problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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[Reverse Process]
A peasant, buying goods, paid the first merchant half of his money and another 1 ruble; then he paid the second merchant half of the remaining money and another 2 rubles, and finally, he paid the third merchant half of what was left and another 1 ruble. After this, the peasant had no money left. How ... | Before coming to the third merchant, the peasant had 1 ruble and as much again - a total of 2 rubles. Before coming to the second merchant, he had $2+2$ = 4 rubles and as much again - a total of 8 rubles. Finally, before coming to the first merchant, the peasant had $8+1=9$ and as much again; so initially, the peasant ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,189 |
[ Dirichlet's Principle (continued).]
In a bag, there are balls of two different colors: black and white. What is the smallest number of balls that need to be taken out of the bag blindly so that among them there are definitely two balls of the same color?
# | We need to draw three balls in total, so the balls are the "rabbits," and the colors are the "cages." Since there are fewer cages than rabbits, by the Pigeonhole Principle, there will be a cage with at least two rabbits. That is, two balls of the same color. It is easy to notice that by drawing two balls, we might get ... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,190 |
[ Dirichlet's Principle (continued).]
In the forest, there are a million fir trees. It is known that each of them has no more than 600,000 needles. Prove that there are two fir trees in the forest with the same number of needles.
# | We have a million "rabbits"-trees and, alas, only 600001 cells numbered from 0 to 600000. Each "rabbit"-tree is planted in a cell with a number equal to the number of needles on that tree. Since there are far more "rabbits" than cells, at least two "rabbits" must be in the same cell—if there were no more than one in ea... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,191 |
a) In the store "Everything for Tea," there are 5 different cups and 3 different saucers. How many ways can you buy a cup and a saucer?
b) The store also has 4 teaspoons. How many ways can you buy a set consisting of a cup, a saucer, and a teaspoon?
c) The store still sells 5 cups, 3 saucers, and 4 teaspoons. How man... | a) Let's choose a cup. To it, we can choose any of the three saucers. Therefore, there are 3 different sets containing the chosen cup. Since there are 5 cups in total, the number of different sets is 5$\cdot$3 = 15.
b) Any of the 15 sets from part a) can be complemented with a spoon in 4 ways. Therefore, the total numb... | 47 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,192 |
[ Dirichlet's Principle (continued).]
20 tourists went on a hike. The oldest of them is 35 years old, and the youngest is 20 years old. Is it true that there are people of the same age among the tourists?
# | The age of the tourist is one of the numbers: $20,21,22, \ldots, 35$ (a total of 16 options). And there are more tourists.
## Answer
Correct. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,193 |
[ Dirichlet's Principle (continued).]
Prove that there will be twenty Muscovites with the same number of hairs on their heads.
(It is known that a person has no more than 400,000 hairs on their head, and there are at least 8 million residents in Moscow.)
# | If it were not so, there would be fewer residents in Moscow.
## Solution
We will divide all Muscovites into 400,001 groups - by the number of hairs. Suppose that in each of these groups there are no more than 19 people. Then the total number of residents in Moscow is no more than $19 \cdot 400001$ < 8,000,000. Contra... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,195 |
Can 12 identical coins be arranged along the walls of a large square box so that along each wall there are exactly
a) 2 coins;
b) 3 coins;
c) 4 coins;
d) 5 coins;
e) 6 coins;
f) 7 coins
(You are allowed to stack coins on top of each other.) | Coins cannot be too many or too few. Understand how many is too many and how many is too few.
## Solution
a) Since all the coins must be placed along the walls, and each wall is touched by exactly two coins, the total number of coins is no more than 8.
b)-d) Examples of the required arrangements are shown in the fig... | ),e)Notpossible;b)-)possible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,196 |
[ Dirichlet's Principle (finite number of points, lines, etc.).)]
In a 4 x 4 meter carpet, a moth has eaten 15 holes. Is it always possible to cut out a 1 x 1 meter carpet that does not contain any holes inside? (Holes are considered point-like). | Let's divide the carpet into 16 squares of 1x1. Since there are 15 holes, at least one 1x1 square will not have a hole inside. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,197 |
Three merchants' wives - Sosipatra Titovna, Olympiada Karpovna, and Polikseina Uvarovna - sat down to drink tea. Olympiada Karpovna and Sosipatra Titovna drank 11 cups together, Polikseina Uvarovna and Olympiada Karpovna - 15, and Sosipatra Titovna and Polikseina Uvarovna - 14. How many cups of tea did all three mercha... | Notice, the cup drunk by each merchant's wife was mentioned twice in the problem's condition.
## Solution
|If we add up all the cups accounted for, we get twice the sum of the cups drunk. Therefore, we need to divide this sum by two.
## Answer
20 cups. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,198 |
[ Divisibility of numbers. General properties ] [ Number and sum of divisors of a number ]
Prime numbers have only two distinct divisors - one and the number itself. What numbers have only three distinct divisors? | Notice that for any divisor $m$, there corresponds a "paired divisor" $-{ }^{M / m}$.
## Solution
Any number $M$ always has divisors 1 and $M$. If $M$ has a divisor $m$, then it also has a divisor ${ }^{M} / m$. Therefore, for $M$ to have exactly three distinct divisors, the following conditions must be met: $m={ }^{... | Theof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,199 |
In the class, there are 38 students. Prove that among them, there will be four who were born in the same month.
# | Recall problem 145.
## Solution
If no more than three students in this class were born in each month, then there could not be more than 312 = 36 students in the class, but according to the condition, there are 38.
## Answer
If no more than $3 \times$ students in this class were born in each month, then there could ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,201 |
There are 20 scarves hanging on a coat rack. 17 girls take turns approaching the coat rack, and each either takes off or hangs up exactly one scarf.
Can there be 10 scarves left on the coat rack after the girls leave? | After the last girl leaves, there will be an odd number of scarves on the hanger.
## Answer
It cannot. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,202 |
Sharygin I.F.
A rectangle is composed of six squares (see the right figure). Find the side of the largest square if the side of the smallest one is 1.
The side of the largest square is equ... | Notice that the side of the largest square is equal to the sum of the sides of two squares: the one following it clockwise and the smallest one. Denoting the side of the largest square as \( x \), we can sequentially express the sides of the other squares: \( x-1, x-2, x-3, x-3 \) (see the figure). Now notice that the ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,203 |
$\underline{\text { Akulich I.F. }}$.
