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Trailer $\underline{\text { P. }}$.
The numbers from 1 to 50 are written on the board. It is allowed to erase any two numbers and write down one number, which is the absolute value of their difference. After repeating this procedure 49 times, one number will remain on the board. What can this number be? | Clearly, any number written on the board will be between 0 and 50. Moreover, it is odd (see the solution to problem $\underline{30303}$).
We will show that any of the odd numbers from 1 to 49 can be obtained. Suppose we want to get the number $2 m+1$ $(m=0,1, \ldots, 24)$. We will divide the numbers from 1 to 50 into ... | 1,3,5,\ldots,49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,326 |
Into how many parts can $n$ planes divide space?
(Any three planes intersect at one point, and no four planes have a common point.) | Let's denote this number as $K_{n}$. Let's see how this number changes when the $(n+1)$-th plane is drawn. This plane intersects the other planes along $n$ lines of "general position" and is divided by them into $1 / 2\left(n^{2}+n+2\right)$ pieces (see problem 60323). Each of these pieces splits one of the "old" parts... | \frac{1}{6}(n^3+5n+6) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,327 |
[The Fundamental Theorem of Arithmetic. Factorization into Prime Factors]
Given the number \( H = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \) (the product of prime numbers). Let \( 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, \ldots, H \) be all its divisors, listed in asc... | We will call a divisor "even" if it can be factored into an even number of prime factors (1 is considered an "even" divisor), and "odd" if it can be factored into an odd number of prime factors. We will prove by induction on $k$ that the number $N_{k}=2 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot p_{k}$ (the product of ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,328 |
Geometric progression Proof by contradiction
[ Proof by contradiction $]$
Given a non-increasing sequence of numbers $1 / 2 k=a_{1} \geq a_{2} \geq \ldots \geq a_{n} \geq \ldots>0, a_{1}+a_{2}+\ldots+a_{n}+\ldots=1$. Prove that there exist $k$ numbers, the smallest of which is greater than half of the largest. | Let, on the contrary, among any $k$ numbers of our sequence, the smallest is not greater than half of the largest. Consider the numbers $a_{1}, a_{2}, \ldots, a_{k}$. By the condition, $a_{1}$ is the largest among them, $a_{k}$ is the smallest; according to the made assumption, $a_{k} \leq 1 / 2 a_{1}$. Considering the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,329 |
4On a chessboard, two cells of the same color are selected.
Prove that a rook, starting from the first, can visit all cells once, and visit the second selected cell twice.
# | Clearly, if the desired path of the rook exists for some choice of marked cells, then it exists for any other choice that is obtained from the original by permuting the columns or permuting the rows of the chessboard. Therefore, it is sufficient to prove the statement of the problem in just two cases: if the first mark... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,330 |
From the set of weights weighing $1, 2, \ldots, 26$, select six weights such that it is impossible to choose two subsets of these weights that have the same total weight.
Prove that it is impossible to select seven weights with the same property. | The set of weights $26, 25, 24, 22, 19$ and 11 satisfies the condition.
Let some seven weights be chosen. Note that it is impossible to take weights 26, 25, 24, 23 simultaneously (among them there are already two piles of equal weight), and therefore the sum of the weights of any four weights out of these seven will b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,333 |
Factorization Induction (etc.)
It is known that $a^{n}-b^{n}$ is divisible by $n$ (where $a, b, n$ are natural numbers, $a \neq b$). Prove that $\frac{a^{n}-b^{n}}{a-b}$ is divisible by $n$. | Let's use the method of mathematical induction.
Base. Let $n$ be a prime number. If $a-b$ is not divisible by $n$, then everything is clear. If $a \equiv b(\bmod n)$, then $\frac{a^{n}-b^{n}}{a-b}=a^{n-1}+a^{n-2} b+\ldots+a b^{n-2}+b^{n-1} \equiv n b^{n-1} \equiv 0(\bmod n)$.
Induction step. Suppose the statement is ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,334 |
One after another, $2^{n-1}$ different sequences of zeros and ones of length $n$ are written down. It is known that for any three of the written sequences, there exists a number $p$ such that in the $p$-th position, all three have a 1. Prove that in some position, all the written sequences have a 1 and such a position ... | Let 0 denote a sequence of all zeros, $xy$ - the sequence obtained by element-wise multiplication of sequences $x$ and $y$, and $\bar{x}$ - the sequence obtained from sequence $x$ by replacing all zeros with ones and all ones with zeros. In these notations, the problem statement can be written as follows.
There are $2... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,335 |
[
[ The Fundamental Theorem of Arithmetic. Prime Factorization]
In the country of Anchuria, where President Miraflores is in power, the time for new presidential elections is approaching. The country has exactly 20 million voters, of which only one percent support Miraflores (the regular army of Anchuria). Miraflores,... | We will divide the voters into groups of 5 people. In 66666 such groups, we can place 3 military personnel. As a result, we get 66666 military electors out of 4 million.
We will divide these 4 million electors into groups of 4 people, placing 3 military personnel in 22222 of them. As a result, we get 22222 military el... | Hecan | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,336 |
Tolpygo A.K.
Prove that if a natural number $k$ is divisible by 10101010101, then in its decimal representation at least six digits are non-zero. | If a number $A$, divisible by 10101010101, has no more than 12 digits, then it has the form $\overline{a b a b a b a b a b a b}$, and the problem is solved. It remains to reduce the problem to this case.
Let the number 10101010101 be denoted by $M$.
Lemma. If $A$ is divisible by $M$ and has more than 12 digits (count... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,337 |
## [Examples and Counterexamples. Constructions]
Prove that it is possible to place real numbers $x_{1}, x_{2}, \ldots, x_{n}$, all different from 0, at the vertices of a regular $n$-gon, such that for any regular $k$-gon, all vertices of which are vertices of the original $n$-gon, the sum of the numbers at its vertic... | Let's draw a line $l$ through the center $O$ of a regular polygon $A_{1} \ldots A_{n}$, which does not pass through its vertices. Let $x_{i}$ be the projection of the vector $\overrightarrow{O A}_{i}$ onto a line perpendicular to $l$. Then all $x_{i}$ are non-zero. The sum of the numbers $x_{i}$ located at the vertices... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,338 |
[ Dirichlet's Principle (continued).] $[\quad$ Backward step $]$
In the country of Mara, there are several castles. From each castle, three roads lead out. A knight set out from one of the castles. Wandering along the roads, he turns either right or left from each castle he encounters, relative to the road he arrived ... | All castles in the country of Mara are connected by a finite number of roads. If a knight travels through the country for a long enough time, he will pass through many roads, so he will travel at least one road $AB$ (where $A$ and $B$ are castles) no less than five times. At least three times he will travel this road i... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,339 |
7,8,9 |
| | Product Rule | |
Let $K(x)$ be the number of irreducible fractions $a / b$ such that $a<x$ and $b<x$ (where $a$ and $b$ are natural numbers).
For example, $K(5 / 2)=3$ (fractions $1, 2, 1 / 2$).
