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3. In triangle $ABC$, the median $BM$ is twice as short as side $AB$ and forms an angle of 40 degrees with it. Find angle $ABC$.
| Answer: $110^{\circ}$. Solution. Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a parallel... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Tom Sawyer took on the task of painting a very long fence, adhering to the condition: any two boards, between which there are exactly two, exactly three, or exactly five boards, must be painted in different colors. What is the smallest number of different colors he can manage with. | Answer: Three. Solution. Note that between the first and fourth, and the fourth and seventh boards, there are two boards each, and between the first and seventh boards, there are five boards. Therefore, the first, fourth, and seventh boards of the fence must be painted in different colors, so Tom will need at least thr... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The length of a rectangle was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$? | Answer: By 18%. Solution. Let the length be $a$, and the width be $b$. According to the condition, $2(0.1 a + 0.2 b) = 0.12(2 a + 2 b)$, from which we get $a = 4 b$. If the length is reduced by $20\%$, and the width by $10\%$, then the perimeter will decrease by $2(0.2 a + 0.1 b) = 1.8 b$, and it was $10 b$. Therefore,... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a coffee shop, 55 Indians and Turks met, each drinking tea or coffee. All Indians tell the truth when drinking tea and lie when drinking coffee, while all Turks do the opposite. When asked "Are you drinking coffee?" 44 people answered "yes," when asked "Are you a Turk?" 33 people answered "yes," and 22 people agr... | Answer: 0. Solution. Let $h_{\psi}, h_{\kappa}$ be the number of Indians drinking tea and coffee, respectively. Similarly, define the quantities $T_{4}$ and $T_{k}$. Then from the first question, it follows that $T_{4}+T_{\kappa}=44$, and from the second question, it follows that $h_{\kappa}+T_{\kappa}=33$. Suppose tha... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 100 identical-looking coins, of which 85 are counterfeit and 15 are genuine. You have a miracle tester, into which you can place two coins and get one of three results - "both coins are genuine," "both coins are counterfeit," and "the coins are different." Can you find all the counterfeit coi... | Answer. Yes. Solution. We will present one of the possible ways to define the fake coins. Divide the coins into 50 pairs and check all pairs except one. We will learn the number of fake coins in each pair. Since the total number of fake coins is known, we will also learn how many are fake in the remaining pair. We need... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Black and white balls are arranged in a circle, with black balls being twice as many as white ones. It is known that among pairs of adjacent balls, there are three times as many monochromatic pairs as polychromatic ones. What is the smallest number of balls that could have been arranged? (B. Trushin) | Answer: 24. Solution: Since the number of black balls is twice the number of white balls, the total number of balls is divisible by three. Let's denote it by $n$. All the balls are divided into alternating groups of consecutive balls of the same color (a group can consist of just one ball). Since the colors of the grou... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural $k$ such that for some natural number $a$, greater than 500,000, and some natural number $b$, the equality $\frac{1}{a}+\frac{1}{a+k}=\frac{1}{b}$ holds. (I. Bogdanov) | Answer. $k=1001$. Solution. Estimation. Let $a+k=c$ and $\text{GCD}(a, c)=d$. Then $a=d a_{1}, c=d c_{1}$ and $\frac{1}{a}+\frac{1}{c}=\frac{a_{1}+c_{1}}{d a_{1} c_{1}}$. Since the numbers $a_{1} c_{1}$ and $a_{1}+c_{1}$ are coprime, $d$ must divide $a_{1}+c_{1}$. Therefore, $d \geq a_{1}+c_{1}$ and $d^{2} \geq d\left(... | 1001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the vertices of a regular 300-gon, numbers from 1 to 300 are placed once each in some order. It turns out that for each number a, among the 15 nearest numbers to it in the clockwise direction, there are as many numbers less than a as there are among the 15 nearest numbers to it in the counterclockwise direction. ... | Answer: 10. Solution: Estimation. To prove that there are no fewer than 10 huge numbers, it is sufficient to prove that among any 30 consecutive numbers, there will be a huge one. Indeed, suppose this is not the case. Then consider the largest of these 30 numbers. On one side of it, all 15 nearest numbers will be part ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Wrote down two numbers - the first and the second. Added the second to the first - got the third, added the third to the second - got the fourth, and so on. The sum of the first six numbers is 2008. What is the fifth number? | Answer: 502. Solution: Let the first number be $a$, the second - $b$. Then the third number is $a+b$, the fourth $-a+2b$, the fifth $-2a+3b$, the sixth $-3a+5b$, and the sum of all six numbers is $8a+12b$. Thus, the fifth number is a quarter of the sum of all six numbers, that is, $2008: 4=502$.
