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V.V. Proizvolov
The ends of $N$ chords divide the circle into $2N$ arcs of unit length. It is known that each chord divides the circle into two arcs of even length.
Prove that the number $N$ is even.
# | We will color the ends of the chords alternately in white and black. Chords connect ends of the same color. Therefore, $N$ white ends are paired, and $N$ is even.
Problem | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,821 |
Shapovalov A.V.
Numbers from 1 to 16 are arranged in a $4 \times 4$ table. In each row, each column, and on each diagonal (including diagonals consisting of one cell), the largest number in it is marked (a number can be marked multiple times). Could it happen that
a) all numbers, except perhaps two, are marked?
b) a... | The numbers in the corners will be marked in any case - these are the numbers on diagonals of length 1. Among the other numbers, there is at least one unmarked number. Indeed, consider the smallest of the numbers not in the corners. It is not marked because on each line with it, there are other "non-corner" numbers. Th... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,822 |
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 | 53,823 |
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 | 53,824 |
Proizvolov V.V.
There are 200 weights with masses of $1, 2, \ldots, 200$ grams. They were distributed on two scales, with 100 weights on each, and the scales balanced. On each weight, the number of weights on the opposite scale that are lighter than it was recorded.
Prove that the sum of the numbers recorded on the w... | For each weight, consider the total number of weights that are lighter than it. If the mass of the weight is $m$, then for it, there will be $m-1$ such weights. Since the scales are in equilibrium, the sums $S$ of such numbers for both sides of the scale are the same: 100 less than the sum of the masses of the weights ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,825 |
[ 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 | 53,827 |
Folklore
There are a thousand tickets numbered $000, 001, \ldots, 999$ and a hundred boxes numbered $00, 01, \ldots, 99$. A ticket is allowed to be placed in a box if the box number can be obtained from the ticket number by erasing one of the digits. Is it possible to distribute all the tickets into 50 boxes? | Let's divide the boxes into two groups: Group $A$ will include all boxes whose numbers contain only the digits $0,1,2,3,$ and $4$; and Group $B$ will include all boxes whose numbers contain only the digits $5, 6, 7, 8,$ and $9$. Each group will have 25 boxes: Group $A$ will include half of the boxes from the first five... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,828 |
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 | 53,831 |
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 | 53,832 |
Tokorev S.I.
Can 100 weights with masses $1,2,3, \ldots, 99,100$ be distributed into 10 piles of different masses so that the condition is met: the heavier the pile, the fewer weights it contains? | Suppose it is possible to distribute the weights according to the condition of the problem. The total mass of all weights is 5050. Therefore, the mass of the heaviest pile is not less than $5050: 10=505$. Since there are no weights in the set with a mass greater than 100, there are no fewer than 6 weights in this pile.... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,833 |
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 | 53,835 |
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 | 53,836 |
[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 | 53,837 |
$\underline{\text { Folklore }}$
On a chessboard, 11 knights are placed such that no two attack each other.
Prove that on the same board, one more knight can be placed while maintaining this property. | The first method. To cover the 12 cells highlighted in the left figure, at least 12 knights are required, as no two cells can be covered by a single knight. Therefore, among them, there is at least one cell that is not attacked by any of the 11 placed knights. Another knight can be placed on this cell.
![](https://cdn.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,838 |
At the central telegraph office, there are change machines that exchange 20 kopecks for 15, 2, 2, and $1 ; 15$ kopecks for 10, 2, 2, and $1 ; 10$ kopecks for 3, 3, 2, and 2. Petya exchanged 1 ruble 25 kopecks in silver for copper. Vasya, looking at the result, said: "I know exactly what coins you have" and named them. ... | Since two 15-kopeck coins can be exchanged for the same combination as a set of one 10-kopeck and one 20-kopeck coin, Petya could not have had more than one 15-kopeck coin in his original set, nor could he have had both a 10-kopeck and a 20-kopeck coin simultaneously.
It is impossible to get 125 kopecks using only 10 ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,840 |
| |
10 magazines lie on a coffee table, completely covering it. Prove that five of them can be removed so that the remaining magazines will cover at least half of the table's area. | Reduce the problem to the case where the magazines do not overlap.
## Solution
Number the magazines from 1 to 10. From the second magazine, cut off the part (if any) that is already covered by the first magazine. From the third magazine, cut off the part that is already covered by the first and second magazines. Cont... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,842 |
On the plane, there are $p$ points, no three of which lie on the same line. Prove that they can be labeled $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{n}$ in such an order that the closed broken line $\mathrm{A}_{1} \mathrm{~A}_{2} \ldots \mathrm{A}_{n}$ is non-self-intersecting. | Connect the points with a closed broken line in some order, and then, if there are self-intersections, replace the pair of intersecting segments with a pair of non-intersecting segments.
## Solution
Let the points be denoted as $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{n}$ in some arbitrary order. If the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,843 |
Prove that the sum of the squares of the distances from an arbitrary point on the plane to two opposite vertices of a rectangle is equal to the sum of the squares of the distances from this point to the other two vertices of the rectangle.
