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A set of several numbers, none of which are the same, has the following property: the arithmetic mean of some two numbers from this set is equal to the arithmetic mean of some three numbers from the set and is equal to the arithmetic mean of some four numbers from the set. What is the smallest possible number of number... | Let $C\left(a_{1}, \ldots, a_{k}\right)$ be the arithmetic mean of the numbers $\left(a_{1}, \ldots, a_{k}\right)$. Note that adding a number different from the arithmetic mean of a set changes the original arithmetic mean of the set.
Suppose that $(a, b, c, d)$ is a set of four numbers satisfying the condition, and $... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
A monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines? | Let's put the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed no more than one tangerine to each monkey, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. Here's how we can satisfy 45 monkeys:
5 monkeys eat a pear, a banana, and ... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berroov S.L.
One hundred integers are written in a circle. Each number is greater than the sum of the two numbers following it in a clockwise direction.
What is the maximum number of positive numbers that can be among those written?
# | Evaluation. Suppose two non-negative numbers stand next to each other. Then the number preceding them is greater than their sum, meaning it is positive. Similarly, the number before it is also positive, and so on. In the end, we get that all numbers are non-negative; but then the smallest of them cannot be greater than... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility of numbers. General properties ] [ Examples and counterexamples. Constructions ]
Natural numbers $A$ and $B$ are divisible by all natural numbers from 1 to 65. What is the smallest natural number that $A+B$ may not be divisible by? | From the condition, it follows that $A+B$ is divisible by all numbers from 1 to 65. This sum is also divisible by $66=2 \cdot 33$. However, $A+B$ is not necessarily divisible by 67. For example,
$A=65!, B=2 \cdot 65!, A+B=3 \cdot 65!$ which is not divisible by 67.
## Answer
Not divisible by 67. | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
P.G. Zhenoarove
In a box, there are 111 balls: red, blue, green, and white. It is known that if you pull out 100 balls from the box without looking, there will definitely be four balls of different colors among them. What is the smallest number of balls that need to be pulled out without looking to ensure that there a... | There are no less than 12 balls of each color (otherwise, they could all end up among the 11 remaining in the box). Therefore, the number of balls of two colors is no more than
$111-24=87$. Hence, among any 88 balls, there will be balls of at least three colors.
87 balls are insufficient, for example, for the color ... | 88 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
In a box, there are 100 balls: white, blue, and red. It is known that if 26 balls are drawn from the box without looking, there will definitely be 10 balls of the same color among them. What is the smallest number of balls that need to be drawn from the box, without looking, to ensure that there are 30 ... | Let's prove that 66 balls are sufficient. Suppose among them there are no more than 29 of each color. Then there are no less than $66-2 \cdot 29=$ 8 balls of each color and no less than $66-29=17$ of any two colors. But then there is a set of 8, 9, 9, which contradicts the condition.
65 balls are insufficient, for exa... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
Forty children were holding hands in a circle. Out of them, 22 were holding hands with a boy and 30 were holding hands with a girl. How many girls were in the circle?
# | $22+30=52$, so $52-40=12$ children held hands with both a boy and a girl. Therefore, $30-12=18$ children held hands only with girls. These 18 children held $18 \cdot 2=36$ girls' hands, and the other 12 held one girl's hand each, so the girls had a total of $36+12=48$ hands. Therefore, there were $48: 2=24$ girls.
## ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The Unified State Exam (USE) in mathematics in the magical country of Oz is organized as follows. Each paper is independently checked by three teachers, and each gives 0 or 1 point for each problem. Then the computer finds the arithmetic mean of the scores for that problem and rounds it to the nearest integer. The poin... | It's easy to come up with a case where you get 4 points. To get 5 or more, there must be at least 10 units in total, but the teachers have only given 9 units in total. Therefore, it's impossible to get 5.
## Answer
4 points. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
A hundred bear cubs found berries in the forest: the youngest managed to grab 1 berry, the next older cub - 2 berries, the next one - 4 berries, and so on, with the oldest getting \(2^{99}\) berries. The fox suggested they divide the berries "fairly." She can approach two bear cubs and distribute their be... | Notice that at each step, if both participating cubs had at least one berry, then each of them will have at least one berry left. Therefore, in the end, each cub will have at least one berry.
We will prove that the fox can leave each cub exactly one berry, that is, transition from the position (1, $2, \ldots, 2^{99}$)... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Pythagorean Theorem (direct and inverse). ] [ Examples and counterexamples. Constructions ]
Is there a right triangle in which the lengths of two sides are integers, and the length of the third side is $\sqrt{2016}$? | Consider, for example, a right-angled triangle with legs $\sqrt{2016}$ and 3. Its hypotenuse is equal to $\sqrt{2016+9}=\sqrt{2025}=45$.
## Answer
It exists. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the table lies a rectangular sheet of paper. Sasha cuts it along a straight line into two pieces and puts the pieces on the table. Then he takes one of the pieces, cuts it again along a straight line into two pieces, and puts the pieces back on the table. Then he takes one piece from the table and cuts it again, and... | Notice that after one cut, the total number of vertices increases by two (if the cut passes through two vertices), by three (if the cut passes through a vertex and a point inside a side), or by four (if the cut passes through the interior points of two sides). Suppose $k$ cuts have been made, resulting in $k+1$ pieces,... | 2015 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{ll}{\left[\begin{array}{l}\text { Common fractions } \\ \text { [Examples and counterexamples. Constructions ] }\end{array}\right]}\end{array}\right]$
Author: Akonn E, Kaminin D.
Can the stars in the equation $\frac{*}{*}+\frac{*}{*}+\frac{*}{*}+\frac{*}{*}=$ * be replaced with the digits from 1 ... | From the possible examples, let's give two: $7 / 4+6 / 8+5 / 1+3 / 2=9$, $5 / 4+6 / 8+9 / 3+2 / 1=7$.
## Answer
It can be done. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Frankin B.R.
There are 100 non-zero numbers written in a circle. Between every two adjacent numbers, their product was written, and the original numbers were erased. The number of positive numbers did not change. What is the minimum number of positive numbers that could have been written initially?
# | Evaluation. Suppose there were no more than 33 positive numbers. Negative numbers in the product can only be formed if one of the factors is positive, and each positive number can participate in no more than two such products. Therefore, there can be no more than 66 negative numbers. But then the total number of number... | 34 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a theater troupe, there are 60 actors. Any two of them have at least once played in the same play. In each play, no more than 30 actors are involved.
What is the minimum number of plays the theater could have staged? | Example. Let's divide the troupe into four groups of 15 people and hold 6 performances, in each of which some two groups are involved. The number of ways to choose two groups out of four is 4$\cdot$3:2=6.
