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# Problem 4. Maximum 20 points
Svetlana is convinced that any feline can be interesting to her for two purposes - catching mice and conducting cat therapy. Based on these considerations, Svetlana has decided to acquire 2 kittens, each of which has an equal probability of being either a male or a female. The weekly pro... | # Solution:
(a) Males catch mice better than females. Two males can catch $40 \cdot 2=80$ mice in a week.
Answer: 160 (2 points).
(b) There are 3 options: one male and one female, two males, two females.
Answer: 3 (2 points).
(c) Two males with individual PPFs: $M=80-4K$. The opportunity costs for males are consta... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. Maximum 15 points
Find the values of the variable $x$ such that the four expressions: $2 x-6, x^{2}-4 x+5, 4 x-8, 3 x^{2}-12 x+11$ differ from each other by the same number. Find all possible integer values of the variable $x$ for any order of the expressions. | # Solution:
From the properties of numbers that differ from the following in a numerical sequence by the same number, we form two equations:
$$
\begin{gathered}
x^{2}-4 x+5-(2 x-6)=4 x-8-\left(x^{2}-4 x+5\right) \\
4 x-8-\left(x^{2}-4 x+5\right)=3 x^{2}-12 x+11-(4 x-8)
\end{gathered}
$$
Only \( x=4 \) satisfies both... | 4 | 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 1. Maximum 20 points
Let's say in country A there are only two firms engaged in research and development in a certain field. Currently, each of them independently decides whether to participate in the development of a new technology. It is known that if a firm develops a new technology, it will bring it $V$ mon... | # Solution and Grading Scheme:
(a) For each firm to decide to participate in the development, it is necessary that the expected profit from participation for each firm is greater than 0 (we use the formula for expected income in the case where both firms are involved in development):
$\alpha(1-\alpha) V+0.5 \alpha^{2... | 3 | Algebra | 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 5. Maximum 15 points
In the treasury of the Magic Kingdom, they would like to replace all old banknotes with new ones. There are a total of 3,628,800 old banknotes in the treasury. Unfortunately, the machine that prints new banknotes requires major repairs and each day it can produce fewer banknotes: on the fir... | # Solution:
(a) If a major repair is to be carried out, it is most effective to do so on the second day, as this will allow the production of new banknotes to increase from 604,800 to 1 million on that day, and the production will also be 1 million in subsequent days, which is more than the possibilities without major... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. Maximum 20 points
Let's say in country A there are only two firms engaged in research and development in a certain field. Currently, each of them independently decides whether to participate in the development of a new technology. It is known that if a firm develops a new technology, it will bring it $V$ mon... | # Solution and Grading Scheme:
(a) For each firm to decide to participate in the development, it is necessary that the expected profit from participation for each firm is greater than 0 (we use the formula for expected income in the case where both firms are involved in development):
$\alpha(1-\alpha) V+0.5 \alpha^{2... | 5 | Algebra | 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 |
# Task 3. Maximum 20 points
Consider a consumer who plans to use a coffee machine. Let's assume the consumer lives for two periods and values the benefit of using the coffee machine at 10 monetary units in each period. Coffee machines can be produced in two types: durable, working for two periods, and low-quality, com... | # Solution and Grading Scheme:
(a) When coffee machines are produced by a monopoly, it sets a selling price that will extract all consumer surplus:
1) when producing a durable machine, the price equals the consumer's benefit from using the coffee machine for 2 periods $\mathrm{p}_{\mathrm{L}}=2 \cdot 10=20$ (1 point)... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. Maximum 20 points
In Moscow, a tennis tournament is being held. Each team consists of 3 players. Each team plays against every other team, with each participant of one team playing against each participant of the other exactly one match. Due to time constraints, a maximum of 150 matches can be played in the ... | # Solution:
Assume there are two teams - then each of the three members of one team plays against each of the other - i.e., $3 * 3 = 9$ games. The number of team pairs can be $150: 9 = 16.6 \ldots$ a maximum of 16. Two teams form only one pair; three teams form three pairs. If there is a fourth team, add 3 more pairs,... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 3. Maximum 20 points