# | Let's denote the distances from an arbitrary point on the plane to the lines containing the sides of a rectangle as \(x, y, z\), and \(t\), and apply the Pythagorean theorem (or use the coordinate method).
## Solution
First method.
Let \(M\) be an arbitrary point on the plane, and \(ABCD\) be a rectangle. Denote the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,844 |
Prove by induction the Binet's formula:
$$
F_{\mathrm{n}}=\frac{\varphi^{n}-\hat{\varphi}^{n}}{\sqrt{5}}
$$
where $\varphi=\frac{1+\sqrt{5}}{2}$ — the "golden ratio" or number of Phidias, and $\widehat{\varphi}=\frac{1-\sqrt{5}}{2}$ ("phi with a hat") — conjugate to it. | Prove that the numbers $F_{\mathrm{n}}$, found using Binet's formula, satisfy the initial conditions $F_{0}=0, F_{1}=1$ and the recurrence relation $F_{n+1}=F_{n}+F_{n-1}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,845 |
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 | 53,846 |
Emelyanov L.A.
Among five visually identical coins, 3 are genuine and two are counterfeit, identical in weight but it is unknown whether they are heavier or lighter than the genuine ones. How can you find at least one genuine coin with the fewest number of weighings? | It will not be possible to find the genuine coin in one weighing if an inequality occurs (this is easy to check in the case when there is one coin on each pan, as well as in the case when there are two). We will indicate how to find the genuine coin in 2 weighings. Weigh 1 and 2, and then 3 and 4. If both weighings res... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,847 |
Bogonov I.I.
Find all pairs of numbers $x, y \in\left(0 ; \frac{\pi}{2}\right)$, satisfying the equation $\sin x+\sin y=\sin (x y)$. | There are no such numbers. Indeed, if $x \leq 1$, then $\sin (x y) \leq \sin y1>\frac{\pi}{6}$, then $\sin x, \sin y>\sin \frac{\pi}{6}=\frac{1}{2}$, i.e., $\sin x+\sin y>\frac{1}{2}+\frac{1}{2}=1 \geq \sin (x y)$.
## Answer
There are no such numbers. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,849 |
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 | 53,852 |
$[\quad$ 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 | 53,854 |
Hang the picture on a string between two nails so that the picture falls when either nail is pulled out.
# | Around each of the nails, an equal number of turns should be made clockwise and counterclockwise.
## Solution
Here is one of the possible options.
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,855 |
Prove that every integer $A$ can be represented as
$$
A=a_{0}+2 a_{1}+2^{2} a_{2}+\ldots+2^{\mathrm{n}} a_{\mathrm{n}}
$$
where each of the numbers $a_{\mathrm{k}}=0,1$ or -1 and $a_{\mathrm{k}} a_{\mathrm{k}+1}=0$ for all $0 \leqslant k \leqslant n-1$, and such a representation is unique. | If $A$ is even, then the representation of the number $A$ is obtained from the representation of a smaller number $m=A / 2$ by a "shift" by one digit. If $A$ is odd, then $a_{0}= \pm 1$ and the number $a_{1}$ must be zero; therefore, the number $A$ - $a_{0}$ is divisible by 4 and the representation of the number $A$ is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,858 |
Peshnin A.
What is the minimum number of colors needed to color the natural numbers so that any two numbers differing by 2 or by a factor of two are colored differently?
# | Example. We will sequentially paint natural numbers in ascending order. 1 and 2 will be painted in two different colors. For the color of each number $k>2$, there are no more than two restrictions: it cannot be the same color as the number $k / 2$ or the number $k-2$. Therefore, for any such $k$, there will definitely ... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,859 |
Frankin B.R.
The hostess baked a square cake and cut several pieces from it. The first cut was made parallel to the side of the original square from edge to edge. The next cut was made in the remaining part from edge to edge perpendicular to the previous cut, and so on (several times). All the cut pieces have equal ar... | The part remaining after the next cutting is called the remainder. The length of the remainder is defined as the size of the side along which it is cut, and the width is the size of the other side. The length of the cut rectangle (piece) is also considered the size of the side along which it is cut, and the width is th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,861 |
Given a trihedral angle with vertex $O$. Can we find a plane section $A B C$ such that the angles $O A B, O B A, O B C$, ОСВ, ОАС, ОСА are acute? | Answer: Yes, it is possible. Let's choose points $A, B, C$ at the same distance from point $O$. Then all the specified angles will be the base angles of isosceles triangles, and the base angle of an isosceles triangle is necessarily acute. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,863 |
$[$ Point Systems ]
On a circle of length 15, $n$ points are chosen such that for each point, there is exactly one chosen point at a distance of 1 and exactly one at a distance of 2 (distance is measured along the circle). Prove that $n$ is divisible by 10. | From the solution of problem $\underline{78064}$, it follows that the minimum number of selected points is 10. In addition, the system of selected points consists of several sets of such tens of points. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,865 |
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 | 53,867 |
There are $m$ points, some of which are connected by segments such that each is connected to $l$ points. What values can $l$ take?