Estimate. In total, the actors played no more than $30 \cdot 5=150$ roles, so if there are five performances, ther... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Lobov A.
Let \( n \) be a natural number. We will call a sequence \( a_1, a_2, \ldots, a_n \) interesting if for each \( i = 1, 2, \ldots, n \), one of the equalities \( a_i = i \) or \( a_i = i + 1 \) holds. We will call an interesting sequence even if the sum of its terms is even, and odd otherwise. For each odd int... | Denoting the sum containing the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$ by $A_n$, and the other by $B_n$, we will prove the equality $A_n - B_n = 1$ by induction.
Base case. $A_1 - B_1 = 2 - 1 = 1$.
Inductive step. Represent the sum $A_n$ as $A' + A''$, where $A'$ contains all terms of the form
$a_1a_2...a_{n-1}(... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { [Quadratic inequalities and systems of inequalities } \\ {[\quad \text { Pairing and grouping; bijections }]}\end{array}\right]$
Given two sets of numbers $a_{1}>a_{2}>\ldots>a_{n}$ and $b_{1}>b_{2}>\ldots>b_{n}$. Prove that $a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}>a_{1} b_{n}+$ $a_{2... | Let's group the terms in both expressions that are equidistant from the ends. As a result, we get that the difference between the expression on the left and the expression on the right is the sum of expressions of the form \(a_{k} b_{k} + a_{n+1-k} b_{n+1-k} - a_{k} b_{n+1-k} - a_{n+1-k} b_{k} = (a_{k} - a_{n+1-k})(b_{... | 26 | Inequalities | proof | Yes | Yes | olympiads | false |
[ Divisibility rules for 3 and 9 $]$
There is a three-digit number $\overline{a b c}$, we take $\overline{c b a}$ and subtract the smaller from the larger. We get the number $\overline{a_{1}^{1}} \overline{b_{1}} \overline{1} \bar{c}_{1}^{-}$, and do the same with it, and so on.
Prove that at some step we will get ei... | 1) $b_{1}=9$
2) $\bar{a}_{1}^{-} \bar{b}_{1}^{-} \bar{c}_{1}^{-}$ is divisible by 9.
Therefore, $\overline{a_{1}} \overline{\overline{1}} \overline{\overline{1}} \bar{c}_{1}^{-}$ is one of the numbers $99,198,297,396,495,594,693,792,891,990$. It remains to note that $594 \rightarrow 99$
$\rightarrow 891 \rightarrow 6... | 495 | Number Theory | proof | Yes | Yes | olympiads | false |
[ Regular Polyhedra. Duality and Relationships ] [ Distance between_two points. Equation of a sphere ]
Authors: Rabzimiotskyl., Giadkih A.
Can an octahedron be inscribed in a cube such that the vertices of the octahedron lie on the edges of the cube?
# | In the figure, an octahedron is inscribed in a cube with an edge length of 4; the vertices of the octahedron divide the edges of the cube in the ratio $1: 3$. The square of the length of each edge of the octahedron is 18 (either $3^{2}+3^{2}$, or $4^{2}+1^{2}+$ $\left.1^{2}\right)$, that is, all its edges are equal.
!... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Evoikinov M.A.
A pirate has five bags of coins, each containing 30 coins. He knows that one bag contains gold coins, another contains silver coins, a third contains bronze coins, and each of the two remaining bags contains an equal number of gold, silver, and bronze coins. You can simultaneously take any number of coi... | Example. Let's take one coin from each bag. Among these five coins, there are coins of all three types, so there is only one coin of a certain type. If it is, for example, a gold coin, then it was taken from the bag with gold coins. Indeed, for each coin from the "mixed" bag, there is a matching one from the correspond... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Prime numbers and their properties $]$ [ Examples and counterexamples. Constructions ]
Find the smallest natural number $n$, for which the following condition is satisfied: if the number $p-$ is prime and $n$ is divisible by $p-1$, then $n$ is divisible by $p$. | Let $n$ satisfy this condition. Since $n$ is divisible by $1=2-1$, it must be divisible by 2, but then it is divisible by $3=2+1$, by
$7=2 \cdot 3+1$ and by $43=2 \cdot 3 \cdot 7+1$. Therefore, $n$ must be divisible by $1806=2 \cdot 3 \cdot 7 \cdot 43$. Hence, the minimum $n$ (if it exists) is no less than 1806.
On t... | 1806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
A cube with a side of 20 is divided into 8000 unit cubes, and a number is written in each cube. It is known that in each column of 20 cubes, parallel to the edge of the cube, the sum of the numbers is 1 (columns in all three directions are considered). In a certain cube, the number 10 is written. Thro... | Through the given cube K, one horizontal layer G and two vertical layers pass. The sum of all numbers in 361 vertical columns, not included in the last two layers, is 361. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with G (there are 361 such cub... | 333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Among the angles of each lateral face of a pentagonal prism, there is an angle $\varphi$. Find all possible values of $\varphi$. | In a right prism, each lateral face is a rectangle, so $\varphi=90^{\circ}$ is suitable.
Suppose $\varphi \neq 90^{\circ}$. We can assume that $\varphi<90^{\circ}$ (if the obtuse angles in the parallelograms are equal, then the acute angles are also equal). Draw in the plane of the base through one of the vertices $V$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Blinkov A.d:
In a football championship, 16 teams participated. Each team played against each of the others once, with 3 points awarded for a win, 1 point for a draw, and 0 for a loss. We will call a team successful if it scored at least half of the maximum possible number of points. What is the maximum number of succ... | Each team played 15 games and therefore could have earned a maximum of $15 \cdot 3=45$ points. Thus, a team is successful if it has at least 23 points.
But one of the teams scored no more than the average possible number of points. Even if all matches were successful, the average is
$15 \cdot 1.5=22.5$.
We will show... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Agakhanov N.K.
The cosines of the angles of one triangle are respectively equal to the sines of the angles of another triangle.
Find the largest of the six angles of these triangles. | From the condition, it follows that the angles $\alpha_{1}, \alpha_{2}, \alpha_{3}$ of the first triangle are acute $\left(\cos \alpha_{i}=\sin \beta_{i}>0\right.$, where $\beta_{1}, \beta_{2}, \beta_{3}$ are the angles of the second triangle). Therefore, $\beta_{i}=90^{\circ} \pm \alpha_{i}$,
$i=1,2$, 3. From the equ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Akopyan A.V.