Consider a consumer who plans to use a coffee machine. Let's assume the consumer lives for two periods and evaluates the benefit from using the coffee machine at 20 monetary units in each period. Coffee machines can be produced in two types: durable, working for two periods, and low-quality... | # Solution and Grading Scheme:
(a) When coffee machines are produced by a monopoly, it sets a selling price that will extract all consumer surplus:
1) when producing a durable machine, the price equals the consumer's benefit from using the coffee machine for 2 periods $\mathrm{p}_{\mathrm{L}}=2 \cdot 20=40$ (1 point)... | 6 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. Maximum 20 points
In Moscow, a tennis tournament is being held. Each team consists of 3 players. Each team plays against every other team, with each participant of one team playing against each participant of the other exactly one match. Due to time constraints, a maximum of 200 matches can be played in the ... | # Solution:
Assume there are two teams - then each of the three members of one team plays against each of the other - i.e., $3 * 3=9$ games. The number of team pairs can be $200: 9=22.2 \ldots$ a maximum of 22. Two teams form only one pair; three teams form three pairs. If there is a fourth team, add 3 more pairs, so ... | 7 | Combinatorics | 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 |
5. In a certain kingdom, the workforce consists only of the clan of dwarves and the clan of elves. Historically, in this kingdom, dwarves and elves have always worked separately, and no enterprise has ever hired both at the same time. The aggregate supply of labor resources of the dwarves is represented by the function... | # Solution:
Before the law was introduced, the wages of elves and gnomes were determined by the condition of equality of supply and demand in each market:
$w_{\text {gnome }}^{S}=w_{\text {gnome }}^{D}$, from which $1+\frac{L}{3}=10-2 L / 3$ and $L_{\text {gnome }}=9$ and $w_{\text {gnome }}=4$
$w_{\text {elf }}^{S}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given a 2015-digit number divisible by 9. Let the sum of its digits be $\boldsymbol{a}$, the sum of the digits of $\boldsymbol{a}$ be $\boldsymbol{b}$, and the sum of the digits of $\boldsymbol{b}$ be $\boldsymbol{c}$. Find the number $\boldsymbol{c}$. (14 points). | Solution:
The sum of the digits of any number gives the same remainder when divided by 9 as the number itself.
The largest 2015-digit number consists of 2015 nines. The sum of its digits is $2015 * 9 = 18135$, i.e., it has 5 digits.
The sum of the digits of the largest 5-digit number is 45 (b).
Numbers less than 45... | 9 | Number Theory | 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 |
3. On the extensions of sides $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{A}$ of the convex quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$, points $\boldsymbol{B}_{1}, \boldsymbol{C}_{1}, \boldsymbol{D}_{1}$ and $\boldsymbol{A}_{1}$ are taken such that $\boldsymbol{B} \boldsymbol{B}_{1}=\... | # Solution:
Let the area of quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ be $\boldsymbol{S}$. The median divides the area of a triangle in half. Therefore, $S_{A B C}=S_{C B B 1}=S_{C B 1 C 1}$. Consequently, $S_{B B 1 C 1}=2 S_{A B C}$. Similarly, we have $S_{C C 1 D 1}=2 S_{B C D}$, $S_{D D 1 A 1}=2 S_{C D A}$,... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The power in the kingdom of gnomes was seized by giants. The giants decided to get rid of the gnomes and told them the following: "Tomorrow we will line you up so that each of you will see those who stand after and not see those who stand before (i.e., the 1st sees everyone, the last sees no one). We will put either... | # Solution:
The gnomes agree as follows: they risk the first gnome, telling him the following: "Denote a white hat as 1 and a black hat as 0, and count the sum of the remaining n-1 gnomes. If the sum is even, say 'white'; if it is odd, say 'black'. Under this condition, the first gnome dies with a probability of $1 / ... | 1 | 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 |
3. On the extensions of sides $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{A}$ of the convex quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$, points $\boldsymbol{B}_{1}, \boldsymbol{C}_{1}, \boldsymbol{D}_{1}$ and $\boldsymbol{A}_{1}$ are taken such that $\boldsymbol{B} \boldsymbol{B}_{1}=\... | # Solution:
Let the area of quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ be $\boldsymbol{S}$. A median divides the area of a triangle in half. Therefore, $S_{A B C}=S_{C B B 1}=S_{C B 1 C 1}$. Consequently, $S_{B B 1 C 1}=2 S_{A B C}$. Similarly, we have $S_{C C 1 D 1}=2 S_{B C D}$, $S_{D D 1 A 1}=2 S_{C D A}$, a... | 5 | Geometry | 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 |