Avant-garde artist Zmey Kletochkin painted several cells on a $7 \times 7$ board, following the rule: each subsequent painted cell must share a side with the previously painted cell, but must not share a side with any other previously painted cell. He managed to paint 31 cells.
B... | Do not try to completely color the sides, leave the middle of the sides uncolored.
## Solution
a, b) If we can color 33 cells, then 32 cells can be colored by stopping in time. Three examples where 33 cells are colored are shown in the figure (in fact, there are many more such examples). It is impossible to color mor... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,204 |
6
A grasshopper jumps along a straight line forward by 80 cm or backward by 50 cm. Can it move away from the starting point by exactly 1 meter 70 cm in fewer than 7 jumps? | According to the condition, it is not necessary to move forward!
## Solution
After several unsuccessful attempts to move 1 m 70 cm forward, it may seem that the answer is "no". Notice that moving 1 m 70 cm does not necessarily mean moving forward. Let's try moving backward. 170 $=50 \cdot 5-80$.
## Answer
Yes, it c... | Yes,itcan.Forexample,5jumpsbackward1forward | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,205 |
[ Dirichlet's Principle (continued).]
Prove that if 21 people collected 200 nuts, then there are two people who collected the same number of nuts.
# | If everyone has a different number of nuts, then a total of no less than $0+1+2+3+\ldots+20=210$ nuts would have been collected, which contradicts the condition of the problem. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,206 |
Baranov d..V.
The hare bought seven drums of different sizes and seven pairs of sticks of different lengths for her seven baby hares. If a baby hare sees that both its drum is larger and its sticks are longer than those of one of its brothers, it starts to drum loudly. What is the maximum number of baby hares that can... | Not all the bunnies can play the drum, as the baby bunny that gets the smallest drum will not play it. On the other hand, if the same baby bunny is also given the shortest drumsticks, all the other bunnies will play the drum.
## Answer
6 bunnies. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,207 |
The numbers $1, 2^{1}, 2^{2}, 2^{3}, 2^{4}, 2^{5}$ are written on the board. It is allowed to erase any two numbers and replace them with their difference - a non-negative number.
Can the number 15 be the only number left on the board after several such operations? | The required sequence of operations is evident from the following record: $15=32-16-(8-4-2-1)$.
## Answer
It can. | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,208 |
[ Examples and counterexamples. Constructions ]
It is known about numbers $a$ and $b$ that $a=b+1$. Can it happen that $a^{4}=b^{4}$? | Let $a=1 / 2, b=-1 / 2$, then $a^{4}=b^{4}=\frac{1}{16}$. It can be proven that this example is unique (this is not required from students). Indeed, $a^{4}=b^{4} \Leftrightarrow|a|=|b|$. The case $a=b$ is impossible, the case $a=-b$ gives the specified example.
## Answer
## Problem | =1/2,b=-1/2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,210 |
[ Dirichlet's Principle (continued).]
A store received 25 boxes of three different types of apples (each box contains apples of only one type). Prove that among them, there are at least 9 boxes of apples of the same type.
# | 25 boxes - "rabbits" will be distributed among 3 cells-sorts. Since $25=3 \cdot 8+1$, we apply the "generalized pigeonhole principle" for $N=3, k=8$ and obtain that in some cell-sort there will be no less than 9 boxes. | 9 | Combinatorics | proof | Yes | Yes | olympiads | false | 53,211 |
[ Pigeonhole Principle (continued).]
What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other
# | Answer: 16 kings. Divide the board into 16 squares, in each of which there can be no more than one king. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,212 |
7,8
What is the maximum number of rooks that can be placed on an 8x8 chessboard so that they do not attack each other
# | Obviously, 8 rooks can be placed, for example, along the diagonal from a1 to h8. Let's prove that it is impossible to place 9 rooks that do not attack each other.
On one horizontal row, there cannot be more than one rook - otherwise, they would attack each other; therefore, the number of rooks that can be placed canno... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,213 |
[ $\underline{\text { Counting in two ways }}$ ]
A carpet covers a staircase with 9 steps. The length and height of the staircase are 2 meters. Will this carpet be enough to cover a staircase with 10 steps, with a length and height of 2 meters? | The length of the horizontal sections of the path is 2 meters, and the length of the vertical sections of the path is also 2 meters.
## Solution
Let the path cover a staircase with a certain number of steps, where the length and height of the staircase are both 2 meters. Mentally project the path onto the horizontal ... | itisenough | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,214 |
[ Dirichlet's Principle (continued).]
In the alphabet of the Ni-Bum-Bum tribe, there are 22 consonants and 11 vowels; a word in this language is any letter combination in which there are no two consecutive consonants and no letter is used more than once. The alphabet is divided into six non-empty groups. Prove that a ... | Let's replace consonant letters with ones and vowels with minus ones. Since the sum of all the resulting numbers is 11, and there are six groups, the sum in one of the groups does not exceed 1. This means that there are enough vowels in it to fill all the gaps between consonants. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,217 |
From a glass of milk, three spoonfuls of its contents are transferred to a glass of tea and stirred carelessly. Then, three spoonfuls of the resulting mixture are scooped and poured back into the glass of milk. What is now more: tea in the glass of milk or milk in the glass of tea
# | After the second pouring, the amount of tea in the glass with milk is exactly the same as the amount of milk that was taken from it: after all, the volume of the liquid has not changed. Therefore, in the end, there is as much tea in the milk as there is milk in the tea.
## Answer
Equally. | Equally | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,218 |
$\begin{aligned} & \text { [ Examples and counterexamples. Constructions ] } \\ & \text { [Numerical inequalities. Comparing numbers.] }\end{aligned}$
The sum of several numbers is 1. Can the sum of their squares be less than 0.1? | The condition of the problem is satisfied by eleven numbers equal to $1 / 11$. The sum of their squares is $11 \cdot (1 / 11)^2 = 1 / 11 < 0.1$.
## Answer
Yes. | 1/11<0.1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 53,220 |
[ Completing the Square. Sums of Squares ]
It is known that the number $a + 1/a$ is an integer. Prove that the number $a^2 + 1/a^2$ is also an integer. | $a^{2}+1 / a^{2}=(a+1 / a)^{2}-2$ | ^{2}+1/^{2}=(+1/)^{2}-2 | Algebra | proof | Yes | Yes | olympiads | false | 53,221 |
[ Dirichlet's Principle (continued).]