Through a terminal, money can be transferred to a mobile phone, with a commission of a natural number of percent. Fedya put an integer amount of rubles on his mobile phone, and his account was credited with 847 rubles. How much money did Fedya put on the account, given that the commission is less than $30... | Let Fedia put $n$ rubles, and the commission is $k \%$. Then $(1-k / 100) n=847$, that is, $84700=$ $(100-k) n .84700=2 \cdot 2 \cdot 5 \cdot 5 \cdot 7 \cdot 11 \cdot 11$. According to the condition $70<100-k<100$, so it is necessary to find all numbers,
dividing 84700, in this range. A small enumeration shows that the... | 1100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
|
| | $[$ Chess Coloring $]$ | |
Author: Raskina I.V.
At the edge of a round rotating table, 30 cups of tea were placed at equal intervals. The March Hare and Alice sat down at the table and started drinking tea from two of the cups (not necessarily adjacent). When they finished their tea, the Hare turned the table... | We will paint every other cup blue and red. Let the March Hare drink from a red cup at first. We will prove that Sonya drank from a blue cup at first. Indeed, if she drank from a red one, then after any rotation of the table, two cups of the same color would be emptied. Since there are 15 of each color, and they are em... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false |
Folklore
In a certain state, the airline system is arranged in such a way that each city is connected by air routes to no more than three other cities, and from any city, you can reach any other city with no more than one layover. What is the maximum number of cities that can be in this state? | Evaluation. From a fixed city $A$, one can directly reach no more than three cities, and with one transfer - no more than $3 \cdot 2=6$ additional cities. Thus, the total number of cities can be no more than ten. An example of a network of 10 cities is shown in the figure.
. Between them lies $k_{i}-i-1$ cards, so the desired sum $S=\left(k_{... | 1260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Class $\mathbf{M}$.
In the class, there are 15 boys and 15 girls. On March 8, some boys called some girls to congratulate them on the holiday (no boy called the same girl twice). It turned out that the children could be uniquely divided into 15 pairs so that in each pair there was a boy who called the girl. What is th... | Let's denote the boys as $M_{1}, M_{2}, \ldots, M_{15}$, and the girls as $-D_{1}, D_{2}, \ldots, D_{15}$, such that $M_{1}-D_{1}, M_{2}-D_{2}, \ldots, M_{15}-D_{15}$ is the only pairing that satisfies the condition of the problem. Suppose each boy called at least two girls. We will draw an arrow from each girl $D_{i}$... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A.Y. Evnin
2000 people registered on a new website. Each of them invited 1000 people to be their friends. Two people are considered friends if and only if each of them invited the other to be a friend. What is the minimum number of pairs of friends that could have formed? | Evaluation. A total of 2,000,000 invitations were sent, while the number of pairs on the site is $1000 \cdot 1999 = 1999000$. There are 1000 more invitations than pairs, so within at least 1000 pairs, two invitations were sent. Therefore, at least 1000 pairs were formed.
Example: Place everyone at the vertices of a re... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-
Can we, by applying the functions sin, cos, tan, cot, arcsin, arccos, arctan, arccot to the number 1 in some order, obtain the number 2010? (Each function can be used any number of times.) | Let $f(x)=\operatorname{ctg}(\operatorname{arctg} x)=1 / x, g(x)=\sin (\operatorname{arctg} x)=\frac{x}{\sqrt{1+x^{2}}}$. By induction, we check that $g^{n}(x)=g\left(g^{n-1}(x)\right)=$ $\frac{x}{\sqrt{1+n x^{2}}}$
Taking $n=2010^{2}-1$, we get that $g^{n}(1)=1 / 2010$, from which $f\left(g^{n}(1)\right)=2010$.
## A... | 2010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Division with remainder ]
[ Evaluation + example ]
Author: Polish A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 222 nuts into two small boxes. After looking at
the distribution, Chichikov names any integer \( N \) from 1 to 222. Then Nozdryov must, if necessary, move one or several nuts to an emp... | Upper bound. By placing 74 and 148 nuts in the boxes, Nozdryov can, for any $N$, move no more than 37. Indeed, $N$ can be written in the form $74 k+r$, where $k=0,1,2,3$, and $-37 \leq r0$, and the numbers 74, 148, and 222 can be collected with one or two boxes, without moving anything. If $r0,74 k$ $=0,74$ or 148. By ... | 37 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Podolsky A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 1001 nuts among three small boxes. After looking at the distribution, Chichikov names any integer \( N \) from 1 to 1001. Then Nozdryov must, if necessary, move one or several nuts to an empty fourth box and show Chichikov one or several boxes... | Upper bound. By placing 143,286 = 2 * 143 and 572 = 4 * 143 nuts in the boxes, Nozdryov can, for any N, move no more than 71. Indeed, N can be represented as 143k + r, where 0 ≤ k ≤ 7, and -71 ≤ r < 71, and the number 143k can be collected with one or several boxes without moving anything. If r < 0, then 1001 - N = 143... | 71 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folklore
In a football championship, 18 teams are participating. As of today, 8 rounds have been played (in each round, all teams are divided into pairs, and the teams in each pair play against each other, with no pairs repeating). Is it true that there will be three teams that have not played a single match against e... | Consider one of the teams, denoted as A. Over 8 rounds, it played against eight teams and did not play against nine teams. If among these nine teams, there are two teams $B$ and $C$ that did not play against each other, then $A, B$, and $C$ form the desired triplet.
Otherwise, these 9 teams played a full round-robin t... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.v.
Fox Alice and Cat Basilio have grown 20 fake banknotes on a tree and are now filling in seven-digit numbers on them. Each banknote has 7 empty cells for digits. Basilio calls out one digit at a time, either "1" or "2" (he doesn't know any other digits), and Alice writes the called digit in any free cell... | Basilio can always get two banknotes: he knows where the last digit should be written and names it so that it differs from the digit in the same position on another banknote. Then the numbers on these two banknotes will be different, and the cat can take them.
Let's show how Alice can ensure that there are no more tha... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Blinkov A.D:
The teams held a football tournament in a round-robin format (each team played one match against every other team, with 3 points for a win, 1 point for a draw, and 0 points for a loss). It turned out that the sole winner scored less than $50 \%$ of the maximum possible points for one participant. What is ... | Let's prove that there could not have been fewer than six teams. If, for example, there were five teams in the tournament, then they played $5 \cdot 4: 2=10$ matches and scored a total of at least 20 points. Therefore, the sole winner scored more than $20: 5=4$ points. However, according to the condition, he scored no ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7,8,9 |
In a deck of 16 cards, numbered from top to bottom. It is allowed to take a portion of the deck from the top, after which the removed and remaining parts of the deck, without flipping, are "interleaved" with each other. Can it happen that after several such operations, the cards end up numbered from bottom to ... | Let's consider a method that allows achieving the required order in four operations. Each time, we will take exactly half of the deck - 8 cards from the top and "interleave" the removed part into the remaining part "one by one". The transformation of the deck during such operations is shown in the diagram:
| Top | | ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On a bookshelf, 30 volumes of an encyclopedia stand in some order. In one operation, it is allowed to swap any two adjacent volumes. What is the minimum number of operations required to guarantee that all volumes can be arranged in the correct order (from the first to the thirtieth from left to right) regardless of the... | Let's call a disorder a pair of volumes where the volume with the larger number is to the left of the volume with the smaller number. In one operation, the number of disorders changes by no more than 1.