# | Since each point can be connected to no more than $m$ - 1 others, $l < m$. Moreover, the total number of pairs of the form (segment, end of this segment) is $l m$, which means the total number of segments is $l m / 2$, from which it follows that the number $l m$ is even. We will prove that for any $l < m$, for which th... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,869 |
$\underline{78255}$ topics: [ vectors (other)
It is known that $Z_{1}+\ldots+Z_{\mathrm{n}}=0$, where $Z_{\mathrm{k}}$ are complex numbers. Prove that among these numbers, there are two such that the difference of their arguments is greater than or equal to $120^{\circ}$. | First, note that each complex number can be associated with a vector on the complex plane. Therefore, we can consider that we have $n$ vectors drawn from point $O$ to points $Z_{1}, \ldots, Z_{\mathrm{n}}$, the sum of which is zero. If $n=2$, then since their sum is zero, they are equal in magnitude and oppositely dire... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,870 |
Prove that any natural number can be represented as the sum of several different terms of the sequence $1,2,3,5,8,13, \ldots, a_{\mathrm{n}}=a_{\mathrm{n}-1}+a_{\mathrm{n}-2}, \ldots$.
# | We will prove the required statement by induction. The base case of the induction is obvious. The sequence $\left\{a_{n}\right\}$ is monotonically increasing, so for any natural number $m$, we can choose $n$ such that $a_{\mathrm{n}} \leq m < a_{\mathrm{n}+1}$. By the induction hypothesis, the number $m - a_{\mathrm{n}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,871 |
$2 n$ people gathered, each of whom is acquainted with at least $n$ of those present. Prove that it is possible to choose four of them and seat them around a round table in such a way that each will be sitting next to their acquaintances ( $n \geq 2$ ). | If all the present people are familiar with each other, then the possibility of seating 4 people in such a way raises no doubts. Let now $A$ and $B$ be unfamiliar with each other. Each of them has among the other $2n$ 2 present people no less than $n$ acquaintances; since $n+n=2n=(2n-2)+2$, then $A$ and $B$ have at lea... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,872 |
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 | 53,873 |
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 | 53,876 |
[Processes and operations] $[\underline{\text { Induction (etc.) }}]$
A deck of punch cards of four colors is laid out in a single row. If two punch cards of the same color lie next to each other or with one card between them, then the leftmost of the two can be discarded. In addition, any number of punch cards from o... | Let's show that if a deck contains punch cards of $k$ different colors, then it is possible to insert and discard punch cards in such a way that only $k$ cards remain. We will only insert colors that are already present in the deck. First, note that the last two cards can be swapped. Indeed, if $a_{1}, \ldots, a_{\math... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,877 |
Two players play the following game. Each player, in turn, crosses out 9 numbers (of their choice) from the sequence $1,2,3, \ldots, 100,101$. After eleven such crossings, two numbers will remain. Then the second player awards the first player as many points as the difference between these remaining numbers. Prove that... | Let's describe the winning strategy for the first player. The first player's first move is to cross out nine numbers from 47 to 55. The remaining numbers are divided into pairs of the form ( $x, 55+x$ ). The first player should then act in such a way that after each of his moves, all the numbers are divided into pairs ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,878 |
There is a grid consisting of $1 \times 1$ squares. Each of its nodes is painted in one of four given colors such that the vertices of any $1 \times 1$ square are painted in different colors. Prove that there is a line belonging to the grid such that the nodes lying on it are painted in two colors. | Assume the opposite. Since adjacent nodes are colored in different colors, there will be three consecutive nodes of different colors on a horizontal line. Let these colors be denoted as 1, 2, and 3. Then, above and below these nodes, there must be nodes 3, 4, 1 (indeed, above and below node 2 there can only be node 4).... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,879 |
In an acute-angled triangle $A B C$, the median $A M$, the bisector $B K$, and the altitude $C H$ are drawn. Let $M^{\prime} K^{\prime} H^{\prime}$ be the triangle with vertices at the points of intersection of the three drawn segments. Can the area of the resulting triangle be greater than 0.499 of the area of triangl... | Answer: Yes, it can. Consider a right triangle $ABC_{1}$ with legs $AB=1$ and $BC_{1}=2n$. In it, draw the median $AM_{1}$, the angle bisector $BK_{1}$, and the altitude $C_{1}H_{1}$. The area of the triangle formed by these segments is greater than $S_{\mathrm{ABM}_{1}} - S_{\mathrm{ABK}_{1}}$. It is clear that $S_{\m... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,881 |
8 Does there exist a natural number $n$ such that the sum of the digits of $n^{2}$ is 100? | 
Prove that the maximum number of sides of a convex polygon whose sides lie on the diagonals of a given convex 100-gon is no more than 100.
## Solution
Let the sides of a convex $n$-gon lie on... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,882 |
Solve the equation $x^{\mathrm{x}^{4}}=4(x>0)$.
# | Answer: $x=\sqrt{2}$.
Obviously, $x=\sqrt{2}$ is a solution to the equation. The function $y=x^{x^{4}}$ is monotonically increasing on $[1 ;+\infty)$, so there are no other solutions on this interval. On the interval $(0 ; 1)$, there are also no solutions, as the values of the function on this interval are less than 1... | \sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,885 |
In March 1987, a teacher decided to hold 11 math club sessions. Prove that if the club does not meet on Saturdays and Sundays, then there will be three consecutive days in March during which there will be no math club sessions.