Evening Mathematical School classes are held in nine classrooms. Among others, 19 students from the same school attend these classes.
a) Prove that no matter how they are seated, at least three such students will end up in one classroom.
b) Is it true that in some classroom ther... | a) Indeed, let's assume that in each of the classrooms, there are no more than two students from this school. Then in all nine classrooms, there are no more than 18 such students - a contradiction. Therefore, in some classroom, there are at least three students from this school.
b) No, this is incorrect. For example, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,222 |
[ $\underline{\text { A Segment Inside a Triangle is Less Than the Longest Side }}$ $[\quad$ Regular Polygons Hexagons
Inside a regular hexagon with side 1, there are 7 points. Prove that among them, there are two points at a distance of no more than 1.
# | Divide a regular hexagon into six equilateral triangles with a side length of 1.
## Solution
A regular hexagon can be divided into six equilateral triangles with a side length of 1. Then, at least one of these triangles will contain two marked points. The distance between them will not exceed the side length of the t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,223 |
[ Plane,_divided by lines ] $[\quad \underline{\text { Coloring }}]$
Several lines divide the plane into parts. Prove that these parts can be colored in 2 colors such that adjacent parts will have different colors.
# | Let's use induction on the number of lines.
## Solution
We will use induction on the number of lines $n$. For $n=1$, the statement of the problem is obvious: it is sufficient to paint the two half-planes, into which one line divides the plane, in different colors. Suppose the statement of the problem is true for $k$ ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,224 |
2+ [ Identical Transformations ]
It is known that $x+1 / x$ is an integer. Prove that $x^{n}+1 / x^{n}$ is also an integer for any integer $n$.
# | Obviously, it is enough to prove the statement for non-negative integers. We will do this by induction.
Base. For $n=0$ the statement is obvious, and for $n=1$ it is given in the condition.
Induction step. Suppose the statement is already proven for all numbers from 0 to $n$. Then the number
$x^{n+1}+\frac{1}{x^{n+1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,225 |
Can an $n \times n$ table be filled with the numbers $-1, 0, 1$ so that the sums in all rows, all columns, and on the main diagonals are different?
# | How many specified sums and possible values for the specified sums?
## Solution
The condition requires that the values of $2 n+2$ sums ( $n$ rows, $n$ columns, and two diagonals) be distinct.
Each of these sums consists of $n$ terms, each of which can take one of the values $-1,0,1$. Therefore, each sum takes an int... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,227 |
[
Several points were marked on a line. After that, a point was added between each pair of adjacent points. This operation was repeated three times, and as a result, there were 65 points on the line. How many points were there initially?
# | If there were $n$ points on a line, then in one operation, $n-1$ points were added.
## Solution
If there were 65 points at the end, then there were 33 (and 32 were added) points before that. Similarly, there were 17 points after the first operation, and 9 before it.
## Answer
9 points. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,228 |
Several sixth-graders and seventh-graders shook hands. It turned out that each sixth-grader shook hands with seven seventh-graders, while each seventh-grader shook hands with six sixth-graders. Who were there more of - sixth-graders or seventh-graders? | How is the number of handshakes between sixth graders and seventh graders related to the number of seventh graders?
## Solution
Note that the total number of handshakes between students of different classes is six times the number of seventh graders and seven times the number of sixth graders.
## Answer
The number ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,229 |
$2+$ $[$ Lines and Planes in Space (other).]
A plane given by the equation $x+2 y+3 z=0$ divides the space into two half-spaces. Determine whether the points ( $1,2,-2$ ) and ( $2,1,-1$ ) lie in the same or in different half-spaces. | How to define a half-space using an inequality
## Solution
It is not difficult to verify that one of the half-spaces (the one in the direction in which the normal vector (1, 2, 3) points) is defined by the inequality $x+2y+3z>0$, and the other half-space by the inequality $x+2y+3z<0$. Thus, they lie in different half... | indifferent | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,230 |
Prove the inequality: $2^{n}>n$.
# | Note that $(a-1)(b-1) \geq 1$ for $a, b \geq 2$. From this, $a b \geq a+b$. Therefore, $2^{n} \geq 2+2+\ldots+2=2 n>n$. Send comment
$\begin{array}{lll}\text { Problem } \underline{60311} \text { Topics: } \\ {\left[\begin{array}{l}\text { Induction (other) }\end{array}\right]} & \text { Difficulty: } 2+ \\ {[\underli... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,232 |
[ Completing the square. Sums of squares ]
Is the number $100^{2}+201$ prime or composite? | $100^{2}+201=100^{2}+2 \cdot 100+1=101^{2}$
## Answer
Composite. | Composite | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,233 |
In the fairy country of Perra-Terra, among other inhabitants, Karabases and Barabases live. Each Karabas is acquainted with six Karabases and nine Barabases. Each Barabas is acquainted with ten Karabases and seven Barabases. Who is more numerous in this country - Karabases or Barabases? | Recall the problem $\underline{88268}$.
## Solution
Let's imagine stretching threads between each Karabas and the Barabas he knows. Then, from each Karabas, 9 threads will stretch, and from each Barabas, 10 threads will stretch. Therefore, the number of threads will simultaneously be 9 times the number of Karabas and... | Karabas | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,234 |
Shapovalov A.V.
Anya, Borya, and Vasya were forming words from given letters. Each of them formed a different number of words: Anya formed the most, and Vasya the least. Then the children summed up the points for their words. If a word is found in two players' lists, it scores 1 point; if it is found in only one playe... | Let's say, for example, a total of 15 words were formed, of which 2 are only with Boris, 4 - only with Vasya, 6 - with Anya and Boris, and 3 - with Anya and Vasya. In total, Anya formed 9 words and scored 9 points, Boris - 8 words and 10 points, and Vasya - 7 words and 11 points.
Answer
It could. | Itcould | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,235 |
[Coordinate method on the plane]
Given points $A(0; -2)$, $B(-2; 1)$, $C(0; 0)$, and $D(2; -9)$. Indicate which of them lie on the line $2x - 3y + 7 = 0$.