## Solution
Suppose we have some arrangement of volumes on a shelf. Consider all possible pairs of volumes (there a... | 435 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
99 children stand in a circle, each initially having a ball. Every minute, each child with a ball throws their ball to one of the two neighbors; if two balls land with the same child, one of these balls is lost irretrievably. What is the minimum time after which only one ball may remain with the children? | Number the children and balls clockwise from 1 to 99.
Example. Suppose children 1 and 2 toss the first ball to each other. The other balls with odd numbers are always thrown counterclockwise until they reach the second child, who discards them (this happens after an odd number of minutes, and at that moment he also re... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Fon-der-Flaass }}$ D:
In the vertices of a cube, numbers $1^2, 2^2, \ldots, 8^2$ are placed (one number in each vertex). For each edge, the product of the numbers at its ends is calculated. Find the maximum possible sum of all these products. | We will color the vertices of a cube in two colors such that the ends of each edge are of different colors. Let the numbers $a_{1}, a_{2}, a_{3}, a_{4}$ be placed in the vertices of one color, and the numbers $b_{1}, b_{2}, b_{3}, b_{4}$ in the vertices of the other color, with numbers having the same indices placed in... | 9420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Linear Inequalities and Systems of Inequalities ] Evaluation + Example
Authors: Bogdanov I.I., Knop K.A.
King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, 2, ..., 11 kg. He also has a bag that will tear if more than 11 kg is placed in i... | Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then ingots weighing 1, 4, and 6 kg. In both cases, the bag does not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_{1}, \ldots, w_{6} \) kg instead of ingots ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Keshukhin } 0 . H .}$
Three cyclists are riding in the same direction on a circular track 300 meters long. Each of them moves at their own constant speed, and all speeds are different. A photographer can take a successful photo of the cyclists if all of them end up on some segment of the track $d$ ... | Without loss of generality, we assume that all cyclists are riding counterclockwise on the track, with the first one being the fastest and the third one being the slowest. Consider the motion of the cyclists in a reference frame attached to the second cyclist. Then the second cyclist is always at some point $A$, while ... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
A grasshopper can jump along a strip of $n$ cells by 8, 9, and 10 cells in either direction. We will call a natural number $n$ jumpable if the grasshopper can, starting from some cell, visit the entire strip, visiting each cell exactly once. Find at least one $n>50$ that is not jumpable. | Suppose a grasshopper has jumped over a strip of 62 cells. We will paint the 8 leftmost cells of the strip white, the next 10 cells black, then 8 cells white again, and so on. In total, there will be 32 white cells and 30 black cells. Since the difference in the number of white and black cells is greater than 1, there ... | 62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $\left.\quad \begin{array}{lc}{\left[\begin{array}{l}\text { Processes and operations }\end{array}\right]} \\ {[} & \text { Semivariants }\end{array}\right]$
Authors: Fadin M. Kovalenko K.
Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it,... | Let $N>1$, and $1=d_{1}<d_{2}<\ldots<d_{k}<d_{k+1}=N-$ be all the divisors of $N$.
Notice that $d_{i} d_{k+2-i}=N$. Therefore,
$d_{1}^{2}+d_{2}^{2}+\ldots+d_{k}^{2}=\frac{N^{2}}{d_{k+1}^{2}}+\frac{N^{2}}{d_{k}^{2}}+\ldots+\frac{N^{2}}{d_{2}^{2}} \leq N^{2}\left(\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{N^{2}}\r... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4 [ Examples and counterexamples. Constructions
In some cells of a $10 \times 10$ board, $k$ rooks were placed, and then all cells that are attacked by at least one rook (a rook attacks the cell it stands on as well) were marked. For what largest $k$ can it happen that after removing any rook from the board, at least ... | Evaluation. Let's consider the placement of $k$ rooks satisfying the condition. There are two possible cases.
1) In each column, there is at least one rook. Then the entire board is under attack, and a rook can be removed from any column that has at least two rooks. Therefore, in this case, there is exactly one rook i... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Golovanov A.S
In an infinite increasing sequence of natural numbers, each number is divisible by at least one of the numbers 1005 and 1006, but none are divisible by 97. Additionally, any two consecutive numbers differ by no more than \( k \). What is the smallest \( k \) for which this is possible? | Let's denote our sequence as $\left(a_{n}\right)$. It is clear that $a_{1}D$ (while $a_{n} \neq D$ by the condition). However, the largest numbers less than $D$ and divisible by 1005 and 1006 are $D-1005$ and $D-1006$, respectively; therefore, $a_{n} \leq D-1005$. Similarly, $a_{n+1} \geq D+1005$; hence, $a_{n+1}-a_{n}... | 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapoveaov A.B.
The plan of the palace is a $6 \times 6$ square, divided into rooms of size $1 \times 1$. There is a door in the middle of each wall between the rooms. The Shah told his architect: "Knock down some walls so that all rooms become $2 \times 1$, no new doors appear, and the path between any two rooms pass... | Consider an arbitrary route from the lower left corner of the palace to the upper right. Since one needs to "climb" 5 horizontal levels and "shift right" 5 vertical levels, one has to pass through at least 10 doors, visiting at least 11 rooms (including the starting and ending rooms).
11 rooms of size $1 \times 1$ cou... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Dmitriev 0.
The mammoth figure moves like a bishop (diagonally), but only in three of the four possible directions (the missing direction can differ for different mammoths). What is the maximum number of non-attacking mammoths that can be placed on an $8 \times 8$ chessboard? | Evaluation. From each mammoth, we will release three arrows in the directions in which it can attack. We will match an arrow to a diagonal (not necessarily the main one) if the mammoth from which the arrow originates is on this diagonal, and the arrow travels along it. Then, no more than two arrows are matched to each ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.V.
There is a board $1 \times 1000$, initially empty, and a pile of $n$ chips. Two players take turns. The first player, on their turn, "places" no more than 17 chips on the board, one on any free cell (they can take all 17 from the pile, or part of them - from the pile, and part - by moving them on the bo... | a) Let's outline the strategy of the first player. Initially, he builds 12 series of 8 chips each, such that adjacent series are separated by one space, sequentially restoring a removed series and adding another. Then, after restoring the configuration following the second player's move, he inserts two chips into the o... | 98 | Combinatorics | proof | Yes | Yes | olympiads | false |
Razin M.