# | Suppose it is possible to conduct exactly 11 club sessions in March such that no sessions are held on Saturdays and Sundays, and in any three consecutive days, at least one session is held. Therefore, sessions must be held before Saturday and Sunday and after Saturday and Sunday.
Thus, sessions were held on the 6th, 9... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,886 |
Prove that from 53 different natural numbers, not exceeding a total sum of 1990, it is always possible to select 2 numbers that sum to 53.
# | Suppose that such a set, from which 2 numbers summing to 53 are chosen, does not exist. Let $k$ be the number of numbers in the set that are less than 53. Since from each pair of the form ( $a, 53$ - a) only one number can be in the set, $k \leq 26$. Since all numbers in the set are distinct, their sum is no less than ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,887 |
The numbers $1,2,3, \ldots, 20$ are written on a board. It is allowed to erase any two numbers $a$ and $b$ and replace them with their sum $ab + a + b$. What number can result after 19 such operations? | Note that the equality $a b+a+b=(a+1)(b+1)-1$ holds. Let's carefully calculate the result after applying two such operations, that is, for three numbers $a, b, c$. After the first operation, the remaining numbers are $(a+1)(b+1)-1, c$. After the second operation, we need to take the product of the new numbers, each inc... | 21!-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,888 |
Fishing. Four friends came back from fishing. Each pair of them counted the sums of their catches. Six numbers were obtained: $7,9,14,14,19,21$. Can you find out what their catches were? | ## Solution
There were four friends, which means each person counted their catch three times. Therefore, if we add up all six sums, we get three times the total catch. From this, we immediately get that the total catch is 28 fish. This, in turn, means that our six sums should break down into pairs of numbers that add ... | 1,6,8,13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,889 |
[ Dirichlet's Principle (angles and lengths). ]
Inside a square with side length 1, several circles are placed, the sum of whose radii is 0.51. Prove that there exists a line that is parallel to one of the sides of the square and intersects at least 2 circles. | Let's project these circles onto one side of the square. The projection of one circle is a segment whose length is equal to the diameter of the circle. Therefore, the sum of the lengths of the projections of the circles onto the side is $2 \times 0.51 = 1.02$, which is greater than the side of the square, so at least t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,890 |
$3+$ Counting in two ways $\quad]$
Cutting a rectangle. A rectangular grid was cut into 1x2 rectangles (dominoes) such that any line running along the grid lines cuts a multiple of four dominoes. Prove that the length of one of the sides is divisible by 4. | For each line running along the grid, we will write down the number of dominoes it cuts, and add up all the numbers written down. Obviously, the sum will be divisible by 4. Since each domino is cut by exactly one line, the resulting sum is equal to the total number of dominoes. Therefore, the number of dominoes is divi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,891 |
The line of recruits stood facing the sergeant. On the command "left," some turned left, some turned right, and the rest turned around. Can the sergeant always find a place in the line so that there are an equal number of recruits facing him on both sides? | Let's agree that in the case when the sergeant stands in the formation, the letter $m$ will denote the number of people standing to the left of the sergeant facing him, and the letter $n$ will denote the number of people standing to the right of the sergeant facing him.
Suppose the sergeant first stands at the left ed... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,892 |
Shapovalov A.V.
In a $10 \times 10$ grid, the centers of all unit squares are marked (a total of 100 points). What is the minimum number of lines, not parallel to the sides of the square, needed to cross out all the marked points? | Let's draw all lines parallel to one of the diagonals of the square and containing more than one of the marked
points - there are 17 such lines. The two corner points will remain unstruck. Th... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,893 |
Guakhnov H.X.
Can it be that for some natural $k$ all natural numbers from 1 to $k$ can be divided into two groups and the numbers in each group can be written in a row in some order so that the two resulting numbers are the same? | Assume the opposite. It is clear that $k>10$, since in the set of digits from 1 to 9 there are no repetitions. Consider the largest power of ten $10^{n}$ that does not exceed $k$. The sequence of digits of the number $10^{n}$ will completely fit into one of the constructed numbers. But then the same sequence of one and... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,897 |
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 | 53,898 |
Does there exist a finite set of distinct vectors $\overrightarrow{a_{1}}, \overrightarrow{a_{2}}, \ldots, \overrightarrow{a_{n}}$ in the plane such that for any pair of distinct vectors from this set, there exists another pair from this set such that the sums of each pair are equal? | Answer: No, it does not exist. Let $\overrightarrow{a_{1}}, \overrightarrow{a_{2}}, \ldots, \overrightarrow{a_{n}}$ be distinct vectors. Choose the $O x$ axis such that the projections of these vectors onto it are distinct. Then the sum of two vectors with the largest coordinates cannot be equal to the sum of two other... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,900 |
On the plane, two parabolas are arranged such that their axes are mutually perpendicular, and the parabolas intersect at four points.