# | The equation of the line is satisfied only by the coordinates of point $B$.
## Answer
B. | B | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,236 |
Igor painted several cells in a $6 \times 6$ square. After this, it turned out that in all $2 \times 2$ squares the number of painted cells is the same and in all $1 \times 3$ strips the number of painted cells is the same. Prove that diligent Igor painted all the cells. | Let in each $2 \times 2$ square, $m$ cells are shaded, and in each strip $1 \times 3-n$ cells are shaded. Since the $6 \times 6$ square is divided into 9 smaller squares, a total of $9 m$ cells are shaded. Similarly, we get that a total of $12 n$ cells are shaded. Therefore, $9 m=12 n$, which means $3 m=4 n$. But $0<m ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,238 |
a) At each vertex of a cube, a number 1 or 0 is written. On each face of the cube, the sum of the four numbers written at the vertices of this face is written. Can it happen that all the numbers written on the faces are different?
b) The same question, if the numbers 1 or -1 are written at the vertices. | a) On each face, one of the five numbers is written: $0,1,2,3$ or 4. But there are only six faces, which means that on some two faces, the same number will be written.
b) The solution is the same, only on each face one of the five numbers is written: $-4,-2,0,2,4$.
Submit a comment | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,239 |
[ Dirichlet's Principle (continued).]
In a class of 25 students, it is known that any two girls in the class have a different number of boy friends from this class. What is the maximum number of girls that can be in this class?
# | If there are 13 girls in the class, the number of their boy friends from this class can be any integer from 0 to 12 (13 different options), which meets the condition. If there are more than 13 girls (at least 14), then the number of boys in the class will not exceed 11, which means the number of different options for t... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,241 |
2+ [Numerical tables and their properties]
Can some number of asterisks be placed in the cells of a $10 \times 10$ square so that in each $2 \times 2$ square there are exactly two asterisks, and in each $3 \times 1$ rectangle - exactly one asterisk? (No more than one asterisk can be placed in each cell.)
# | The first method. Suppose the stars can be arranged in this way. A $10 \times 10$ square can be divided into 25 non-overlapping $2 \times 2$ squares. Since each of them contains two stars, there are a total of 50 stars. On the other hand, 99 cells of the original square can be divided into 33 non-overlapping $3 \times ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,242 |
Avor: Gorinov E.N.
In 10 identical jugs, milk was distributed - not necessarily evenly, but each turned out to be filled to no more than $10 \%$. In one operation, you can choose a jug and pour any part of its contents equally into the other jugs. Prove that in no more than 10 such operations, it is possible to achiev... | Pour from each jug into all the others $1 / 10$ of the initial amount of milk in that jug. Then the milk from each jug will be evenly distributed, meaning there will be an equal amount in each jug. Send comment
Problem | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,246 |
In the country of Courland, there are $m$ football teams (each with 11 players). All the players gathered at the airport to travel to another country for an important match. The plane made 10 flights, each time carrying
$t$ passengers. One more player arrived at the location of the upcoming match by helicopter. Prove t... | Since a total of $10 m+1$ football players have been transported, when distributed into the cell-teams, we get that in some cell, 11 football players are sitting. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,247 |
On the sides of a hexagon, six numbers were written, and at each vertex - a number equal to the sum of the two numbers on the adjacent sides. Then all the numbers on the sides and one number at a vertex were erased. Can the number that was at the vertex be restored?
# | Let $A B C D E F$ be the given hexagon, and we need to restore the number at vertex $A$. Note that the sum of the numbers at vertices $A, C$ and $E$ is equal to the sum of the numbers at vertices $B, D$ and $F$: both these sums are equal to the sum of the numbers originally on all sides of the hexagon. Therefore, to re... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,248 |
[Dirichlet's Principle (angles and lengths).]
On a plane, 12 lines are drawn passing through point O. Prove that it is possible to choose two of them such that the angle between them is less than 17 degrees. | Let's use proof by contradiction. Suppose that the angle between any two of these lines is not less than 17 degrees. These lines divide the plane into 24 angles, so their sum is not less than $24 * 17=408$ degrees, but on the other hand, their sum is 360 degrees, - a contradiction. Therefore, there will be two lines, t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,249 |
[ Semivariants ]
In a country, there are several cities, the pairwise distances between which are all different. A traveler set out from city A to the farthest city B from it, then from there to the farthest city C, and so on. Prove that if C does not coincide with A, then the traveler will never return to A.
# | Each subsequent trip is not shorter than the previous one.
## Solution
Assume that on the second step, the traveler did not return to A, i.e., city C is different from city A. Then the route from A to B is shorter than the route from B to C (since C is the farthest city from B). In the future, each subsequent route w... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 53,251 |
[The Pigeonhole Principle (finite number of points, lines, etc.).]
On a chessboard, more than a quarter of the fields are occupied by chess pieces. Prove that at least two adjacent (side by side or diagonally) cells are occupied.
# | Divide the board into 16 squares of $2 * 2$. If in each of these squares no more than one cell is occupied, then in total no more than 16 cells are occupied by chips, which does not exceed ( $1 / 4$ )*64 contrary to the condition. Thus, in some of the squares, chips occupy no less than two cells. But this means that th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,254 |
On each of the cells of a $9 \times 9$ board, there is a chip. Petya wants to move each chip to an adjacent cell so that there is again one chip in each of the cells. Can Petya do this? | Consider the chessboard coloring of the board. A piece can only be moved from a black square to a white one, and from a white square to a black one.
## Solution
We will color the squares of the board in a checkerboard pattern so that all the corner squares are black. Then, there will be 41 black squares and 40 white ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,255 |
[ Dirichlet's Principle (angles and lengths).]
On a plane, there are $n$ pairwise non-parallel lines. Prove that the angle between some two of them is no more than $180^{\circ} / n$.
# | Let's take an arbitrary point on the plane and draw lines through it parallel to the given ones. They will divide the plane into $2 n$ angles, which sum up to $360^{\circ}$. Therefore, one of these angles does not exceed $180^{\circ} / n$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,256 |
| |
In the bag, there are 70 balls that differ only in color: 20 red, 20 blue, 20 yellow, and the rest are black and white.
What is the smallest number of balls that need to be drawn from the bag, without seeing them, to ensure that among them there are at least 10 balls of the same color? | By drawing 37 balls, we risk getting 9 red, blue, and yellow balls each, and there will be no ten balls of one color. If we draw 38 balls, however, the total number of red, blue, and yellow balls among them will be no less than 28, and the number of balls of one of these colors will be no less than ten (since $28>3 \cd... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,257 |
[ Dirichlet's Principle (continued).]
Some points from a given finite set are connected by segments. Prove that there are two points from which the same number of segments emanate.
