problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Problem 8.4. The distances from a point $P$, lying inside an equilateral triangle, to its vertices are 3, 4, and 5. Find the area of the triangle. | Answer: $9+\frac{25 \sqrt{3}}{4}$.
Solution. Let the vertices of the triangle be denoted by $A, B$, and $C$ such that $PA = 3$, $PB = 4$, and $PC = 5$. Construct point $C_1$ such that $ACC_1$ is an equilateral triangle and mark a point $P_1$ inside it such that $P_1A = 3$, $P_1C = 4$, and $P_1C_1 = 5$ (Fig. 1). Triang... | 9+\frac{25\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,516 |
Problem 8.5. At a dance evening, $n$ pairs of partners arrived, each pair consisting of a girl and a boy. The evening consists of at least $n$ dances, each of which involves all the attendees. Initially, the boys are seated around a round table. For the first dance, each girl invites one of the boys (not necessarily he... | Answer: For odd $n$.
Solution. We will show how to meet the condition for odd $n=2 k+1$. Number the boys around the table and the corresponding girls with numbers $1,2, \ldots, n$. Let the girl with number $j$ invite the boy with number $2 j$ for the first dance if $j \leqslant k$, and the boy with number $2 j-n$ if $... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,517 |
4. Find all natural $n$ such that the number $8^{n}+n$ is divisible by $2^{n}+n$.
---
The provided text has been translated from Russian to English, maintaining the original formatting and structure. | Answer: $n=1,2,4,6$.
By the formula of abbreviated multiplication, the number $8^{n}+n^{3}$ is divisible by $2^{n}+n$, so the condition of the problem is equivalent to the condition $n^{3}-n \vdots 2^{n}+n$. But for $n \geqslant 10$ the inequality $n^{3}<2^{n}$ holds. It remains to check 9 options and get the answer.
... | 1,2,4,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,518 |
5. A cube with a side of 5 is made up of 125 smaller cubes with a side of 1. How many small cubes does a plane perpendicular to one of the cube's diagonals and passing through its midpoint intersect? | Answer: 55.
Let's introduce a coordinate system such that the cube is located in the first octant (the set of points with non-negative coordinates) and the mentioned diagonal extends from the origin $O$. The midpoint of the cube's diagonal has coordinates $(5 / 2, 5 / 2, 5 / 2)$, so the specified plane is given by the... | 55 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,519 |
Problem 1. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all pos... | Answer: 30.
Solution. Consider any arbitrary quartet of consecutive people. If there were at least 3 knights in it, the leftmost of them would definitely lie, which is impossible. If there were at least 3 liars, the leftmost of them would definitely tell the truth, which is also impossible. Therefore, in every quartet... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,520 |
Problem 2. Given an acute scalene triangle $A B C$, point $O$ is the center of its circumscribed circle. The extension of the altitude $B H$ of triangle $A B C$ intersects its circumscribed circle at point $N$. Points $X$ and $Y$ are marked on sides $A B$ and $B C$ respectively such that $O X \| A N$ and $O Y \| C N$. ... | Solution. Since $O X \| A N$ and $O Y \| C N$, we have $\angle X O Y=\angle A N C$. Therefore,
$$
180^{\circ}=\angle A B C+\angle A N C=\angle X B Y+\angle X O Y
$$
which means that the five points $O, X, B, Y, Z$ lie on the same circle.
On one hand,
$$
\angle Y O Z=\angle Y B Z=90^{\circ}-\angle A C B
$$
On the o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,521 |
Problem 3. Let's call small all natural numbers not exceeding 150. Does there exist a natural number $N$ that is not divisible by some 2 consecutive small numbers, but is divisible by the remaining 148 small numbers?
Answer: Yes, it exists. | Solution. Let $150!=2^{k} \cdot a$, where $a$ is odd. We will prove that the number $N=2^{6} \cdot \frac{a}{127}$ satisfies the condition of the problem, namely that it is not divisible by 127 and 128, but is divisible by all other small numbers.
It is obvious that the number 127 is prime, and $a$ is divisible by 127 ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,522 |
Problem 4. A buyer came to an antique shop. The trader laid out 2022 coins on the table, among which there are genuine and counterfeit coins, and warned the buyer that the genuine coins among them are more than half. To the buyer, all the coins look indistinguishable, while the trader knows exactly which coins are genu... | Answer: Yes, it can.
Solution. We will present the buyer's strategy. The first move will be to ask the trader about any two coins. In the following actions, we will always ask about the coin that remained on the table from the previous move and any other coin that has not been involved yet.
We will show the rule by w... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,523 |
Problem 5. Real numbers $x, y, z$ are such that $x+y+z=2$ and $x y+y z+z x=1$. Find the greatest possible value of the quantity $x-y$.
---
The original text has been translated into English, maintaining the original formatting and line breaks. | Answer: $\frac{2 \sqrt{3}}{3}$.
Solution. Eliminate the variable $z$:
$$
\begin{gathered}
1=x y+z(x+y)=x y+(2-x-y)(x+y)=2 x+2 y-x^{2}-y^{2}-x y \Rightarrow \\
x^{2}+y^{2}+x y+1=2 x+2 y .
\end{gathered}
$$
Let $a=x+y$ and $b=x-y$, express everything in terms of $a$ and $b$:
$$
\begin{gathered}
x^{2}+y^{2}+x y+1=2 x+... | \frac{2\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,524 |
1. On an island, there live knights who always tell the truth and liars who always lie. The population of the island is 1000 people, distributed across 10 villages (with no fewer than two people in each village). One day, every islander claimed that all their fellow villagers are liars. How many liars live on the islan... | Answer: 990.
In one village, at least two knights cannot live, because otherwise the knights would lie. Also, in the village, they cannot all be liars, since then these liars would tell the truth. Therefore, in each village there is exactly one knight, and there are 10 knights in total, and 990 liars. | 990 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,525 |
2. Petya climbed up a moving upward escalator, counting 75 steps, and then descended the same escalator (i.e., moving against the direction of the escalator), counting 150 steps. During the descent, Petya walked three times faster than during the ascent. How many steps are there on the stopped escalator? | Answer: 120.
For convenience, let's introduce a unit of time during which Petya took one step while ascending the escalator. We will measure all speeds in steps per unit of time. Petya's speed while ascending is 1 step per unit of time, and his speed while descending is 3 steps per unit of time. Let the speed of the e... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,526 |
3. For a convex quadrilateral $A B C D$, it is known that $A B=B C=C A=$ $C D, \angle A C D=10^{\circ}$. A circle $\omega$ is circumscribed around triangle $B C D$ with center $O$. Line $D A$ intersects circle $\omega$ at points $D$ and $E$. Find the measure of angle $E O A$, express your answer in degrees. | Answer: 65.