Calculate the sum $K(100)+K\left(100 / 2\right)+K\left(100 / 3\right)+\ldots+K\left(100 / 99\right)+K\left(1... | The irreducible fraction $a / b$ is counted several times in our sum. Indeed, it is counted once in all $K\left({ }^{100} /\right)$ where $l a<100$ and $l b<100$. Thus, we obtain that the desired sum is equal to the number of all fractions $a / b$ (including reducible ones) or, equivalently, pairs $(a, b)$ that satisfy... | 9801 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,341 |
Eлиссев C.Л.
In a certain state, cities are connected by roads. The length of each road is less than 500 km, and from any city, one can travel to any other city by driving less than 500 km on the roads. When one road was closed for repair, it turned out that from any city, one could still travel to any other city by t... | Note that:
1) if some path $A-\ldots-B-\ldots-C-$ is the shortest between cities $A$ and $C$, then a part of it, for example, from $A$ to $B$ - is the shortest path between the corresponding cities;
2) the shortest path does not pass through the same road twice.
Suppose the road connecting cities $A$ and $B$ is close... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,342 |
Induction (continued).
## [Sums of numerical sequences and series of differences]
Given the formula $1^{3}+2^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$, prove that for different natural numbers $a_{1}, a_{2}, \ldots, a_{n}$, the inequality
$\left(a_{1}^{7}+a_{2}^{7}+\ldots+a_{n}^{7}\right)+\left(a_{1}^{5}+a... | We will prove this by induction. Base: $a^{7}+a^{5} \geq 2 a^{6} \Leftrightarrow a^{5}(a-1)^{2} \geq 0$.
Induction step. Let $a_{1}<\ldots<a_{n+1}$ be arbitrary natural numbers. By the induction hypothesis
$\left(a_{1}^{7}+a_{2}^{7}+\ldots+a_{n}^{7}\right)+\left(a_{1}^{5}+a_{1}^{5}+\ldots+a_{n}^{5}\right) \geq 2\left... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,345 |
Is the number of all 64-digit natural numbers, not containing zeros in their notation and divisible by $101$, even?
# | Consider the number $A=\underbrace{111 \ldots 11}_{64} 0$. It is divisible by $1111=11 \cdot 101$, so $A$ is divisible by 101. We will associate with each 64-digit number $n=\overline{a_{1}^{-}} \overline{\overline{2}} \overline{2}-\ldots \bar{a}_{64}$, which does not contain zeros in its representation and is divisibl... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,346 |
In a certain country, there are 1985 airfields. From each of them, an airplane took off and landed at the most distant airfield from the starting point. Could it happen that as a result, all 1985 airplanes ended up on 50 airfields? (The Earth can be considered flat, and the routes are straight; the pairwise distances b... | Let's first place 50 airfields at the vertices of a regular 50-gon, and the remaining airfields at its center. Then all planes from the center will fly to the vertices, and from the vertices, they will fly to the diametrically opposite vertices. Thus, all planes will gather at all 50 vertices of the 50-gon. Considering... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,347 |
Prove that a rhombus with half the area can be placed inside a centrally symmetric convex polygon.
# | Let $O$ be the center of symmetry, $A$ one of the farthest vertices from $O$ in the polygon, and $B$ the vertex symmetric to $A$ with respect to $O$. Construct the perpendicular $l_{\mathrm{A}}$ to the line $A B$ at point $A$, the perpendicular $l_{\mathrm{O}}$ at point $O$, and the perpendicular $l_{\mathrm{B}}$ at po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,349 |
Koganev I.
In Shvambrania, there are $N$ cities, each pair of which is connected by a road. At the same time, roads intersect only at the cities (there are no intersections, one road is elevated over the other). An evil wizard sets up one-way traffic on all roads in such a way that if you can leave a city, you cannot ... | a) Number the cities in an arbitrary order and for each pair of cities, keep the direction of movement from the lower number to the higher one.
b) Assign to city A a number equal to the number of cities from which a path leads to A. Obviously, all numbers are different, so they cover all values from 0 to $N-1$.
c) Th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,352 |
$\underline{\text { Folklore }}$
In a class, there are 32 students. 33 clubs were organized, each consisting of three people, and no two clubs have the same members. Prove that there will be two clubs that intersect exactly at one student. | Let's solve a more general problem: suppose $k$ students attend $n$ clubs (each consisting of three people), $k \leq n$. Assume the opposite: any two clubs either do not intersect or intersect exactly by two students. Note that if clubs $K$ and $L$ intersect with club $M$, then they also intersect with each other (thei... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,353 |
Bona M.
In a round-robin football tournament, 28 teams participated. By the end of the tournament, it turned out that more than $3 / 4$ of all the games ended in a draw.
Prove that some two teams scored the same number of points.
# | Less than $1 / 4 \cdot 1 / 2 \cdot 28 \cdot 27=94.5$ games were productive. Subtract one point from all results, that is, we will consider that 1 point is given for a win, and -1 for a loss. If a team scored $n$ points, then it won (or lost, if $n<0$) no less than $|n|$ matches. Let $n_{k}$ be the number of points scor... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,354 |
Razborov A.
In the country, there are 1988 cities and 4000 roads.
Prove that one can specify a circular route passing through no more than 20 cities (each road connects two cities).
# | Let's call a city "backwater" if no more than two roads lead out of it. We will erase any backwater city from the map of the country, along with the roads leading out of it. We will clean up the new map in the same way and continue erasing backwater cities until they disappear. The number of roads we can erase in this ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,355 |
In a competition, 32 boxers are participating. Each boxer can only have one match per day. It is known that all boxers have different strengths, and the strongest always wins. Prove that in 15 days, the place of each boxer can be determined.
(Schedules for each day of the competition are set in the evening before and ... | Lemma. Let $2^{n+1}$ participants in a competition be divided into two groups of $2^{n}$ boxers, with the boxers in each group already ordered by strength. Then all boxers can be ordered in $n+1$ days.
Proof. Induction on $n$. Base case. For $n=1$, the statement is obvious.
Inductive step. By the inductive hypothesis... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,358 |
Vasinivev N. 5.
Can natural numbers be arranged in a $4 \times 4$ table such that the following conditions are simultaneously satisfied:
1) the products of the numbers in each row are the same for all rows;
2) the products of the numbers in each column are the same for all columns;
3) there are no equal numbers among... | Consider a Graeco-Latin square (the table on the left). In it, each letter appears exactly once in each row and each column, and all 64 pairs of letters in the cells are different.
| $A a$ | $B b$ | $C c$ | $D d$ |
| :--- | :--- | :--- | :--- |
| $B d$ | $A c$ | $D b$ | $C a$ |
| $C b$ | $D a$ | $A d$ | $B c$ |
| $D c... | Itispossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,359 |
A 9×9 square is divided into 81 unit cells. Some cells are shaded, and the distance between the centers of any two shaded cells is greater than 2.
a) Provide an example of a coloring where there are 17 shaded cells.
b) Prove that there cannot be more than 17 shaded cells. | b) Suppose it was possible to color 18 cells. Divide the $9 \times 9$ square into nine $3 \times 3$ squares. Label the cells of the square with horizontal coordinates as letters from $a$ to $i$, and vertical coordinates as numbers from 1 to 9. Denote the $3 \times 3$ squares with Greek letters from $\alpha$ to $\iota$,... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,360 |
Andjassans $A$.
A numerical sequence is defined by the conditions: $a_{1}=1, a_{n+1}=a_{n}+\left[\sqrt{a_{n}}\right]$.