# | If there are $n$ points in total, then from each point, there can be from 0 to $n-1$ lines. But there cannot be two points such that from one, $n-1$ lines come out, and from the other, 0. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,258 |
[ Induction in geometry ] [Convex polygons ]
Given two convex polygons $A_{1} A_{2} A_{3} A_{4} \ldots A_{\mathrm{n}}$ and $B_{1} B_{2} B_{3} B_{4} \ldots B_{\mathrm{n}}$. It is known that $A_{1} A_{2}=B_{1} B_{2}, A_{2} A_{3}=B_{2} B_{3}, \ldots$, $A_{\mathrm{n}} A_{1}=B_{\mathrm{n}} B_{1}$ and $n-3$ angles of one po... | Answer: Yes, they will. We apply induction on $n$. For $n=3$, we have two triangles with corresponding sides equal. Now consider two $n$-gons, where $n \geq 4$. By the condition, they have a pair of equal corresponding angles. Cut off a triangle from each polygon, the two sides of which enclose the given angle. These t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,260 |
[ Coordinate method on the plane $]$
Find the equation of the line passing through the intersection point of the lines $3 x+2 y-5=0$ and $x-3 y+2=0$ and parallel to the y-axis. | Solving the system of equations
$$
\left\{\begin{array}{l}
3 x+2 y-5=0 \\
x-3 y+2=0
\end{array}\right.
$$
we find the coordinates of the point $B\left(x_{0} ; y_{0}\right)$ of intersection of these lines: $x_{0}=1, y_{0}=1$.
Since the desired line is parallel to the y-axis and passes through the point $B\left(x_{0} ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,261 |
3 [ Examples and counterexamples. Constructions ]
Is it true that all numbers of the form $p_{1} p_{2} \ldots p_{n}+1$ are prime? ( $p_{k}-k$-th prime number.) | $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13+1=30031=59 \cdot 509$
## Answer
Incorrect.
## Problem | 30031=59\cdot509 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,262 |
$\left[\begin{array}{ll}\text { Pairings and Groupings; Bijections } \\ {[\quad \underline{\text { Modular Arithmetic (other) }}]} \\ {[\quad \text { Factorization }}\end{array}\right]$
Prove that the number $1^{1999}+2^{1999}+\ldots+16^{1999}$ is divisible by 17. | $k^{1999}+(17-k)^{1999}$ is divisible by 17. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,264 |
3 [ Invariants $]
A disappointed investor in the "OilDiamondInvest" fund tore a share into 8 pieces. Not satisfied with this, he tore one of the pieces into 8 more, and so on.
Could he have ended up with 2002 pieces? | Each tear adds 7 pieces, so the number of pieces always gives a remainder of 1 when divided by 7. And 2002 is divisible by 7.
## Answer
It could not. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,265 |
$\left.\begin{array}{ll}{\left[\begin{array}{l}\text { Pairing and groupings; bijections }\end{array}\right]}\end{array}\right]$
Prove that if $n>2$, then the number of all irreducible fractions with denominator $n$ is even. | All such fractions can be split into pairs $k / n, \,^{k-k} / n$. The numbers in such a pair cannot coincide. Indeed, from the equality $k / n=n-k / n$ it follows that $n$ is even, $k=n / 2$ and the fraction $k / n$ can be reduced by $n / 2$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,267 |
$[$ P $[\quad$ Case Analysis $\quad]$
Find all three-digit numbers that are 12 times the sum of their digits.
# | According to the condition, the number is divisible by 3. Therefore, the sum of its digits is also divisible by 3, which means the number itself is divisible by 9.
In addition, it is divisible by 4. Therefore, we need to look for numbers that are divisible by 36. The sum of the digits of a three-digit number does not ... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,268 |
Iterations $\quad]$ [ Binary Number System ]
A flock of geese was flying. On each lake, half of the geese and another half-goose landed. The rest continued flying. All the geese landed on $n$ lakes.
How many geese were there in the flock? | ## Answer
$2^{n}-1$. | 2^{n}-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,270 |
Let $f(x)=x^{2}+p x+q$. For which $p$ and $q$ are the equalities $f(p)=f(q)=0$ satisfied? | The first method. From the condition $2 p^{2}+q=q^{2}+p q+q=0$. Hence, $2 p^{2}-p q-q^{2}=(2 p+q)(p-q)=0$.
For $p=q$, we get $2 p^{2}+p=0$, that is, $p=0$ or $p=-\frac{1}{2}$.
For $q=-2 p$, we get $2 p^{2}-2 p=0$, that is, $p=0$ or $p=1$.
The second method. Consider two cases.
1) $p$ and $q$ are different roots of ... | (0,0),(-\frac{1}{2},-\frac{1}{2}),(1,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,272 |
For what $p$ and $q$ does the equation $x^{2}+p x+q=0$ satisfy two distinct numbers $2 p$ and $p+q$? | According to Vieta's theorem, $3 p+q=-p, 2 p(p+q)=q$. From the first equation, $q=-4 p$. Substituting into the second, we get $6 p^{2}=4 p$, from which $p=0$ or $p=2 / 3$. The first case does not satisfy the condition.
## Answer
$p=2 / 3, q=-8 / 3$. | p=2/3,q=-8/3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,273 |
Solve the system
$x^{2}+y^{2}=1$
$4 x y\left(2 y^{2}-1\right)=1$ | Making the substitution $x=\sin \varphi, y=\cos \varphi$, we obtain the equation $4 \sin \varphi \cos \varphi\left(2 \cos ^{2} \varphi-1\right)=1$, that is, $\sin 4 \varphi=1$.
From this, $\varphi=\pi / 8+k \pi / 2$.
## Answer
$( \pm \sin \pi / 8, \pm \cos \pi / 8),( \pm \sin 5 \pi / 8, \pm \cos 5 \pi / 8)$, or
$\l... | (\\frac{\sqrt{2-\sqrt{2}}}{2},\\frac{\sqrt{2+\sqrt{2}}}{2}),(\\frac{\sqrt{2+\sqrt{2}}}{2},\\frac{\sqrt{2-\sqrt{2}}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,274 |
Prove the inequality for positive values of the variables: $x^{4}+y^{4}+z^{2}+1 \geq 2 x\left(x y^{2}-x+z+1\right)$. | $x^{4}+y^{4}+z^{2}+1-2 x\left(x y^{2}-x+z+1\right)=x^{4}-2 x^{2} y^{2}+y^{4}+z^{2}-2 x z+x^{2}+x^{2}-2 x+1=(x^{2}-y^{2})^{2}+(x-z)^{2}+(x-1)^{2} \geq 0$.