Fig. 2: to the solution of problem 3.
Fig. 2. Since triangle $A D C$ is isosceles, and we know the angle at its vertex, the angles at its base are $\angle D A C = \angle C D A =... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,527 |
4. Anya writes a natural number, and Boris replaces one of its digits with a digit differing by 1. What is the smallest number Anya should write to ensure that the resulting number is guaranteed to be divisible by 11? | Answer: 909090909.
According to the divisibility rule for 11, the remainder of a number when divided by 11 is the same as the remainder of the alternating sum of its digits when divided by 11. Therefore, changing a digit by 1 also changes the remainder by 1 when divided by 11. This means the original number gives a re... | 909090909 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,528 |
5. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles. | Answer: 28.
Fig. 3: Solution to problem 5.
Example: The arrangement of 28 tiles can be seen in Fig. 3.
Estimation. Cells not covered by tiles will be called empty. First, let's prove that no... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,529 |
6. On the parade ground, 2018 soldiers are lined up in one row. The commander can order either all soldiers standing in even positions or all soldiers standing in odd positions to leave the formation. After this order, the remaining soldiers close up into one row. In how many ways can the commander issue a series of 8 ... | Answer: 30.
Let's add 30 imaginary people to the end of the line. We will number all the people in the line from 0 to 2047. We will write all these numbers in binary using 11 digits. This will result in sequences from 00000000000 to 11111111111. The imaginary soldiers correspond to numbers from 11111100010 to 11111111... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,530 |
2. When a five-digit number is multiplied by 9, the result is a number composed of the same digits but in reverse order. Find the original number. | Answer: 10989.
Let $\overline{a b c d e}$ be the original number. The condition is written as the equation $9 \cdot \overline{a b c d e}=\overline{e d c b a}$. Note that $a=1$, because if $a \geqslant 2$, then $9 \cdot \overline{a b c d e} \geqslant 9 \cdot 20000>100000>\overline{e d c b a}$.
We have $9 \cdot \overli... | 10989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,531 |
3. We will call a pair of numbers magical if the numbers in the pair add up to a multiple of 7. What is the maximum number of magical pairs of adjacent numbers that can be obtained by writing down all the numbers from 1 to 30 in a row in some order? | Answer: 26.
Example: $1,6,8,13,15,20,22,27,29,2,5,9,12,16,19,23,26,30,3,4$, $10,11,17,18,24,25,7,14,21,28$. It is not hard to see that in this sequence, only the pairs $(29,2),(30,3),(25,7)$ are not magical.
Evaluation: Suppose it is possible to make no less than 27 pairs magical. Then there would be no more than two... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,532 |
5. Dima has 25 identical bricks of size $5 \times 14 \times 17$. Dima wants to build a tower from all his bricks, each time adding one more brick on top (each new brick adds 5, 14, or 17 to the current height of the tower). We will call a number $n$ constructible if Dima can build a tower of height exactly $n$. How man... | Answer: 98.
In essence, we need to find the number of different towers by height that can be built from the given set of bricks.
Mentally reduce the length, width, and height of each brick by 5: from $5 \times 14 \times 17$ to $0 \times 9 \times 12$. Then the total height of a potential tower will decrease by $25 \cd... | 98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,533 |
6. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles. | Answer: 28.
Fig. 1: to the solution of problem 6.
Example: the arrangement of 28 tiles can be seen in Fig. 1.
Estimate. Cells not covered by tiles will be called empty. To begin with, let's... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,534 |
10.1. The sum of the fourth and fifth power of some non-integer number is known. Can the sign of the original number always be determined
保留源文本的换行和格式，直接输出翻译结果。
(Note: The note at the end is not part of the translation but is provided to clarify that the format has been preserved.) | Answer: No. The study of the graph of $f(x)=x^{4}+x^{5}$ shows that it is continuous, increasing for $x>0$, and decreasing for $-0.8<x<0$. Accordingly, $f(-0.8)=0.08192$ is a local maximum, $f(0)=0$ is a local minimum. For $0 \leq x \leq 1$, the function takes all values from 0 to 2, including the value 0.08192. It is ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,535 |
10.2. A natural number is called a palindrome if it remains unchanged when its digits are written in reverse order (for example, the numbers 4, 55, 626 are palindromes, while 20, 201, 2016 are not). Prove that any number of the form 2016... 2016 (the digit group 2016 is repeated several times) can be represented as the... | Solution. $2016 \ldots 2016=2016 \cdot 100010001 \ldots 01$, where the ones are interspersed with triplets of zeros, and the number of ones equals the number of 2016 groups in the original number. The right number is a palindrome, while the left one is not. However, 2016=252*8 is the product of two palindromes. Multipl... | 800080\ldots8.252 | Number Theory | proof | Yes | Yes | olympiads | false | 4,536 |
10.3. Given a triangle $A B C$. From a point $P$ inside it, perpendiculars $P A^{\prime}, P B^{\prime}, P C^{\prime}$ are dropped to the sides $B C$, $C A$, $A B$ respectively. Then from the point $P$, perpendiculars $P A^{\prime \prime}, P B^{\prime \prime}$ are dropped to the sides $B^{\prime} C^{\prime}$ and $C^{\pr... | Solution. Since angles $P B^{\prime} A$ and $P C^{\prime} A$ are right angles, the quadrilateral
$P B^{\prime} A C^{\prime}$ is inscribed in a circle with diameter $P A$, from which $\angle P B^{\prime} C^{\prime} = \angle P A C^{\prime}$. Therefore, the right triangles $P B^{\prime} A^{\prime \prime}$ and $P A C^{\pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,537 |
10.4. Through the point with coordinates $(10,9)$, lines (including those parallel to the coordinate axes) have been drawn, dividing the plane into angles of $10^{\circ}$. Find the sum of the x-coordinates of the points of intersection of these lines with the line $y=101-x$. | Answer: 867. Solution: Shift the entire picture to the left by 1. We get a set of lines passing through the point $(9,9)$ and intersecting the line $y=100-x$. The picture will become symmetric with respect to the line $y=x$, so the sum of the abscissas on it is equal to the sum of the ordinates. Through the point $(9,9... | 867 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,538 |
10.5. There are 64 checkers of several colors, paired such that the colors of the checkers in each pair are different. Prove that all the checkers can be arranged on a chessboard so that the checkers in each two-square rectangle are of different colors. | Solution. Let's distribute the checkers into piles by color.
