problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
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 |
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 |
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 |
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 |
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 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 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 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 |
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 |
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 |
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.
$$
\begin{aligned}
& \frac{a^{12}+729^{2}}{729 a^{6}}=\frac{a^{6}}{729}+\frac{729}{a^{6}}=\frac{a^{6}}{729}-2+\frac{729}{a^{6}}+2=\left(\frac{a^{3}}{27}-\frac{27}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{27}-3 \cdot \frac{a}{3}+3 \cdot \frac{3}{a}-\frac{27}{a^{3}}+3\left(\frac{a}{3}-\frac{3}{a}\right)\r... | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{12}+4096}{64 a^{6}}$, if $\frac{a}{2}-\frac{2}{a}=3$.
# | # Solution.
$$
\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 |
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 1. (5 points) Calculate
$$
\left(\frac{10001}{20232023}-\frac{10001}{20222022}\right) \cdot 4090506+\sqrt{4092529}
$$ | # Solution.
$$
\begin{aligned}
& \left(\frac{10001}{20232023}-\frac{10001}{20222022}\right) \cdot 4090506+\sqrt{4092529} \\
& =\left(\frac{10001}{2023 \cdot 10001}-\frac{10001}{2022 \cdot 10001}\right) \cdot 4090506+\sqrt{2023^{2}}= \\
& \quad=\left(\frac{1}{2023}-\frac{1}{2022}\right) \cdot 2022 \cdot 2023+2023=\frac... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) At the research institute, a scientific employee, Tatyana Vasilyevna, received an object for research containing about 150 oil samples (a container designed for 150 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or... | # Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{2}{11}$, and the number closest to 150 that is divisible by $11-143$. Therefore, the total number of samples in the container is 143. Samples of high-sulfur oil consist of... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}+1296}{36 a^{4}}$, if $\frac{a}{\sqrt{6}}+\frac{\sqrt{6}}{a}=5$.
# | # Solution.
$$
\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 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,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_... | -2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (5 points) Find $\frac{a^{8}-6561}{81 a^{4}} \cdot \frac{3 a}{a^{2}+9}$, if $\frac{a}{3}-\frac{3}{a}=4$. | Solution.
$\frac{a^{8}-6561}{81 a^{4}} \cdot \frac{3 a}{a^{2}+9}=\left(\frac{a^{4}}{81}-\frac{81}{a^{4}}\right) \cdot \frac{3 a}{a^{2}+9}=\left(\frac{a^{2}}{9}+\frac{9}{a^{2}}\right)\left(\frac{a^{2}}{9}-\frac{9}{a^{2}}\right) \cdot \frac{3 a}{a^{2}+9}=$
$=\left(\frac{a^{2}}{9}-2+\frac{9}{a^{2}}+2\right)\left(\frac{a... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) At the quality control department of an oil refinery, Engineer Valentina Ivanovna received a research object consisting of about 200 oil samples (a container designed for 200 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low... | # Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{1}{9}$. The number closest to 200 that is divisible by 9 is 198. Therefore, the total number of samples in the container is 198. The samples of high-sulfur oil consist of ... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+398 x_{n}+x_{n+1}}{400}$ holds. Find $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}+2021$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\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.
$$
\begin{aligned}
& \frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{4}}{16}-\frac{16}{a^{4}}\right) \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)\left(\frac{a^{2}}{16}-\frac{16}{a^{2}}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\frac{a^{2}}{4}-2+\frac{4}{a^{2}... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) In the educational center "Young Geologist," an object consisting of about 150 monoliths (a container designed for 150 monoliths, which was almost completely filled) was delivered. Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relati... | # Solution.
Let's determine the exact number of monoliths. It is known that the relative frequency of a monolith being loamy sand is $\frac{2}{11}$. The number closest to 150 that is divisible by 11 is 143. Therefore, there are 143 monoliths in total. Monoliths of lacustrine-glacial origin make up all loamy sands ( $1... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+98 x_{n}+x_{n+1}}{100}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2021$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\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.
$$
\begin{aligned}
& \frac{a^{8}+256}{16 a^{4}}=\frac{a^{4}}{16}+\frac{16}{a^{4}}=\frac{a^{4}}{16}+2+\frac{16}{a^{4}}-2=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{4}+2+\frac{4}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{2}+\frac{2}{a}\right)^{2}-2\right)^{2}-2=\left(3^{2}-2\... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (15 points) An educational center "Young Geologist" received an object for research consisting of about 300 monoliths (a container designed for 300 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The rel... | # Solution.