Prove that these four points lie on a single circle. | The equation of the desired circle is a linear combination of the equations of the two given parabolas.
## Solution
We introduce coordinates such that the axis \$ \$ \$ is the axis of symmetry of the first parabola, and the axis \$ $\$$ is the axis of symmetry of the second parabola. The equations of the parabolas th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,902 |
Does there exist a continuous function that takes each real value exactly 3 times? | An example can be found as the sum of a periodic and a linear function.
## Solution
One possible option is the function \( f(x) = \frac{\mathrm{x}}{3} + \arcsin (\sin \mathrm{x}) \), the graph of which is a "sawtooth" as shown in the image.
.
$$
1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}
$$ | Induction on $m$ easily proves that $1^{3}+2^{3}+3^{3}+\ldots+m^{3}=\left(\frac{m(m+1)}{2}\right)^{2}$. Indeed, the base case of induction
is obvious, so we only need to verify the equality
$$
\frac{m^{2}(m+1)^{2}}{2}+(m+1)^{3}=\frac{(m+1)^{2}(m+2)^{2}}{2}
$$
After canceling out $m+1$ and multiplying by 4, we get th... | n^{2}(2n^{2}-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,905 |
The 200-digit number 89252525... 2525 is multiplied by the number 444x18y27 (x and y are unknown digits). It turned out that the 53rd digit of the resulting number (counting from the right) is 1, and the 54th is 0. Find x and y.
# | Answer: $x=4, y=6$. Let $A=89252525 \ldots 2525, B=444 x 18 y 27$. Represent the number $B$ as a sum
$B=444 x 00 y 00+18027$. The product $A \cdot 18027$ has the form $3 . .272727272727268175$, in particular, its 53rd digit is 7, and the 54th digit is 2. Let the number $C \cdot(B-18027) / 4$ be written with the digits... | 4,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,907 |
In the city of "Diversity," there live $n$ residents, any two of whom are either friends or enemies. Each day, no more than one resident can start a new life: fall out with all of their friends and become friends with all of their enemies. Prove that all residents can become friends.
Note. If $A$ is a friend of $B$, a... | Let $A, B$ and $C$ be any three residents of the city.
It is clear that it is possible for all of them to be friends with each other; it is also possible that one of them (say, $A$) is not friends with either $B$ or $C$, while $B$ and $C$ are friends with each other: in this case, for $A, B$, and $C$ to all become fri... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,908 |
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 | 53,909 |
A.Y. Evnin
A $10 \times 10$ table is filled according to the rules of the game "Minesweeper": some cells are marked with a mine, and in each of the remaining cells, the number of mines in the neighboring cells (adjacent by side or corner) is recorded. Can the sum of all numbers in the table increase if all the "old" m... | The sum of all numbers in the table is equal to the number of pairs consisting of adjacent mined and unmined cells. When the specified operation is performed, these pairs are preserved, so the sum does not change.
## Answer
It cannot.
Task | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,910 |
Around a round table, 33 representatives of four tribes are sitting: humans, dwarves, elves, and goblins. It is known that humans do not sit next to goblins, and elves do not sit next to dwarves. Prove that some two representatives of the same tribe are sitting next to each other.
# | 1) Let's see who is more at the table: humans with goblins or elves with dwarves. Without loss of generality, we can assume that there are more humans with goblins; then there are at least 17 of them, since there are 33 beings sitting at the table in total.
2) Now let the elves and dwarves stand up from their seats. Am... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,911 |
Each voter in the election enters the surnames of 10 candidates into the ballot. There are 11 ballot boxes at the polling station. After the election, it turned out that each ballot box contains at least one ballot, and for any selection of 11 ballots, one from each box, there is a candidate whose surname appears in ea... | Take an arbitrary ballot from the 11th urn and reason by contradiction.
## Solution
Take an arbitrary ballot from the $11$-th urn. Number the candidates whose surnames appear in this ballot. Suppose that the required condition is not met. Then in the $\mathrm{k}$-th urn ( $k=1,2, \ldots 10$ ) there will be a ballot t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,912 |
[ Sequences (other) $]$ ]
2000 trees - oaks and baobabs - are planted in a row. A sign is attached to each tree indicating the number of oaks among the tree with the sign and its neighbors. Can the numbers on the signs be used to determine which trees are oaks? | ## Solution
Consider two such sequences of trees. In the first sequence, oaks grow at positions of the form $3k$ and $3k+2$, where $k$ is an integer, and baobabs grow at all other positions. In the second sequence, oaks grow at positions of the form $3k$ and $3k+1$, where $k$ is an integer, and baobabs grow at all oth... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,913 |
[ Invariants and Semi-invariants ]
In the Martian alphabet, there are two letters - U and Y, and if any adjacent letters UY are removed from any word, the meaning of the word does not change. Similarly, the meaning does not change when the letter combinations YU or UUYU are added to any position in the word. Is it tru... | When transitioning from any word to a word with the same meaning, the difference between the number of letters У and Ы does not change.