4. In all other cases, $-\mathbf{0}$ points.
## Task 4
Maxim Andreevich, a former university lecturer, gives math lessons to groups of schoolchildren. The cost of one hour (60 minutes) of Maxim Andreevich's lesson with a group of schoolchildren is 3 thousand rubles (after all taxes). In addition to income from tutori... | # Solution and Grading Scheme:
(a) Maxim Andreevich works $\left(24-8-2 L_{i}-k_{i}\right)=16-2 L_{i}-k_{i}$ hours a day, where $L_{i}$ is the number of hours he works as a tutor on the $i$-th working day, and $k_{i}$ is the number of hours he spends on rest and household chores on the $i$-th working day. The number o... | 11 | 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 |
# Task 4.
## Maximum 15 points
Find the value of the expression under the condition that the summation is performed infinitely
$$
\sqrt{20+\sqrt{20+\sqrt{20+\cdots}}}
$$
# | # Solution:
Let $x=\sqrt{20+\sqrt{20+\sqrt{20+\cdots}}}, x>0$.
Square both sides of the obtained equation. We get:
$x^{2}=20+x$
$x^{2}-x-20=0 ;$
$D=1+80=81 ; x_{1}=\frac{1-9}{2}=-4-$ does not satisfy the condition $x>0$;
$x_{1}=\frac{1+9}{2}=5-$ fits.
Answer: 5.
## Criteria
0 points - If calculated approximate... | 5 | Algebra | 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 |
1. Consider the vectors $\vec{a}=(3, x), \vec{b}=(\sqrt{x-2}, 4)$, then $\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|$, which is equivalent to
$$
3 \sqrt{x-2}+4 x \leq \sqrt{\left(9+x^{2}\right)(x+14)}
$$
Equality is possible if and only if the vectors are collinear
$$
\frac{\sqrt{x-2}}{3}=\frac{4}{x}>0 \Leftrightar... | Answer: 1. 6. 2.6 3.74. 7.
## Checking Criteria: | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In all other cases - 0 points
## *Important: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.
## Assignment 2 (12 points)
Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest... | # Solution:
1) It is clear that the expression $4 x_{1}-3 p_{1}-44=0 \Leftrightarrow x=\frac{3}{4} p+11$ defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression
$$
\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 |
6. Maximum 15 points. A school economics teacher believes it is extremely important to know the basics of both microeconomics and macroeconomics. Therefore, for his subject, he has introduced the following final grading system. First, the student's study of the basics of microeconomics ( $O_{\text {micro }}$ ) is asses... | # Solution:
(a) If the minimum value of the two is determined by the expression $0.75 * O_{\text {мИкро }} + 0.25 * O_{\text {мАкро }}$, i.e., $O_{\text {мИкро }}O_{\text {мАкро }}$, then it is more advantageous for Ivanov to spend an additional unit of time studying macroeconomics.
Thus, to achieve the highest grade... | 4 | 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 |
# 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 |
1. A necklace consists of 30 blue and a certain number of red beads. It is known that on both sides of each blue bead there are beads of different colors, and one bead away from each red bead there are also beads of different colors. How many red beads can be in this necklace? (The beads in the necklace are arranged cy... | Answer: 60.
Solution. It is obvious that blue beads appear in the necklace in pairs, separated by at least one red bead. Let there be $n$ red beads between two nearest pairs of blue beads. We will prove that $n=4$. Clearly, $n \leqslant 4$, since the middle one of five consecutive red beads does not satisfy the condit... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the segment $A B$ of length 10, a circle $\omega$ is constructed with $A B$ as its diameter. A tangent to $\omega$ is drawn through point $A$, and a point $K$ is chosen on this tangent. A line through point $K$, different from $A K$, is tangent to the circle $\omega$ at point $C$. The height $C H$ of triangle $A ... | Answer: 8.
Solution. Let $O$ be the center of $\omega$. Note that
$$
B H=\frac{1}{5} A B=2, \quad A H=8, \quad O H=\frac{1}{2} A B-B H=3, \quad C H=\sqrt{O C^{2}-O H^{2}}=4
$$
Right triang... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A knight is placed in each cell of a chessboard. What is the smallest number of knights that can be removed from the board so that no knight remains that attacks exactly three other knights? (A knight attacks the squares that are one square away horizontally and two squares away vertically, or vice versa.) | Answer: 8 knights.
