a railway between each pair of them. A tourist wants to travel around the country, visiting each city exactly once, and return to the city where he started. Prove that the tourist can choose the starting city and the route in such a way that he wil... | Let's rephrase the problem in the language of graphs. We are given a complete graph with $n$ vertices, the edges of which are colored in two colors. We need to prove that we can select a cycle in this graph that passes through all vertices and consists of no more than two monochromatic parts.
We will prove this by ind... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,801 |
Pevzner I. A set of cells on a grid plane is called rook-connected if from any of its cells you can get to any other, moving along the cells of this set by the moves of a rook (the rook is allowed to fly over fields not belonging to our set). Prove that a rook-connected set of 100 cells can be divided into pairs of cel... | We will prove the statement of the problem for any rook-connected set $X$ consisting of $2n$ cells by induction on $n$. Cells will henceforth refer to the cells of the set $X$. The base case ($n=1$) is obvious.
Inductive step. We will call pairs of cells lying in the same row or column dominoes. Remove some domino, co... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,802 |
Volchenkov S.G.
On a grid paper, a rectangle is drawn, the sides of which form angles of $45^{\circ}$ with the grid lines, and the vertices do not lie on the grid lines.
Can each side of the rectangle intersect an odd number of grid lines? | Suppose it is possible for rectangle $ABCD$. Let $AB$ be its smallest side. Choose the origin of the coordinate system at a grid node and direct the coordinate axes along the grid lines so that among the vertices of the rectangle, vertex $A$ has the smallest abscissa, and vertex $B$ has the smallest ordinate. Let $A_x,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,803 |
Senderov V.A.
Does there exist an infinite increasing arithmetic progression $\left\{a_{n}\right\}$ of natural numbers such that the product $a_{n} \ldots a_{n+9}$ is divisible by the sum
$a_{n}+\ldots+a_{n+9}$ for any natural $n ?$ | Suppose such a progression exists. Then the number $A_{n}=\left(2 a_{n}\right) \ldots\left(2 a_{n+9}\right)$ is divisible by $B_{n}=a_{n+4}=a_{n+5}$ for any natural $n$. On the other hand, denoting the common difference of the progression by $d$, we have $A_{n}=\left(B_{n}-9 d\right)\left(B_{n}-\right.$
$7 d) \ldots\le... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,804 |
a) All vertices of the pyramid lie on the faces of the cube, but not on its edges, and at least one vertex lies on each face. What is the maximum number of vertices that the pyramid can have?
b) All vertices of the pyramid lie in the planes of the faces of the cube, but not on the lines containing its edges, and at l... | a) The section of the cube by the plane of the pyramid's base intersects all its faces and, therefore, is a convex hexagon. The vertices of the base lie on the sides of this hexagon, but not at its vertices. It is easy to see that if more than two vertices of the base lie on any one side, it is impossible to connect th... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,807 |
Kozhevnikov P.A.
Petya and Vasya were given identical sets of $N$ weights, in which the masses of any two weights differ by no more than 1.25 times. Petya managed to divide all the weights of his set into 10 equal mass groups, and Vasya managed to divide all the weights of his set into 11 equal mass groups. Find the s... | Example. Let the set consist of 20 weights of 50 g and 30 weights of 40 g. In this case, Petya can divide all the weights into ten groups, each containing two weights of 50 g and three weights of 40 g; and Vasya can divide all the weights into five groups, each containing four weights of 50 g, and six groups, each cont... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,809 |
Berroov S.L. Do there exist three pairwise coprime natural numbers such that the square of each of them is divisible by the sum of the two remaining ones? | Suppose such numbers $a, b, c$ are found. Note that the numbers $a+b, b+c, c+a$ are pairwise coprime. Indeed, suppose, for example, that the numbers
$a+b, b+c$ are divisible by some prime $p$. Since $c^{2}$ is divisible by $a+b$, and $a^{2}$ is divisible by $b+c$, the numbers $c$ and $a$ are also divisible by $p$, and... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,811 |
Bogdanov I.I.
Rational numbers $x, y$, and $z$ are such that the numbers $x+y^{2}+z^{2}, x^{2}+y+z^{2}$, and $x^{2}+y^{2}+z$ are integers. Prove that the number $2 x$ is an integer. | Let's bring the fractions $x, y$ and $z$ to a form with the least common denominator: $x=a / D, y=b / D, z=c / D$. Then $\gcd(a, b, c, D)=1$.