## Solution
Assume the opposite - the words ЫУЫУЫ and УЫУЫУ have the same meaning. Then, from the word ЫУЫУЫ to the word УЫУЫУ, one can transition through several transformations des... | nottrue | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,914 |
Authors: Kleptsyn V.A. Raskin M.A.
In the 7th grade parallel, there are 100 students, some of whom are friends with each other. On September 1, they organized several clubs, each founded by three students (each club had its own). Every day, students who were friends with at least three members of a club joined that cl... | Let's divide the seventh graders into two groups of 50 students each. In each group, all students are friends with everyone, and three students A, B, and C from the first group have three different friends each in the second group, and no one else is friends with anyone outside their own group.
 and (2) does not exceed $n-1+1 / 2 n(n-1)<n^{2}$.
Therefore, there will be ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,916 |
Compose the equation of the line passing through the point $A(0 ; 7)$ and tangent to the circle $(x-15)^{2}+(y-2)^{2}=25$.
# | Let $k$ be the slope of the desired tangent line. Then the equation of the tangent line is $y-7=k(x-0)$, or $y=k x+7$. This problem reduces to finding all such numbers $k$ for which the system of equations
$$
\left\{\begin{array}{l}
(x-15)^{2}+(y-2)^{2}=25 \\
y=k x+7
\end{array}\right.
$$
has exactly one solution.
S... | 7or-\frac{3}{4}x+7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,917 |
[ Coordinate method on the plane [ [MT - circle or arc of a circle $]$
Given points $A, B$ and a positive number $d$. Find the geometric locus of points $M$, for which $A M^{2}+B M^{2}=d$.
# | Choose the coordinate system $X O Y$ so that point $A$ is its origin, and point $B$ lies on the positive half-axis $O X$.
## Solution
Let $A B=b$. Choose the coordinate system $X O Y$ so that point $A$ is its origin, and point $B$ lies on the positive half-axis $O X$. Then the coordinates of points $A$ and $B$ are $(... | (x-\frac{b}{2})^{2}+y^{2}=\frac{2-b^{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,918 |
Berroov S.L.
Each integer point on the plane is colored in one of three colors, and all three colors are present. Prove that there exists a right-angled triangle with vertices of three different colors. | Let's call an integer point a node.
If on each vertical line all nodes are of the same color, then we choose any node (let it be of color 1). We draw two perpendicular lines through it, going at an angle of $45^{\circ}$ to the vertical, and choose points of colors 2 and 3 on these lines (this is possible since there a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,919 |
Setting out on the route at 4 a.m., mountaineer Jeff Lowe reached the peak "Freedom of Korea" by evening. After spending the night at the summit, the next day he set out at the same time and quickly descended back along the same path he had climbed. Prove that there is a point on the route that Lowe passed at the same ... | Let's imagine that at the moment Jeff began to descend, his doppelgänger started to exactly repeat Jeff's path from yesterday. Since Jeff is going down and his doppelgänger is going up, they will inevitably meet at some point. The place where they meet will be the point we are looking for.
Send a comment | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 53,921 |
n bandits are dividing the loot. Each of them has their own opinion about the value of a particular share of the loot, and each wants to receive no less than $1 / \mathrm{n}$ of the loot (from their point of view). Come up with a way to divide the loot among the bandits.
# | Use induction on the number of bandits.
## Solution
For two bandits, the problem is easily solved - one divides the loot into two equal parts as he sees fit, and the other chooses the largest part in his opinion. We will solve the problem using induction on the number of bandits, i.e., assume that $k$ bandits already... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,922 |
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 | 53,923 |
(
Diameters $A B$ and $C D$ of circle $S$ are perpendicular. Chord $E A$ intersects diameter $C D$ at point $K$, and chord $E C$ intersects diameter $A B$ at point $L$. Prove that if $C K: K D=2: 1$, then $A L: L B=3: 1$.
# | Let's take points $K$ and $L$ on segments $A B$ and $C D$, respectively, dividing them in the given ratios. It is sufficient to prove that the intersection point of lines $A K$ and $C L$ lies on the circle $S$. Introduce a coordinate system with the origin at the center $O$ of the circle $S$ and axes $O x$ and $O y$ di... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,924 |
[ Smallest or Largest Angle ]
Inside a circle of radius 1, there are eight points. Prove that the distance between some two of them is less than 1.
# | At least seven points are distinct from the center $O$ of the circle. Therefore, the smallest of the angles $A_{\mathrm{i}} O A_{\mathrm{j}}$, where $A_{\mathrm{i}}$ and $A_{\mathrm{j}}$ are the given points, does not exceed $360^{\circ} / 7 < 60^{\circ}$. If $A$ and $B$ are the points corresponding to the smallest ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,925 |
$\left.\begin{array}{l}{[\underline{\text { Induction in Geometry }}]} \\ {[\underline{\text { Colorings }}}\end{array}\right]$
Prove that if a plane is divided into regions by straight lines and circles, then the resulting map can be colored with two colors in such a way that regions sharing a segment or arc will be ... | We will prove this by induction on the total number of lines and circles. For one line or circle, the statement is obvious. Now suppose that any map defined by $n$ lines and circles can be colored in the required manner, and we will show how to then color a map defined by $n+1$ lines and circles. Remove one of these li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,926 |
[Examples and counterexamples. Constructions]
Can a $3 \times 3$ table be filled with distinct natural numbers such that the sums in the rows are equal to each other and the products in the columns are also equal to each other (but the sums are not necessarily equal to the products)? | For example, see the figure. The sum in each row is 45, and the product in each column is 720.
| 1 | 4 | 40 |
| :---: | :---: | :---: |
| 24 | 15 | 6 |
| 30 | 12 | 3 |
It is better to start selecting the numbers with the condition of equal products in the columns. It makes sense to choose a number that can be factore... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,928 |
Evdokimov M.A.