Case 1) There are two piles. Then the number of checkers in each pile is equal. We can place one color on the white squares and the other on the black squares.
Case 2) There are 3 piles. In each pile, there will be no more than 32 checkers. We will place th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,539 |
Problem 1. Masha wrote a positive number on the board. It turned out that its integer part is $43\%$ less than the number itself. What number did Masha write? Find all possible options and prove that there are no others.
The integer part of a number is the greatest integer not exceeding the given number. | Solution. Answer: $1 \frac{43}{57}$.
Let $n \geqslant 0$ be the integer part of Masha's number, and $0 \leqslant \epsilon < 1$, which is impossible.
Thus, the only possible option is the number $1 \frac{43}{57}$.
## Criteria
One of the largest suitable criteria is used:
7 p. Any complete solution to the problem.
... | 1\frac{43}{57} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,540 |
Problem 3. Given a rectangular trapezoid $A B C D$ with a right angle at $A$ ($B C \|$ $A D$). It is known that $B C=1, A D=4$. On side $A B$, point $X$ is marked, and on side $C D$, point $Y$ is marked such that $X Y=2, X Y \perp C D$. Prove that the circumcircle of triangle $X C D$ is tangent to $A B$. | Solution. Let lines $A B$ and $C D$ intersect at point $Z$. It is not hard to notice the similarity of three triangles
$$
\triangle Z B C \sim \triangle Z Y X \sim \triangle Z A D
$$
from which it follows that
$$
\frac{Z C}{B C}=\frac{Z X}{X Y}=\frac{Z D}{A D}
$$
Therefore,
$$
\begin{gathered}
\frac{Z X}{Z C}=\fra... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,541 |
Task 4. A pair of natural numbers is called good if one of the numbers is divisible by the other. Numbers from 1 to 30 were divided into 15 pairs. What is the maximum number of good pairs that could result? | Solution. Answer: 13.
First, let's prove that more than 13 good pairs could not have been formed. We will call a number bad if it is prime and not less than 15. There are only four bad numbers: \(17, 19, 23, 29\). None of the other numbers from 1 to 30 can be divisible by any bad number, and the bad numbers themselves... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,542 |
Problem 5. Petya and Vasya are playing the following game. They have a grid rectangle $1000 \times 2020$, and Petya moves first. On their turn, the first player divides the rectangle into two smaller ones with a single cut along a grid line. Then the second player chooses one of the two resulting rectangles, on which t... | Solution. Answer: Petya.
We will prove a more general fact: Petya has a winning strategy only for such $m$ and $n$ where the powers of two in their prime factorizations differ (respectively, if the powers of two in $m$ and $n$ are the same, then Vasya has a winning strategy). Since 1000 is divisible by 8, and 2020 is ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,543 |
Problem 6.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Answer: 4.
Solution. Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,544 |
Problem 6.2. A square with a side of 1 m is cut into three rectangles with equal perimeters. What can these perimeters be? List all possible options and explain why there are no others. | Answer: $8 / 3$ or $5 / 2$.
Solution. Note that, up to rotation, a square can be cut into three rectangles in only the following two ways:
First cutting. Note that if the rectangles have the... | \frac{8}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,545 |
Problem 6.3. A country has the shape of a square and is divided into 25 identical square counties. In each county, either a knight-count, who always tells the truth, or a liar-count, who always lies, rules. One day, each count said: "Among my neighbors, there are an equal number of knights and liars." What is the maxim... | Answer: 8.
Solution. First, note that counties with exactly three neighboring counties must be governed by lying counts (Fig. ??). Therefore, the corner counties are also governed by liars, since both of their neighbors are liars.
| | $L$ | $L$ | $L$ | |
| :--- | :--- | :--- | :--- | :--- |
| $L$ | | | | $L$ |
|... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,546 |
Problem 6.4. All divisors of a natural number $n$ were written in a row in ascending order from 1 to $n$ itself. It turned out that the second-to-last number in the row is 101 times greater than the second. For what largest $n$ is this possible? | Answer: $101^{3}$.
Solution. Let $p$ be the smallest divisor of the number $n$ different from 1, then $p$ is a prime number, otherwise there would be smaller divisors. Thus, the first two numbers in the row will be equal to 1 and $p$, and the last two - equal to $\frac{n}{p}$ and $n$. According to the problem, we have... | 101^3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,547 |
Problem 6.5. An equilateral triangle is divided into 36 identical equilateral triangles. For which $k$ can it be cut along the grid lines into $k$ identical polygons? | Answer: $1,3,4,9,12,36$.
Solution. First, note that $k$ must be a divisor of the number 36. In Fig. ??, the divisions into $1,3,4,9,12,36$ equal parts are shown. It remains to prove that the triangle cannot be divided into 2,
 t-$ $100-b=0$. Therefore, $b+100=(a+200) t>a+200$, so $b-a>100$. | b->100 | Algebra | proof | Yes | Yes | olympiads | false | 4,549 |
1. If two right triangles have the same area and perimeter, must they be congruent? | Answer. Yes. Solution. Let a and $\mathrm{b}$ be the legs of the triangle, $\mathrm{P}$ be its perimeter, and $\mathrm{S}$ be its area. Then $\mathrm{ab} / 2=\mathrm{S}$ and $a+b+\sqrt{a^{2}+b^{2}}=P$. By moving a and $\mathrm{b}$ to the right side in the second equation and squaring, we get $a+b=\frac{P^{2}+4 \mathrm{... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,550 |
3. There are 288 visually identical coins weighing 7 and 8 grams (both types are present). On the scales, 144 coins were placed on each pan such that the scales are in balance. In one operation, you can take any two groups of the same number of coins from the pans and swap their places. Prove that it is possible to mak... | Solution. We will change groups of coins from different scales. Let's assume that with each of the following replacements, the balance is maintained. We will swap one coin. They are the same. We will swap one of these coins with a new one. Now three coins are the same: a pair on one scale and one on the other. We will ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,552 |
4. We will call a natural number fashionable if its record contains the group of digits 2016 (for example, the numbers 32016, 1120165 are fashionable, while 3, 216, 20516 are not). Prove that any natural number can be obtained as the quotient of dividing a fashionable number by a fashionable number. | Solution. Let's say we need to obtain a $\mathrm{k}-3$ digit number $N$. Among the $10^{k}$ numbers from 20160...0 (k zeros) to 20169...9 (k nines), at least one is divisible by $N$. Let's denote it as $A$, and $A/N=B$. Let the number $\mathrm{C}=2016 \mathrm{~N}-\mathrm{m}-3$ digits. By appending $\mathrm{m}-4$ zeros ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,553 |
6. Given a quadratic trinomial $x^{2}+b x+c$. Prove that there exists an irrational $x$ such that the value of $\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ is rational. | Solution. Let $\mathrm{P}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$. Choose a sufficiently large rational number $\mathrm{r}$ so that $\mathrm{P}(\mathrm{x})-\mathrm{r}$ has two roots: $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$. Then, by Vieta's theorem, $\mathrm{P}(\mathrm{x})-\mathrm{r}=\left(\mathrm{x}-\mathrm{x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,554 |
Task 1. Find such two numbers $a$ and $b$, that $b$ is obtained from $a$ by permuting its digits, and $a-b$ consists only of the digit 1. | Answer: for example, 234567809 and 345678920.
