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Yamenniko i.v.
The numbers 2, 3, 4, ..., 29, 30 are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? | Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$) the even number $2n$ is already m... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
16 cards with integers from 1 to 16 are laid face down in a $4 \times 4$ table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the c... | Evaluation. Let's number the cells as shown in Figure 1.
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | 1 | 4 | 3 |
| 5 | 6 | 7 | 8 |
| 6 | 5 | 8 | 7 |
Fig. 1
Fig. 2
 measure the distance in centimeters between two given points; b) compare two given numbers. What is the minimum number of operations this device needs to perform to definitely determi... | To determine whether $A B C D$ is a rectangle, it is sufficient to check the equalities $A B=C D$, $B C=A D$, and $A C=B D$ - a total of 9 operations (3 operations for each equality: two measurements and one comparison). The rectangle $A B C D$ will be a square if $A B=B C$ - for this, one more, the 10th, operation of ... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a single-round-robin tournament, 10 chess players are participating. What is the minimum number of rounds after which a sole winner can be determined prematurely? (In each round, the participants are paired.
Win - 1 point, draw - 0.5 points, loss - 0).
# | Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one who has no less than 3 points. Since there are still 3 rounds ahead, the winner is stil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In a store, 21 white and 21 purple shirts are hanging in a row. Find the smallest $k$ such that for any initial order of the shirts, it is possible to remove $k$ white and $k$ purple shirts so that the remaining white shirts hang together and the remaining purple shirts also hang together. | First, let's show that \( k \), equal to 10, is sufficient.
First method. We will walk along the row of shirts and count the white and purple shirts separately. As soon as we count 11 of one color - let's assume, without loss of generality, purple - shirts, we will stop. Now we will remove all the white shirts that we... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Gooovanovo A.C. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation... | By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
Then $a=\frac{m}{10^{30}-1}, b=\frac{n}{10^{30}-1}$. We also know t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10,11 |
Let $x, y, z$ be positive numbers and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$. | Make the substitutions $x=p-a, y=p-b, z=p-c$.
## Solution
Let $a=y+z, b=x+z, c=x+y, p=x+y+z$. Consider a triangle with sides $a, b, c$ (the triangle inequalities are obviously satisfied). The perimeter of this triangle is $2 p$, and let the area be denoted by $S$. By Heron's formula,
$S^{2}=p(p-a)(p-b)(p-c)=(x+y+z) ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.i.
Initially, there are 111 pieces of plasticine of the same mass on the table. In one operation, you can choose several groups (possibly one) with the same number of pieces and in each group, combine all the plasticine into one piece. What is the minimum number of operations required to get exactly 11 piec... | Let the mass of one original piece be 1. If in the first operation in each group there are $k$ pieces, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to get 11 pieces of different masses in one operation.
We will show that the required result can be achieved in two operations. For t... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.B.
A boy and a girl were sitting on a long bench. Twenty more children approached them one by one, and each of them sat between two of the already seated children. We will call a girl brave if she sat between two neighboring boys, and a boy brave if he sat between two neighboring girls. When everyone was sea... | The first method. Let's look at the number of pairs of adjacent boys and girls. Initially, it is equal to 1. Notice that if a boy sits between two boys, the number of such pairs does not change. If he sits between a boy and a girl, he "destroys" one such pair and "creates" one, so the number of such pairs does not chan... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
For what value of $a$ does the polynomial $P(x)=x^{1000}+a x^{2}+9$ divide by $x+1$?
# | $P(-1)=1+a+9=a+10$. By the theorem of Bezout, this number should equal zero.
## Answer
For $a=-10$. | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The principle of the extreme (etc.). $\quad]$ Symmetry helps solve the task_ $\quad]$
There are thirty cards, each with a number: on ten cards - $a$, on ten others - $b$, and on the remaining ten - $c$ (the numbers $a, b, c$ are all different). It is known that for any five cards, it is possible to find another five s... | Let $a<b<c$. We will mark on the number line all possible sums of the numbers on five cards. For each of them, the opposite is also marked, so the marked points are symmetrically located relative to zero. In particular, the largest (5c) and the smallest (5a) sums are opposite, so $5 a+5 c=0$, which means $c=-a$. The su... | 0 | Combinatorics | proof | Yes | Yes | olympiads | false |
Bakayev E.v.
10 children of different heights are standing in a circle. From time to time, one of them runs to another place (between some two children). The children want to stand in order of increasing height as quickly as possible, in a clockwise direction (from the shortest to the tallest). What is the minimum num... | Number the children in ascending order of height $-1, 2, \ldots, 10$.