The teacher wrote a two-digit number on the board and asked Dima, in turn, whether it is divisible by 2? by 3? by 4? ... by 9? Dima answered all eight questions correctly, and the number of "yes" and "no" answers was equal.
a) Can you now answer at least one of the teacher's questions correctly without... | a) Let's show that the written number is even. If it were odd, then Dima would have answered "no" to the questions about divisibility by 2, 4, 6, and 8, and then, accordingly, he would have answered "yes" to the questions about divisibility by 3, 5, 7, and 9. But if the number is divisible by 5, 7, and 9, then it is di... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,929 |
Evdokimov M.A.
The king summoned two sages and announced a task to them: the first sage thinks of 7 different natural numbers with a sum of 100, secretly informs them to the king, and tells the second sage only the fourth largest of these numbers. After that, the second sage must guess the numbers thought of. The sage... | If the first sage named the number 22, then the second can uniquely determine all the numbers, as the sum of 100 can only be achieved in one way—by taking the smallest possible numbers: 1, 2, 3, 22, 23, 24, and 25.
## Answer
They can. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,933 |
Bakayev E.V.
There is a set of coins with radii $\$ 1,2,3, \ldots, 10 \$$ cm. You can place two of them on a table so that they touch each other, and add coins one by one so that the next one touches at least two of the already placed ones. A new coin cannot be placed on top of an old one. Is it possible to place seve... | We will look for a construction with four coins (obviously, three coins are insufficient). Let the coins with
touch. Denote by \(O_r\) the center of the circle with radius \(r\). Applying the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,935 |
[ Pairing and Grouping; Bijections ]
Given 12 numbers, \(a_{1}, a_{2}, \ldots a_{12}\), and the following inequalities hold:
\[
\begin{aligned}
& a_{2}\left(a_{1}-a_{2}+a_{3}\right)<0 \\
& a_{3}\left(a_{2}-a_{3}+a_{4}\right)<0
\end{aligned}
\]
\[
a_{11}\left(a_{10}-a_{11}+a_{12}\right)<0
\]
Prove that among these n... | Let's first prove that among these numbers, there will be at least 3 negative ones. For this, we will divide the given numbers into three sets of four consecutive numbers and prove that in each set of four, there will be at least one negative number. Suppose that in some set of four consecutive numbers \(a, b, c, d\) a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,939 |
From any six points on a plane (none of which are collinear), one can select three such that the triangle formed by these points has at least one angle not greater than $30^{\circ}$. Prove this. | Let's take a circle on the plane with such a large radius that it contains all the given points inside it, and consider any line $l$ lying outside this circle. We will now approach the line to our points until it passes through one of them; for example, it will pass through point $A$. Let's connect point $A$ with all t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,940 |
Prove that from eleven arbitrary infinite decimal fractions, one can choose two fractions such that their difference has in its decimal representation either an infinite number of zeros or an infinite number of nines.
# | According to the Pigeonhole Principle, for each decimal place, there will be a pair of fractions that have the same digit in that place. Let's mark such a pair of sequences for each decimal place. Since the total number of possible pairs is finite, some pair will occur infinitely many times, i.e., there will be a pair ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,941 |
[ Pairing and Grouping; Bijections ]
Prove that any even number $2 n \geq 0$ can be uniquely represented in the form $2 n=(x+y)^{2}+$ $3 x+y$, where $x$ and $y$ are non-negative integers. | Let's number the points with non-negative integer coordinates $(x, y)$ in the following order: $(0,1),(1,0),(0,2),(1,1),(2,0),(0,3),(1,2),(2,1),(3,0),(0,4)$, and so on. We will prove that the point with coordinates $(x, y)$ has the number $n = \frac{(x+y)^{2}+3 x+y}{2}$. For the first point, this is obvious. Assume tha... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,943 |
[ 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 | 53,944 |
[ $\underline{\text { Counting in two ways }}$ ]
Seven schoolchildren decided to visit seven cinemas over the weekend. In all of them, the sessions start at 9:00, 10:40, 12:20, 14:00, 15:40, 17:20, 19:00, and 20:40 (8 sessions). For each session, six of them went together, while one (not necessarily the same person ea... | Assume the opposite. Let in some cinema, someone was at every session. Then during each session, one of the groups (one person or six people) sat in this cinema, while the other group was in one of the remaining ones. Thus, in total, they attended no more than eight sessions in other cinemas. On the other hand, each gr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,945 |
[ [ 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 | 53,948 |
$\left[\begin{array}{l}\text { Convex hull and supporting lines (planes). } \\ {[\underline{\text { Inequalities with angles }}]}\end{array}\right]$
Let there be five points in general position on a plane, that is, no three of them lie on the same line and no four lie on the same circle. Prove that among these points,... | Among the given points, let's choose points $A$ and $B$ such that all other points lie on one side of the line $A B$. The remaining three points will be denoted as $C, D, E$ such that $\angle A C B > \angle A D B > \angle A E B$. Then points $C$ and $E$ lie on opposite sides of the circle passing through points $A, B$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,951 |
[ [ $\left[\begin{array}{l}\text { Examples and counterexamples. Constructions } \\ {[\quad \underline{\text { Prime numbers and their properties }}]}\end{array}\right]$
Does there exist 1000000 such distinct natural numbers that no sum of several of these numbers is a perfect square?