Grading Guidelines. An... | 502 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a right triangle, the height dropped to the hypotenuse is four times shorter than the hypotenuse. Find the acute angles of the triangle. | Answer: 15 and 75 degrees. Solution. Let $A B C$ be the given right-angled triangle with the right angle at vertex $C$. As is known, the median $C M$ is half the hypotenuse $A B$. The height $C H$ is given to be $A B / 4$. Therefore, in the right-angled triangle $C H M$, the hypotenuse $C M$ is twice the length of the ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. The decimal representation of a natural number $N$ consists only of ones and twos. It is known that by erasing digits from this number, any of the 10000 numbers consisting of 9999 ones and one two can be obtained. Find the smallest possible number of digits in the representation of $N$. (G. Chelnokov) | Answer: 10198. Solution: Example. The number 1...121...12...21...121...1, where there are 100 twos, 99 ones at the beginning and end, and 100 ones between adjacent twos. The number consisting of 9999 ones and a two, where before the two there are $100 m+n$ ones ($0 \leq m, n \leq 99$), is obtained by deleting all twos ... | 10198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On a $100 \times 100$ chessboard, 1975 rooks were placed (each rook occupies one cell, different rooks stand on different cells). What is the maximum number of pairs of rooks that could be attacking each other? Recall that a rook can attack any number of cells along a row or column, but does not attack a rook that i... | Answer: 3861. Solution: Sequentially remove from the $100 \times 100$ board the verticals and horizontals that do not contain rooks, each time gluing the edges of the removed strip. We will get a rectangle $\pi$, in each vertical and each horizontal of which there is at least one rook (obviously, the number of pairs of... | 3861 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Zeus has scales that allow him to find out the weight of the load placed on them, and a bag with 100 coins, among which there are 10-gram and 9-gram coins. Zeus knows the total number $N$ of 10-gram coins in the bag, but it is unknown which ones weigh how much. He would like to make four weighings on the scales and ... | Answer. For $N=15$. Solution. First, let's outline Zeus's algorithm for $N=15$. By weighing a certain number of coins, he immediately determines the number of heavy coins among the weighed ones. Since he only needs to identify one light coin, he can weigh just 8 coins in the first weighing. If there are light coins amo... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. What is the maximum number of white and black pawns that can be placed on a 9x9 grid (a pawn, regardless of its color, can be placed on any cell of the board) so that no pawn attacks any other (including those of the same color)? A white pawn attacks two diagonally adjacent cells on the next higher horizontal row, w... | Answer: 56. Solution: An example with 56 pawns is shown in the figure. Evaluation: Note that in each rectangle of three rows and two columns, there are no more than 4 pawns. Indeed, if there are at least 5, then on one of the colors, all three cells are occupied, and the pawn in the middle row attacks one of the two re... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In each cell of a $2 \times 2$ table, one number is written. All numbers are distinct, the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second column. Find the sum of all four... | Answer: 0. Solution: Let the numbers in the top row of the table be (from left to right) $a$ and $b$, and the numbers in the bottom row (from left to right) be $c$ and $d$. By the condition, $a+b=c+d$ and $a c=b d$. Expressing $c$ from the first equation and substituting into the second, we get $a(a+b-d)=b d \Leftright... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. 200 people are standing in a circle. Each of them is either a liar or a conformist. Liars always lie. A conformist who stands next to two conformists always tells the truth. A conformist who stands next to at least one liar can either tell the truth or lie. 100 of those standing said: "I am a liar," and the other 10... | Answer: 150. Solution: A liar cannot say, "I am a liar." Therefore, 100 people who said, "I am a liar," are conformists. All of them lied, so next to each of them stands a liar. Since next to a liar there can be a maximum of two conformists, there are no fewer than 50 liars. Thus, there are no more than 150 conformists... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In pentagon $A B C D E A B=B C=C D=D E, \angle B=96^\circ$ and $\angle C=\angle D=108^\circ$. Find angle $E$. | Answer: $102^{\circ}$. Solution. Draw segments $B D$ and $C E$. Let them intersect at point $O$. Note that triangles $B C D$ and $C D E$ are isosceles with an angle of $108^{\circ}$ at the vertex, so the base angles are $36^{\circ}$ (they are marked on the diagram with one arc). Then $\angle B C E = \angle B D E = 72^{... | 102 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In triangle $ABC$, angle $C$ is three times larger than angle $A$, and side $AB$ is twice as long as side $BC$. Prove that angle $ABC$ is 60 degrees. | Solution. Let $D$ be the midpoint of side $A B$. Since $B D=B C$, triangle $B C D$ is isosceles. Let $\angle C A D=x, \angle A C D=y$. Then $\angle D C B=3 x-y$, and $\angle C D B=x+y$. Since $\angle D C B=\angle C D B$, we have $3 x-y=x+y$, from which $y=x$. Therefore, $D C=D A=D B=B C$, which means triangle $B C D$ i... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
6. In the Thirtieth Kingdom, there are 100 cities, and no more than one road connects any two cities. One day, the tsar ordered that one-way traffic be introduced on each road, and at the same time, each road should be painted either white or black. The Minister of Transport proudly reported that after the order was ca... | Answer: 150. Solution: Example. Arrange the cities on a circle so that they divide it into equal arcs, and declare these arcs to be roads directed clockwise. Paint these 100 arcs in white and black colors so that the colors alternate on the circle. Direct another 50 white roads along the chords from the cities where bl... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. At the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip is ... | Answer: 50. Solution: Example: The chip 50 is sequentially exchanged 99 times with the next one counterclockwise.