Estimation. Suppose they initially stood in reverse order.
First method. If there were fewer than eight sprints, then some three children remained in their places, and their order is opposite to the required one.
Second method. Let's call the comp... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
The weight of each weight in the set is a non-integer number of grams. They can balance any integer weight from 1 g to 40 g (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the smallest number of weights in such a set? | Example 1. Let's take weights of 1, 1, 3, 5, 11, 21, 43 g. The first two can measure any integer weight up to 2 g. Therefore, the first three can measure up to 5 g, the first four up to 10 g, the first five up to 21 g, the first six up to 42 g, and all seven up to 85 g. If we reduce the weight of each weight by half, a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The hostess made a pirog (a type of Russian pie) and wants to pre-cut it into such (not necessarily equal) pieces so that the pie can be evenly divided both among five and seven people. What is the minimum number of pieces she can manage with?
# | This task is a particular case of problem $\underline{98057 .}$
## Answer
11 pieces. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the minimum number of weights needed to be able to weigh any number of grams from 1 to 100 on a balance scale, if the weights can be placed on either pan of the scale?
# | When solving this problem, we need the following interesting property of the ternary numeral system: any natural number can be represented as the difference of two numbers, the ternary representation of which contains only 0 and 1.
To prove this, we need to write the original number in ternary notation and construct t... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Ionin Yu.I.
a) Does there exist an infinite sequence of natural numbers with the following property: no number in the sequence divides another, but among any three numbers, one can choose two whose sum is divisible by the third?
b) If not, how many numbers can be in a set with such a property?
c) Solve the same prob... | b) To the sequence of four numbers $3,5,7,107$ given in the condition, we can add the fifth number 10693: 10693 + 5 is divisible by $3,10693+3$ is divisible by $7,10693+7$ is divisible by $5,10693+107$ is divisible by 3 and by $5,10693+7$ is divisible by 107.
We will show that it is impossible to form a sequence satis... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ CaseAnalysis ] $[$ Divisibility Rules (etc.) $]$ Author: Fomin S.B. A natural number $n$ is written in the decimal system. It is known that if a digit appears in this representation, then $n$ is divisible by this digit (0 does not appear in the representation). What is the maximum number of different digits that t... | If the digit 5 is included in the representation of a number, then the number must end in 5. Therefore, it is odd and, consequently, contains only odd digits. Thus, it cannot have more than five digits. If 5 does not appear in the decimal representation of the number, then it can include all other 8 digits. For example... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems ] $[$ Case enumeration ]
In a box, there are blue, red, and green pencils. In total, there are 20. The number of blue pencils is 6 times the number of green pencils, and the number of red pencils is less than the number of blue pencils. How many red pencils are in the box? | Think about how many blue pencils there can be.
## Solution
Since there are 20 pencils in total, and blue and green pencils together make up 7 parts. This means there can be 6 or 12 blue pencils, and green and red pencils would then be 1 and 13 or 2 and 6, respectively. Since there are fewer red pencils than blue one... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
On a circular road, there are four gas stations: $A, B, C$, and $D$. The distance between $A$ and $B$ is 50 km, between $A$ and $C$ is 40 km, between $C$ and $D$ is 25 km, and between $D$ and $A$ is 35 km (all distances are measured along the circular road in the shortest direction).
a) Provide an exa... | First, determine the arrangement of gas stations $A, C$, and $D$.
## Solution
The problem provides all three distances between $A, C$, and $D$. First, let's determine the arrangement of these three gas stations. Gas stations $A$ and $C$ divide the circular road into two arcs. If gas station $D$ were on the shorter ar... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pairing and grouping; bijections $]$ [ Decimal number system ]
Find the last digit of the number $1^{2}+2^{2}+\ldots+99^{2}$. | $1^{2}+2^{2}+\ldots+99^{2} \equiv 10\left(1^{2}+2^{2}+\ldots+9^{2}\right)=0(\bmod 10)$.
## Answer
0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Investigation of a quadratic trinomial ]
A quadratic trinomial $y=a x^{2}+b x+c$ has no roots and $a+b+c>0$. Determine the sign of the coefficient $c$.