Prove that there are infinitely many perfect squares among the terms of this sequence. | Note that if $a_{n}=m^{2}+k$, where $1 \leq k \leq m$, then $a_{n+1}=m^{2}+m+k$, and $a_{n+2}=m^{2}+2 m+k=(m+1)^{2}+k-1$, which means the "excess" over the square decreases by 1.
If now $a_{s}=m^{2}$, then $a_{s+1}=m^{2}+m, a_{s+3}=(m+1)^{2}+m-1, \ldots, a_{s+2 m+1}=(2 m)^{2}=4 m^{2}$, that is, the next square is 4 ti... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,361 |
Prof. V.V. Proizvolov
In an isosceles triangle $ABC (AB = AC)$, the angle $A$ is equal to $\alpha$. On the side $AB$, a point $D$ is taken such that $AD = AB / n$. Find the sum of $n-1$ angles under which the segment $AD$ is seen from the points dividing the side $BC$ into $n$ equal parts:
a) for $n=3$
b) for an arb... | b) Let's construct point $E$ on side $AC = AB$, symmetric to $D$ with respect to the height dropped from $A$ - the axis of symmetry of triangle $ABC$. The sum of the angles we need to find is obviously half the sum of $\angle D K_{1} E + \angle D K_{2} E + \ldots + \angle D K_{n-1} E (*)$
angles under which the segmen... | \alpha/2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,363 |
Tokoreve C.
Does there exist a six-digit number $A$ such that among the numbers $A, 2A, \ldots, 500000A$ there is not a single number ending in six identical digits? | Let $A$ be coprime with 10. Then the numbers $A, 2A, \ldots, 10^{6}A$ give all possible remainders modulo $10^{6}$ exactly once (see problem 6 0733). Therefore, it is sufficient to find a number $A$ such that the remainders 111111, 222222, ..., 999999 appear in the last nine digits, that is, among the numbers from $\le... | There\exists | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,364 |
Tokorev S.i.
In each cell of a $4 \times 4$ table, there is a sign "+" or "-". It is allowed to simultaneously change the signs to the opposite in any cell and in all cells that share a side with it. How many different tables can be obtained by repeatedly applying such operations? | Let's assume that each cell contains a button, pressing which we change the sign in this cell and in all adjacent cells.
Lower bound. Note that by pressing the buttons in the second row, we can bring the first row to any state (each button changes the sign in the cell above it and does not change the state of the othe... | 2^{12} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,371 |
Several thin matches of the same length are lying on the table. Can their ends always be painted a) in 2, b) in 3 colors so that the two ends of each match are of different colors, and any two touching ends (of different matches) are of the same color? | a) It is sufficient to take three matches and form an equilateral triangle from them. Suppose we manage to color its vertices in two colors. Some two vertices will be of the same color, which means that the match connecting these vertices will have both ends of the same color. Contradiction.
b) Let's form a structure ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,372 |
Galperin G.A.
A natural number $N$ is 999... 99 ( $k$ nines) times greater than the sum of its digits. Specify all possible values of $k$ and for each of them, provide an example of such a number. | Answer: such a number exists for any $k: N_{k}=9 k \cdot\left(10^{k}-1\right)$. Let $9 k=\overline{s_{1}^{1}} \overline{-\ldots} S_{t}^{-} \overline{0}-.0$ ( $s_{t}$ is not equal to 0, there may be no zeros at the end). We will check that the sum of the digits of the number $N_{k}$ is equal to $9 k$. Write the differen... | N_{k}=9k\cdot(10^{k}-1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,373 |
Kanel-Belov A.Y.
A billiard table has the shape of a polygon (not necessarily convex), in which adjacent sides are perpendicular to each other. The vertices of this polygon are pockets, into which the ball remains if it falls. A ball is shot from vertex $A$ with an (internal) angle of $90^{\circ}$, and it reflects off... | If the ball flies out of A along the side, it will fall into the nearest pocket. Let the ball fly out at an acute angle to the side $A B$. It is easy to see that upon reflection from both the side parallel and the side perpendicular to the line $A B$, the smallest angle between $A B$ and the segment of the trajectory d... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,374 |
Given a paper circle. Is it possible to cut it into several pieces using scissors, from which a square of the same area can be assembled? (Cutting along straight lines and arcs of circles is allowed).
# | Suppose this can be done. Then the edge of each piece consists of segments and arcs of circles. For each arc, we define its "angular measure" as the ratio of the length of this arc to the radius, taken with a positive sign if the radius vector drawn from the center of the circle points "inside" our paper circle, and wi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,375 |
$\underline{\text { Ignatiev K.A. }}$
In a $10 \times 10$ square grid, one needs to place one ship of size $1 \times 4$, two ships of size $1 \times 3$, three ships of size $1 \times 2$, and four ships of size $1 \times 1$. The ships must not share any points (even vertices) with each other, but they can touch the bou... | b) An example of a "non-continuable" arrangement of nine ships is shown in the figure.
a) A ship $1 \times 4$ can be placed. We will prove that the next ship $1 \times 3$ will fit. For this... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,376 |
$\underline{\text { Frankin B.R. }}$ B.
In a round-robin tournament, there were no draws; 1 point was awarded for a win, and 0 for a loss. Then, a coefficient for each participant was determined. It was equal to the sum of the points scored by those whom the given athlete defeated. It turned out that all participants ... | Suppose not all participants scored the same number of points. Let those who took first place (the firsts) have scored $K$ points, and those who took last place (the lasts) have scored $L$ points. (Places are determined by points, not by coefficients.)
The coefficient of the firsts is the sum of $K$ numbers, each of w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,377 |
Konagin S.
Find $x_{1000}$, if $x_{1}=4, x_{2}=6$, and for any natural $n \geq 3$, $x_{n}$ is the smallest composite number greater than $2 x_{n-1}-x_{n-2}$. | We will prove by induction that $x_{n}=1 / 2 n(n+3)$.
Base. For $n=3,4$, the formula is correct: $2 x_{2}-x_{1}=8$, so $x_{3}=9$; $2 x_{3}-x_{2}=12$, so $x_{4}=14$.
Induction step. $2 x_{n}-x_{n-1}=2 \cdot 1 / 2 n(n+3)-1 / 2(n-1)(n+2)=1 / 2(n+1)(n+4)-1$. By the condition, $x_{n+1}$ is the first composite number great... | 501500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,378 |
[ [MT - circle or arc of a circle] [Coordinate method on the plane]
On the plane, points $A$ and $B$ are given. Prove that the set of all points $M$, which are 3 times farther from $A$ than from $B$, is a circle.
# | Let's solve the problem using the coordinate method. Introduce a coordinate system such that $A$ is at the origin and $B$ has coordinates $(1,0)$. Let the point $M(x, y)$ be the desired point. Then $1 / 3 = M B / M A = \sqrt{\frac{(x-1)^{2}+y^{2}}{x^{2}+y^{2}}}$, and $B$ has coordinates $(1,0)$. Let the point $M(x, y)$... | (x-9/8)^{2}+y^{2}=(9/8)^{2}-9/8 | Geometry | proof | Yes | Yes | olympiads | false | 52,380 |
}
Authors: [Orran R..., Sereeev P..