Evaluation. We reason by contradiction. Let $k<50$. First proof. We will consider the shifts of the chips relative to their initial positions, with shifts clockwise counted as positive and counterclockwise... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some natural number. Then the first said: “My number is greater than 1”, the second said: “My number is greater than 2”, ..., the tenth said: “My number is greater than 10”. After that, they, in ... | Answer: 8. Solution: Those who in the first series of answers said that their numbers are greater than 9 and 10 are definitely liars, because these answers are incompatible with any of the answers in the second series. Therefore, there are no more than eight knights. An example when there are exactly 8 knights: the fir... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There is a cube, each face of which is divided into 4 identical square cells. Oleg wants to mark 8 cells with invisible ink so that no two marked cells share a side. Rustem has detectors. If a detector is placed in a cell, the ink on it becomes visible. What is the minimum number of detectors Rustem can place in the... | Answer: 16. Solution: Example. Let's divide all 24 cells into eight triples, where each triple consists of three cells adjacent to one vertex of the cube. Any two cells in the same triple share a common side. Since the number of marked cells is the same as the number of triples, there must be exactly one marked cell in... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Given an equilateral triangle ABC. Point $D$ is chosen on the extension of side $A B$ beyond point $A$, point $E$ is on the extension of $B C$ beyond point $C$, and point $F$ is on the extension of $A C$ beyond point $C$ such that $C F=A D$ and $A C+E F=D E$. Find the angle BDE. (A. Kuznetsov) | Answer: 60 - . Solution: Complete triangle $A C E$ to parallelogram $A C E G$. Since $C F=A D$, $C E=A G$ and $\cdot F C E=\cdot D A G=60 \cdot$, triangles $D A G$ and $F C E$ are equal, from which $G D=E F$. Therefore, $D E=A C+E F=G E+G D$. This means that point $G$ lies on segment $D E$, and thus $D E \| A C$, from ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that among 100 balls, exactly 51 are radioactive. There is a device into which two balls can be placed, and if both are radioactive, a light will turn on (if at least one of the two balls is not radioactive, the light will not turn on). Can all the radioactive balls be found using the device no more than... | Answer. Yes. Solution. Let's divide the balls into 50 pairs and test them. Consider two possible cases.
1) Exactly one of these tests revealed two radioactive balls. Then in each of the remaining 49 pairs, there is exactly one radioactive ball. Testing one of the found radioactive balls with one ball from each of the ... | 145 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The numbers 1, 2, 3, 4, 5, 6, 7 were written in a circle in some order. We will call a written number good if it is equal to the sum of the two numbers written next to it. What is the maximum possible number of good numbers among the written ones? (E. Bakayev) | Answer: 3. Solution: If the numbers are written, for example, in the order 2, 7, 5, 6, 1, 4, 3, then the numbers 7, 6, and 4 will be good. It remains to show that there cannot be more than three good numbers.
Note that a good number is greater than both of its neighbors, so two good numbers cannot stand next to each o... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. On a white checkered board of size $25 \times 25$ cells, several cells are painted black, with exactly 9 cells painted black in each row and each column. What is the smallest $k$ such that it is always possible to repaint $k$ cells to white in such a way that it is impossible to cut out a black $2 \times 2$ square? ... | Solution. Evaluation. Note that if 9 cells are shaded in a row, then four of them can be repainted so that no two shaded cells are adjacent: it is enough to renumber the shaded cells from left to right and repaint the cells with even numbers. If such repainting is done with all even rows, then 48 cells will be repainte... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. From the natural numbers from 1 to 25, Dasha chose six such that the difference between any two chosen numbers is divisible by 4. What is the maximum number of prime numbers that Dasha could have chosen? | Answer: Five. First solution. The difference of two numbers is divisible by 4 if and only if these numbers have the same remainder when divided by 4. Let's list all prime numbers less than 25 and their remainders when divided by 4: 2-2, 3-3, 5-1, 7-3, 11-3, 13-1, 17-1, 19-3, 23-3. The most common remainder is 3, which ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle ABC, the median BM is drawn. It is known that $\angle A B M=40^{\circ}$, and $\angle C B M=70^{\circ}$. Find the ratio $A B: B M$. | Answer: 2. First solution. Complete the triangle $A B C$ to a parallelogram $A B C D$. Since the diagonals of a parallelogram bisect each other, point $M$ is the point of their intersection and $B D=2 B M$. On the other hand, $\angle B D A=\angle C B D=70^{\circ}$, and $\angle B A D=180^{\circ}-\angle B D A-\angle A B ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The cells of an $n \times n$ square are colored black and white with the condition that no four cells, located at the intersection of two rows and two columns, can all be the same color. What is the largest possible value of $n$? | Solution. An example for $n=4$ is shown in the figure on the right. We will prove by contradiction that it is impossible to color a $5 \times 5$ square in this way. Let's call a row predominantly black if it has more black squares than white, and predominantly white otherwise. Among the five rows, there will be either ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A two-digit number $N$ was multiplied by 2, the digits of the result were swapped, and then the number was divided by 2. The result was the same number $N$. How many such numbers $N$ exist?
Answers:
A) none (-) B) exactly 4 (-) C) at least 10 (+) D) at least 14 (+) E) at least 15 (-) | Solution. Since the result turned out to be the same number, two identical digits were swapped. This means that $2 \mathrm{~N}$ should have two such digits. Let's consider several cases:
1) When multiplying by 2, there was no carry-over to the next place value. Obviously, as $N$, the numbers 11, 22, 33, 44 fit and onl... | 14 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Task 2. (10 points) A numerical sequence is given:
$x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1$.
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2021$. | # Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2021)$, if $f(1)=5, f(4)=2$. | # Solution.
Substituting the pairs of numbers \(a=4, b=1\) and \(a=1, b=4\) into the given equation, respectively, we get
If \(a=4, b=1\), then \(f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{2+2 \cdot 5}{3}=4, f(2)=4\).
If \(a=1, b=4\), then \(f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3... | -2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A numerical sequence is given:
$$
x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1
$$
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2022$. | # Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2021)$, if $f(1)=1, f(4)=7$.
# | # Solution.
Substituting the pairs of numbers $a=4, b=1$ and $a=1, b=4$ into the given equation, respectively, we get
If $a=4, b=1$, then $f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{7+2 \cdot 1}{3}=3, f(2)=3$.
If $a=1, b=4$, then $f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\fr... | 4041 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere. Segment $C D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $A A_{1}$. Find the volume of the prism if $C K=2 \sqrt{6}, D K=4$. | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (Fi... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 55 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | # Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similar... | 1705 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 41 and 24, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Solution.