## Criteria
Any correct example is worth 7 points. | 234567809345678920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,555 |
Problem 2. In a $3 \times 3$ table, natural numbers (not necessarily distinct) are placed such that the sums in all rows and columns are different. What is the minimum value that the sum of the numbers in the table can take? | Answer: 17.
Solution. Note that in each row and column, the sum of the numbers is no less than 3. Then, the doubled sum of all numbers in the table, which is equal to the sum of the sums of the numbers in the rows and columns, is no less than \(3+4+\ldots+8=33\), so the simple sum of the numbers in the table is no les... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,556 |
Problem 3. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, cont... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,557 |
Problem 4. Polycarp has 2 boxes, the first of which contains $n$ coins, and the second is empty. In one move, he can either transfer one coin from the first box to the second, or remove exactly $k$ coins from the first box, where $k$ is the number of coins in the second box. For which $n$ can Polycarp make the first bo... | Solution. Let's call a move where one coin is moved from the first box to the second box a move of the first type, and a move where coins are removed from the first box a move of the second type. Suppose a total of $x$ moves of the first type and $y$ moves of the second type were made. Then, the second box contains no ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,558 |
Problem 1. A train consists of 20 cars, numbered from 1 to 20, starting from the beginning of the train. Some of the cars are postal cars. It is known that
- the total number of postal cars is even;
- the number of the nearest postal car to the beginning of the train is equal to the total number of postal cars;
- the ... | Answer: $4,5,15,16$.
Solution. The number of the last postal car cannot exceed 20; the number of postal cars is four times less than this number and therefore does not exceed 5. Since the number of postal cars is even, there are 2 or 4 of them.
Suppose there are two. Then the first one has the number 2, and the last ... | 4,5,15,16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,559 |
Task 2. Lёnya has cards with digits from 1 to 7. How many ways are there to glue them into two three-digit numbers (one card will not be used) so that their product is divisible by 81, and their sum is divisible by 9? | Answer: 36.
Solution. If one of the numbers is not divisible by 9, then the other is not either, since their sum is divisible by 9. But then the product cannot be divisible by 81, a contradiction. Therefore, both numbers are divisible by 9.
Then the sum of the digits in each number is divisible by 9, and thus the tot... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,560 |
Problem 3. In a $5 \times 5$ square, there are 5 columns, 5 rows, and 18 diagonals, including diagonals of length one. In each cell of this square, Vova wrote the number $1, 3, 5$ or 7, and Lesha calculated the sum of the numbers in each column, row, and diagonal. Prove that among the sums obtained by Lesha, there are ... | Solution. We will call a row, column, or diagonal along which Andrey summed the numbers a line. Note that there are a total of 20 lines consisting of an odd number of cells (5 lines in each direction). Since all numbers in the table are odd, the sums in these lines are also odd. At the same time, they cannot exceed \(5... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,561 |
Problem 5. Six boys and six girls stood in a circle, alternating. Each of them wrote a non-zero number in their notebook. It is known that each number written by a boy is equal to the sum of the numbers written by the adjacent girls, and each number written by a girl is equal to the product of the numbers written by th... | Answer: 4.5
Solution. Let's choose any five people standing in a circle in a row, with the outermost being boys. Let their numbers be $x, x y, y, y z, z$. Then $y=x y+y z$. Since $y \neq 0$, we can divide by it, obtaining $x+z=1$. Thus, the sum of any two numbers said by boys standing three apart is 1.
Let the boys s... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,562 |
2. Petya and Vasya each thought of two real numbers and told them to Masha. It turned out that the sum of the numbers thought of by Petya is equal to the product of the numbers thought of by Vasya, and that the product of the numbers thought of by Petya is equal to the sum of the numbers thought of by Vasya. Masha adde... | 2. Petya and Vasya each thought of two real numbers and told them to Masha. It turned out that the sum of the numbers thought of by Petya is equal to the product of the numbers thought of by Vasya, and that the product of the numbers thought of by Petya is equal to the sum of the numbers thought of by Vasya. Masha adde... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,563 |
3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\le... | 3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\le... | 4080401 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,564 |
4. In an acute-angled triangle $ABC$, a line $\ell$ is drawn through vertex $A$, perpendicular to the median from vertex $A$. The extensions of the altitudes $BD$ and $CE$ of the triangle intersect the line $\ell$ at points $M$ and $N$. Prove that $AM = AN$. | 4. In an acute-angled triangle $ABC$, a line $\ell$ is drawn through vertex $A$, perpendicular to the median from vertex $A$. The extensions of the altitudes $BD$ and $CE$ of the triangle intersect the line $\ell$ at points $M$ and $N$. Prove that $AM = AN$.
Let $\overrightarrow{AB} = \vec{b}, \overrightarrow{AC} = \v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,565 |
5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal? | 5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal?
Answer: 432.
Estimate. We color the cells of the table ... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,566 |
Problem 2. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all pos... | Answer: 30.