On the sides $BC$, $AC$, and $AB$ of an acute-angled triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are taken such that the rays $A_1A$, $B_1B$, and $C_1C$ are the angle bisectors of triangle $A_1B_1C_1$. Prove that $AA_1$, $BB_1$, and $CC_1$ are the altitudes of triangle $ABC$.
# | Let's draw the bisectors of the external angles of triangle $A_{1} B_{1} C_{1}$. Suppose the bisectors of the external angles at $B_{1}$ and $C_{1}$ intersect at point $A_{2}$, and so on. The bisector of angle $A_{1}$ also passes through point $A_{2}$ (since point $A_{2}$ is equidistant from the lines $A_{1} B_{1}$, $B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,381 |
Teresen D..
In a line, natural numbers from 1 to 1993 are written in some order. An operation is performed over the line: if the number $k$ stands at the first place, then the first $k$ numbers in the line are reversed. Prove that after several such operations, the number 1 will be at the first place.
# | We will prove this statement for a string containing numbers from 1 to $n$ by induction on $n$. Base case. When $n=1$, the number 1 is already in the first position.
Inductive step. If, as a result of applying the described operations to a string of $n$ numbers, the number $n$ ends up in the last position, then we can... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,382 |
Zlobin S.A.
For some positive numbers $x$ and $y$, the inequality $x^{2}+y^{3} \geq x^{3}+y^{4}$ holds. Prove that $x^{3}+y^{3} \leq 2$.
# | Suppose that $x+y^{2}<x^{2}+y^{3}$. Then, adding this inequality to the inequality $x^{3}+y^{4} \leq x^{2}+y^{3}$, we get ( $x$ $+x^{3})+(y^{2}+y^{4})<2 x^{2}+2 y^{3}$, which contradicts the inequalities $x+x^{3} \geq 2 x^{2}$ and $y^{2}+y^{4} \geq 2 y^{3}$.
Therefore, $x+y^{2} \geq x^{2}+y^{3} \geq x^{3}+y^{4}$, from... | x^{3}+y^{3}\leq2 | Inequalities | proof | Yes | Yes | olympiads | false | 52,385 |
Berlov S.L.
Let $a, b, c$ be positive numbers whose sum is 1. Prove the inequality:
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$ | Note that $\frac{1}{2}\left(\frac{1}{1-a}+\frac{1}{1-b}\right) \geq \frac{2}{2-a-b}=\frac{2}{1+c}$ (we used the inequality $\frac{x+y}{2} \geq \frac{2}{\frac{1}{x}+\frac{1}{y}}$ between the arithmetic mean and the harmonic mean for positive $x, y$). It remains to add three similar inequalities. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,386 |
Senderovv B.A.
Find the smallest natural number that cannot be represented in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are natural numbers. | $1=\frac{4-2}{4-2}, \quad 2=\frac{8-4}{4-2}, 3=\frac{8-2}{4-2}, 4=\frac{16-8}{4-2}, 5=\frac{32-2}{8-2}$
$6=\frac{16-4}{4-2}, 7=\frac{16-2}{4-2}, 8=\frac{32-16}{4-2}, 9=\frac{128-2}{16-2}, 10=\frac{64-4}{8-2}$.
Assume that $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$. Without loss of generality, let $a>b, c>d$. Denote $m=a-b,... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,387 |
Authors: Sadykov R., Cherelánov E.
The cells of a 200×200 table are colored black and white such that there are 404 more black cells than white ones. Prove that there exists a $2 \times 2$ s... | Suppose that in any $2 \times 2$ square there is an even number of white cells. Then if the upper cells of the square are colored the same, the lower cells are also colored the same, and if the upper cells are colored differently, the lower cells are also colored differently.
Consider the top row of the table and the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,391 |
Let the polynomial $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$ have at least one real root and $a_{0} \neq 0$. Prove that, by sequentially erasing monomials in the expression of $P(x)$, one can obtain the number $a_{0}$ from it, such that each intermediate polynomial also has at least one real root. | Let's outline a scheme for crossing out monomials that yields polynomials with roots at each step.
Let the polynomial $P(x) = a x^{n} + b x^{m} + \ldots + c (a, b, c \neq 0)$ contain no fewer than three terms ($x^{n}$ and $x^{m}$ being the two highest degrees of the variable $x$ in $P$). If $n$ or $m$ is odd, by cross... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,395 |
M. Murraikin
In a racing tournament, there are 12 stages and $n$ participants. After each stage, all participants receive points $a_{k}$ depending on the place $k$ they occupy (the numbers $a_{k}$ are natural, and $a_{1}>a_{2}>\ldots>a_{n}$). For what smallest $n$ can the tournament organizer choose the numbers $a_{1}... | Evaluation. Suppose there are no more than 12 participants.
Let one of the participants (call him $A$) win all 11 stages, and each of the remaining participants at least once took the last place.
Then participant $A$ after 12 stages will score no less than $11 a_{1} + a_{n}$ points, while each of the remaining partic... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,396 |
Given a sequence of numbers $F_{1}, F_{2}, \ldots ; F_{1}=F_{2}=1$ and $F_{n+2}=F_{n}+F_{n+1}$. Prove that $F_{5 k}$ is divisible by 5 for $k=1,2, \ldots$. | Induction on $k$. Base: $F_{5}=5$.
Induction step. $F_{5 k+2} \equiv F_{5 k+1} \pmod{F_{5 k}}, F_{5 k+3} \equiv 2 F_{5 k+2} \pmod{F_{5 k}}, F_{5 k+4} \equiv 3 F_{5 k+1} \pmod{F_{5 k}}, F_{5 k+5} \equiv 5 F_{5 k+1}$ $\pmod{F_{5 k}}$. By the induction hypothesis, $F_{5 k}$ is divisible by 5. Therefore, $F_{5 k+5}$ is al... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,397 |
$a, b, p$ - any integers. Prove that there exist such coprime $k, l$ that $a k+b l$ is divisible by $p$.
# | Since the signs of the numbers $k$ and $l$ can be changed, we can assume that $a \geq 0$ and $b \geq 0$.
We will prove the statement by induction on $a+b$. The case when $a=0$ or $b=0$ is obvious.
Induction step. Let $0 < b \leq a$. By the induction hypothesis, there exist coprime numbers $m$ and $n$ such that $(a-b)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,398 |
In an $8 \times 8$ table, all integers from 1 to 64 are written. Prove that in this case, there will be two adjacent numbers, the difference between which is not less than 5. (Adjacent numbers are those that are in cells sharing a common side.)
# | Consider a horizontal row of a table containing the number 1, and a vertical column containing the number 64. We can move from the cell with the number 1 to the cell with the number 64 by first moving along the row and then along the column, and our path will consist of no more than 14 moves (a move is a transition fro... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,399 |
In a $9 \times 9$ table, all integers from 1 to 81 are written. Prove that there will be two adjacent numbers, the difference between which is not less than 6.
# | Somewhere in the table, there is a 1, and somewhere there is an 81. Let's connect these numbers with a chain, moving from one number to another (located in an adjacent cell), making the minimum number of moves (no more than 16). If the difference between adjacent numbers in the table is less than 6, then the number of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,400 |
Solve the system in positive numbers:
$$
\begin{cases}x^{y} & =z \\ y^{z} & =x \\ z^{x} & =y\end{cases}
$$ | First, note that if one of the unknowns is equal to one, then the others are also equal to one.