# | 
The quadratic polynomial has no roots, which means its graph does not intersect the x-axis. Since \( y(1) = a + b + c > 0 \), the graph is located in the upper half-plane (see the figure), th... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In a circle, there are boys and girls (both are present), a total of 20 children. It is known that for each boy, the neighbor in the clockwise direction is a child in a blue T-shirt, and for each girl, the neighbor in the counterclockwise direction is a child in a red T-shirt. Can the number of boys in th... | The MHD (clockwise) arrangement cannot be due to the color of child X's T-shirt. Therefore, one boy should stand clockwise from the boy, one boy from him, and so on. This means that there are no fewer than half of all the children in the circle who are boys. By similar considerations, there are no fewer than half of th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V. A sequence of several natural numbers is written, with a sum of 20. No number and no sum of several consecutive numbers equals 3. Could there be more than 10 numbers written? | Example with 11 numbers: $1,1,4,1,1,4,1,1,4,1,1$.
## Answer
It could.
Send a comment | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ [Reverse Process ]
Lёnya thought of a number. He added 5 to it, then divided the sum by 3, multiplied the result by 4, subtracted 6, divided by 7, and got 2. What number did Lёnya think of?
Try to form an equation to determine the unknown number.
# | Let's denote the number thought of by Lёna as $x$. Then we can form the equation
$$
\{[((x+5): 3)-4]-6\}: 7=2
$$
By sequentially moving all numbers from the left side to the right, we get a new equation
$$
x=\{[((27)+6): 4] 3\}-5
$$
from which it is easy to determine that $x=10$. From this, it is also clear that to... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ 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 |
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 |
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 |
[ 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 |
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 |
[
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 |
[ 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 |
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 |
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 |
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 |
[ 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 |
$\begin{aligned} & {\left[\begin{array}{l}\text { Irrational Equations } \\ \text { [ Completing the Square. Sums of Squares }\end{array}\right]}\end{aligned}$
Solve the equation
$$
\left(x^{2}+x\right)^{2}+\sqrt{x^{2}-1}=0
$$ | Since the numbers $\left(x^{2}+x\right)^{2}$ and $\sqrt{x^{2}-1}$ are non-negative, and their sum is zero, then both these numbers are equal to zero. On the other hand, if both these numbers are equal to zero, then their sum is zero. Therefore, the original equation is equivalent to the following system:
$$
\left\{\be... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameter $]
For the quadratic trinomial $f(x)=a x^{2}-a x+1$, it is known that $|f(x)| \leq 1$ for $0 \leq x \leq 1$. Find the greatest possible value of $a$. | Since $f(0)=f(1)=1$, the graph of the quadratic function is a parabola symmetric about the line $x=0.5$. From the condition $|f(x)| \leq 1$ for
$0 \leq x \leq 1$, it follows that the branches of the parabola are directed upwards. The minimum value of $f(x)$ is $f(0.5)=1-\frac{a}{4}$.
The maximum possible value of $a$... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8,9
The areas of the projections of a certain triangle onto the coordinate planes Oxy and Oyz are $\sqrt{6}$ and $\sqrt{7}$, respectively, and the area of the projection onto the plane $O x z$ is an integer. Find the area of the triangle itself, given that it is also an integer. | Let the vector perpendicular to the plane of the original triangle form angles $\alpha, \beta$, and $\gamma$ with the coordinate axes $O x, O y$, and $O z$ respectively. Then
$$
\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=1
$$
Let the area of the original triangle be denoted by $S$, and the areas of the projections on t... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
At the beginning of the school year, Andrey started recording his math grades. Upon receiving a new grade (2, 3, 4, or 5), he called it unexpected if, up to that point, it had occurred less frequently than each of the other possible grades. (For example, if he had received the grades 3, 4, 2, 5, 5, 5, 2, ... | The first unexpected grade will be the last one received for the first time. The second unexpected grade will be the last one received for the second time, and so on. Therefore, there will be a total of 10 unexpected grades.
## Answer
It can be. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).] Case analysis $\quad]$
After a hockey game, Anton said that he scored 3 goals, and Ilya only one. Ilya said that he scored 4 goals, and Seryozha as many as 5. Seryozha said that he scored 6 goals, and Anton only two. Could it be that together they scored 10 goals, given that each ... | There are two cases.
1) Anton told the truth about himself, that is, he scored 3 goals. Then Seryozha lied about Anton, so he told the truth about himself, that is, he scored 6 goals. Therefore, Ilya lied about Seryozha and told the truth about himself, that is, he scored 4 goals. In this case, the boys scored a total... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
$\left[\begin{array}{ll}{\left[\begin{array}{l}\text { Common fractions } \\ \text { [Examples and counterexamples. Constructions ] }\end{array}\right]}\end{array}\right]$
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 |
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 |
[ 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 |
|
| | $[$ 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.