Let's determine the exact number of monoliths. It is known that the relative frequency of a monolith being loamy sand is $\frac{1}{8}$. The number closest to 300 that is divisible by $8-296$. Therefore, there are 296 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands $(29... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2022$. | # Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\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 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 |
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 |
# Problem 5. Maximum 15 points
The great-grandfather-banker left a legacy to his newborn great-grandson. According to the agreement with the bank, the amount in the great-grandson's account increases. Every year, on the day after the birthday, the current amount is increased by 1 million rubles more than in the previo... | # Solution:
Since the number consists of identical digits, it can be represented as 111 multiplied by a. According to the problem, the same number should be obtained as the sum of an arithmetic progression. The first element of the progression is 1, the last is \( \mathrm{n} \), and the number of elements in the progr... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 3. Maximum 20 points
At the conference "Economics of the Present," an intellectual tournament was held, in which more than 198 but fewer than 230 scientists, including doctors and candidates of sciences, participated. Within one match, participants had to ask each other questions and record correct answers with... | Solution: Let there be $\mathrm{n}$ scientists participating in the tournament, of which $\mathrm{m}$ are doctors and $\mathrm{n}-\mathrm{m}$ are candidates of science. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the doctors of science ... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 3. Maximum 20 points
At the "Economics and Law" congress, a "Tournament of the Best" was held, in which more than 220 but fewer than 254 delegates—economists and lawyers—participated. Within one match, participants had to ask each other questions within a limited time and record the correct answers. Each partic... | Solution: Let there be $\mathrm{n}$ delegates participating in the tournament, of which $\mathrm{m}$ are economists and $\mathrm{n}-\mathrm{m}$ are lawyers. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the economists competed in $\mathrm... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the minimum loss, which is EXPENSE - INCOME, where the letters $\boldsymbol{P}, \boldsymbol{A}, \boldsymbol{C}, \boldsymbol{X}, \boldsymbol{O}, \boldsymbol{D}$ represent digits forming an arithmetic progression in the given order. (2 points).
# | # Solution:
The difference in the progression is 1; otherwise, 6 digits will not fit (if the "step" is 2, then 6 digits will exceed the field of digits). The smaller the first digit, the smaller the loss. Therefore, $\boldsymbol{P}=1, \boldsymbol{A}=2, \boldsymbol{C}=3, \boldsymbol{X}=4, \boldsymbol{O}=5, D=6$. Thus, ... | 58000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a certain state, only liars and economists live (liars always lie, while economists tell the truth). At a certain moment, the state decided to carry out monetary and credit, as well as budgetary and tax reforms. Since it was unknown what the residents expected, everyone was asked several questions (with only "yes... | # Solution:
Let $\boldsymbol{x}$ be the proportion of liars in the country, then (1-x) is the proportion of economists. Each economist answers affirmatively to one question, and each liar answers affirmatively to three. Therefore, we can set up the equation:
$3 x+1-x=0.4+0.3+0.5+0 ; 2 x=0.2 ; x=0.1$. Thus, in the cou... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There are 2015 coins on the table. Two players play the following game: they take turns; on a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. How many coins should the first player ta... | # Solution:
The strategy of the first player: he takes 95 coins, and then on each move, he takes (101-x) coins, where $\boldsymbol{x}$ is the number of coins taken by the second player. Since $\boldsymbol{x}$ is even (by the condition), 101-x is odd. Then $2015-95=1920$, since 101-x+x=101 coins will be taken per move,... | 95 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. The bank issued a loan to citizen $N$ on September 9 in the amount of 200 mln rubles. The repayment date is November 22 of the same year. The interest rate on the loan is $25 \%$ per annum. Determine the amount (in thousands of rubles) that citizen N will have to return to the bank. Assume that there are 365 days in... | # Solution:
Number of days of the loan: September - 21 days, October - 31 days, November - 21 days, i.e., $21+31+21=73$ days. The accrued debt amount is calculated using the formula:
$\boldsymbol{F V}=\boldsymbol{P V} \cdot(\boldsymbol{1}+\boldsymbol{t} \cdot \mathbf{Y}$ ), where $\boldsymbol{F} \boldsymbol{V}$ - the... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. In the Tumba-Yumba tribe with a population of 30 people, a trader arrives. After studying the customs of the tribe, the trader proposes to play a game. For each natural exchange of goods conducted in the market by two tribespeople, the trader gives each participant one gold coin. If at the end of the day, two diffe... | Solution: In any company, there are at least two people who have the same number of acquaintances. Therefore, the natives had no chance. The chief proposed to distribute 270 coins, which means he knew that some natives would be removed. Let's say x people were removed. To distribute a different number of coins to every... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can two knights, two bishops, two rooks, a queen, and a king be arranged on the first row of a chessboard so that the following conditions are met:
1) The bishops stand on squares of the same color;
2) The queen and the king stand on adjacent squares. (20 points). | # Solution:
Let's number the cells of the first row of the chessboard in order from left to right with numbers from **1** to **8** ( **1** - the first white cell, **8** - the last black cell). Since the queen and king are standing next to each other, they can occupy one of 7 positions: 1-2, 2-3, ..., 7-8. Additionally... | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In his country of Milnlandia, Winnie-the-Pooh decided to open a company that produces honey. Winnie-the-Pooh sells honey only in pots, and it costs him 10 milnovs (the monetary units of Milnlandia) to produce any pot of honey. The inverse demand function for honey is given by $\boldsymbol{P}=310-3 \boldsymbol{Q}$ (w... | # Solution:
a) Profit $=P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q = 310Q - 3Q^2 - 10Q = 300Q - 3Q^2$.
Since the graph of the function $300Q - 3Q^2$ is a parabola opening downwards, its maximum is achieved at the vertex: $Q = -b / 2a = -300 / (-6) = 50$.
b) Winnie-the-Pooh maximizes the quantity $P(Q) \c... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. To climb from the valley to the mountain top, one must walk 4 hours on the road, and then -4 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 17 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour,... | # Solution:
The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the seco... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. To climb from the valley to the mountain top, one must walk 6 hours on the road, and then - 6 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 25 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour... | # Solution:
The path along the road and the trail (there and back) takes 24 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 12 hours. Therefore, if you start moving along the trail immediately after the sec... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In all other cases - o points.
## Task 2
Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$.
How many solutions of this equation satisfy the condition:
$0 \leq x \leq 2020 ?$ | Solution. Let $t=\frac{\pi x}{4}$. Then the equation takes the form
$2 \sqrt{2} \sin ^{3} t=\cos \left(\frac{\pi}{4}-t\right)$.
$2 \sqrt{2} \sin ^{3} t=\cos \frac{\pi}{4} \cos t+\sin \frac{\pi}{4} \sin t$
$2 \sqrt{2} \sin ^{3} t=\frac{\sqrt{2}}{2} \cos t+\frac{\sqrt{2}}{2} \sin t$
$4 \sin ^{3} t=\cos t+\sin t ; 4 \... | 505 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In all other cases - $\mathbf{0}$ points.
## Task 2
## Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi}{4}(1+x)\right)$.
How many solutions of this equation satisfy the condition: $2000 \leq x \leq 3000$? | # Solution:
$\sin \left(\frac{\pi}{4}(1+x)\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$. The equation becomes
$2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$, i.e., we get problem 2 from option 1, the solution of which is:
$x=1+4 n, n \in Z . \quad 2000 \leq x \leq 3000,2000 \... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 3.
## Maximum 10 points
In the Country of Wonders, a pre-election campaign is being held for the position of the best tea lover, in which the Mad Hatter, March Hare, and Dormouse are participating. According to a survey, $20 \%$ of the residents plan to vote for the Mad Hatter, $25 \%$ for the March Hare, and $3... | # Solution:
Let the number of residents in Wonderland be $N$, then $0.2 N$ residents are going to vote for Dum, $0.25 N$ residents for the Rabbit, and $0.3 N$ residents for Sonya. The undecided voters are $0.25 N$ residents. Let $\alpha$ be the fraction of the undecided voters who are going to vote for Dum. Dum will n... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Assignment 7.