Send a comment
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$x^{4}+y^{4}... | Inequalities | proof | Yes | Yes | olympiads | false | 53,275 | |
Prove the inequality for positive values of the variables: $a^{2}\left(1+b^{4}\right)+b^{2}\left(1+a^{4}\right) \leq\left(1+a^{4}\right)\left(1+b^{4}\right)$. | $2\left(1+a^{4}\right)\left(1+b^{4}\right)-2\left(a^{2}\left(1+b^{4}\right)+b^{2}\left(1+a^{4}\right)\right)=\left(a^{4} b^{4}-2 a^{4} b^{2}+a^{4}\right)+\left(a^{4} b^{4}-2 a^{2} b^{4}+b^{4}\right)+\left(a^{4}-2 a^{2}+1\right)+\left(b^{4}-2 b^{2}+\right.$ 1) $=$
## Problem
Translate the text above into English, pl... | Inequalities | proof | Yes | Yes | olympiads | false | 53,276 | |
[Sum of angles in a triangle. Theorem about the exterior angle.] Proof by contradiction
Given two triangles. The sum of two angles of the first triangle is equal to a certain angle of the second. The sum of another pair of angles of the first triangle is also equal to a certain angle of the second. Is it true that the... | Suppose the first triangle is scalene, $\alpha, \beta$ and $\gamma$ are its angles, and the sums $\alpha+\beta$ and $\beta+\gamma$ are equal to the angles of the second triangle. Since these sums are not equal, they are different angles of the second triangle. Therefore, the sum of the angles of this triangle is greate... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,277 |
3 [ Examples and counterexamples. Constructions ]
Can integers be written in the cells of a $4 \times 4$ table so that the sum of all the numbers in the table is positive, while the sum of the numbers in each $3 \times 3$ square is negative? | The central square of size $2 \times 2$ is contained in each square of size $3 \times 3$. If we place the number -9 in one of the cells of the central square, and fill the rest of the cells of this table with ones, then the sum of all numbers in the table is $15+(-9)=6$, and the sum of the numbers inside any $3 \times ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,278 |
Multiplied several natural numbers and got 224, and the smallest number was exactly half of the largest. How many numbers were multiplied
# | $224=2^{5} \cdot 7$. Consider the two numbers mentioned in the condition: the smallest and the largest. If one of them is divisible by 7, then the other must also be divisible by 7. But 224 is not divisible by 7², so both of these numbers must be powers of two. From the condition, it also follows that these are two con... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,279 |
Masha has two-ruble and five-ruble coins. If she takes all her two-ruble coins, she will be 60 rubles short to buy four pies. If all five-ruble coins - she will be 60 rubles short for five pies. And in total, she is 60 rubles short to buy six pies. How much does a pie cost? | If Masha takes all her two-ruble and five-ruble coins, she will be short of $60+60=120$ rubles for $4+5=9$ pies. On the other hand, she will be short of 60 rubles for 6 pies. That is, three pies cost 60 rubles.
## Answer
20 rubles. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,280 |
A circle intersects the coordinate axes at points $A(a, 0), B(b, 0) C(0, c)$ and $D(0, d)$. Find the coordinates of its center. | The center $O(x, y)$ of the given circle is the intersection of the perpendicular bisectors of the chords $A B$ and $C D$. The equations of these perpendiculars are:
$x=1 / 2(a+b)$ and $y=1 / 2(c+d)$ respectively.
## Answer
$\left(\frac{a+b}{2} ; \frac{c+d}{2}\right)$ | (\frac{+b}{2};\frac{+}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,282 |
[ Quadratic equations and systems of equations ] [Completing the square. Sums of squares] Solve the system: $\left\{\begin{array}{c}x-y \geq z \\ x^{2}+4 y^{2}+5=4 z\end{array}\right.$. | Multiply both sides of the inequality by 4 and substitute the value of $4z$ from the equation. We get:
$x^{2}+4 y^{2}+5 \leq 4 x-4 y \Leftrightarrow(x-2)^{2}+(2 y+1)^{2} \leq 0 \Leftrightarrow x=2, y=-0.5$. Substituting the found values into the equation, we get $z=2.5$.
## Answer
$(2; -0.5; 2.5)$. | (2;-0.5;2.5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,288 |
| $\left[\begin{array}{l}{\left[\begin{array}{l}\text { Decimal numeral system }\end{array}\right]} \\ {[\text { Examples and counterexamples. Constructions }]}\end{array}\right.$ |
| :---: | :---: |
Does there exist a digit $a$ such that $\overline{\operatorname{aaa(a-1)}}=(a-1)^{a-2}$. | When $a=7$, we get the correct equality $7776=6^{5}$.
## Answer
It exists. | 7776=6^{5} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,289 |
At the end of the chess tournament, Nезнайка said: "I scored 3.5 points more than I lost." Could his words be true?
(Victory - 1 point, draw - $1 / 2$ point, loss - 0.) | Before the tournament, the difference between the points scored and lost by Nезнайка was zero. If a game is decisive, this difference changes by 1, and if the game ends in a draw, the difference does not change. Therefore, such a difference can only take integer values, that is, it cannot be equal to 3.5.
## Otвет
Ca... | Cannot | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,290 |
Folklore
To repair the propeller, Karlson needs to buy three blades and one screw. In the store, blades cost 120 tugriks each and screws cost 9 tugriks each. However, after a purchase of at least 250 tugriks, a 20% discount is given on all subsequent purchases. Will Karlson be able to repair the propeller if he has on... | Let the first purchase be two blades and two screws, costing $2(120+9)=258$ tugriks. Since the cost of the purchase is more than 250 tugriks, Carlson can buy the third blade with a $20\%$ discount, spending $120 \cdot 0.8=96$ tugriks. Therefore, in total, Carlson will spend $258+96=354$ tugriks.
## Answer
He can. | 354 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,292 |
It is known that ЖЖ + Ж = МЁД. What is the last digit of the product: В $\cdot И \cdot H \cdot H \cdot U \cdot \Pi \cdot У \cdot X$ (different letters represent different digits, the same letters represent the same digits)?
# | Since a two-digit number ЖЖ was added to a one-digit number Ж to get a three-digit number, then Ж $=9$, and МЁД = 108. Four digits have already been used. In the product $\mathrm{B} \cdot \mathrm{U}^{\prime} \cdot \mathrm{H} \cdot \mathrm{H} \cdot И \cdot П \cdot У \cdot \mathrm{X}$, six other digits are used.