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of tria... | 984 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds. The sequence must contain a term $a_{k}=2021$. Determine the maximum number of three-digit numbers, divisible by 25, that ... | # Solution.
The final sequence can contain all three-digit numbers, as it can consist of a given number of natural numbers starting from the chosen number $a_{i}$.
We will prove that for any term of the arithmetic progression $1,2,3, \ldots$ defined by the formula for the $n$-th term $a_{n}=n$, the equality $a_{k+2}=... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2022)$, if $f(1)=1, f(4)=7$.
# | # Solution.
Substituting the pairs of numbers \(a=4, b=1\) and \(a=1, b=4\) into the given equation, respectively, we get
If \(a=4, b=1\), then \(f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{7+2 \cdot 1}{3}=3, f(2)=3\).
If \(a=1, b=4\), then \(f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3... | 4043 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 5. (20 points) At the first deposit, equipment of the highest class was used, and at the second deposit, equipment of the first class was used, with the highest class being less than the first. Initially, $40 \%$ of the equipment from the first deposit was transferred to the second. Then, $20 \%$ of the equipment ... | # Solution.
Let there initially be $x$ units of top-class equipment at the first deposit and $y$ units of first-class equipment at the second deposit $(x1.05 y$, from which $y48 \frac{34}{67} .\end{array}\right.\right.$
This double inequality and the condition "x is divisible by 5" is satisfied by the unique value $x... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1} \mathrm{c}$ is inscribed in a sphere with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$. Segment $C D$ is the diameter of this sphere, point $K$ and $L$ are the midpoints of edge $A A_{1}$ and $A B$ respectively. Find the volume of the... | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$, with their centers at points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (F... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this se... | # Solution.
Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that an arithmetic progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equation $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.
In... | 225 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2022)$, if $f(1)=5, f(4)=2$. | # Solution.
Substituting the pairs of numbers $a=4, b=1$ and $a=1, b=4$ into the given equation, respectively, we get
If $a=4, b=1$, then $f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{2+2 \cdot 5}{3}=4, f(2)=4$.
If $a=1, b=4$, then $f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\fr... | -2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ is inscribed in a sphere with the base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$. The segment $C_{1} D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $C C_{1}$. Find the volume of the prism if $D K=2, D A=\sq... | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$, with their centers at points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (F... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 367 and 6, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | # Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the simila... | 2202 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 101 and 20, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similari... | 2020 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. If we now take into account that after reflections in the mirrors, the ray EC' has entered the eyepiece, that is, the direction of the ray after two deflections coincides with the direction of DA, it follows that the total deflection of the ray EC' is equal to the angle between the rays EC' and DA
$$
\mathrm{psi}_{... | Answer: There is no need to rotate mirror C relative to point C. The angle of rotation of mirror C relative to the platform is $0^{\circ}$.
## Solution Method 2. | 0 | Geometry | proof | Yes | Yes | olympiads | false |
4. (6 points) In an ideal gas, a thermodynamic cycle consisting of two isochoric and two adiabatic processes is carried out. The ratio of the initial and final absolute temperatures in the isochoric cooling process is \( k = 1.5 \). Determine the efficiency of this cycle, given that the efficiency of the Carnot cycle w... | Answer: $\eta=1-k\left(1-\eta_{C}\right)=0.25=25 \%$.
## Evaluation Criteria
| Performance | Score |
| :--- | :---: |
| Participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for work in the cycle is written | $\mathbf{1}$ |
| Expression for the amount of heat ... | 25 | Other | math-word-problem | Yes | Yes | olympiads | false |
8. Finally, let's calculate the change in resistance when the switch is closed:
$$
\Delta R=R_{p}-R_{3}=33[\text { Ohms] }-30[\text { Ohms] }=3[\text { Ohms] }
$$ | Answer: 3.0 Ohms
Criteria (maximum 10 points) | 3 | Other | math-word-problem | Yes | Yes | olympiads | false |
Task 2. By how many units can the city's fleet of natural gas vehicles be increased in 2022, assuming that the capacity of each of the old CNG stations in the city is equal to the capacity of the new station on Narodnaya Street, and that the city's fleet constitutes only $70 \%$ of all vehicles refueling at CNG station... | Task 2. There are a total of 15 stations: 4 new ones and 11 old ones.
The throughput capacity of the old stations is 11 x $200=2200$ vehicles per day, and for the new ones: $200+700=900$ vehicles per day. In total, the stations can refuel: $2200+900=3100$ vehicles per day.
Vehicles from the city fleet account for onl... | 1170 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}+4096}{64 a^{6}}$, if $\frac{a}{2}-\frac{2}{a}=5$.
# | # Solution.
$$
\begin{aligned}
& \frac{a^{12}+4096}{64 a^{6}}=\frac{a^{6}}{64}+\frac{64}{a^{6}}=\frac{a^{6}}{64}-2+\frac{64}{a^{6}}+2=\left(\frac{a^{3}}{8}-\frac{8}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{8}-3 \cdot \frac{a}{2}+3 \cdot \frac{2}{a}-\frac{8}{a^{3}}+3\left(\frac{a}{2}-\frac{2}{a}\right)\right)^{2}+... | 19602 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}+729^{2}}{729 a^{6}}$, if $\frac{a}{3}-\frac{3}{a}=4$.