Indeed, let $x=1$. Then $z=1^{\mathrm{y}}=1, y=z^{\mathrm{x}}=1^{1}=1$. Suppose there exists another solution besides $(1,1,1)$. Let's first consider $x>1$. Then $z=x^{\mathrm{y}}>1, y=z^{\mathrm{x}}>1$. Therefore, $z=x^{\m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,401 |
Induction (miscellaneous).
[Properties of the modulus. Triangle inequality]
[Algebraic inequalities (miscellaneous). ..... ]
Given a sequence of positive numbers $q_{1}, \ldots, q_{n}, \ldots$, a sequence of polynomials is constructed as follows:
$$
\begin{aligned}
& f_{0}(x)=1 \\
& f_{1}(x)=x \\
& \cdots \\
& f_{n+1... | We will prove by induction on $n$ that if $|x|>1$, then $\left|f_{n+1}(x)\right|>\left|f_{n}(x)\right|$. For $n=0$, this is obvious.
Induction step. If $|x|>1$, then $\left|f_{n+1}(x)\right| \geq\left(1+q_{n}\right)\left|x f_{n}(x)\right|-q_{n}\left|f_{n-1}(x)\right|>\left(1+q_{n}\right)\left|f_{n}(x)\right|-q_{n}\lef... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,403 |
[ Numerical tables and their properties ]
In table $A$ of size $10 \times 10$, some numbers are written. Let the sum of all numbers in the first row be denoted by $s_{1}$, in the second row by $s_{2}$, and so on. Similarly, the sum of numbers in the first column is denoted by $t_{1}$, in the second column by $t_{2}$, ... | Example of table $A$, for which the sum of all numbers is 955.
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 91 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 92 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 93 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 94 |
| 0 | 0 | 0 ... | 955 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,404 |
[Tournaments and tournament tables]
12 tennis players participated in a tournament. It is known that every two players played against each other exactly once and there was no player who lost all their matches. Prove that there exist players $A, B$, $C$ such that $A$ won against $B$, $B$ won against $C$, and $C$ won ag... | Consider the tennis player $B$, who has lost the most matches. By the condition, there is a player $C$, whom he has beaten. If $C$ had lost to all the participants who won against $B$, then $C$ would have lost more matches than $B$, which contradicts the selection of $B$. Therefore, there will be a player $A$, who won ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,405 |
$3+$ Proof by contradiction $\quad]$
Given a polynomial with integer coefficients. At three integer points, it takes the value 2.
Prove that it does not take the value 3 at any integer point.
# | Let $P(x)$ be our polynomial. Then for integers $a$ and $b$, the number $P(a)-P(b)$ is divisible by $a-b$ (see the solution of problem 35562). Suppose that
$P\left(a_{1}\right)=P\left(a_{2}\right)=P\left(a_{3}\right)=2$ and $P(b)=3$, where $a_{1}, a_{2}, a_{3}$ and $b$ are distinct integers. Then $\left|a_{1}-b\right|... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,407 |
Konyagin S.V.
At the congress, scientists gathered, among whom there are friends. It turned out that any two of them, having an
equal number of friends at the congress, do not have common friends. Prove that there will be a scientist who has exactly
one friend among the participants of the congress. | Let's take a scientist $A$, the number of friends of whom is maximal (if there are several such scientists, we take any of them); let's denote this number by $N$. Each of the $N$ friends of scientist $A$ has at least one friend (A), does not have more than $N$ friends, and no two of them have the same number of friends... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,408 |
[Arithmetic operations. Numerical identities] [Examples and counterexamples. Constructions]
Prove that it is possible to find more than a thousand triples of natural numbers $a, b, c$, for which the equality $a^{15}+b^{15}=c^{16}$ holds. | The given equality can be written as
$$
\left(\frac{a}{c}\right)^{15}+\left(\frac{b}{c}\right)^{15}=c
$$
Choose arbitrary natural numbers $n$ and $m$ and set $c=n^{15}+m^{15}, a=c n, b=c m$.
Send a comment | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,410 |
[ Quadratic inequalities (several variables).] The Pigeonhole Principle (other)
The numbers $a_{1}, a_{2}, \ldots, a_{1985}$ are the numbers $1,2, \ldots, 1985$ rearranged in some order. Each number $a_{k}$ is multiplied by its index $k$, and then the largest of the 1985 resulting products is chosen. Prove that it is ... | The numbers $a_{k}$, not less than 993, are exactly 993. Therefore, at least one of them has a number not less than 993, and for it $k a_{k} \geq 993^{2}$. | 993^2 | Inequalities | proof | Yes | Yes | olympiads | false | 52,413 |
Let's perform the following operation: between any two adjacent numbers, insert the number that results from subtracting the left number from the right one. We will perform the same operation on the new row, and so on. Find the sum of the numbers in the row that results after one hundred such operations.
# | Let's see how the sum of the numbers in a row changes after one operation. Let $a_{1}, a_{2}, \ldots, a_{\mathrm{n}}$ be the row to which the operation is applied. Then the new row has the form $a_{1}, a_{2}-a_{1}, a_{2}, a_{3}-a_{2}, \ldots, a_{\mathrm{n}-1}, a_{\mathrm{n}}-a_{\mathrm{n}-1}, a_{\mathrm{n}}$. The sum o... | 726 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,414 |
$\left[\begin{array}{l}\text { Equations in integers } \\ {[\text { Case enumeration }}\end{array}\right]$
Let $S(x)$ be the sum of the digits of a natural number $x$. Solve the equation $x+S(x)=2001$. | If $x>1999$, then $x+S(x) \geq 2002$, if $x<1000$, then $x+S(x)<1000+3 \cdot 9=1027$. Therefore, $x-$ is a four-digit number with the first digit 1.
Let $a$ be the hundreds digit, $b$ be the tens digit, and $c$ be the units digit, then $(1000+100a+10b+c)+1=2001$, or $101a+11b+2c=1000$. If $a<9$ the last sum is less th... | 1977 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,415 |
[ Dirichlet's Principle (continued). ] [ GCD and LCM. Mutual simplicity ]
Prove that there exists a number of the form
a) 1989...19890...0 (the number 1989 is repeated several times, followed by several zeros), divisible by 1988;
b) $1988 . . .1988$, divisible by 1989. | a) Consider the numbers 1989, 19891989, ..., 19891989... 1989 (in the last number, 1989 is repeated 1989 times).
Consider the remainders of these numbers when divided by 1988. Since there are fewer remainders than the numbers written, there must be two numbers with the same remainder (by the Pigeonhole Principle), and... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,416 |
[ $[$ Examples and counterexamples. Constructions ] [ Recurrent relations (miscellaneous). ]
Does there exist 2005 such distinct natural numbers that the sum of any 2004 of them is divisible by the remaining number?