# | Let's specify such numbers.
First method. Take a prime number $p$ greater than $10^{12}$. The desired numbers are $p, 2 p, \ldots, 1000000 p$. Indeed, denote the sum of some $n$ of them by $S$. Then $S=k p$, where $k<10^{6} \cdot 10^{6}<p$. Thus, $S$ is divisible by $p$, but not by $p^{2}$, meaning it cannot be a perf... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,954 |
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 | 53,955 |
Prove that the lines given by the equations $y=k_{1} x+l_{1}$ and $y=k_{2} x+l_{2}$ and not parallel to the coordinate axes are perpendicular if and only if $k_{1} k_{2}=-1$.
# | Let's first consider the case when both lines pass through the origin, i.e., when their equations have the form $y=k_{1} x$ and $y=k_{2} x$. By setting $x=1$, we find the ordinates of the points $M_{1}\left(1 ; y_{1}\right)$ and $M_{2}\left(1 ; y_{2}\right)$ lying on these lines:
$$
y_{1}=k_{1} \text { and } y_{2}=k_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,957 |
Ribamko A.V.
Both Vintik and Shpuntik have three sticks with a total length of 1 meter each. Both Vintik and Shpuntik can form a triangle with their three sticks. At night, Neznaika sneaked into their house, took one stick from Vintik and one from Shpuntik, and swapped them. In the morning, it turned out that Vintik c... | Let $x_{1} \geq x_{2} \geq x_{3}$ be the lengths of the sticks that Vintik ended up with in the morning, and $y_{1} \geq y_{2} \geq y_{3}$ be the lengths of the sticks that Shpuntik ended up with. Vintik cannot form a triangle with his sticks, so $x_{1} \geq x_{2} + x_{3}$. Suppose that Shpuntik also cannot form a tria... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,958 |
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 | 53,959 |
[ $\begin{array}{ccc}\text { Induction (other) } & ] \\ \text { Fibonacci numbers } & ]\end{array}$ [Number systems (other) ]
Prove that any natural number can be represented as the sum of several different terms of the Fibonacci sequence. (The Fibonacci sequence $\left\{\mathrm{a}_{n}\right\}$ is defined by the condi... | Subtract from the number the largest Fibonacci number not exceeding it.
## Solution
We will use induction on p. The base case of induction is trivial - the number 1 is itself a Fibonacci number. Next, suppose that all natural numbers less than some number $k$ can be represented as the sum of several different members... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,960 |
Shnol D.Z.
In quadrilaterals \$ABCD\$ and \$A_1B_1C_1D_1\$, the corresponding angles are equal. In addition, \$AB=A_1B_1\$, \$AC=A_1C_1\$, \$BD=B_1D_1\$. Are the quadrilaterals \$ABCD\$ and \$A_1B_1C_1D_1\$ necessarily equal? | Let \$A=A_1\$, \$B=B_1\$, \$ on side $\$ \mathrm{~B} X$ \$ and $\$ \mathrm{D} \$$, \$ \$AC=AC_1=BD=BD_1\$ and isosceles trapezoids \$ABCD\$, \$A_1B_1C_1D_1\$ satisfy all conditions, but are not equal.
$ -> $... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,963 |
[ Invariants $]$
In a test tube, there are Martian amoebas of three types: $A, B$, and $C$. Two amoebas of any two different types can merge into one amoeba of the third type. After several such mergers, only one amoeba remained in the test tube. What is its type, if initially there were 20 amoebas of type $A$, 21 of ... | ## Task
Translate the text above into English, preserving the original text's line breaks and formatting. Output the translation directly. | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,964 | |
[ $\underline{\text { invariants }}$ ]
At the vertices of a cube, numbers are arranged: seven zeros and one one. In one move, it is allowed to add one to the numbers at the ends of any edge of the cube. Is it possible to achieve that all numbers are equal? Is it possible to achieve that all numbers are divisible by 3? | It is not possible. You need to mark four vertices of a cube such that no two of them are connected by an edge, and then consider the difference between the sum of the numbers in the marked vertices and the sum of the numbers in the unmarked vertices.
## Problem
Translate the text above into English, preserving the o... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,965 | |
[ 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 | 53,967 |
[ 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 | 53,968 |
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