# | # Solution.
$$
\begin{aligned}
& \frac{a^{12}+729^{2}}{729 a^{6}}=\frac{a^{6}}{729}+\frac{729}{a^{6}}=\frac{a^{6}}{729}-2+\frac{729}{a^{6}}+2=\left(\frac{a^{3}}{27}-\frac{27}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{27}-3 \cdot \frac{a}{3}+3 \cdot \frac{3}{a}-\frac{27}{a^{3}}+3\left(\frac{a}{3}-\frac{3}{a}\right)... | 5778 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}+729^{2}}{729 a^{6}}$, if $\frac{a}{3}-\frac{3}{a}=2$. | Solution.
$$
\begin{aligned}
& \frac{a^{12}+729^{2}}{729 a^{6}}=\frac{a^{6}}{729}+\frac{729}{a^{6}}=\frac{a^{6}}{729}-2+\frac{729}{a^{6}}+2=\left(\frac{a^{3}}{27}-\frac{27}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{27}-3 \cdot \frac{a}{3}+3 \cdot \frac{3}{a}-\frac{27}{a^{3}}+3\left(\frac{a}{3}-\frac{3}{a}\right)\r... | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A circle touches the extensions of two sides $A B$ and $A D$ of square $A B C D$ with side $2 \sqrt{3} \mathrm{~cm}$. Two tangents are drawn from point $C$ to this circle. Find the radius of the circle if the angle between the tangents is $30^{\circ}$, and it is known that $\sin 15^{\circ}=\frac{\sq... | # Solution.
Fig. 1
The segment cut off from vertex $A$ by the point of tangency of the circle is equal to the radius of this circle. The diagonal of the square $A B C D A C=2 \sqrt{6}$. If... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}+4096}{64 a^{6}}$, if $\frac{a}{2}-\frac{2}{a}=3$.
# | # Solution.
$$
\begin{aligned}
& \frac{a^{12}+4096}{64 a^{6}}=\frac{a^{6}}{64}+\frac{64}{a^{6}}=\frac{a^{6}}{64}-2+\frac{64}{a^{6}}+2=\left(\frac{a^{3}}{8}-\frac{8}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{8}-3 \cdot \frac{a}{2}+3 \cdot \frac{2}{a}-\frac{8}{a^{3}}+3\left(\frac{a}{2}-\frac{2}{a}\right)\right)^{2}+... | 1298 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}-729}{27 a^{6}}$, if $\frac{a^{2}}{3}-\frac{3}{a^{2}}=4$.
# | # Solution.
$$
\begin{aligned}
& \frac{a^{12}-729}{27 a^{6}}=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}+1+\frac{9}{a^{4}}\right)=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}-2+\frac{9}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\left(\frac{a^{2}... | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}-4096}{64 a^{6}}$, if $\frac{a^{2}}{4}-\frac{4}{a^{2}}=3$. | # Solution.
$$
\begin{aligned}
& \frac{a^{12}-4096}{64 a^{6}}=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}+1+\frac{16}{a^{4}}\right)=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}-2+\frac{16}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\left(\frac{... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}-729}{27 a^{6}}$, if $\frac{a^{2}}{3}-\frac{3}{a^{2}}=6$. | Solution.
$$
\begin{aligned}
& \frac{a^{12}-729}{27 a^{6}}=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}+1+\frac{9}{a^{4}}\right)=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}-2+\frac{9}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\left(\frac{a^{2}}{... | 234 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}-4096}{64 a^{6}}$, if $\frac{a^{2}}{4}-\frac{4}{a^{2}}=5$. | # Solution.
$$
\begin{aligned}
& \frac{a^{12}-4096}{64 a^{6}}=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}+1+\frac{16}{a^{4}}\right)=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}-2+\frac{16}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\left(\frac{... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (6 points) In an ideal gas, a thermodynamic cycle consisting of two isochoric and two adiabatic processes is carried out. The ratio of the initial and final absolute temperatures in the isochoric cooling process is \( k = 1.5 \). Determine the efficiency of this cycle, given that the efficiency of the Carnot cycle w... | Answer: $\eta=1-k\left(1-\eta_{C}\right)=0.25=25 \%$.
## Evaluation Criteria
| Performance | Score |
| :--- | :---: |
| Participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for work in the cycle is written | $\mathbf{1}$ |
| Expression for the amount of heat ... | 25 | Other | math-word-problem | Yes | Yes | olympiads | false |
18. Determine the amount of substance of sodium carbonate
According to the chemical reaction equation:
$$
\mathrm{n}\left(\mathrm{CO}_{2}\right)=\mathrm{n}\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)=0.125 \text { mol }
$$
Determination of the amount of substance - 2 points
19. Calculate the mass of sodium carbonat... | Answer: $10 \%$
## TOTAL 10 points
## TASK № 2
Aluminum oxide was melted with potash. The resulting product was dissolved in hydrochloric acid and treated with excess ammonia water. The precipitate that formed was dissolved in excess sodium hydroxide solution. Write the equations for the four reactions described.
#... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
28. Let's determine the amount of sulfur substance
According to the chemical reaction equation:
$$
\mathrm{n}(\mathrm{S})=\mathrm{n}\left(\mathrm{SO}_{2}\right)=1 \text { mole }
$$
29. Let's calculate the mass of pure sulfur:
$$
m(S)=n \cdot M=1 \cdot 32=32 \text { g }
$$
Determination of the mass of the substance... | Answer: 6 g.
## TOTAL 10 points
## TASK № 2
Potassium phosphate was calcined with coke in the presence of river sand. The simple substance formed reacted with excess chlorine. The resulting product was added to an excess of potassium hydroxide solution. The formed solution was treated with lime water. Write the equa... | 6 | Other | math-word-problem | Yes | Yes | olympiads | false |
39. Let's calculate the volume of carbon dioxide.
According to Avogadro's law, the following rule applies: in equal volumes of different gases taken at the same temperatures and pressures, the same number of molecules is contained.