There is a set of 20 weights with which any integer weight from 1 to 1997 g can be measured (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the minimum possible weight of the heaviest weight in such a set, if:
a) the weights in the set are all integers
b) the... | Let's order the weights of the weights (in grams) in ascending order: $p_{0} \leq p_{1} \leq \ldots \leq p_{19}$. Clearly, $p_{0} \leq 1$ (otherwise, it's impossible to weigh a load of 1 g). Similarly, $p_{1} \leq 2$ (to weigh 2 g). Since these two weights are insufficient to weigh 4 g, $p_{2} \leq 4=2^{2}$. Continuing... | 146 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
a) All vertices of the pyramid lie on the faces of the cube, but not on its edges, and at least one vertex lies on each face. What is the maximum number of vertices that the pyramid can have?
b) All vertices of the pyramid lie in the planes of the faces of the cube, but not on the lines containing its edges, and at l... | a) The section of the cube by the plane of the pyramid's base intersects all its faces and, therefore, is a convex hexagon. The vertices of the base lie on the sides of this hexagon, but not at its vertices. It is easy to see that if more than two vertices of the base lie on any one side, it is impossible to connect th... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Capacity }}$..
On a line, 100 sets $A_{1}, A_{2}, . ., A_{100}$ are chosen, each of which is the union of 100 pairwise non-intersecting segments. Prove that the intersection of the sets $A_{1}, A_{2}, . ., A_{100}$ is the union of no more than 9901 pairwise non-intersecting segments (a point is als... | Let sets $A$ and $B$ on the line be unions of $m$ and $n$ segments, respectively. Then $A \cap B$ is a union of no more than $m+n-1$ segments. It is clear that $A \cap B$ is also a union of segments. Let the number of these segments be $k$. The endpoints of the segments in $A \cap B$ are endpoints of segments in $A$ or... | 9901 | Combinatorics | proof | Yes | Yes | olympiads | false |
Shapovadov A.V.
In a set of several weights, all of which have different masses. It is known that if any pair of weights is placed on the left pan, the scales can be balanced by placing one or several weights from the remaining ones on the right pan. Find the smallest possible number of weights in the set. | To balance a pair of the heaviest weights, at least three weights are needed, which means there are at least five weights in total. Suppose there are exactly five weights, and their weights are
$P_{1}2$, then a pair
$(m+1, n-1)$ weighs the same. If $m>3$ and $n<8$, then a pair ( $m-1, n+1$ ) weighs the same. The case... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A.K.
On a circle, 999 numbers are arranged, each being 1 or -1, and not all numbers are the same. We will take the product of every 10 consecutive numbers and sum them up.
a) What is the smallest sum that can be obtained?
b) And what is the largest? | a) Evaluation. If two adjacent products are equal, then the first number of the left is equal to the last number of the right, meaning the numbers 10 places apart are equal. Since 10 and 999 are coprime, stepping by 10 will cover all numbers. But among the numbers, there are different ones, so among the products, there... | -997 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Examples and counterexamples. Constructions] [ Symmetry helps solve the task. ]
What is the minimum number of cells that need to be marked on a chessboard so that each cell of the board (marked or unmarked) shares a side with at least one marked cell? | Let's highlight 20 white cells (on the diagram, they are marked with the sign "x").
Any black cell is adjacent to no more than two highlighted white cells. Therefore, to ensure that these w... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the value of the expression $\sqrt{1+2011^{2}+\left(\frac{2011}{2012}\right)^{2}}+\frac{2011}{2012}$. | The expression under the square root is $\quad\left(2011^{2}+2 \cdot 2011+1\right)-2 \cdot 2011+\left({ }^{2011} / 2012\right)^{2}=2012^{2}-2 \cdot 2012 \cdot 2011 / 2012+$
$\left({ }^{2011} / 2012\right)^{2}=(2012-2011 / 2012)^{2}$, so the original expression equals $(2012-2011 / 2012)+{ }^{2011} / 2012=2012$.
## An... | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Schön D.…
On the island of knights and liars, a traveler came to visit his acquaintance, a knight, and saw him at a round table with five guests.
- I wonder, how many of you are knights? - he asked.
- Why don't you ask each of us a question and find out yourself, - one of the guests suggested.
- Alright. Tell me each... | If everyone said, "Both of my neighbors are knights," it would be immediately clear that everyone sitting at the table is a knight. Indeed, the traveler's acquaintance, who is a knight, told the truth, meaning that both of his neighbors also told the truth, and so on, which means everyone told the truth.
If everyone s... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogosnov I.I.
On the plane, the curves $y=\cos x$ and $x=100 \cos (100 y)$ were drawn, and all points of their intersection with positive coordinates were marked. Let $a$ be the sum of the abscissas, and $b$ be the sum of the ordinates of these points. Find $a / b$. | After replacing $x=100 u$, the equations will become: $y=\cos (100 u), u=\cos (100 y)$. The ordinates of the corresponding intersection points of the new curves will be the same, while the abscissas will be reduced by a factor of 100. Let $c$ be the sum of the abscissas of the new intersection points (with positive coo... | 100 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Rice grains were placed on the cells of a chessboard. The number of grains on any two adjacent cells differed by exactly
1. At the same time, there were three grains on one of the cells of the board, and 17 grains on another. A rooster pecked all the grains from one of the main diagonals of the board, and a hen - from... | Let, for example, cell $A$, which contains three grains, is $k$ cells to the left and $n$ cells below cell $B$, which contains 17 grains. Consider the shortest paths leading from cell $A$ to cell $B$. Each such path consists of $k$ steps to the neighboring cell to the right and $n$ steps to the neighboring cell upwards... | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Quadratic equations. Vieta's theorem]
Can all roots of the equations $x^{2}-p x+q=0$ and $x^{2}-(p+1) x+q=0$ be integers if:
a) $q>0$
b) $q<0$ ?
# | a) For example, the roots of the equations $x^{2}-7 x+12=0$ and $x^{2}-8 x+12=0$ are integers (3 and 4, 2 and 6, respectively).
b) Each of these equations has roots of different signs. Let $x_{1}>0$ and $-x_{2}<0$, and $x_{3}>0$ and $x_{4}<0$ (the case when $x_{1}>x_{3}$ is considered similarly). Since all roots are i... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kuznetsov
All natural numbers from 1 to $N, N \geq 2$ are written in some order around a circle. For any pair of adjacent numbers, there is at least one digit that appears in the decimal representation of each of them. Find the smallest possible value of $N$.