![](https://cdn.mathpix.com/cropped/2024_05_0... | 10 | 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 |
[ 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 |
[ $\left.\quad \begin{array}{lc}{\left[\begin{array}{l}\text { Processes and operations }\end{array}\right]} \\ {[} & \text { Semivariants }\end{array}\right]$
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 |
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 |
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 |
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 |
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 |
[ [ 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 |
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 |
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 |
[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 |
[ 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 |
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.
![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-16.jpg?he... | 6 | Logic and Puzzles | 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 |
Agakhanov N.X.
At a round table, 30 people are sitting - knights and liars (knights always tell the truth, while liars always lie). It is known that each of them has exactly one friend at the same table, and a knight's friend is a liar, while a liar's friend is a knight (friendship is always mutual). When asked, "Is y... | All those sitting at the table are paired as friends, which means there are an equal number of knights and liars. Consider any pair of friends. If they are sitting next to each other, the knight will answer "Yes" to the given question, and the liar will answer "No." If they are not sitting next to each other, their ans... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Solve the equation: $2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}=\frac{10}{x-4}$. | Since the left, and therefore the right, part of the equation takes only positive values, then $x>4$. On the interval $(4,+\infty)$, the function
$f(x)=2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}$ is increasing, while the function $g(x)=\frac{10}{x-4}$ is decreasing, so the equation $f(x)=g(x)$ has no more than one root.
It rem... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kosukhin O.n.
Sasha found that there were exactly $n$ working digit buttons left on the calculator. It turned out that any natural number from 1 to 99999999 can either be entered using only the working buttons, or obtained as the sum of two natural numbers, each of which can be entered using only the working buttons. ... | Let's show that the conditions of the problem are met if the buttons with digits $0,1,3,4,5$ remain functional.
Indeed, any digit from 0 to 9 can be represented as the sum of some two "functional" digits. Let the number from 1 to 99999999 that we want to obtain consist of digits $a_{1}, a_{2}, \ldots, a_{8}$ (some of ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Authors: Shapovalov A.V., Yatsenno I.V.
At the "Come on, creatures!" competition, 15 dragons are standing in a row. The number of heads of neighboring dragons differs by 1. If a dragon has m... | It is convenient to represent a row of dragons as a graph: instead of each dragon, we draw a point at a height corresponding to the number of heads the dragon has, and connect these points.
a) See the figure.
b) First, note that somewhere between every two cunning dragons stands a strong one. Indeed, if we walk along... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Among the actors of Karabas Barabas theater, a chess tournament was held. Each participant played exactly one game with each of the others. One solido was given for a win, half a solido for a draw, and nothing for a loss. It turned out that among any three participants, there would be ... | Example. Let's denote the participants with letters A, B, V, G, D. Suppose A won against B, B won against V, V won against G, G won against D, D won against A, and all other matches ended in a draw. The condition of the problem is satisfied.
Evaluation. From the condition, it follows that for this tournament, two stat... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
On a chessboard, 32 white and 32 black pawns are placed in all squares. A pawn can capture pawns of the opposite color by moving diagonally one square and taking the place of the captured pawn (white pawns can only capture to the right-up and left-up, while black pawns can only capture to the left-down a... | Note that a pawn that stood on a black square will always move only on black squares. Then after each move (on black squares) there will always be at least one pawn on black squares - the one that made the move. Similarly, at least one pawn will remain on white squares, and there will be no less than two pawns in total... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
The distance between two cells on an infinite chessboard is defined as the minimum number of moves in the path of a king between these cells. On the board, three cells are marked, the pairwise distances between which are 100. How many cells exist such that the distances from them to all three marked cell... | Consider two arbitrary cells $A$ and $B$. Let the difference in the abscissas of their centers be $x \geq 0$, and the difference in the ordinates be $y \geq 0$. Then the distance $\rho(A, B)$ between these cells is $\max \{x, y\}$.
Let cells $A, B, C$ be marked. Then for each pair of cells, there exists a coordinate i... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underset{\text { Tokarev } C . \text {. }}{ }$
Numbers $a, b$ and $c$ are such that $(a+b)(b+c)(c+a)=a b c,\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)=a^{3} b^{3} c^{3}$. Prove that $a b c=0$. | First, note that $x^{2}-x y+y^{2}>|x y|$ for any distinct numbers $x$ and $y$.