From the chemical reaction equation, it follows that the ratio of the amount of substa... | Answer: 7 l.
## TOTAL 10 points
## TASK № 2
Chromium (III) oxide was melted with potassium sulfite. The resulting product was added to water. To the precipitate formed, a mixture of bromine and sodium hydroxide was added, resulting in a yellow solution. Upon adding hydrogen sulfide water to the obtained solution, a ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}+256}{16 a^{4}}$, if $\frac{a}{2}+\frac{2}{a}=5$. | Solution.
$$
\begin{aligned}
& \frac{a^{8}+256}{16 a^{4}}=\frac{a^{4}}{16}+\frac{16}{a^{4}}=\frac{a^{4}}{16}+2+\frac{16}{a^{4}}-2=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{4}+2+\frac{4}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{2}+\frac{2}{a}\right)^{2}-2\right)^{2}-2=\left(5^{2}-2\... | 527 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) Laboratory engineer Sergei received an object for research consisting of about 200 monoliths (a container designed for 200 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequen... | # Solution.
Let's determine the exact number of monoliths. It is known that the probability of a monolith being loamy sand is $\frac{1}{9}$. The number closest to 200 that is divisible by 9 is 198. Therefore, there are 198 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands $(198: 9=2... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 5. (20 points)
Find $x_{0}-y_{0}$, if $x_{0}$ and $y_{0}$ are the solutions to the system of equations:
$$
\left\{\begin{array}{l}
x^{3}-2023 x=y^{3}-2023 y+2020 \\
x^{2}+x y+y^{2}=2022
\end{array}\right.
$$ | # Solution.
Rewrite the system as
$$
\left\{\begin{array}{l}
x^{3}-y^{3}+2023 y-2023 x=2020 \\
x^{2}+x y+y^{2}=2022
\end{array}\right.
$$
Let $x_{0}$ and $y_{0}$ be the solution to the system of equations. Then
$\left\{\begin{array}{l}x_{0}{ }^{3}-y_{0}{ }^{3}+2023 y_{0}-2023 x_{0}=2020, \\ x_{0}{ }^{2}+x_{0} y_{0}... | -2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}$, if $\frac{a}{2}-\frac{2}{a}=3$.
# | # Solution.
$$
\begin{aligned}
& \frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{4}}{16}-\frac{16}{a^{4}}\right) \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\frac{a^{2}}{4}-2+\frac{4}{a^{2}... | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) Lab engineer Dasha received an object for research consisting of about 100 monoliths (a container designed for 100 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequency (stat... | # Solution.
Let's determine the exact number of monoliths. It is known that the probability of a monolith being loamy sand is $\frac{1}{7}$. The number closest to 100 that is divisible by 7 is 98. Therefore, there are 98 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands ($98: 7=14$)... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) At the research institute, a scientific employee, Ivan Ivanovich, received an object for research containing about 300 oil samples (a container designed for 300 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or hig... | # Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a selected sample being a heavy oil sample is $\frac{1}{8}$, and the number closest to 300 that is divisible by $8-296$. Therefore, the total number of samples in the container is 296. The samples of high-sulfur oil... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) Find the smallest natural solution of the inequality $\left(\frac{2023}{2022}\right)^{27+18+12+8+\ldots+27 \cdot\left(\frac{2}{3}\right)^{n}}>\left(\frac{2023}{2022}\right)^{72}$. | Solution.
$\left(\frac{2023}{2022}\right)^{27+18+12+8+\ldots+27 \cdot\left(\frac{2}{3}\right)^{n}}>\left(\frac{2023}{2022}\right)^{72}$
We transition to an equivalent inequality
$27+18+12+8+\ldots+27 \cdot\left(\frac{2}{3}\right)^{n}>72$
$27\left(1+\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\left(\frac{2}{3}\right)^{... | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Calculate
$$
\left(\frac{10001}{20232023}-\frac{10001}{20222022}\right) \cdot 4090506+\sqrt{4092529}
$$ | # Solution.
$$
\begin{aligned}
& \left(\frac{10001}{20232023}-\frac{10001}{20222022}\right) \cdot 4090506+\sqrt{4092529} \\
& =\left(\frac{10001}{2023 \cdot 10001}-\frac{10001}{2022 \cdot 10001}\right) \cdot 4090506+\sqrt{2023^{2}}= \\
& \quad=\left(\frac{1}{2023}-\frac{1}{2022}\right) \cdot 2022 \cdot 2023+2023=\frac... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) At the research institute, a scientific employee, Tatyana Vasilyevna, received an object for research containing about 150 oil samples (a container designed for 150 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or... | # Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{2}{11}$, and the number closest to 150 that is divisible by $11-143$. Therefore, the total number of samples in the container is 143. Samples of high-sulfur oil consist of... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) Find the smallest natural solution of the inequality $\left(\frac{2023}{2022}\right)^{36+24+16+\ldots+36\left(\frac{2}{3}\right)^{n}}>\left(\frac{2023}{2022}\right)^{96}$. | Solution.
$\left(\frac{2023}{2022}\right)^{36+24+16+\ldots+36\left(\frac{2}{3}\right)^{n}}>\left(\frac{2023}{2022}\right)^{96}$.
$36+24+16+. .+36 \cdot\left(\frac{2}{3}\right)^{n}>96$
$36\left(1+\frac{2}{3}+\left(\frac{2}{3}\right)^{2}+\ldots+\left(\frac{2}{3}\right)^{n}\right)>96$
$\left(1+\frac{2}{3}+\left(\frac{... | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}+1296}{36 a^{4}}$, if $\frac{a}{\sqrt{6}}+\frac{\sqrt{6}}{a}=5$.