The number $x^{2}+y^{2}+z=\frac{a^{2}+b^{2}+c D}{D^{2}}$ is an integer, so $a^{2}+b^{2}$ is divisible by $D$. Similarly, $D$ divides the sums $b^{2}+c^{2}$ and $a^{2}+c^{2}$. Th... | 2x | Number Theory | proof | Yes | Yes | olympiads | false | 53,812 |
Pov V. A.
On the segment [0; 1], a function $f$ is defined. This function is non-negative at all points, $f(1)=1$, and for any two non-negative numbers $x_{1}$ and $x_{2}$, the sum of which does not exceed 1, the value $f\left(x_{1}+x_{2}\right)$ does not exceed the sum of the values $f\left(x_{1}\right)$ and $f\left(... | a) The function satisfying the condition of the problem is monotonically increasing. Indeed, if $x_{2} \geq x_{1}$ and $x_{2} \leq 1$, then $f\left(x_{2}\right) \geq f\left(x_{1}\right)+f\left(x_{2}-x_{1}\right)$ and $f\left(x_{2}-x_{1}\right) \geq 0$; therefore, $f\left(x_{2}\right) \geq f\left(x_{1}\right)$. Therefor... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,814 |
[ [Examples and counterexamples. Constructions] [ Auxiliary coloring $]$
A schoolboy wants to cut out the largest number of rectangles of size $1 \times(n$ $+1)$ from a square of size $2 n \times 2 n$. Find this number for each natural value of $n$.
# | The area of a square $2 n \times 2 n$ is $4 n^{2}$, and the area of a rectangle $1 \times (n+1)$ is $n+1$. Therefore, the number of cut rectangles does not exceed $\left[\frac{4 n^{2}}{n+1}\right]$, where $[n]$ is the integer part of the number $n$, i.e., the largest integer not exceeding $n$. Note that
$$
\left[\frac... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,815 | |
$\underline{\text { Capacity }}$..
On a line, 100 sets $A_{1}, A_{2}, . ., A_{100}$ are chosen, each of which is the union of 100 pairwise non-intersecting segments. Prove that the intersection of the sets $A_{1}, A_{2}, . ., A_{100}$ is the union of no more than 9901 pairwise non-intersecting segments (a point is als... | Let sets $A$ and $B$ on the line be unions of $m$ and $n$ segments, respectively. Then $A \cap B$ is a union of no more than $m+n-1$ segments. It is clear that $A \cap B$ is also a union of segments. Let the number of these segments be $k$. The endpoints of the segments in $A \cap B$ are endpoints of segments in $A$ or... | 9901 | Combinatorics | proof | Yes | Yes | olympiads | false | 53,816 |
Authors: Tolovanov A.S., $\underline{\text { Sopkina E. }}$.
The cells of a 50×50 square are colored in four colors. Prove that there exists a cell that has cells of the same color on all four sides (i.e., above, below, to the left, and to the right) of it (not necessarily adjacent to this cell). | Suppose the cells of an $n \times n$ square have been colored in such a way that for any cell, there is no cell of the same color on any of its sides.
Consider then all the cells of one color and draw an arrow in each of them in the direction where there is no cell of the same color. Then, for each cell on the edge of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,818 |
[ Tangent Circles ] [ Analytic Method in Geometry ]
Inside a convex quadrilateral, there are four circles, each of which is tangent to two adjacent sides of the quadrilateral and to two other circles (externally). It is known that a circle can be inscribed in the quadrilateral. Prove that at least two of the given cir... | The segment of the common external tangent to touching circles of radii $r$ and $R$, enclosed between the points of tangency, is equal to $2 \sqrt{r R}$.
## Solution
Let $x, y, z$ and $t$ be the radii of the circles inscribed in the angles $A, B, C$ and $D$ of quadrilateral $ABCD$. The distance between the points of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,819 |
Folklore
In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge 1, points $T, P$, and $Q$ are the centers of the faces $A A^{\prime} B^{\prime} B, A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, and $B B^{\prime} C^{\prime} C$ respectively. Find the distance from point $P$ to the plane $A T Q$. | Vertices $B^{\prime}$ and $C$ of the cube lie in the plane $A T Q$, so the planes $A T Q$ and $A B^{\prime} C$ coincide (see figure). By symmetry, the base $H$ of the perpendicular $P H$ dropped onto the plane $A B^{\prime} C$ lies on the line $B^{\prime} O$. Thus, $P H$ is the height of the right triangle $B^{\prime} ... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,820 |
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