Assume that $a b c \neq 0$. Then, dividing the second equality by the first, we get $\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}\right.$
$\left.-c a+a^{2}\right)=|a b| \cdot |b c| \cdot |a c|$.
All the parentheses o... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
B.R.'s Problem
The segments connecting an inner point of a convex non-equilateral $n$-gon with its vertices divide the $n$-gon into $n$ equal triangles.
For what smallest $n$ is this possible? | Let's prove that the specified situation is impossible for $n=3,4$.
The first method. For $n=3$, the angles of the triangles in the partition that meet at an internal point are equal, since the sum of any two different angles in them is less than $180^{\circ}$. But then the sides opposite to them, which are sides of t... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Berdonikov A.
Let's call a natural number good if all its digits are non-zero. A good number is called special if it has at least $k$ digits and the digits are in strictly increasing order (from left to right).
Suppose we have some good number. In one move, it is allowed to append a special number to either end or in... | Obviously, a special number does not have more than nine digits. If $k=9$, then with each operation, the number of digits changes by exactly 9, meaning the remainder of the number of its digits divided by 9 does not change, and a two-digit number cannot be made from a one-digit number.
Let $k=8$. Since all operations ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Senderov V.A.
Find all natural numbers $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first power).
# | Let $n \geq 2$, and $2=p_{1}1$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation (*) would be divisible by $q$, which is not the case. Therefore, $a>p_{k}$.
Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, th... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$2+$
John had a full basket of trempons. First, he met Anna and gave her half of his trempons and another half-trempon. Then he met Banna and gave her half of the remaining trempons and another half-trempon. After meeting Vanna and giving her half of the trempons and another half-trempon, the basket was empty. How man... | Notice that before meeting Vanna, John had one tremponch left, as half of this amount was half a tremponch. Before meeting Banna, he had 3 tremponchs, because half of this amount was one and a half tremponchs, that is, one and a half. Similarly, we get that initially there were 7 tremponchs.
## Answer
7 tremponchs. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $M$ be a finite set of numbers. It is known that among any three of its elements, there are two whose sum belongs to $M$.
What is the maximum number of elements that can be in $M$? | Consider either the four largest or the four smallest numbers.
## Solution
Example of a set of 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that the set $M=\left\{a_{1}, a_{1}, \ldots, a_{n}\right\}$ of $n>7$ numbers does not have the required property. We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9,10,11 |
Author: S $\underline{\text { Saghafian M. }}$.
In the plane, five points are marked. Find the maximum possible number of similar triangles with vertices at these points. | Example. Vertices and the center of a square.
Evaluation. Let's describe all configurations of four points forming four similar triangles. Let $\$ \mathrm{~A} \$$, $\$ \mathrm{~B} \$$, \$C \$ \$ \$
1. Point $\$ \mathrm{~A} \$$ lies inside triangle $\$ B C D \$$. Let $\$ B \$$ be the largest angle of triangle $\$ B C ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Chessboards and chess pieces ] [ Examples and counterexamples. Constructions ]
What is the maximum number of queens that can be placed on an $8 \times 8$ chessboard without attacking each other? | You can place no more than eight queens that do not attack each other, as each queen controls one row.
Let's show how you can place eight queens:
Send a comment | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3[ Examples and counterexamples. Constructions]
In the cells of a $5 \times 5$ square table, the numbers 1 and -1 are arranged. It is known that the number of rows with a positive sum is greater than the number of rows with a negative sum.
What is the maximum number of columns in this table that can have a negative s... | Let's consider one of the possible examples (see the table). The sum of the numbers in each of the three upper rows is positive, while the sum of the numbers in each column is negative.
| 1 | 1 | 1 | -1 | -1 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | -1 | 1 | -1 | 1 |
| -1 | 1 | -1 | 1 | 1 |
| -1 | -1 | -1 | -1... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the minimum number of sportlotto cards (6 out of 49) you need to buy to ensure that at least one number is guessed correctly in at least one of them?
# | Let's fill eight cards as follows: in the first one, we will strike out numbers from 1 to 6, in the second one - from 7 to 12, and so on, in the last one - from 43 to 48. The number 49 will remain unstruck in any card. Therefore, at least five of the winning numbers will be struck out.
We will prove that seven cards m... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kostrikina I.A.