# | # Solution.
$$
\begin{aligned}
& \frac{a^{8}+1296}{36 a^{4}}=\frac{a^{4}}{36}+\frac{36}{a^{4}}=\frac{a^{4}}{36}+2+\frac{36}{a^{4}}-2=\left(\frac{a^{2}}{6}+\frac{6}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{6}+2+\frac{6}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{\sqrt{6}}+\frac{\sqrt{6}}{a}\right)^{2}-2\right)^{2}... | 527 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A circle with a radius of 15 touches two adjacent sides $AB$ and $AD$ of the square $ABCD$. On the other two sides, the circle intersects at points, cutting off segments of 6 cm and 3 cm from the vertices, respectively. Find the length of the segment that the circle cuts off from vertex $B$ at the p... | # Solution.
Fig. 1
Let $X$ be the desired segment, then $X+15$ is the side of the square. The segment $K L=15+X-6-3=X+6$. Consider $\triangle O N L$. By the Pythagorean theorem, the follow... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) At the quality control department of an oil refinery, Engineer Pavel Pavlovich received a research object consisting of about 100 oil samples (a container designed for 100 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-su... | # Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{1}{7}$. The number closest to 100 that is divisible by $7$ is $98$. Therefore, there are 98 samples in total in the container. The samples of high-sulfur oil include all t... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300}$ holds. Find $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}-2023$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_... | -2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}-6561}{81 a^{4}} \cdot \frac{3 a}{a^{2}+9}$, if $\frac{a}{3}-\frac{3}{a}=4$. | Solution.
$\frac{a^{8}-6561}{81 a^{4}} \cdot \frac{3 a}{a^{2}+9}=\left(\frac{a^{4}}{81}-\frac{81}{a^{4}}\right) \cdot \frac{3 a}{a^{2}+9}=\left(\frac{a^{2}}{9}+\frac{9}{a^{2}}\right)\left(\frac{a^{2}}{9}-\frac{9}{a^{2}}\right) \cdot \frac{3 a}{a^{2}+9}=$
$=\left(\frac{a^{2}}{9}-2+\frac{9}{a^{2}}+2\right)\left(\frac{a... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A circle with a radius of 10 touches two adjacent sides $AB$ and $AD$ of the square $ABCD$. On the other two sides, the circle intersects at points, cutting off segments of 4 cm and 2 cm from the vertices, respectively. Find the length of the segment that the circle cuts off from vertex $B$ at the p... | Solution.
Fig. 1
Let $X$ be the desired segment, then $X+10$ is the side of the square. The segment $K L=10+X-4-2=X+4$. Consider $\triangle O N L$. By the Pythagorean theorem, the followin... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) At the quality control department of an oil refinery, Engineer Valentina Ivanovna received a research object consisting of about 200 oil samples (a container designed for 200 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low... | # Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{1}{9}$. The number closest to 200 that is divisible by 9 is 198. Therefore, the total number of samples in the container is 198. The samples of high-sulfur oil consist of ... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+398 x_{n}+x_{n+1}}{400}$ holds. Find $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}+2021$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+398 x_{n}+x_{n+1}}{400} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}$, if $\frac{a}{2}-\frac{2}{a}=5$. | Solution.
$$
\begin{aligned}
& \frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{4}}{16}-\frac{16}{a^{4}}\right) \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)\left(\frac{a^{2}}{16}-\frac{16}{a^{2}}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\frac{a^{2}}{4}-2+\frac{4}{a^{2}... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A circle touches two adjacent sides $AB$ and $AD$ of square $ABCD$ and cuts off segments of length 4 cm from vertices $B$ and $D$ at the points of tangency. On the other two sides, the circle intersects and cuts off segments of 2 cm and 1 cm from the vertices, respectively. Find the radius of the ci... | # Solution.
Fig. 1
Let $R$ be the radius of the circle, then $R+4$ is the side of the square. The segment $K L=4+R-2-1=R+1$. Consider the triangle $O N L$. By the Pythagorean theorem, the ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) In the educational center "Young Geologist," an object consisting of about 150 monoliths (a container designed for 150 monoliths, which was almost completely filled) was delivered. Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relati... | # Solution.
Let's determine the exact number of monoliths. It is known that the relative frequency of a monolith being loamy sand is $\frac{2}{11}$. The number closest to 150 that is divisible by 11 is 143. Therefore, there are 143 monoliths in total. Monoliths of lacustrine-glacial origin make up all loamy sands ( $1... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+98 x_{n}+x_{n+1}}{100}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2021$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+98 x_{n}+x_{n+1}}{100} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}+256}{16 a^{4}}$, if $\frac{a}{2}+\frac{2}{a}=3$. | Solution.
$$
\begin{aligned}
& \frac{a^{8}+256}{16 a^{4}}=\frac{a^{4}}{16}+\frac{16}{a^{4}}=\frac{a^{4}}{16}+2+\frac{16}{a^{4}}-2=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{4}+2+\frac{4}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{2}+\frac{2}{a}\right)^{2}-2\right)^{2}-2=\left(3^{2}-2\... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. (10 points) A circle touches two adjacent sides $AB$ and $AD$ of square $ABCD$ and cuts off segments of length 8 cm from vertices $B$ and $D$ at the points of tangency. On the other two sides, the circle intersects and cuts off segments of 4 cm and 2 cm from the vertices, respectively. Find the radius of the ci... | # Solution.
Fig. 1
Let $R$ be the radius of the circle, then $R+8$ is the side of the square. The segment $K L=8+R-4-2=R+2$. Consider $\triangle O N L$. By the Pythagorean theorem, the fol... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) An educational center "Young Geologist" received an object for research consisting of about 300 monoliths (a container designed for 300 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The rel... | # Solution.