# | Since single-digit numbers do not have common digits, then $N>9$. And since the numbers adjacent to the number 9 must contain a nine in their notation, the smaller one cannot be less than 19, and the larger one cannot be less than 29. Therefore, $N \geq 29$.
Equality $N=29$ is possible, since the conditions of the pro... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
There are 20 people - 10 boys and 10 girls. How many ways are there to form a company where the number of boys and girls is equal?
# | For each such company, associate a set of 10 people, which includes all the girls who joined the company and all the boys who did not join it.
## Solution
Let there be some company consisting of $k$ boys and $k$ girls. We will associate it with a set of 10 people, which includes $k$ girls who joined the company and $... | 184756 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[\quad$ Evaluation + example $\quad]$
The hostess baked a pie for her guests. She may have either 10 or 11 guests. Into what smallest number of slices should she cut the pie in advance so that it can be evenly divided among either 10 or 11 guests? | If 10 guests arrive, each should receive no less than two pieces.
## Solution
If 10 guests arrive, each should receive no less than two pieces. Indeed, otherwise one of the 10 guests would receive one piece, which is $\frac{1}{10}$ of the pie, and if 11 guests arrived, this piece would need to be further divided. Thu... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A regular 100-gon is placed on a table, with the numbers $1,2, \ldots, 100$ written at its vertices. Then these numbers are rewritten in the order of their distance from the front edge of the table. If two vertices are at the same distance from the edge, the left number is written first, followed by the right one. All ... | Due to the symmetry of the regular 100-gon, each number appears in the sets at the 13th position the same number of times, which means the number of sets is a multiple of 100. If the regular 100-gon is continuously rotated counterclockwise around its center, then, first, all sets will appear, and second, the change of ... | 10100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum difference between adjacent numbers among those whose sum of digits is divisible by 7?
# | Let's first prove that this difference cannot be greater than 13. Indeed, among 13 consecutive natural numbers, the sum of the digits of at least one of them is divisible by 7, since among 13 consecutive natural numbers, at least 7 lie in the same decade. Then, the sums of the digits of these seven numbers are consecut... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On a circle of radius 1, a point $O$ is marked, and from it, a notch is made to the right with a radius of $l$. From the resulting point $O_{1}$, another notch is made in the same direction with the same radius, and this is repeated 1968 times. After this, the circle is cut at all 1968 notches, resulting in 1968 arcs. ... | We will prove by induction on $n$ that the number of different arcs after $n$ cuts does not exceed 3. For $n=2$, this is obvious. Let $A_{k}$ denote the cut with number $k$. Suppose $n$ cuts have been made and the point $A_{n}$ falls on the ARC $A_{k} A_{l}$. Then the point $A_{n-1}$ falls on the arc $A_{k-1} A_{l-1}$.... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
To glaze 15 windows of various sizes and shapes, 15 pieces of glass have been prepared exactly to fit the windows (the windows are such that each window should have one piece of glass). The glazier, not knowing that the glasses are matched, works as follows: he approaches the next window and tries the unused glasses un... | First, let's show that if at any moment there are no fewer than 8 windows (and, accordingly, no fewer than 8 panes), then a pane for one of the remaining windows can be found. Indeed, no more than seven panes have been used, so at least one of the eight panes, intended for the eight remaining windows, remains. This one... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Malkin M.I.
On the board, 101 numbers are written: $1^{2}, 2^{2}, \ldots, 101^{2}$. In one operation, it is allowed to erase any two numbers and write down the absolute value of their difference instead.
What is the smallest number that can result from 100 operations? | From four consecutive squares (in three operations), you can get the number 4: $(n+3)^{2}-(n+2)^{2}-((n+$ $\left.1)^{2}-n^{2}\right)=(2 n+5)-(2 n+1)=4$.
We can get 24 such fours from the numbers $6^{2}, 7^{2}, \ldots, 101^{2}$. 20 fours can be turned into zeros by pairwise subtraction. From the numbers $4,9,16,25$ we ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Solve the equation $2^{\mathrm{x}}+3^{\mathrm{x}}=5^{\mathrm{x}}$.
#
The above text has been translated into English, preserving the original text's line breaks and format. | One solution is obvious - x=1. Prove that there are no other solutions using the monotonic increase of some functions.
## Solution
One of the solutions can be guessed immediately: $\mathrm{x}=1$. Now it is enough to show that this equation has no more than one solution. Transform the equation to the form $(2 / 5)^{\m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Various cutting tasks [Cutting into parts with special properties]
A paper square was punctured at 1965 points. From the puncture points and the vertices of the square, no three lie on the same straight line. Then several straight, non-intersecting cuts were made, each starting and ending only at puncture points or ... | Answer: 5896 cuts, 3932 triangles. Let's solve the problem in the general case when the square is pierced in $n$ points. Let the number of resulting triangles be $x$. On the one hand, the sum of the angles of all these triangles is $x \cdot 180^{\circ}$. On the other hand, it is equal to $360^{\circ} + n \cdot 360^{\ci... | 5896 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [ Evaluation + example $\quad]$
In a corridor 100 meters long, 20 carpet strips with a total length of 1000 meters are laid. What is the maximum number of uncovered segments (the width of the strip is equal to the width of the corridor)?
# | First, let's provide an example: take eleven long paths, each 90.5 meters long, and the remaining nine short paths, each 0.5 meters long. Place the eleven long paths on top of each other, leaving a 0.5-meter gap from the edge of the corridor, and in the remaining 9 meters of the corridor, leave the first half empty and... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The surface of a 3 x 3 x 3 Rubik's cube consists of 54 cells. What is the maximum number of cells that can be marked so that the marked cells do not share any vertices?
# | In Fig. 1, it is shown how to mark 7 cells on three adjacent faces of a cube. On the three "invisible" faces, seven cells symmetrical to these need to be marked.
Now let's prove that it is impossible to mark more than 14 cells in the required manner. We will do this in two ways.
The first way. Let's count the total n... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kazitsyna T.
Baba Yaga was given large sandglasses for 5 minutes and small ones for 2 minutes. The potion must boil continuously for exactly 8 minutes. When it started boiling, all the sand in the large sandglasses was in the lower half, and in the small sandglasses, some (unknown) part of the sand was in the upper ha... | Let at the beginning in the upper half of the small hourglass there was sand for $x$ minutes.