Solution. Consider any four consecutive people. If there were at least 3 knights among them, the leftmost knight would definitely lie, which is impossible. If there were at least 3 liars, the leftmost liar would definitely tell the truth, which is also impossible. Therefore, in every group of four consecut... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,567 |
Problem 4. In each cell of a $7 \times 7$ table, one of the five integers: $-2, -1, 0, 1, 2$ is written, such that the sum of all numbers in the table is 0. Prove that there exists a $3 \times 3$ square in which the absolute value of the sum of all nine numbers does not exceed 6. | Solution. Consider two $3 \times 3$ squares intersecting in a $2 \times 3$ rectangle. Let the cells of these squares contain the numbers $a, b, c, d, e, f, g, h, i, j, k, l$.
| a | b | c | d |
| :---: | :---: | :---: | :---: |
| e | f | g | h |
| $\mathbf{i}$ | $\mathbf{j}$ | k | l |
## Let
$S_{1}=|a+b+c+e+f+g+i+j+k... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,568 |
Problem 5. The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$. Points $P$ and $Q$ are the midpoints of segments $AC$ and $BD$ respectively. Points $A_1, B_1, C_1, D_1$ are marked on segments $OA, OB, OC, OD$ respectively such that $AA_1 = CC_1, BB_1 = DD_1$.
- The circumcircles of triangles $AOB$ an... | Solution. We will prove that points $O, K, P, Q$ lie on the same circle.
Notice that triangles $A K C$ and $B K D$ are similar. Indeed, $\angle O C K = \angle O D K$, since points $O, C, D, K$ lie on the same circle, and $\angle O B K = \angle O A K$, since points $O, B, A, K$ lie on the same circle.
In similar figur... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,569 |
7.1. Masha and the Bear ate a basket of raspberries and 60 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate raspberries 6 times faster than Masha, and pies only 3 times faster. How many pies did the Bear... | Answer: 54 pies. Solution: Divide the raspberries into 3 equal parts. The bear ate each part 6 times faster than Masha, but there are two parts, so he spent only 3 times less time on the raspberries than Masha. This means Masha ate pies in one-third of the time the bear did. Since she eats three times slower, she ate 9... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,570 |
7.2. If the width of a rectangle is increased by 3 cm and the height is decreased by 3 cm, its area will not change. How will the area change if, instead, the width of the original rectangle is decreased by 4 cm and the height is increased by 4 cm? | Answer. It will decrease by $28 \mathrm{~cm}^{2}$. Solution. Let the width of the original rectangle be $s$, and the height $h$. The area of the original rectangle is $sh$, and the area of the modified one is $(s+3)(h-3)$, which equals $sh+3(h-s)-9$. Given the equality of areas, $3(h-s)-9=0$, from which $h-s=3$. In the... | 28\mathrm{~}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,571 |
7.3. A natural number is called a palindrome if it remains unchanged when its digits are written in reverse order (for example, the numbers 4, 55, 626 are palindromes, while 20, 201, 2016 are not). Represent the number 2016 as a product of three palindromes greater than 1 (find all possible options and explain why ther... | Answer: 2-4.252. Solution. The number 2016 is not divisible by 11. Therefore, none of the palindromes are two-digit (they all are divisible by 11). This means that two of the palindromes are one-digit (if the two factors have no less than 3 digits, then the product is no less than $100 \cdot 100 > 2016$). The product o... | 2\cdot4\cdot252 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,572 |
7.4. Segments $\mathrm{KL}$ and $\mathrm{MN}$ intersect at point $\mathrm{T}$. It is known that triangle KNT is equilateral and $\mathrm{KL}=\mathrm{MT}$. Prove that triangle $\mathrm{LMN}$ is isosceles. | Solution. On the ray $\mathrm{TM}$, we mark the segment $\mathrm{TD}=\mathrm{TL}$. Since $\mathrm{TD}=\mathrm{TL}<\mathrm{KL}=\mathrm{TM}$, the point $\mathrm{D}$ lies on the segment TM. Since $\angle \mathrm{DTL}=\angle \mathrm{MTK}=60^{\circ}$ and $\mathrm{DT}=\mathrm{TL}$, the triangle $\mathrm{DTL}$ is equilateral.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,573 |
7.5. At the Olympiad, 300 seventh-graders from 4 schools arrived. Prove that they can be divided into teams of 3 people each in such a way that in each team either all three students are from the same school, or all three are from different schools. | Solution. We will gather students from one school and separate them into teams of 3 until 0, 1, or 2 are left. Since the initial number of students is a multiple of 3, the total number of remaining students will also be a multiple of 3. But there are no more than $2+2+2+2=8$, so there are only 0, 3, or 6 left. Let's fo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,574 |
1. Buses from Moscow to Oryol depart at the beginning of each hour (at 00 minutes). Buses from Oryol to Moscow depart in the middle of each hour (at 30 minutes). The journey between the cities takes 5 hours. How many buses from Oryol will the bus that left from Moscow meet on its way? | Answer: 10.
It is clear that all buses from Moscow will meet the same number of buses from Orel, and we can assume that a bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Orel at $7:30, 8:30, \ldots, 15:30, 16:30$ and only them. There are 10 such buses.
$\pm$ Correct reaso... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,575 |
2. Three generous friends, each of whom has candies, redistribute them as follows: Vasya gives some of his candies to Petya and Kolya, doubling the number of candies they have. After this, Petya gives some of his candies to Kolya and Vasya, doubling the number of candies they have as well. Finally, Kolya gives some of ... | Answer: 252.
Let's track the number of candies Kolya has. After the first redistribution, he has 72, and after the second - 144. Therefore, he gave away $144-36=108$ candies, and during this time, the number of candies Vasya and Petya had doubled. So, the total number of candies Vasya, Petya, and Kolya have together i... | 252 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,576 |
1. Nезнayka, Doctor Pilulkin, Knopochka, Vintik, and Znayka participated in a math competition. Each problem in the competition was solved by exactly four of them. Znayka solved strictly more than each of the others - 10 problems, while Nезнayka solved strictly fewer than each of the others - 6 problems. How many probl... | Answer: 10.
Each of Dr. Pill, Knopochka, and Vintik, according to the condition, solved from 7 to 9 problems. Therefore, the total number of solved problems ranges from $10+6+3 \cdot 7=37$ to $10+6+3 \cdot 9=43$. Note that this number should be equal to four times the number of problems. Among the numbers from 37 to 4... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,577 |
5. On the table, there are 2018 playing cards (2018 stacks, each with one card). Petya wants to combine them into one deck of 2018 cards in 2017 operations. Each operation consists of merging two stacks. When Petya merges stacks of $a$ and $b$ cards, Vasily Ivanovich pays Petya $a \cdot b$ rubles. What is the maximum a... | Answer: $\frac{2017 \cdot 2018}{2}=2035153$
Let's mentally connect the cards in one stack with invisible threads, each with each. Then the operation of combining two stacks with $a$ and $b$ cards adds exactly $a \cdot b$ threads. In the end, each card will be connected by a thread to each other, and the number of thre... | 2035153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,578 |
Problem 10.1. A bus leaves city $A$ for city $B$, which is 240 km away from $A$, at a speed of 40 km/h. At the same time, a car leaves $B$ for $A$ at a speed of $v$ km/h. Half an hour after meeting the bus, the car, without reaching city $A$, turns around and heads back towards $B$ at the same speed. Determine all valu... | Answer: $v \in(56 ; 120)$.