# | It's not difficult to choose three natural numbers such that the sum of any two of them is divisible by the third: 1, 2, and 3. Note that one of these numbers is equal to the sum of the other two ($3 = 2 + 1$). Adding their sum to these numbers, we get the set $(1, 2, 3, 6)$, which has the property that the sum of any ... | (1+\sqrt{2}) | Number Theory | proof | Yes | Yes | olympiads | false | 52,417 |
\left[\begin{array}{l}\text { Mathematical logic (miscellaneous). } \\ {[\quad \text { Processes and operations }}\end{array}\right]
In the dense Murom forest, ten sources of dead water spring from the ground, numbered from N 1 to N 10. Anyone can take dead water from the first nine sources, but source N 10 is in Kosc... | Ivanushka, though a fool, prepared for the duel: he drank in advance from the first source (he could have from any other except the tenth). And that is why the poison from Koschei's source No. 10 acted as an antidote on him, not as a poison. To Koschei, Ivanushka gave ordinary water. After that, Koschei went to his cav... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,418 |
}
Twenty knights put on twenty cloaks, and each cloak turned out to be too short. Then the knights, taking off their cloaks, lined up by height. The tallest knight took the longest cloak, the second took the longest cloak from the remaining ones, and so on. The knight of the smallest height took the shortest cloak. Pr... | First, try to line up the knights by height, and then distribute the cloaks "by height."
## Solution
At first, each knight's cloak was too short. Let's start lining up the knights by height and redistributing the cloaks simultaneously. We will swap the cloaks of the tallest knight and the knight with the longest cloa... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,419 |
Lissicicy A. d $_{5}$
Natural numbers $M$ and $K$ differ by a permutation of digits.
Prove that
a) the sum of the digits of the number $2 M$ is equal to the sum of the digits of the number $2 K$;
b) the sum of the digits of the number ${ }^{M} / 2$ is equal to the sum of the digits of the number $K / 2$ (if $M$ and... | First, let's prove an auxiliary statement.
Let \( S(x) \) be the sum of the digits of a natural number \( x \), and \( N(x) \) be the number of its digits greater than 4. Then \( S(2x) = 2S(x) - 9N(x). \quad(*) \)
Suppose we add the number \( x \) to itself column by column. A carry of one to the next, \((k+1)\)-th, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,421 |
Pechkovssiy A.n.
In a dance class, 15 boys and 15 girls formed two parallel columns, creating 15 pairs. In each pair, the difference in height between the boy and the girl was measured (the difference is taken as an absolute value, meaning the larger height minus the smaller height). The maximum difference was 10 cm. ... | Let boys and girls be paired in descending order of height. Suppose that in one of the pairs, say the $k$-th, the height of the boy differs from the height of the girl by more than 10 cm, for example, the boy (denote him as $B$) is taller than the girl ($G$). Then $B$, and thus all the first $k$ boys, are taller than t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,422 |
There are many cubes of the same size, painted in six colors. Each cube is painted in all six colors, each face in its own color, but the arrangement of colors on different cubes can be different. The cubes are laid out on a table to form a rectangle. It is allowed to take any column of this rectangle, rotate it around... | Let $\varphi$ be a $90^{\circ}$ rotation of some row, $\psi$ be a $90^{\circ}$ rotation of some column. The commutators $\varphi \psi \varphi^{-1} \psi^{-1}$ and $\psi \varphi \psi^{-1} \varphi^{-1}$ rotate only the cube lying at their intersection, returning all other cubes to their original positions. Now it is clear... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,424 |
Two players play on an $8 \times 8$ chessboard. The player who starts the game makes the first move - placing a knight on the board. Then they take turns moving it (according to the usual rules), and it is not allowed to place the knight on a square where it has already been. The player who has no legal moves loses. Wh... | Let's divide all the squares into 32 pairs so that the squares in one pair are connected by a knight's move (see the figure).
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| 2 | 1 | 4 | 3 | 6 | 5 | 8 | 7 |
| 10 | 9 | 12 | 11... | Partner | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,425 |
Fomin D.
Find six different natural numbers such that the product of any two of them is divisible by the sum of these two numbers.
# | Let's take the numbers $1,2,3,4,5$ and 6. We will compute the least common multiple of 15 sums $1+2,1+3, \ldots, 5+6$. It equals $2^{3 \cdot 3} \cdot 3^{2 \cdot 5 \cdot 7 \cdot 11}=27720$. Multiply the numbers $1,2,3,4,5,6$ by 27720 and we get the desired set.
## Answer
For example, 27720, 55440, 83160, 110880, 13860... | 27720,55440,83160,110880,138600,166320 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,427 |
Fomin S.B.
The ladder has 100 steps. Kolya wants to go down the ladder, starting from the top, jumping down and up alternately. Jumps can be of three types - six steps (over five to the sixth), seven, and eight. He does not stand on the same step twice. Will he be able to get down? | In two jumps, Kolya can descend, at most, two steps, and only in the case where he always jumps down 8 and up 6. But such jumping can only be done $3+3$ times, otherwise he will have to step on a previously traversed step again. Thus, in 8 jumps, Kolya can descend, at most, 7 steps; in 72 jumps - 63 steps, and he did n... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,428 |
$3+$ [ Rectangles and Squares. Properties and Characteristics ]
Author: Anjanis $A$.
Consider a finite set $M$ of unit squares on the plane. Their sides are parallel to the coordinate axes (it is allowed for the squares to intersect). It is known that for any pair of squares, the distance between their centers does ... | Let's draw the x-axis through the lowest, and the y-axis through the leftmost center of the square. Then the centers of all squares are located in the square
$[0,2] \times [0,2]$. It is clear that the square with the center at the point $(1,1)$ intersects with all of them. | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,429 |
Fomin D.
30 boots are lined up in a row: 15 left and 15 right. Prove that among some ten consecutive boots, there are an equal number of left and right boots.
# | Let $L_{i}$ denote the number of left boots among ten, occupying positions from $i$ to $(i+9)$. Then $L_{1}+L_{11}+L_{21}=15$. If one of the numbers $L_{1}, L_{11}, L_{21}$ is equal to 5, then we have found the desired ten. Otherwise, one of them is greater than 5, and one is less than 5. But $L_{i+1}$ can differ from ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,430 |
Leibniz Gottfried Wilhelm
Consider the numerical triangle:
$$
\begin{array}{cccccc}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \ldots & \frac{1}{1993} \\
\frac{1}{2} & \frac{1}{6} & \frac{1}{12} & \ldots & \frac{1}{1992 \cdot 1993} \\
\frac{1}{3} & \frac{1}{12} & \ldots & &
\end{array}
$$
(The first row is given,... | On the $j$-th place in the $i$-th row, the number $\frac{(i-1)!(j-1)!}{(i+j-1)!}$ stands (this can be easily proven by induction on $\left.i\right)$. Substituting $j$ $=1, i=1993$, we get the desired number.
## Answer
$1 / 1993$. | \frac{1}{1993} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,431 |
$\underline{\text { R ubin }}$ A.
Three chess players $A, B$ and $C$ played a match-tournament (each played the same number of games with each other). Could it happen that in terms of points, A took first place, $C$ - last, and in terms of wins, on the contrary, $A$ took last place, $C$ - first (one point is awarded f... | Let the tournament have proceeded in six rounds, $C$ won four games (three against $B$ and one against $A$) and lost five, $B$ won three (all against $C$) and lost three, $A$ won two (against $C$) and lost one. Then $A$ scored 6.5 points, $B-6$, and $C-5.5$.
## Answer
It is possible. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,433 |
$\underline{\text { Anjance }}$.