Solution 1. We will say that a knight controls a square on the board if it attacks this square or stands on it. First, we will prove that it is impossible to remove fewer than 8 knights. It is sufficient to check that at least 4 knights must be removed from each half of the board. Consider, for defi... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 175 beads of red, blue, and green colors. It is known that each red bead has neighbors of different colors, and on any segment of the necklace between two green beads, there is at least one blue bead. What is the minimum number of blue beads that can be in this necklace? (The beads in the neck... | Answer: 30.
Solution 1. We will show that any block of six consecutive beads contains a blue bead. We can assume that there is no more than one green bead in it, otherwise there is nothing to prove. If the block contains 5 red beads, then at least 3 of them are consecutive, and the middle one does not satisfy the prob... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given a right triangle $ABC$ with a right angle at $C$. On its leg $BC$ of length 26, a circle is constructed with $BC$ as its diameter. A tangent $AP$ is drawn from point $A$ to this circle, different from $AC$. The perpendicular $PH$, dropped from point $P$ to segment $BC$, intersects segment $AB$ at point $Q$. Fi... | Answer: 24.
Solution. Let $O$ be the center of $\omega$. Note that
$$
B H=\frac{4}{13} B C=8, \quad C H=18, \quad O H=\frac{1}{2} B C-B H=5, \quad P H=\sqrt{O P^{2}-O H^{2}}=12
$$
Right tr... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A knight is placed in each cell of a chessboard. What is the minimum number of knights that can be removed from the board so that no knight remains that attacks exactly four other knights? (A knight attacks the squares that are one square away horizontally and two squares away vertically, or vice versa.) | Answer: 8 knights.
Solution. First, we will show that no fewer than 8 knights need to be removed. On the left diagram, all knights that attack exactly 4 squares of the board are marked (for convenience, they are highlighted in different colors). Let's call such knights bad. To stop a knight from attacking four others,... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 100 beads of red, blue, and green colors. It is known that among any five consecutive beads, there is at least one blue one, and among any seven consecutive beads, there is at least one red one. What is the maximum number of green beads that can be in this necklace? (The beads in the necklace ... | Answer: 65.
Solution. Let there be a set of beads $A$ such that in every set of $n$ consecutive beads, there is at least one from $A$. We will show that $A$ contains no fewer than $\frac{100}{n}$ elements. Indeed, between any two adjacent beads from $A$, there are no more than $n-1$ beads. If the set $A$ contains $m$ ... | 65 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. For a natural number ending not in zero, one of its digits (not the most significant) was erased. As a result, the number decreased by 9 times. How many numbers exist for which this is possible? | Answer: 28.
Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m>0$. By erasing the digit $a$, we get the number $m+10^{k} n$. According to the condition,
$$
m+10^{k} a+10^{k+1} n=9\left(m+10^{k} n\right) \Lon... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A square $4 \times 4$ is divided into 16 squares $1 \times 1$. We will call a path a movement along the sides of the unit squares, in which no side is traversed more than once. What is the maximum length that a path connecting two opposite vertices of the large square can have? | Answer: 32.
Solution. Let's call the sides of the $1 \times 1$ squares edges, the vertices of these squares nodes, and the number of edges adjacent to a node the multiplicity of the node. Notice that the $1 \times 1$ squares generate 40 distinct edges. If a path passes through a node of multiplicity 3, it enters the n... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A string is threaded with 150 beads of red, blue, and green. It is known that among any six consecutive beads, there is at least one green, and among any eleven consecutive beads, there is at least one blue. What is the maximum number of red beads that can be on the string? | Answer: 112.
Solution. We can choose $\left[\frac{150}{11}\right]=13$ consecutive blocks of 11 beads each. Since each block contains at least one blue bead, there are at least 13 blue beads on the string. In addition, we can group all the beads into 25 consecutive blocks of 6 beads each. Each block contains at least o... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. For a natural number ending not in zero, one of its digits was replaced by zero (if it is the leading digit, it was simply erased). As a result, the number decreased by 9 times. How many numbers exist for which this is possible? | Answer: 7.
Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m>0$, otherwise $m=a=0$. Then the number $8 m$ is a multiple of 10 and therefore ends in 0. By the condition, the number $m$ does not end in 0. Thus... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A rectangle $3 \times 5$ is divided into 15 squares $1 \times 1$. We will call a path a movement along the sides of the unit squares, such that no side is traversed more than once. What is the maximum length that a path connecting two opposite vertices of the rectangle can have? | Answer: 30.
Solution. Let the rectangle be denoted as $A B C D$, and let the path connect its vertices $A$ and $C$. We will call the sides of the $1 \times 1$ squares edges, the vertices of these squares - nodes, and the number of edges adjacent to a node - the multiplicity of the node. Note that the $1 \times 1$ squa... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 50 blue and a certain number of red beads. It is known that in any segment of the necklace containing 8 blue beads, there are at least 4 red ones. What is the minimum number of red beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is a... | Answer: 29.
Solution. Note that any segment of the necklace consisting of 11 beads contains no more than 7 blue and no fewer than 4 red beads (otherwise, it would contain 8 blue beads and no more than 3 red ones). Fix a red bead in the necklace. The 7 consecutive segments of 11 beads adjacent to it do not cover the en... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 6, 24, and 24. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the radius of the smaller base of the truncated cone. | Answer: 2.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $r$ be the radius of this base. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 100 red and a certain number of blue beads. It is known that in any segment of the necklace containing 10 red beads, there are at least 7 blue ones. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one i... | Answer: 78.
Solution. Note that any segment of the necklace containing 16 beads has no more than 9 red and no fewer than 7 blue beads (otherwise, it would contain 10 red beads and no more than 6 blue ones). Fix a blue bead in the necklace. The 11 consecutive segments of 16 beads adjacent to it do not cover the entire ... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 50 blue, 100 red, and 100 green beads. We will call a sequence of four consecutive beads good if it contains exactly 2 blue beads and one each of red and green. What is the maximum number of good quartets that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning... | Answer: 99.
Solution. Blue beads make up one fifth of the total. Therefore, there will be two consecutive blue beads (let's call them $a$ and $b$), separated by at least three beads. Note that $a$ and $b$ are part of no more than three good quartets, while the other blue beads are part of no more than four. If we add ... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 23, 46, and 69. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the radius of the smaller base of the truncated cone.
# | # Answer: 6.
Solution. Let $C$ be the center of the smaller base of the truncated cone, $R$ be the radius of this base, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other c... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A thread is strung with 75 blue, 75 red, and 75 green beads. We will call a sequence of five consecutive beads good if it contains exactly 3 green beads and one each of red and blue. What is the maximum number of good quintets that can be on this thread? | Answer: 123.
Solution. Note that the first and last green beads are included in no more than three good fives, the second and second-to-last - in no more than four fives, and the rest - in no more than five fives. If we add these inequalities, we get $2 \cdot 3 + 2 \cdot 4 + 71 \cdot 5 = 369$ on the right side, and th... | 123 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are $2 r, 3 r$, and $10 r$. A frustum of a cone is placed on the table with its smaller base down, and it shares a common generatrix with each of the other cones. Find $r$ if the radius of the smaller base of t... | Answer: 29.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones, with $R=15$. Denote by $\math... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in ... | Answer: 27.
Solution. If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that there are no fewer green beads tha... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the cells of an $80 \times 80$ table, pairwise distinct natural numbers are placed. Each of them is either a prime number or a product of two prime numbers (possibly the same). It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. ... | Answer: 4266.
Solution. We will say that a composite number $a$ serves a prime number $p$ if $a$ and $p$ are not coprime (i.e., $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than two distinct prime divisors, it serves n... | 4266 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are 1, 2, and 3. A sphere is placed on the table, touching all the cones. It turns out that the center of the sphere is equidistant from all points of contact with the... | Answer: 1.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $R$ be the radius of the sphere, $C$ be the point of contact of the... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Around a round table, 50 schoolchildren are sitting: blondes, brunettes, and redheads. It is known that in any group of schoolchildren sitting in a row, between any two blondes there is at least one brunette, and between any two brunettes - at least one redhead. What is the minimum number of redheads that can sit at... | Answer: 17.
Solution. If only blondes were sitting at the table, the number of pairs of neighbors would be equal to the number of blondes. Since there is a brunette between any two blondes, there are no fewer brunettes than blondes sitting at the table. Similarly, it can be proven that there are no fewer redheads than... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Inside an angle of $30^{\circ}$ with vertex $A$, a point $K$ is chosen, the distances from which to the sides of the angle are 1 and 2. Through point $K$, all possible lines are drawn, intersecting the sides of the angle. Find the minimum area of the triangle cut off by the line from the angle. | Answer: 8.