At a round table, pastries are placed at equal intervals. Igor walks around the table and eats every third pastry he encounters (each pastry can be encountered several times). When there were no pastries left on the table, he noticed that the last pastry he took was the first one he encountered, and he... | Since Igor passed an integer number of circles, he met the first pastry at the moment of approaching the table. In addition, the sequence of eating pastries will not change if we remove the requirement of "equal intervals". If there were four pastries, Igor would have walked exactly five circles (see the figure).
.]
What is the minimum number of shots in the game "Battleship" on a 7*7 board needed to definitely hit a four-deck ship (a four-deck ship consists of four cells arranged in a row)?
# | If on a $7 * 7$ board, n non-overlapping four-deck ships can fit, this means that (n-1) shots would not be enough.
## Solution
In the first picture, an example sequence of 12 shots is provided, with which any four-deck ship would be hit. In the second picture, an example of the placement of twelve non-overlapping fou... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the maximum value that the expression $a e k-a f h+b f g-b d k+c d h-c e g$ can take if each of the numbers $a, b, c, d, e, f, g, h, k$ is equal to $\pm 1$.
# | Evaluation. The product of all addends equals - $(a b c d e f g h k)^{2}$, which is negative. Therefore, the number of minus ones among the addends is odd and thus not equal to zero. Consequently, the sum does not exceed 4.
Example. Let $a=c=d=e=g=h=k=1, b=f=-1$. Then the given expression equals 4.
## Answer
4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Ordering in ascending (descending) order. ] [ Classical combinatorics (other). $\quad]$
a) A traveler stopped at an inn, and the owner agreed to accept rings from a golden chain the traveler wore on his wrist as payment for lodging. However, he set a condition that the payment should be daily: each day the owner sh... | a) It is enough to cut two rings so that pieces of three and six rings are separated. On the third day, the traveler gives the piece of three rings and receives two rings as change, and on the sixth day, the piece of six rings and receives five rings as change.
b) Arrange the resulting pieces of the chain (not countin... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
Two numbers are written on the board in the laboratory. Every day, the senior researcher Petya erases both numbers on the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbe... | The product of the numbers on the board does not change. Indeed, $\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b$. Therefore, the desired product is 2.
## Answer
2.
Send a comment | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kuuyggin A.K.
Among any five nodes of a regular square grid, there will always be two nodes such that the midpoint of the segment between them is also a node of the grid. What is the minimum number of nodes of a regular hexagonal grid that must be taken so that among them there will always be two nodes such that the m... | Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof of the lemma. Introduce the o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Coordinate method on the plane ] [Triangle inequality (other).]
Point $M$ lies on the line $3 x-4 y+34=0$, and point $N-$ lies on the circle $x^{2}+y^{2}-8 x+2 y-8=0$. Find the minimum distance between points $M$ and $N$. | Notice that
$$
x^{2}+y^{2}-8 x+2 y-8=0 \Leftrightarrow x^{2}-8 x+16+y^{2}+2 y+1=25 \Leftrightarrow(x-4)^{2}+(y+1)^{2}=5^{2}
$$
This means the center of the circle is the point $Q(4, -1)$, and the radius is 5.
Let $d$ be the distance from the point $Q$ to the line $3 x-4 y+34=0$. Then
$$
d=\frac{|3 \cdot 4-4 \cdot(-... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Klepitsyn V.A.
Cells of a $5 \times 5$ board are painted in a checkerboard pattern (corner cells are black). A figure, a mini-bishop, moves along the black cells of this board, leaving a trail on each cell it visits and not returning to that cell again. The mini-bishop can move either to free (diagonally) adjacent cel... | Let's first provide an example of the mini-elephant's route that ensures it visits twelve cells (see the figure, numbers from 1 to 12 show the order of cell visits).
| 1 | | 4 | | 6 |
| :--- | :--- | :--- | :--- | :--- |
| | 2 | | 5 | |
| 3 | | 7 | | 9 |
| | 11 | | 8 | |
| 12 | | 10 | | |
We will prove t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Examples and counterexamples. Constructions]
A snail woke up, crawled from the mushroom to the spring, and fell asleep. The journey took six hours. The snail moved sometimes faster, sometimes slower, and stopped. Several scientists observed the snail. It is known that:
1) At every moment of the journey, the snail wa... | Example. Let's divide the entire time of the snail's movement into 30 intervals of 12 minutes each. Suppose the snail crawls 1 m in each of the 10 intervals numbered $1, 6, 7, 12, 13, 18, 19, 24, 25$, and 30, and rests the rest of the time. Scientists observe it in intervals 1-5 (from 1st to 5th - exactly one hour), 2-... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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