Therefo... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,293 |
[ Higher Order Roots (Miscellaneous). ] [ Examples and Counterexamples. Constructions ]
Is there a natural number $n$, greater than 1, such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number? | For example, $n=2^{8}=256$. Indeed, $\sqrt{n \sqrt{n \sqrt{n}}}=n^{7 / 8}$. For $n=2^{8}$, the value of this expression is $2^{7}=$ 128.
## Answer
It exists. | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,294 |
[ Research on quadratic trinomials ] [ Examples and counterexamples. Constructions ]
It is known that $b-c>a$ and $a \neq 0$. Is it necessarily true that the equation $a x^{2}+b x+c=0$ has two roots? | Counterexample: $a=c=-1, b=1.1-(-1)>1$, but the equation $-x^{2}+x-1$ has no roots.
## Answer
Not necessarily. | Notnecessarily | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,295 |
A square with a side of 9 cells was cut along the grid lines into 14 rectangles in such a way that the length of each side of any rectangle is no less than two cells. Could it be that there was not a single square among these rectangles? | The area of a rectangle, different from a square, with the given sides is no less than $2 \cdot 3=6$ cells. Since $6 \cdot 14>9^{2}$, the area of 14 such rectangles is definitely greater than the area of the original square.
Therefore, it is impossible to do without squares.
## Answer
It could not. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,296 |
$\underline{\text { Folklore }}$
Malvina asked Buratino to cut a square into 7 rectangles (not necessarily different), each of which has one side twice as long as the other. Is it possible to complete this task? | In the figure, two examples are shown: for square grids with sides of 6 and 8 cells, respectively.
## Answer
Feasible. | Feasible | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,297 |
Authors: Berrov S.L, Petrov F.
Inside a convex 100-gon, a point $X$ is chosen, not lying on any of its sides or diagonals. Initially, the vertices of the polygon are not marked. Petya and Vasya take turns marking unmarked vertices of the 100-gon, with Petya starting and marking two vertices on his first move, and then... | We will color the sides of a 100-gon in black and white such that any two adjacent sides have different colors. Consider two monochromatic sides $AB$ and $CD$ forming a convex quadrilateral $ABCD$; let the diagonals $AC$ and $BD$ intersect at point $K$. Suppose point $X$ lies inside triangle $KBC$ (see the left figure)... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,299 |
[ Tournaments and tournament tables [ Pigeonhole Principle (other). ]
Authors: Antropov A., Frenkin B.R.
In a sports club, a tennis championship is taking place. The loser of a match is eliminated from the competition (there are no draws in tennis). The pair for the next match is determined by lot. The first match wa... | Let by some moment $n$ games have taken place. This means that $n$ people have been eliminated. Since the first game was judged by an invited judge, no more than $n-1$ eliminated people participated in the judging. Thus, there is someone to judge the next game.
## Answer
It could not.
Send a comment | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,300 |
Evochkimov M.
The quadratic trinomial $x^{2}+b x+c$ has two real roots. Each of its three coefficients was increased by 1.
Could it have turned out that both roots of the trinomial also increased by 1? | Suppose this happened. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}+b x+c=0$. Then $x_{1}+x_{2}=-b, x_{1} x_{2}=c$, $x_{1}+x_{2}+2=-{ }^{b+1} / 2$
$\left(x_{1}+1\right)\left(x_{2}+1\right)={ }^{c+1} / 2$. From here, $b=5, c=9$.
Thus, the desired quadratic polynomial is $x^{2}+5 x+9$. However, the discrimina... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,302 |
[ Irrational Equations $]$ [ Monotonicity and Boundedness
Solve the equation $2017 x^{2017}-2017+x=\sqrt[2017]{2018-2017 x}$.
# | The function $f(x)=2017 x^{2017}-2017+x$ is increasing, while the function $g(x)=\sqrt[2017]{2018-2017 x}$ is decreasing. Therefore, the equation $f(x)=g(x)$ has no more than one root. However, it is obvious that $f(1)=g(1)$.
## Answer
$x=1$.
Author: Volienkov S.G.
A sheet of paper has the shape of a circle. Can fi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,303 |
Shapovalov A.V.
Given four sticks. It turns out that from any three of them, a triangle can be formed, and the areas of all four triangles are equal. Is it necessary that all sticks are of the same length? | Consider 3 sticks of length 1 and a stick of length $a$. Then, from the sticks, one can either form an equilateral triangle with side 1 or a triangle with sides $1,1, a$. To find $a$ such that these triangles have equal area, it is sufficient for the heights to the sides of length 1 in these triangles to coincide, whic... | \sqrt{3} | Geometry | proof | Yes | Yes | olympiads | false | 53,304 |
$\underline{\text { Folklore }}$
On a contour map of Russia, there are 85 regions. Vovochka wants to paint each region in white, blue, or red so that white and red do not share a common border. At the same time, one or even two colors can be unused. Prove that the number of such coloring options is odd. | For each coloring option that includes at least one region in white or red, we can assign a pair: a coloring option where all white regions are repainted red and all red regions are repainted white. In this case, the option will change, but white and red colors will still not be adjacent.
Thus, the number of coloring ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,306 |
On a $6 \times 6$ board, six non-attacking rooks were placed. Then, each unoccupied square was painted according to the following rule: if the rooks threatening this square are at the same distance from it, the square is painted red; if they are at different distances, the square is painted blue. Could all unoccupied s... | A cell will be red if the rooks attacking it are on the same diagonal, and blue otherwise.
a) Place the rooks on the main diagonal.
b) The rooks must not attack each other like queens. See the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_93ee318ee1723cbdd653g-19.jpg?height=326&width=349&top_left_y=609&top_... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,307 |
Evdokimov M.A.
Does there exist a number divisible by 2020 in which all digits $0,1,2, \ldots, 9$ are present in equal amounts? | $2020=20 \cdot 101$, so, for example, the number 10198987676545432320 works.
## Answer
It exists. | 10198987676545432320 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,308 |
Sstrechkova N.P.
Nезнayka (which translates to "Don't-know" in English) does not know about the operations of multiplication and exponentiation. However, he has well mastered addition, subtraction, division, and square root extraction, and he also knows how to use parentheses.
Practicing, Nезнayka chose three numbers... | $\$|frac{20}{2-\\ sqrt{2}}=| frac{20(2+\\ operatorname{sqrt}{2})}{2}=20+10 \\$ sqrt ${2}>20+10 . \\$ There are other solutions.