Solution 1. Choose points $P$ and $Q$ on the sides of the angle such that point $K$ is the midpoint of segment $P Q$. We will show that line $P Q$ cuts off a triangle of the small... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the cells of a $75 \times 75$ table, pairwise distinct natural numbers are placed. Each of them has no more than three distinct prime divisors. It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. What is the maximum number of pri... | Answer: 4218.
Solution. We will say that a composite number $a$ serves a prime number $p$ if the numbers $a$ and $p$ are not coprime (that is, $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than three distinct prime divi... | 4218 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are $2r$, $3r$, and $10r$. A sphere with a radius of 2 is placed on the table, touching all the cones. It turns out that the center of the sphere is equidistant from a... | Answer: 1.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $R$ be the radius of the sphere, $C$ be the point of contact of the... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second must be different. For example, 101 and 292 have this form, while 222 and 123 do not. Similarly, we define the form of the number $\overline{a b c a b d}$. How many odd numbers of the form $\o... | Answer: 448.
Solution. Odd numbers divisible by 5 are numbers ending in 5, so for $d$ we have only one option. For $a$ we have 8 options, as the number cannot start with zero, and $a$ cannot be equal to $d$. The digit $b$ cannot be equal to $a$ or $d$, and there are no other restrictions on it - we get 8 possible valu... | 448 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) The numbers $s_{1}, s_{2}, \ldots, s_{1008}$ are such that their sum is $2016^{2}$. It is known that
$$
\frac{s_{1}}{s_{1}+1}=\frac{s_{2}}{s_{2}+3}=\frac{s_{3}}{s_{3}+5}=\ldots=\frac{s_{1008}}{s_{1008}+2015}
$$
Find $s_{17}$. | Answer: 132.
Solution. Note that none of the $s_{i}$ is equal to zero (otherwise, all the fractions $\frac{s_{i}}{s_{i}+2 i-1}$ would be equal to zero, and, consequently, all $s_{i}$ would have to be equal to zero, which contradicts the fact that their sum is $2016^{2}$). Therefore, the original condition is equivalen... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Cells of an infinite grid paper are painted in $k$ colors (each cell is painted entirely in one color). What is the largest $k$ such that in every grid rectangle with sides 3 and 4, cells of all these colors will be found? | Answer: 10.
Solution. Divide the infinite grid paper into ten-cell figures as shown in the figure.
The required coloring with 10 colors can be achieved by coloring one such figure in 10 co... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell. | Answer: 416.
Solution. A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will be no fewer than 416 such cells.
We will sh... | 416 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In some cells of a $1 \times 2021$ strip, one chip is placed in each. For each empty cell, the number equal to the absolute difference between the number of chips to the left and to the right of this cell is written. It is known that all the written numbers are distinct and non-zero. What is the minimum number of ch... | Answer: 1347.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 1347 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In some cells of a $1 \times 2100$ strip, one chip is placed. In each of the empty cells, a number is written that is equal to the absolute difference between the number of chips to the left and to the right of this cell. It is known that all the written numbers are distinct and non-zero. What is the minimum number ... | Answer: 1400.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 1400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression
$$
A=a^{7} b+b^{7} c+c^{7} a+a b^{3}+b c^{3}+c a^{3}
$$ | Answer: 6.
Solution. By the Cauchy-Schwarz inequality
$$
A=\left(a^{7} b+a b^{3}\right)+\left(b^{7} c+b c^{3}\right)+\left(c^{7} a+c a^{3}\right) \geqslant 2\left(a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}\right) \geqslant \frac{2}{3}\left(a^{2} b+b^{2} c+c^{2} a\right)^{2}=6
$$
Equality is achieved when $a=b=c=1$. | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Given an acute scalene triangle $A B C$. In it, the altitudes $B B_{1}$ and $C C_{1}$ are drawn, intersecting at point $H$. Circles $\omega_{1}$ and $\omega_{2}$ with centers $H$ and $C$ respectively touch the line $A B$. From point $A$ to $\omega_{1}$ and $\omega_{2}$, tangents other than $A B$ are drawn. Denote th... | Answer: $180^{\circ}$.