From the start of the process, $x+x+(2-x)+(2-x)=4$ minutes have passed, and the sand in both hou... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and Operations ]
Between neighboring camps, it takes 1 day to travel. The expedition needs to transfer 1 can of food to the camp located 5 days away from the base camp and return. At the same time:
- each member of the expedition can carry no more than 3 cans of food;
- in 1 day, he consumes 1 can of food... | Suppose that each camp has a trading tent where delivered canned goods are sold. Let the price of one can in the base camp be one ruble, and in each subsequent camp, it is three times greater than in the previous one. In this case, the price of cans delivered to any camp is no less than the price of cans taken from the... | 243 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and Operations ]
When organizing an expedition to Everest, the participants set up four high camps (not counting the base camp), each a day's journey apart from each other, after which everyone descended. After recounting the supplies, the leader decided that one more oxygen cylinder needed to be brought t... | Note that to move one cylinder from one high-altitude camp to the next, it requires 3 cylinders: one to be moved, and another two that the expedition member uses on the way there and back. Therefore, to accomplish the task, no less than three cylinders need to be delivered to the third camp; for this, no less than $3 *... | 81 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Ageev C.M.
A road 1 km long is fully illuminated by street lamps, each of which illuminates a section of the road 1 m long. What is the maximum number of street lamps that can be on the road, given that after turning off any street lamp, the road will no longer be fully illuminated? # | Let's number the street lamps with natural numbers in the order of their placement along the road. If the segments illuminated by the $n$-th and $(n+2)$-th lamps intersect (at least at one point), then the $(n+1)$-th lamp can be turned off.
Therefore, segments with different odd numbers do not intersect. On a segment ... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the number of permutations $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{10}$ of the numbers $1,2, \ldots, 10$, such that $\mathrm{a}_{\mathrm{i}+1}$ is not less than $\mathrm{a}_{\mathrm{i}}-1$ (for $\mathrm{i}=1,2, \ldots, 9)$. | We can encode the required permutations as follows: each $\mathrm{a}_{\mathrm{i}}, \mathrm{i}=1,2, \ldots, 9$, will be marked with a "+" if $\mathrm{a}_{\mathrm{i}+1}>\mathrm{a}_{\mathrm{i}}$, and with a "-", if $\mathrm{a}_{\mathrm{i}+1}=\mathrm{a}_{\mathrm{i}}-1$. Prove that the permutation is uniquely restored by it... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Indicate a six-digit number $N$, consisting of different digits, such that the numbers $2 N, 3 N, 4 N, 5 N, 6 N$ differ from it by a permutation of digits. | Consider the period of the fraction $1 / 7$.
## Solution
Let's take $N=142857$. A direct check shows that $2N=285714, 3N=428571, 4N=571428, 5N$ $=714285, 6N=857142$.
## Answer
$N=142857$. | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Izmeystiev I.V.
The network of bus routes in the suburb of Amsterdam is organized in such a way that:
a) each route has exactly three stops;
b) any two routes either have no common stops at all or have only one common stop. What is the maximum number of routes that can be in this suburb if there are a total of 9 sto... | Evaluation. Consider some stop A. Determine the maximum number of routes passing through it. Besides A, there are 8 other stops in the city. On each route passing through A, there are two more stops. Since no two of these routes can have common stops other than A, a total of no more than 8 / 2 = 4 routes can pass throu... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$4-$ $[$ Formulas for abbreviated multiplication (other) $]$
Twenty-five coins are distributed into piles as follows. First, they are arbitrarily divided into two groups. Then any of the existing groups is again divided into two groups, and so on until each group consists of one coin. With each division of any group i... | The first method. Let's represent the coins as points and connect each pair of points with a segment. We will get 25(25 - 1) : 2 $=300$ segments. Each time we divide one group of coins into two, we will erase all segments connecting points corresponding to coins that ended up in different groups. Suppose at some step w... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $[$ Invariants $]$
On the board, the numbers $1,2,3, \ldots, 19,20$ are written. It is allowed to erase any two numbers $a$ and $b$ and write the number $a+b-1$ instead.
What number can remain on the board after 19 such operations?
# | For any set of $n$ numbers on the board, consider the following quantity $X$: the sum of all numbers, decreased by $n$. It is not hard to verify that this is an invariant. In the set from the condition, $X=(1+2+\ldots+20)-20=190$. After 19 operations, when there is one number $p$ left on the board, $X=p-1$. Therefore, ... | 191 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
If a class of 30 people is seated in a movie theater, then in any case, at least two classmates will end up in the same row. If the same is done with a class of 26 people, then at least three rows will be empty. How many rows are there in the hall?
# | The first condition means that there are no more than 29 rows in the hall. Indeed, if the number of rows were not less than 30, then obviously, a class of 30 students could be seated with no more than one student per row. The second condition means that the number of rows in the hall is not less than 29. Indeed, if the... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Auxiliary coloring (other).]
On a grid paper, 2000 cells are marked arbitrarily. Prove that among them, it is always possible to select at least 500 cells that do not touch each other pairwise (cells are considered touching if they share at least one vertex). | Color the cells in 4 colors so that no two cells of the same color touch.
## Solution
Consider some (infinite) row and all rows that are one cell away from it. We will color the cells of each of these rows alternately red and yellow. Consider the remaining rows. We will color the cells of each of these rows alternate... | 500 | Combinatorics | proof | Yes | Yes | olympiads | false |
[The Pigeonhole Principle (continued).]
In a photo studio, 20 birds flew in - 8 sparrows, 7 wagtails, and 5 woodpeckers. Each time the photographer clicks the camera shutter, one of the birds flies away (permanently). How many shots can the photographer take to be sure: he will have at least four birds of one species ... | 8 snapshots are dangerous: in this time, 3 woodpeckers and 5 wagtails may fly away, leaving only 2 of each.
Let's show that 7 snapshots can be made. Then, in the studio, there will be $20-7=13$ birds left. This means that the number of birds of one species is at least $13: 3$, that is, at least 5. On the other hand, t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Two pirates were playing for gold coins. First, the first pirate lost half of his coins (gave them to the second), then the second pirate lost half of his, then the first pirate lost half of his again. As a result, the first pirate ended up with 15 coins, and the second with 33. How many coins did the first pirate have... | Let's try to trace "from the end" how many coins each pirate had after each game. In the last game, the first pirate lost half of his coins to the second pirate, after which he was left with 15 coins. But this is exactly the amount he just gave to the second pirate! This means that before this, the first pirate had $15... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and Operations ]
In one step, you can replace any letter in any of these words with any other letter (for example, in one step, you can get the word ЗКНОЗА from the word ЗАНОЗА. What is the minimum number of steps needed to make all the words the same (nonsensical words are allowed)?