Decimal representations of natural numbers are written in a row, starting from one, up to some $n$ inclusive: $12345678910111213 \ldots(n)$
Does there exist such an $n$ that in this sequence, all ten digits occur the same number of times? | Let's prove that the number of ones in such a sequence is always greater than the number of zeros.
First method. Induction. The base case (the first few digits are listed) is obvious: there are no zeros in the sequence, but there is a one.
Induction step. Suppose the number $n$ is $(k+1)$-digit. We will divide the se... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,434 |
Folkior
60 students came to the circle. It turned out that among any ten of them, there are at least three classmates.
Prove that among the circle members, there will be at least 15 students who study in the same class.
# | Assume the opposite: no more than 14 students came from each class. Let $m$ be the number of classes from which one student came, and $n$ be the number of classes from which 2 to 14 students came. Then
$m+2 n \leq 9$ (otherwise, taking one student from each of the $m$ classes and two from the rest, we would get a cont... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,435 |
Sazin A.p.
In the table
$0123 \ldots 9$
$9012 \ldots 8$
$8901 \ldots 7$
$1234 \ldots 0$
10 elements are marked such that in each row and each column, one element is marked.
Prove that among the marked elements, there are at least two equal elements. | Let the number marked in the $i$-th row be $k_{i}$. Then the sum of these numbers modulo 10 is $\left(k_{1}-1\right)+\left(k_{2}-2\right)$ $+\ldots+\left(k_{10}-10\right)=55-55=0 \neq 45(\bmod 10)$. Therefore, it cannot be the sum of the numbers from 0 to 9.
Send a comment | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,436 |
Kanel-Belov A.Y.
During the ball, each young man danced a waltz with a girl who was either more beautiful than the previous dance partner or more intelligent, but the majority (no less than $80 \%$) - with a girl who was simultaneously more beautiful and more intelligent. Could this have been possible? (There were an ... | Here is one possible example. At the ball, there were 10 girls and 10 young men (let's assign them numbers $1, 2, \ldots, 10$). The beauty of the girls increased with the number. Girl No. 10 was the least intelligent of all, while the intelligence of the others increased from the first girl to the ninth. In the first d... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,438 |
Shapovalov A.V.
There are tiles (rectangles cut out of cardboard) of size $2 \times 1$. On each tile, one diagonal is drawn. There are two types of tiles, as the diagonal can be placed in two ways, and there are enough tiles of each type. Can 18 tiles be chosen and arranged to form a $6 \times 6$ square such that the ... | See the figure:
## Answer
It is possible.
## [ [Examples and counterexamples. Constructions]
Author: Senderov V.A.
Prove that there are infinitely many odd $n$ for which the number $2^{... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,441 |
Emelyanov L.A.
Prove the inequality $1^{k}+2^{k}+\ldots+n^{k} \leq \frac{n^{2 k}-(n-1)^{k}}{n^{k}-(n-1)^{k}}$ for any natural numbers $n$ and $k$.
# | After multiplying both sides of the inequality by the denominator of the right part and combining like terms, we get:
$\left(1^{k}+2^{k}+\ldots+(n-1)^{k}\right) n^{k} \leq\left(2^{k}+3^{k}+\ldots+n^{k}\right)(n-1)^{k}$
Thus, it is sufficient to verify the inequality $(m-1)^{k} n^{k} \leq m^{k}(n-1)^{k}$ for $m=2,3, \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,443 |
Does there exist such an infinite sequence consisting of
a) real
b) integer
numbers, that the sum of any ten consecutive numbers is positive, and the sum of any first consecutive $10 n+1$ numbers is negative for any natural $n$? | a) Let $a_{10 n}=1+2^{-n}(n>0), a_{10 n+1}=-1(n>0)$, and place zeros in the rest of the sequence. Then, among any ten consecutive terms of the sequence, there are eight zeros, one minus one, and one number greater than one. Thus, their sum is positive. The sum of the first $10 n+1$ terms is $-2^{-n}$.
b) Consider any ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,444 |
Klepitsyn V.A.
Given an $n \times n$ table, each cell contains a number, and all numbers are distinct. In each row, the smallest number is marked, and all marked numbers turned out to be in different columns. Then, in each column, the smallest number is marked, and all marked numbers turned out to be in different rows... | The smallest number in the entire table was obviously marked both times. By the condition, none of the numbers in the same row (or column) with it were marked even once. Therefore, the smallest number in the table obtained from the given one by crossing out this row and column was also marked both times. And so on. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,445 |
Zaslavsky A.A.
Can a triangle be cut into four convex figures: a triangle, a quadrilateral, a pentagon, and a hexagon? | See example in the figure.
## Answer
It is possible.
Send a comment | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,447 |
Zhendarov R.G.
In a bank, there are 2002 employees. All employees came to the anniversary and were seated at one round table. It is known that the salaries of those sitting next to each other differ by 2 or 3 dollars. What is the greatest possible difference between two salaries of the employees of this bank, given th... | Between the employees with the highest and lowest salaries, there are no more than 1000 people. Therefore, their salaries differ by no more than 3003 dollars. A difference of exactly 3003 dollars is not possible: in this case, there would be 1000 people on each side of the table, and the salaries of any two adjacent pe... | 3002 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,448 |
Shaovalov A.V.
All plant species in Russia were numbered consecutively with numbers from 2 to 20000 (the numbers are consecutive without gaps or repetitions). For each pair of plant species, the greatest common divisor of their numbers was recorded, while the numbers themselves were forgotten (due to a computer failur... | Plants with numbers 101 and $10201=101^{2}$ (as well as with numbers $8192=2^{13}$ and $16384=2^{14}$) cannot be distinguished. They have the same greatest common divisors with any number not exceeding 20000. Therefore, it will not be possible to restore their numbers.
## Answer
It is not possible. | notpossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,449 |
Folklore
There are 100 sticks, from which a 100-gon can be formed.
Can it happen that no smaller number of these sticks can form a polygon?
# | Consider, for example, sticks of lengths $1, 2, 2^{2}, 2^{3}, \ldots, 2^{98}$ and $2^{99}-2$. The length of the longest stick is less than the sum of the lengths of the others, so a 100-gon can be formed. But in any subset, the length of the longest stick is not less than the sum of the lengths of the others, so a poly... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,450 |
[ Mixture and Concentration Problems ]
There are two containers with a capacity of 1 liter and 2 liters. From their contents, 0.5 liters of a mixture containing $40\%$ apple juice and 2.5 liters of a mixture containing $88\%$ apple juice were prepared. What is the percentage content of apple juice in the containers?
... | It is clear that in at least one of the vessels, the juice content does not exceed $40 \%$. Under this condition, the maximum amount of juice in 2.5 liters will be if we mix half a liter of 40-percent juice with two liters of pure juice. In this and only this case, in 2.5 liters, there will be $0.5 \cdot 0.4 + 2 = 2.2$... | 40,100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,451 |
In the vertices of a regular nonagon, numbers $1,2,3,4,5,6,7,8,9$ are placed, after which the product of the numbers at the ends of each diagonal is written on it. Can the numbers be arranged at the vertices in such a way that all the numbers on the diagonals are different?