Let's determine the exact number of monoliths. It is known that the relative frequency of a monolith being loamy sand is $\frac{1}{8}$. The number closest to 300 that is divisible by $8-296$. Therefore, there are 296 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands $(29... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2022$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_... | 2023 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Development of a Method for Optimizing a Mathematical Problem
Problem Statement. At Unit 3 of an oil and gas company, the task is to optimize the operation of the navigation system elements. The task of this unit is to receive encrypted signals from Unit 1 and Unit 2, synthesize the incoming data packets, and ... | Solution to the problem. Solving the problem "head-on" by raising one number to the power of another is bound to exceed the computational power not only of a single computer but even a data center would require a certain amount of time to perform the calculations. Since we only need to send the last digit, let's focus ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) In a sequence of natural numbers, each subsequent number, starting from the third, is equal to the absolute difference of the two preceding ones. Determine the maximum number of elements such a sequence can contain if the value of each of them does not exceed 2022.
# | # Solution.
To maximize the length of the sequence, the largest elements should be at the beginning of the sequence. Let's consider the options:
1) $n, n-1,1, n-1, n-2, n-3,1, n-4, n-5,1, \ldots, 2,1,1$;
2) $n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
3) $1, n, n-1,1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
4) $n-1, n, ... | 3034 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) In a sequence of natural numbers, each subsequent number, starting from the third, is equal to the absolute difference of the two preceding ones. Determine the maximum number of elements such a sequence can contain if the value of each of them does not exceed 2021.
# | # Solution.
To maximize the length of the sequence, the largest elements should be at the beginning of the sequence. Let's consider the options:
1) $n, n-1,1, n-1, n-2, n-3,1, n-4, n-5,1, \ldots, 2,1,1$;
2) $n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
3) $1, n, n-1,1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
4) $n-1, n, ... | 3033 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. (20 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds. The sequence must contain a term $a_{k}=2021$. Determine the maximum number of three-digit numbers, divisible by 25, t... | # Solution.
The final sequence can contain all three-digit numbers, as it can consist of a given number of natural numbers starting from the chosen number $a_{i}$.
We will prove that for any term of the arithmetic progression $1,2,3, \ldots$ defined by the formula for the $n$-th term $a_{n}=n$, the equality $a_{k+2}=... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 6. (30 points) At the first deposit, equipment of the highest class was used, and at the second deposit, equipment of the first class was used, with the highest class being less than the first. Initially, $40 \%$ of the equipment from the first deposit was transferred to the second. Then, $20 \%$ of the equipment ... | # Solution.
Let there initially be $x$ units of top-class equipment at the first deposit and $y$ units of first-class equipment at the second deposit $(x1.05 y$, from which $y48 \frac{34}{67} .\end{array}\right.\right.$
This double inequality and the condition “x is divisible by 5” is satisfied by the unique value $x... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) A finite increasing sequence of natural numbers $a_{1}, a_{2}, \ldots, a_{n}(n \geq 3)$ is given, and for all $\kappa \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this seq... | # Solution.
Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that the arithmetic
progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.
I... | 225 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are 41 and 24, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | # Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similar... | 984 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Development of a Method for Optimizing a Mathematical Problem
Problem Statement. At Unit 3 of an oil and gas company, the task is to optimize the operation of the navigation system elements. The task of this unit is to receive encrypted signals from Unit 1 and Unit 2, synthesize the incoming data packets, and ... | Solution to the problem. Solving the problem "head-on" by raising one number to the power of another is bound to exceed the computational power not only of a single computer but even a data center would require a certain amount of time to perform the calculations. Since we only need to send the last digit, let's focus ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Does there exist a natural number that, when divided by the sum of its digits, gives both a quotient and a remainder of 2014? If there is more than one such number, write their sum as the answer. If no such numbers exist, write 0. (12 points)
# | # Solution:
Suppose there exists a natural number $\boldsymbol{n}$ with the sum of its digits $\boldsymbol{s}$, such that $n=2014 s+2014$, from which we get $n-s=2013 s+2014$. By the divisibility rule, $n-s$ is divisible by 3. However, the number $2013 s+2014$ is not divisible by 3, since the number $2013s$ is a multi... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Calculating Samson loves to have lunch at the Italian restaurant "At Pablo's". During his next visit to the restaurant, Samson was offered to purchase a loyalty card for a period of 1 year at a price of 30000 rubles, which gives the client a $30 \%$ discount on the bill amount.
(a) Suppose that during the week Sams... | # Solution:
(a) After purchasing the card, one lunch will cost the client $900 * 0.3 = 270$ rubles less. Since there are 52 full weeks in a year $(365 / 7 = 52.14)$, Samson will save $270 * 3 * 52 = 42120$ rubles on discounts over the year, which is more than the cost of the card. Therefore, it is beneficial for him t... | 167 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. Maximum 15 points
A company that produces educational materials for exam preparation incurs average costs per textbook of $100+\frac{100000}{Q}$, where $Q$ is the number of textbooks produced annually. What should be the annual production volume of the textbook to reach the break-even point if the planned... | # Solution
At the break-even point $\mathrm{P}=\mathrm{ATC}=\mathrm{MC}$
Form the equation $100+10000 / Q=300$
$100 \mathrm{Q}+100000=300 \mathrm{Q}$
$100000=200 \mathrm{Q}$
$\mathrm{Q}=100000 / 200=500$
## Evaluation Criteria
1. The correct answer is justified: 15 points | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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