# | After all the letter replacements in each column, the letters should become the same. The number of replacements will be the smallest if the most frequent letter (any of them if there are several) is kept in each column. For example, in the first column, letters 3 or К can be kept, both requiring four replacements. The... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the perimeter of triangle $ABC$, if the coordinates of its vertices $A(-3,5), B(3,-3)$ and point $M(6$, 1), which is the midpoint of side $BC$, are known.
# | Use the formulas for the coordinates of the midpoint of a segment and the formulas for the distance between two points.
## Solution
Let $(x, y)$ be the coordinates of vertex $C$. According to the condition, $1 / 2(x+3)=6, 1 / 2(y-3)=1$, from which $x=9, y=5$.
Using the distance formula between two points, we find th... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A road 1 km long is fully illuminated by street lamps, with each lamp lighting a section of the road 1 m long. What is the maximum number of lamps that can be on the road, given that after turning off any lamp, the road will no longer be fully illuminated? | If the segments illuminated by the n-th and (n+2)-th street lamps intersect, then the (n+1)-th street lamp can be turned off.
## Solution
We will number the street lamps with natural numbers in the order along the road. If the segments illuminated by the n-th and (n+2)-th street lamps intersect, then the (n+1)-th str... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Combinations and Permutations ] [ Directed Graphs ]
In the discussion, 15 deputies participated. Each of them, in their speech, criticized exactly $k$ of the remaining 14 deputies.
For what smallest $k$ can we assert that there will be two deputies who criticized each other? | Consider a directed graph where the vertices correspond to deputies, and an edge leading from $A$ to $B$ means that deputy $A$ has criticized deputy $B$.
## Solution
If each deputy has criticized 8 others, then the number of edges in the graph is $15 \cdot 8 = 120$, which is greater than the number of pairs $C_{15}^{... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10,11
In the cube $A B C D A 1 B 1 C 1 D 1$, where $A A 1, B B 1, C C 1$ and $D D 1$ are parallel edges, the plane $P$ passes through point $D$ and the midpoints of edges $A 1 D 1$ and $C 1 D 1$. Find the distance from the midpoint of edge $A A 1$ to the plane $P$, if the edge of the cube is 2. | Let $M, N, K$ and $L$ be the midpoints of edges $A1D1, C1D1$, $AA1$ and $CC1$ respectively (Fig.1). The line $KL$ is parallel to the line $MN$, so the line $KL$ is parallel to the plane $P$. Moreover, the line $KL$ passes through the center $O$ of the cube $ABCD A1B1C1D1$. Therefore, the distance from the midpoint $K$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
48 blacksmiths need to shoe 60 horses. Each blacksmith spends 5 minutes on one horseshoe. What is the least amount of time they should spend on the work? (Note, a horse cannot stand on two legs.)
# | Note that it is impossible to shoe all the horses in less than 25 minutes. Why?
## Solution
Let's show how to proceed. First, 48 blacksmiths take 48 horses and shoe each one with one shoe, which takes 5 minutes. At this point, 48 horses have one shoe, and 12 have none. Then, 12 blacksmiths shoe the horses that still ... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Proizvolov V.V.
There are 19 weights of $1, 2, 3, \ldots, 19$ grams: nine iron, nine bronze, and one gold. It is known that the total weight of all iron weights is 90 grams more than the total weight of the bronze weights. Find the weight of the gold weight. | Prove that the nine lightest weights are bronze, and the nine heaviest are iron.
## Solution
The difference between the total weight of the nine heaviest weights and the total weight of the nine lightest weights is $(19+18+\ldots+11)-(9+8+\ldots+1)=90$ grams. Therefore, the iron weights are the heaviest, and the bron... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
In the cells of a $4 \times 4$ table, numbers are written such that the sum of the neighbors of each number is 1 (cells are considered neighbors if they share a side).
Find the sum of all the numbers in the table.
# | Let's divide all cells into 6 groups (in the figure, cells of each group are marked with their own symbol). Each group consists of all neighbors of some one cell, so the sum of the numbers in it is 1. Therefore, the sum of all numbers is 6.
 Find this rule.
b) Find all natural numbers that transform into themselves (according to this rule).
c) Prove that the number $2^{1991}$ will become a single... | a) The law can be guessed, noticing, for example, that while the number is a single digit, it doubles, and then - apparently not. And the fact that 10 turns into 2 suggests that it is not the number itself that doubles, but the sum of its digits. Thus, the sought law has been discovered: "Double the sum of the digits."... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
Place 32 knights on a chessboard so that each of them attacks exactly two others.
# | In a $3 \times 3$ square, it is possible to visit all cells except the central one with a knight's move and return to the starting cell.
## Solution
If knights are placed on all cells of a $3 \times 3$ square except the central one, each knight will attack exactly two others. Now, place four such "squares" on the boa... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Substitution of variables (other) }\end{array}\right]$
$\left[\begin{array}{l}\text { Completing the square. Sums of squares }\end{array}\right]$
$\left[\begin{array}{l}\text { Polynomials (other) }\end{array}\right]$
Given the polynomial $x(x+1)(x+2)(x+3)$. Find its minimum value. | $x(x+3)(x+1)(x+2)=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)$. Let's denote $x^{2}+3 x$ by $z$. Then $\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)=z(z+2)=(z+1)^{2}-1$. The minimum value -1 of this function is reached when $z=-1$. The equation $x^{2}+3 x+1=0$ has solutions (the discriminant is greater than zero), ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3 [ Coloring $\quad]$
Tom Sawyer has taken on the task of painting a very long fence, adhering to the condition: any two boards, between which there are exactly two, exactly three, or exactly five boards, must be painted in different colors. What is the minimum number of paints Tom will need for this job? | Two colors (let's say white and red) are not enough: painting board number 1 in white, Tom will be forced to paint boards with numbers 4, 5, and 7 in red. Then between the red boards numbered 4 and 7, there will be exactly two boards, which violates the condition.
Three colors are sufficient: Tom can paint three board... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The function $f(x)$ is defined for all $x$, except 1, and satisfies the equation: $(x-1) f\left(\frac{x+1}{x-1}\right)=x+f(x)$. Find $f(-1)$.
# | Substitute the values $x=0$ and $x=-1$ into the given equation. We get: $\left\{\begin{array}{c}-f(-1)=f(0), \\ -2 f(0)=-1+f(-1)\end{array}\right.$. Therefore, $2 f(-1)=-1+f(-1)$, which means $f(-1)=-1$.
## Answer
$-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
There are 100 boxes, numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all of t... | Since the order of the answers to his questions is unknown to the audience, he must make a mistake-free choice, knowing only the number of "no" answers. If $N$ notes are sent, the number of "no" answers heard can take any integer value from 0 to $N$, meaning there are $N+1$ possible outcomes. This number must determine... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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