# | For example, the numbers can be arranged in the following order: $3-8-1-6-2-9-4-5-7$.
## Answer
It is possible. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,453 |
Vyalyy M.N.
In a certain country, the total salary of the $10 \%$ highest-paid workers constitutes $90 \%$ of the salary of all workers. Could it be that in each of the regions into which this country is divided, the salary of any $10 \%$ of the workers does not exceed $11 \%$ of the total salary paid in that region? | Let's say, for example, a country consists of two regions: the capital district and the province. In the capital district, 1 million people work, and each earns 81 dollars per hour. In the province, 9 million people work, and each earns 1 dollar per hour.
## Answer
It can. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,455 |
## Diameter, main properties of s. [Perpendicular is shorter than oblique. Inequalities for right triangles] [Chords and secants (other).
In a circle, several (a finite number) different chords were drawn such that each of them passes through the midpoint of some other of the drawn chords. Prove that all these chords ... | Note that the shorter the distance from the center $O$ of the circle to a chord, the longer the length of the chord. Since there are a finite number of chords, there is a shortest one, say $A B$. By the condition, it passes through the midpoint $K$ of some other chord, say $C D$. If the intersection point of $A B$ and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,457 |
Gapochinn A.i.
A paper triangle with angles $20^{\circ}, 20^{\circ}, 140^{\circ}$ is cut along one of its angle bisectors into two triangles, one of which is also cut along its angle bisector, and so on. Can a triangle similar to the original one be obtained after several cuts? | Let's call a triangle bad if one of its angles is not a multiple of $20 \|$. On the first step, both obtained
triangles are bad. Note that in a bad triangle, there are at least two angles that are not multiples of $20 \|$.
From this, it is clear that when a bad triangle is cut along the bisector, two bad triangles ar... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,460 |
What should be the values of $a$ and $b$ so that the polynomial $x^{4}+x^{3}+2 x^{2}+a x+b$ is a perfect square?
# | The given polynomial of the fourth degree can only be the square of a reduced quadratic trinomial. Therefore, $x^{4}+x^{3}+2 x^{2}+a x+b=\left(x^{2}+p x+q\right)^{2}$. By equating the coefficients of like powers of the argument in both parts of the identity, we get
$2 p=1, p^{2}+2 q=2,2 p q=a, q^{2}=b$. Solving this s... | =\frac{7}{8},b=\frac{49}{64} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,461 |
$\underline{\text { Vavilov } B .}$.
Three right triangles are located in the same half-plane relative to a given line $l$ such that one of the legs of each triangle lies on this line. It is known that there exists a line parallel to $l$ that intersects the triangles at equal segments. Prove that if the triangles are ... | In the figure on the left, segments marked with two strokes are equal by condition, and segments marked with one stroke are also equal, since $l_{1} \| l$.
Now, let's arrange the triangles ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,462 |
: Asahanov $H . X$.
It is known that the equation $a x^{5}+b x^{4}+c=0$ has three distinct roots. Prove that the equation $c x^{5}+b x+a=0$ also has three distinct roots. | The number $x=0$ cannot be a root of the equation $a x^{5}+b x^{4}+c=0$, because otherwise $c=0$, and the equation has no more than two distinct roots, which contradicts the condition. Dividing both sides of this equation by $x^{5}$, we get $a+b / x^{+}+x^{5}=0$. Therefore, if $x_{1}, x_{2}$, and $x_{3}$ are distinct r... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,463 |
Kuznetsov D.Yu.
A regular hexagon with side 5 is divided by lines parallel to its sides into regular triangles with side 1 (see figure).
We will call nodes the vertices of all such triangl... | The total number of nodes is 2$\cdot$(6+ $7+8+9+10)=91$. Each node, except the central one, belongs to one of 11 concentric circles centered at the center of the hexagon (see figure).
Suppo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,464 |
Can the cells of a $5 \times 5$ board be painted in 4 colors so that the cells at the intersection of any two rows and any two columns are painted in at least three colors? | Suppose there exists a coloring of a 5×5 table that satisfies the condition.
In each column, there will be a color in which at least two cells of this column are painted. Let's call such a color dominant for the given column (possibly, some column will have two dominant colors).
Similarly, some color (let's call it 1... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,465 |
Zhendarov R.G.
On the board, the product $a_{1} a_{2} \ldots a_{100}$ is written, where $a_{1}, \ldots, a_{100}$ are natural numbers. Consider 99 expressions, each of which is obtained by replacing one of the multiplication signs with an addition sign. It is known that the values of exactly 32 of these expressions are... | Consider the leftmost even number $a_{i}$ and the rightmost even number $a_{k}$. Note that the sums with indices from $i$ to $k-1$ are even and only they are (in sums with smaller indices, the first addend is odd and the second is even; in sums with larger indices, it is the opposite).
Thus, $k-i=32$. The number of ev... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,469 |
Buretov A.i.
Alyosha has pastries distributed in several boxes. Alyosha recorded how many pastries are in each box. Seryozha took one pastry from each box and placed them on the first tray. Then he again took one pastry from each non-empty box and placed them on the second tray - and so on, until all the pastries were... | Let Alyosha write down $k$ different numbers $n_{1}<n_{2}<\ldots<n_{k}$. The first number written by Seryozha is the number of boxes $N_{1}$. It will be written exactly $n_{1}$ times. After this, some of the boxes will become empty, and the next largest number $N_{2}$, written by Seryozha, is the number of boxes that i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,471 |
Senderov V.A.
For real $x>y>0$ and natural $n>k$ prove the inequality $\left(x^{k}-y^{k}\right)^{n}<\left(x^{n}-y^{n}\right)^{k}$. | Dividing by $x^{n k}$ and denoting $\alpha=y / x$, we rewrite our inequality as $\left(1-\alpha^{k}\right)^{n}<\left(1-\alpha^{n}\right)^{k}$.
However, $0<1-\alpha^{k}<1-\alpha^{n}<1$ (since $0<\alpha<1$), so $\left(1-\alpha^{k}\right)^{n}<\left(1-\alpha^{n}\right)^{n}<\left(1-\alpha^{n}\right)^{k}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,472 |
Bogdanov I.I. The angles of a triangle $\alpha, \beta, \gamma$ satisfy the inequalities $\sin \alpha>\cos \beta, \sin \beta>\cos \gamma, \sin \gamma>\cos \alpha$. Prove that the triangle is acute. | Assume the opposite; let $\gamma \geq 90^{\circ}$ for definiteness. Then $\alpha+\beta \leq 90^{\circ}$, and the angles $\alpha$ and $\beta$ are acute. Therefore, $0 < \beta \leq 90^{\circ}-\alpha < 90^{\circ}$, from which $\cos (\beta) \geq \cos \left(90^{\circ}-\alpha\right)=\sin (\alpha)$, which contradicts the cond... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,475 |
A six-digit odometer in a car shows how many kilometers the car has traveled since it was purchased. Currently, it displays a number that contains four "sevens". Could it be that after another 900 km, the odometer will display a number with exactly one "seven"? | To end up with only one "seven", we need to get rid of the other three. Note that when adding 900 to a six-digit number, the digit in the hundreds place will definitely change. Moreover, if the digit in the hundreds place was not zero, the digit in the thousands place will also change. Thus, we can eliminate two "seven... | no,itcannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,476 |
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