Solution. Let $t$ be the time of the meeting, measured in hours. Then, by the condition, $t=\frac{240}{v+40}$. Further, the condition states that the car, having driven for half an hour, did not reach point $A$, which leads to the inequality $v(t+0.5) < 240 - 0.5v$.
Considering the positivi... | v\in(56;120) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,579 |
Task 10.3. Determine all natural numbers $n$ that have exactly $\sqrt{n}$ natural divisors (including 1 and the number $n$ itself). | Answer: 1 and 9.
Solution. First, note that $n=1$ works, and we will assume that $n \geqslant 2$. Second, from the condition it follows that $n=k^{2}$ for some natural number $k$. The squares of natural numbers have an odd number of divisors, since all divisors, except the base of the square, are paired as $d$ and $n ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,581 |
Problem 10.5. In each cell of a square table of size $200 \times 200$, a real number not exceeding 1 in absolute value was written. It turned out that the sum of all the numbers is zero. For what smallest $S$ can we assert that in some row or some column, the sum of the numbers will definitely not exceed $S$ in absolut... | Answer: 100.
Solution. First, we show that $S<40000$.
$$
This means that one of the numbers $A$ or $D$ in absolute value exceeds 10000. However, each of the corresponding squares contains only 10000 cells, and the numbers in them do not exceed 1 in absolute value. Contradiction.
## Criteria
Any correct solution to ... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,583 |
8.1. If the width of a rectangle is increased by $30 \%$, and the height is decreased by $20 \%$, its perimeter will not change. Will the perimeter decrease or increase, and by what percentage, if instead the width of the original rectangle is decreased by $20 \%$, and the height is increased by $30 \%$? | Answer. It will increase by $10 \%$. Solution. Let the width of the original rectangle be $s$, and the height be $h$. In the first case, the modified width and height will be 1.3 and 0.8 $s$ respectively. According to the condition, $2(s+h)=2(1.3 s+0.8 h)$, from which $h=1.5 s$. This means that the original perimeter i... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,584 |
8.2. The diagonals of parallelogram $A B C D$ intersect at point $O$. In triangles $O A B$, $OBC$, $OCD$, medians $OM$, $OM'$, $OM''$ and angle bisectors $OL$, $OL'$, $OL''$ are drawn respectively. Prove that angles $MM'M''$ and $LL'L''$ are equal. | Solution. Let $\mathrm{AB}=\mathrm{a}, \mathrm{BC}=\mathrm{b}, \mathrm{OB}=\mathrm{c}, \mathrm{AO}=\mathrm{OC}=\mathrm{d}$. By the angle bisector theorem $\mathrm{BL}: \mathrm{AL}=\mathrm{BO}: \mathrm{AO}$ $B L:(a-B L)=c: dB L=a c /(c+d)$. Similarly, $B^{\prime}=b c /(c+d)$. Therefore, $\mathrm{BL}: \mathrm{BL}^{\prime... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,585 |
8.3. A natural number is called a palindrome if it remains unchanged when its digits are written in reverse order (for example, the numbers 4, 55, 626 are palindromes, while 20, 201, 2016 are not). Prove that any number of the form 2016... 2016 (the digit group 2016 is repeated several times) can be represented as the ... | Solution. $2016 \ldots . .2016=2016 \cdot 100010001 \ldots 01$, where ones are interspersed with triple zeros, and the number of ones equals the number of 2016 groups in the original number. The right number is a palindrome, while the left one is not. However, 2016=252*8 — the product of two palindromes. Multiplying th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,586 |
8.4. At the Olympiad, 300 students from no fewer than 4 schools arrived. Prove that they can be divided into teams of 3 people each, so that in each team either all three students are from the same school, or all three are from different schools. | Solution. If more than three students come from a school, we separate them into teams of three as long as possible, until 1, 2, or 3 students remain. We do this for all schools. In each school, the remainder will be 0, 1, or 2 students. Since the total number of students is a multiple of 3, and we have separated them i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,587 |
8.5. Through the point with coordinates $(2,2)$, lines (including two parallel to the coordinate axes) are drawn, dividing the plane into angles of $18^{\circ}$. Find the sum of the abscissas of the points of intersection of these lines with the line $y=2016-x$. | Answer: 10080. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(2,2)$, 10 lines are drawn, and the line $y=2016-x$ intersects all of them. For each point on the line $y=2016-x$, the sum of the coordinates is 2016, so... | 10080 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,588 |
Task 1. On a plate, there are various candies of three types: 2 lollipops, 3 chocolate candies, and 5 jelly candies. Svetlana sequentially ate all of them, choosing each subsequent candy at random. Find the probability that the first and last candies eaten were of the same type. | Answer: $14 / 45$.
Solution. Two candies of the same type can be either lollipops, chocolates, or jellies. We will calculate the probabilities of each of these events and add them up.
Arrange the candies in the order of their consumption. The probability that the first candy is a lollipop is $2 / 10$. The probability... | \frac{14}{45} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,589 |
Problem 2. In a school tic-tac-toe tournament, 16 students participated, each playing one game against every other student. A win was worth 5 points, a draw -2 points, and a loss -0 points. After the tournament, it was found that the participants collectively scored 550 points. What is the maximum number of participant... | Solution. A total of $\frac{16 \cdot 15}{2}=120$ games were played in the tournament. In each game, either 5 points were awarded (in case of a win-loss) or 4 points (in case of a draw). If all games had ended in a draw, the total number of points for all participants would have been $120 \cdot 4=480$, which is 70 point... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,590 |
Problem 3. A natural number $A$ is called interesting if there exists a natural number $B$ such that:
- $A>B$;
- the difference between the numbers $A$ and $B$ is a prime number;
- the product of the numbers $A$ and $B$ is a perfect square.
Find all interesting numbers greater than 200 and less than 400. | Answer: $225,256,361$.
Solution. Let $A-B=p-$ a prime number. By the condition $A B=B(B+p)=n^{2}$ for some natural $n$. Note that the GCD of the numbers $B$ and $B+p$ divides their difference, which is $p$, so it is either $p$ or 1. Let's consider two cases.