## Answer
It can.
| 20+10\sqrt{2}>30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,309 |
Bakayev E.V.
Can numbers, among which there are no identical ones, be placed in the cells of a $6 \times 6$ table so that in each $1 \times 5$ rectangle (both vertical and horizontal) the sum of the numbers is equal to 2022 or 2023?
# | Let it be possible. The numbers in adjacent corners differ by 1, since each of them complements the four cells between them to a rectangle $\$ 1$ lime $5 \$$ . Let $\$ a \$$ be the smallest number of the corner ones. Then in the adjacent corners there are numbers $\$ a+1 \$$ . Contradiction.
## Answer
it is impossibl... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,312 |
Evochkimov M.A.
Can all natural numbers greater than 1 be colored in three colors (each number in one color, all three colors must be used) so that the color of the product of any two numbers of different colors differs from the color of each of the factors? | Let it be possible to color the numbers in blue, red, and green. We can assume that the number 2 is blue, and 4 is not red. Take a red number $\$ \mathrm{k} \$$. Then the number $\$ 2 \mathrm{k} \$$ is green, and $\$ 2 \cdot (2 \mathrm{k}) \$$ is red. On the other hand, $\$ 4 \cdot \mathrm{k} \$$ is not red. Contradict... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,313 |
8,9,10 |
| :---: | :---: | :---: |
| | [ Formulas for abbreviated multiplication (other)] | |
| | Evenness and oddness | |
Prove that in the product $\left(1-x+x^{2}-x^{3}+\ldots-x^{99}+x^{100}\right)\left(1+x+x^{2}+x^{3}+\ldots+x^{99}+x^{100}\right)$, after expanding the brackets and combining like terms, there a... | The function $P(x)$ under consideration is an even function: $P(-x)=P(x)$.
## Solution
The given product is equal to $\frac{x^{101}+1}{x+1} \cdot \frac{x^{101}-1}{x-1}=\frac{x^{202}-1}{x^{2}-1}=x^{200}+x^{198}+\ldots+x^{2}+1$.
Submit a comment | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,314 |
There is a closed self-intersecting broken line. It is known that it intersects each of its segments exactly once. Prove that the number of segments is even.
# | Through the point of self-intersection, exactly two segments of the broken line pass (if three segments passed, then each of them would intersect the broken line at least twice). In addition, on each segment lies exactly one point of self-intersection. Therefore, by associating the point of self-intersection with the p... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,317 |
For any numbers $a_{1}$ and $a_{2}$, satisfying the conditions $a_{1} \geq 0, a_{2} \geq 0, a_{1}+a_{2}=1$, there exist numbers $b_{1}$ and $b_{2}$ such that $b_{1} \geq 0, b_{2} \geq 0, b_{1}+b_{2}=1$, $\left(5 / 4-a_{1}\right) b_{1}+3\left(5_{4}-a_{2}\right) b_{2}>1$. Prove. | Suppose that $5 / 4-a_{1} \leq 1$ and $3\left(\frac{5}{4}-a_{2}\right) \leq 1$. Multiplying the first inequality by 3 and adding it to the second, we get $3\left(5 / 2-a_{1}-a_{2}\right) \leq 4$, from which $a_{1}+a_{2} \geq 5 / 2-4 / 3=7 / 6>1$, which contradicts the condition.
Thus, one of the numbers $5 / 4-a_{1} \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,319 |
[ $\underline{\text { Induction (other) }}$. ] $[$ Division with remainder $]$
Prove that any proper fraction can be represented as a (finite) sum of reciprocals of pairwise distinct integers. | Let $m / n$ be a proper fraction. We will prove the required statement by induction on $m$.
Base. For $m=1$, the statement is obvious.
Induction step. Let $m>1$. Write $n$ as $n=q m-r$, where $0 \leq r<m$. Then $m / n=1 / q+r / q n$. If $r=0$, everything is fine. Otherwise, the proper fraction ${ }^{r} /$ qn can be r... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,320 |
[ GCD and LCM. Mutual simplicity ]
Prove that the product of two consecutive natural numbers is not a power of any integer.
# | Suppose that $n(n+1)=a^{m}$, where $m \geq 2$ ( $a, m$ and $n-$ natural numbers). The numbers $n$ and $n+1$ have no common divisors, so $n=b^{m}$ and $n+1=c^{m}$. But this cannot be, because the difference between two consecutive $m$-th powers is greater than 1. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,322 |
\[
\left[\begin{array}{l}
\text { Equations in integers } \\
{[\text { Case enumeration }}]
\end{array}\right]
\]
Solve the system in natural numbers
\[
\begin{aligned}
& x+y=z t, \\
& z+t=x y .
\end{aligned}
\] | Let's consider two cases.
1) $x=1$. Then $z+t=y=z t-1$, from which $(z-1)(t-1)=z t-z-t+1=2$. Therefore, $\{z, t\}=\{2,3\}, y=6-$ $1=5$.
2) $x, y, z, t>1$. Then $(x-1)(y-1) \geq 1$, that is, $x y \geq x+y$. Similarly, $z t \geq z+t$. Adding these, we get $x y+z t$ $\geq x+y+z+t$. On the other hand, adding the equations... | (1,5,2,3),(5,1,2,3),(1,5,3,2),(5,1,3,2),(2,3,1,5),(2,3,5,1),(3,2,1,5),(3,2,5,1),(2,2,2,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,323 |
$\left[\begin{array}{l}\text { Coloring } \\ {[\underline{\text { Induction in Geometry }}]}\end{array}\right]$
The island of Tolpygo has the shape of a polygon. Several countries are located on it, each of which has the shape of a triangle, and any two adjacent countries have a common whole side (i.e., a vertex of on... | We will call such a division of the island into triangles a triangulation. Let's prove the required statement by induction on the number of triangles in the triangulation. For one triangle, the required coloring exists. Now suppose that any triangulation consisting of fewer than $n$ triangles can be colored in the requ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,325 |
[ Principle of the Extreme ] [ Problems with Constraints ]
Prove that in an arbitrary convex $2 n$-gon, there exists a diagonal that is not parallel to any of its sides. | Diagonals parallel to a given side do not exceed $n-2$ (through the farthest vertex from the side, a diagonal parallel to this side cannot pass). Therefore, the total number of diagonals parallel to the sides does not exceed $2 n(n-2)$, which is less than the number $n(2 n-3)$ of all diagonals of a $2 n$-gon (see probl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,327 |
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