## Maximum 10 points
In the modern world, every consumer often has to make decisions about replacing old equipment with more energy-efficient alternatives. Consider a city dweller who uses a 60 W incandescent lamp for 100 hours each month. The electricity tariff is 5 rubles/kWh.
The city dweller can ... | # Solution and Grading Scheme:
a) Expenses for 10 months when installing an energy-saving lamp independently:
$$
120 \text { rub. }+12 \text { (W) * } 100 \text { (hours) / } 1000 \text { * } 5 \text { (rub./kW*hour) * } 10 \text { (months) = } 180 \text { rub. }
$$
Expenses for 10 months when turning to an energy s... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. 25 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that a larger amount of ta... | # Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. For linear demand functions, using the definition of elasticity, we get: $1.5 \cdot \frac{b P_{e}}{Q_{e}}=\frac{6 P_{e}}{Q_{e}}$. From this, we find that $b=4$.
If a per-unit tax $t=30$ is int... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4. 25 points
A novice economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that it was impossible to colle... | # Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. Using the definition of price elasticity for linear demand functions, $1.5 \cdot$ $\frac{4 P_{e}}{Q_{e}}=\frac{d P_{e}}{Q_{e}}$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. Maximum 16 points
Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from point $B$ to the road connecting $A$ and $C$, and from point $C$ to the road connecting $A$ and $B$ is... | # Solution
The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of trian... | 307 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. Maximum 20 points
The commander of a tank battalion, in celebration of being awarded a new military rank, decided to invite soldiers to a tank festival, where the main delicacy is buckwheat porridge. The commander discovered that if the soldiers are lined up by height, there is a certain pattern in the ch... | # Solution
(a) The commander tries to feed as many soldiers as possible, which means he will invite relatively short soldiers first - all other things being equal, their consumption of porridge is less.
Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n- (5+0.1 n) P$, where $n$ i... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The number 2458710411 was written 98 times in a row, resulting in a 980-digit number. From this number, it is required to erase 4 digits. What is the number of ways this can be done so that the newly obtained 976-digit number is divisible by 6? | Solution. If a number is divisible by 6, then it is divisible by 3 and 2. A number is divisible by 2 if and only if its last digit is even. The 980-digit number given in the condition ends with 2458710411, i.e., it has the form ...2458710411. For the number to be divisible by 2, it is necessary to strike out the last t... | 90894 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue c... | # Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 b=6$. We find that $b=$ 4. If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118 ; a=688$. The market demand function is $Q_{d}=688-4 P$. (8 points).
2) It is known t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue c... | # Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot 4=d$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64.688-4\left(P_{s}+90\right)=6 P_{s}+c$; $0.1 c+32.8=P_{s}=64 ; c=-312$. The market supply function is $Q_{s}=6 P-312$. (8 points).
2) It is known t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In all other cases - 0 points
## *Important: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.
## Assignment 2 (12 points)
Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest... | # Solution:
1) It is clear that the expression $4 x_{1}-3 p_{1}-44=0 \Leftrightarrow x=\frac{3}{4} p+11$ defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression
$$
\begin{aligned}
p^{2}-12 p+x^{2}-8 x+4 & =0 \Leftrightarrow(p-6)^{2}+(x-4)^{2}-36-16+43=0 \\
\Leftrightarrow & (p-6)^{2}+... | 13080 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. Maximum 14 points
Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from settlement $B$ to the road connecting $A$ and $C$, and from settlement $C$ to the road connecting $A$ ... | # Solution
The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of trian... | 307 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. Maximum 20 points
The commander of a tank battalion, in celebration of being awarded a new military rank, decided to organize a mass celebration, inviting subordinate soldiers. Only the soldiers whom the commander personally invites can attend. The main delicacy at the celebration is buckwheat porridge. H... | # Solution
(a) The commander tries to feed as many soldiers as possible, which means he will primarily invite relatively short soldiers - all other things being equal, their porridge consumption is less.
Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n-(5+0.1 n) P$, where $n$ i... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that it was impossible to collect a l... | # Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 b=6$. We find that $b=$ 4. If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118 ; a=688$. The market demand function is $Q_{d}=688-4 P$. (8 points).
2) It is known t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue c... | # Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot 4=d$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64.688-4\left(P_{s}+90\right)=6 P_{s}+c$; $0.1 c+32.8=P_{s}=64 ; c=-312$. The market supply function is $Q_{s}=6 P-312$. (8 points).
2) It is known t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Maximum 15 points. On side AB of an equilateral triangle $\mathrm{ABC}$, a right triangle $\mathrm{A} H \mathrm{~B}$ is constructed ( $\mathrm{H}$ - the vertex of the right angle), such that $\angle \mathrm{HBA}=60^{\circ}$. Let point K lie on ray $\mathrm{BC}$ beyond point $\mathrm{C}$ and $\angle \mathrm{CAK}=15^{... | # Solution:
Extend NB and NA beyond points B and A respectively (H-B-B1, H-A-A1)
$\angle \mathrm{B} 1 \mathrm{BC}=60^{\circ}$
$\angle$ KAA1 $=75^{\circ}$, so BK is the bisector of $\angle \... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Maximum 15 points. Masha was given a chest with multicolored beads (each bead has a unique color, there are a total of $\mathrm{n}$ beads in the chest). Masha chose seven beads for her dress and decided to try all possible combinations of them on the dress (thus, Masha selects from a set of options to sew one, two, ... | # Solution:
1) Consider one bead. Before Masha sews it onto the dress, there are two options: to take the bead or not. If we now choose two beads, the number of options becomes four, which can be obtained by multiplying the first option by two. By increasing the number of beads, we conclude that the total number of al... | 127 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Maximum 15 points. The company "Intelligence, Inc" has developed a robot with artificial intelligence. To manufacture it, a special machine is required, which can produce 1 robot in 1 hour. The company owns a large number of such machines, but the created robot is so intelligent that it can produce an exact copy of ... | # Solution:
To minimize the company's costs, it is necessary to find the minimum number of machines that will allow the company to complete the order within the specified time frame.
Let $x$ be the number of machines. Then, in the first hour of operation, they will produce $x$ robots. These $x$ robots will start manu... | 328710 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 4. Maximum 20 points
## Option 1
At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved problem. A complete solution to a planimetry problem is worth 7 points, and a problem in stereometry is... | # Solution:
Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve.
Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points ... | 326 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. Maximum 20 points
In the city of Eifyadl, runic stones are sold. It is known that the first merchant offers a fixed discount of $\mathrm{n} \%$ for every 5th stone purchased, while the second merchant increases the discount by $1 \%$ for each subsequent stone purchased (0% for the 1st stone, 3% for the 4th s... | # Solution and Evaluation Criteria:
a) Let's assume the cost of one rune stone without a discount is 1 unit of currency. We find the average cost of a rune stone from the first merchant:
$(20(1-n)+80) / 100=(100-20 n) / 100$
For the second merchant:
$(1+0.99+0.98+\ldots+0.81+0.8 * 80) / 100=82.1 / 100$
Then:
$100... | 104 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4. Maximum 20 points
## Option 1
At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved problem. A complete solution to a planimetry problem is worth 7 points, and a problem in stereometry is... | # Solution:
Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve.
Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points ... | 326 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. Maximum 20 points
In the city of Eifyadl, runic stones are sold. It is known that the first merchant offers a fixed discount of $\mathrm{n} \%$ for every 5th stone purchased, while the second merchant increases the discount by $1 \%$ for each subsequent stone purchased (0% for the 1st stone, 3% for the 4th s... | # Solution and Evaluation Criteria:
a) Let's assume the cost of one rune stone without a discount is 1 unit of currency. We find the average cost of a rune stone from the first merchant:
$(20(1-n)+80) / 100=(100-20 n) / 100$
For the second merchant:
$(1+0.99+0.98+\ldots+0.81+0.8 * 80) / 100=82.1 / 100$
Then:
$100... | 104 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.