- Suppose, GCD $(B, B+p)=p$. Then $B=p s$ and $B+p=p(s+1)$ ... | 225,256,361 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,591 |
Problem 4. Positive numbers $a, b, c, d$ are greater than 1. Find the smallest possible value of the expression
$$
\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)
$$ | Answer: 67.
Solution. From the properties of logarithms, it follows that $\log _{a} b \cdot \log _{b} c \cdot \log _{c} d \cdot \log _{d} a=1$. Also, all these four factors are positive, since all numbers $a, b, c, d$ are greater than 1.
Transform and estimate the given expression
$$
\begin{gathered}
S=\log _{a}\lef... | 67 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,592 |
Problem 5. Point $P$ inside an acute-angled triangle $A B C$ is such that $\angle B A P=$ $\angle C A P$. Point $M$ is the midpoint of side $B C$. Line $M P$ intersects the circumcircles of triangles $A B P$ and $A C P$ at points $D$ and $E$ respectively (point $P$ lies between points $M$ and $E$, point $E$ lies betwee... | Answer: 35.
Solution. Quadrilateral $A E P C$ is inscribed, so $\angle C A P = \angle C E P$. Similarly, quadrilateral $B P A D$ is inscribed, so $\angle B D P = \angle B A P = \angle C A P = \angle C E P$.
Drop perpendiculars $B X$ and $C Y$ to line $M P$. Note that right triangles $B M X$ and $C M Y$ are congruent ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,593 |
Problem 6. We will call a function $f$ good if
- $f$ is defined on the interval $[0,1]$ and takes real values;
- for all $x, y \in[0,1]$ it holds that $|x-y|^{2} \leqslant|f(x)-f(y)| \leqslant|x-y|$.
Find all good functions. | Answer: $f(x)=x+c, f(x)=-x+c$, where $c \in \mathbb{R}$.
Solution. It is obvious that all the functions listed above are suitable. Note that along with each function $f(x)$ that satisfies the condition, all functions of the form $f(x)+c$ and $-f(x)$ also satisfy it. We will prove that if $f(0)=0$ and $f(1) \geqslant 0... | f(x)=x+,f(x)=-x+ | Inequalities | proof | Yes | Yes | olympiads | false | 4,594 |
1. Several numbers were written on the board, their arithmetic mean was equal to $M$. They added the number 15, after which the arithmetic mean increased to $M+2$. After that, they added the number 1, and the arithmetic mean decreased to $M+1$. How many numbers were on the board initially? (Find all options and prove t... | Answer: 4.
Let there be $k$ numbers in the original list with a sum of $S$. Then, by the condition,
$$
\frac{S+15}{k+1}-\frac{S}{k}=2, \quad \frac{S+15}{k+1}-\frac{S+16}{k+2}=1.
$$
By bringing to a common denominator and transforming in an obvious way, we get that these equations are equivalent to the following two:... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,595 |
5. Let $d_{1}, d_{2}, \ldots, d_{n}$ be all the natural divisors of the number $10!=1 \cdot 2 \cdot \ldots \cdot 10$. Find the sum
$$
\frac{1}{d_{1}+\sqrt{10!}}+\frac{1}{d_{2}+\sqrt{10!}}+\ldots+\frac{1}{d_{n}+\sqrt{10!}}
$$ | Answer: $3 /(16 \sqrt{7})$.
Let's denote the given sum as $S$. Then, since for each $d_{j}$ the number $10!/ d_{j}$ is also a divisor,
$$
S=\sum_{j=1}^{n} \frac{1}{10!/ d_{j}+\sqrt{10!}}=\frac{1}{\sqrt{10!}} \sum_{j=1}^{n} \frac{d_{j}}{\sqrt{10!}+d_{j}}
$$
Therefore, adding the original and the last expression for $... | \frac{270}{2\sqrt{10!}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,596 |
6. Tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to the plane $ABC$. The line $DE$ intersects the plane $ABC$ at point $F$, whi... | Answer: $40^{\circ}$.
Note that point $E$ is equidistant from points $A, B, C$, so its projection onto the plane $A B C$ coincides with the projection of point $O$ onto this plane and is the center of the circumscribed circle of triangle $A B C$.
Consider triangles $A D E$ and $B D E$. They have a pair of equal sides... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,597 |
1. Buses from Moscow to Voronezh depart every hour, at 00 minutes. Buses from Voronezh to Moscow depart every hour, at 30 minutes. The trip between the cities takes 8 hours. How many buses from Voronezh will the bus that left from Moscow meet on its way? | Answer: 16.
It is clear that all buses from Moscow will meet the same number of buses from Voronezh, and we can assume that the bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Oryol at 4:30, 5:30, ..., 18:30, 19:30 and only them. There are 16 such buses.
$\pm$ Correct rea... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,598 |
3. How many solutions in natural numbers does the equation
$$
(2 x+y)(2 y+x)=2017^{2017} ?
$$ | Answer: 0.
Note that the sum of the numbers $A=2x+y$ and $B=2y+x$ is divisible by 3. Since the number on the right side is not divisible by 3, neither $A$ nor $B$ are divisible by 3. Therefore, one of these two numbers gives a remainder of 2 when divided by 3, and the other gives a remainder of 1. Thus, their product ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,599 |
4. The teacher gave Vasya and Petya two identical cardboard $n$-gons. Vasya cut his polygon into 33-gons along non-intersecting diagonals, while Petya cut his polygon into 67-gons along non-intersecting diagonals. Find the smallest possible value of $n$. | Answer: 2017.
The sum of the angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$. If it is cut into $k$ 33-sided polygons, then $(n-2) \cdot 180^{\circ}=k \cdot (33-2) \cdot 180^{\circ}$, hence $n-2 \vdots 31$. Similarly, from the second condition, it follows that $n-2 \vdots 65$. Since 31 and 65 are coprime, ... | 2017 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,600 |
Problem 2. Dodson, Williams, and their horse Bolivar want to get from city A to city B as quickly as possible. Along the road, there are 27 telegraph poles that divide the entire journey into 28 equal segments. Dodson takes 9 minutes to walk one segment, Williams takes 11 minutes, and either of them can ride Bolivar to... | Solution. Let the distance from A to B be taken as a unit, and time will be measured in minutes. Then Dodson's speed is $1 / 9$, Williams' speed is $1 / 11$, and Bolivar's speed is $-1 / 3$.
Let the desired post have the number $k$ (i.e., the distance from city A is $k / 28$). Then Dodson will arrive in time
$$
\frac... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,601 |
Problem 4. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, cont... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,602 |
Problem 5. At the court, 50 musketeers serve. Every day they split into pairs and conduct training duels. Is it true that after 24 days, there will be three musketeers who have not participated in training duels with each other? | Answer: Yes, they will be found.
Solution. Suppose the opposite - let there be no such trio of musketeers.
Let's single out one musketeer, let's call him Petya. According to the condition, he fought duels with no more than 24 others. Consider a set of 25 musketeers with whom he did not fight (if there are more, we wi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,603 |
1. Prove that for any non-negative integer n, the expression $3^{6 n}-2^{6 n}$ is divisible by 35. | Solution. Indeed, for any non-negative integers $m$ and any numbers $a$ and $b$
$$
a^{m}-b^{m}=(a-b)\left(a^{m-1}+a^{m-2} b+\ldots+a b^{m-2}+b^{m-1}\right)
$$
In particular, for even $m=2 k$
$$
a^{2 k}-b^{2 k}=\left(a^{2}-b^{2}\right)\left(a^{2(k-1)}+a^{2(k-2)} b^{2}+\ldots+a^{2} b^{2(k-2)}+b^{2(k-1)}\right) .
$$
T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,604 |
3. Prove that
$$
\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{1}{2}
$$ | Solution. Multiply and divide the expression on the left side of the identity to be proven by $2 \cos \frac{\pi}{14}$, which is not equal to 0. Apply the formula
$$
2 \cos x \cos y=\cos (x+y)+\cos (x-y) .
$$
We get
$$
\begin{array}{r}
\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{2 \cos \frac{\p... | \frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 4,606 |
4. On an $8 \times 8$ chessboard, 8 rooks are placed such that no rook attacks another. Passing by the board, Vitya noticed three rooks standing on white squares. Prove that there is at least one more rook, also standing on a white square. | Solution. Let's number the rows and columns from 1 to 8 from bottom to top and from left to right. Notice that the sum of the indices of any black cell is even, while that of any white cell is odd. On each row and each column, there is exactly one rook. Therefore, the sum of the indices of all cells where the rooks are... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,607 |
5. For which negative integer values of $n$ is the function $f$, defined by the equation
$$
f(x)=\cos 7 x \cdot \sin \frac{25 x}{n^{2}}
$$
a periodic function with period $T=7 \pi$? | Solution. By definition of the period, for any value of $x$ the equality must hold
$$
\cos 7(x+7 \pi) \cdot \sin \frac{25}{n^{2}}(x+7 \pi)=\cos 7 x \cdot \sin \frac{25 x}{n^{2}}
$$
Therefore, it must also hold for $x=0$. In this case, we arrive at the equation
$$
\cos 49 \pi \cdot \sin \frac{175 \pi}{n^{2}}=0
$$
Co... | n=-1,\quadn=-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,608 |
2. Does there exist a natural number $n>3$ such that the decimal representation of the number $n^{2}$ consists only of odd digits? (20 points) | Solution: Suppose such a natural number $n>3$ exists. Then the number $n^{2}$ consists of at least two digits. Take the number formed by the last two digits of the number $n^{2}$, and call it $m$. According to the assumption, both digits of this number are odd. Therefore, the remainder when this number is divided by 4 ... | doesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,609 |
3. Let's call a natural number an almost palindrome if it can be transformed into a palindrome by changing one of its digits. How many nine-digit almost palindromes exist? (20 points) | Solution: After replacing one digit in the almost palindrome, we get a number of the form $\overline{a b c d e d c b a}$. Let's divide the digits into pairs by place value: 1 and 9, 2 and 8, 3 and 7, 4 and 6. In any almost palindrome, the digits in three of the specified pairs of place values must be the same, and in e... | 3240000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,610 |
5. If the angle at the vertex of a triangle is $40^{\circ}$, then the bisectors of the other two angles of the triangle intersect at an angle of $80^{\circ}$. | Write the answer in digits in ascending order, without spaces (for example, 12345). | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,614 |
5. If in an acute scalene triangle three medians, three angle bisectors, and three altitudes are drawn, they will divide it into 34 parts. | Write the answer in digits in ascending order, without spaces (for example, 12345).
## Mathematics $8-9$ grade
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,615 |
5. There exists a quadrilateral that is not a parallelogram, in which the point of intersection of the diagonals bisects one of its diagonals. | Write the answer in digits in ascending order, without spaces (for example, 12345). | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,619 |
5. If the angle at the vertex of a triangle is $40^{\circ}$, then the bisectors of the other two angles of the triangle intersect at an angle of $70^{\circ}$. | Write the answer in digits in ascending order, without spaces (for example, 12345). | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,620 |
5. In a triangle with sides 3 and 5 and an angle of $120^{\circ}$, the third side is 7. | Write the answer in digits in ascending order, without spaces (for example, 12345). | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,622 |
# Task 1. (20 points)
Given six pencils in the form of identical right circular cylinders. Arrange them in space so that each pencil has a common boundary point with any other pencil.
# | # Solution.
Let's highlight a horizontal plane and place three cylinders vertically on it so that they touch each other (obviously, the centers of the circles of the bases form an equilateral triangle with a side equal to the diameter of the pencils). Place three more pencils symmetrically below this plane so that the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,625 |
# Task 2. (20 points)
Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.
# | # Solution.
Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,626 |
# Task 3. (20 points)
In the memory of a supercomputer, there is an infinite string of numbers in both directions. Initially, one number in the string is one, and all others are zeros. In one step, the supercomputer adds to each number in the string the sum of its two neighboring numbers (all additions occur simultane... | # Solution.
Let's write down the four leftmost non-zero numbers in each row, starting from the third, and place the letter "ч" in the place of an even number, and the letter "н" in the place of an odd number. The first letter is always "н", as it results from adding the previous extreme number (one) with zero. In the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,627 |
# Task 4. (20 points)
Can the expression $1+x^{2016} y^{2016}$ be represented as a product $f(x) \cdot g(y)$? Justify your answer.
# | # Solution.
Let's set the variable $x$ to 0 in the equation $x^{2016} y^{2016} + 1 = f(x) \cdot g(y)$. Then $1 = f(0) g(y)$, i.e., $g(y)$ is some constant equal to $\frac{1}{f(0)}$ for all values of $y$. Similarly, when $y=0$, we get that $f(x) = \frac{1}{g(0)}$ for all values of $x$. Clearly, $f(x) g(y) = \frac{1}{f(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,628 |
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