Solution. Since right triangles $A C C_{1}$ and $A C E$ are equal by leg and hypotenuse, line $C A$ is the bisector of isosceles triangle $C_{1} C E$. Therefore, it i... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A circle is circumscribed around an acute-angled triangle $A B C$. Point $K$ is the midpoint of the smaller arc $A C$ of this circle, and point $L$ is the midpoint of the smaller arc $A K$ of this circle. Segments $B K$ and $A C$ intersect at point $P$. Find the angle between the lines $B C$ and $L P$, given that $B... | Answer: $90^{\circ}$.
Solution. Let $\angle A B L=\varphi$. Then $\angle K B L=\varphi$ and $\angle K B C=2 \varphi$. Since $B K=B C$, we get $\angle B K C=\angle B C K=90^{\circ}-\varphi$.... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)^{3}-26 a b c}
$$ | Answer: 3.
Solution. Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$
(we used the Cauchy inequality). Then
$$
a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c
$$
Let $t=\frac{(a+b+c)^{3... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$, and a circle with center at point $O$ encompasses circles $\omega_{1}$ and $\omega_{2}$, touching them at points $C$ and $D$ respectively. It turns out that points $A, C$, and $D$ lie on the same line. Find the angle $A B O$. | Answer: $90^{\circ}$.
Solution. Let $O_{1}$ and $O_{2}$ be the centers of circles $\omega_{1}$ and $\omega_{2}$. Triangles $C O D$, $C O_{1} A$, and $A O_{2} D$ are isosceles. Since points ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}
$$ | Answer: 3.
Solution. Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where
\[
\begin{aligned}
& B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\
& \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{a^{3}+b^{3}+c^{3}-2 a b c}
$$ | Answer: 6.
Solution. By the Cauchy-Schwarz inequality,
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)=\frac{1}{3}\left((a+b+c)^{3}-\left(a^{3}+b^{3}+c^{3}+6 a b c\right)\right) \leqslant \frac{1}{3}\left((a+b+c)^{3}-9 a b c\right)
$$
Note that
$$
a^{3}+b^{3}+c^{3}-2 a b c \geqslant \frac{1}{9}(a+b+c)^{3}-2 a b c=\frac{1}{9}\l... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $A$. Segment $O_{2} A$ intersects circle $\omega_{1}$ again at point $K$, and segment $O_{1} A$ intersects circle $\omega_{2}$ again at point $L$. The line passing through point $A$ parallel to $K L$ intersects cir... | Answer: $90^{\circ}$.
Solution. Triangles $O_{1} A K$ and $O_{2} A L$ are isosceles and have a common base angle. Therefore, their vertex angles are equal. Let the common value of these ang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)}{(a+b+c)^{4}-79(a b c)^{4 / 3}}
$$ | Answer: 3.
Solution. By the Cauchy-Schwarz inequality,
$a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)=(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-\left(a^{4}+b^{4}+c^{4}\right) \leqslant(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-3(a b c)^{4 / 3}$.
Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $B$. The extension of segment $O_{2} B$ beyond point $B$ intersects circle $\omega_{1}$ at point $K$, and the extension of segment $O_{1} B$ beyond point $B$ intersects circle $\omega_{2}$ at point $L$. The line pa... | Answer: $90^{\circ}$.
Solution. Since $\angle O_{1} B K = \angle O_{2} B L$, the isosceles triangles $O_{1} B K$ and $O_{2} B L$ are similar. Therefore, their vertex angles are equal. Let th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on a $10 \times 11$ board such that any placement of a three-cell corner on the board touches at least one marked cell? | Answer: 50.
Solution. It is not hard to notice that in any $2 \times 2$ square, there are at least two marked cells. Since 25 such squares can be cut out from a $10 \times 11$ board, there must be no fewer than 50 marked cells in it. An example with 50 marked cells is obtained if the cells with an even first coordinat... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A warehouse stores 400 tons of cargo, with the weight of each being a multiple of a centner and not exceeding 10 tons. It is known that any two cargos have different weights. What is the minimum number of trips that need to be made with a 10-ton truck to guarantee the transportation of these cargos from the warehous... | Answer: 51.
Solution. We will show that it is always possible to transport the goods in 51 trips, even if the warehouse contains all weights from 1 to 100 tons. Indeed, we can divide all the goods, except for the 50-ton and 100-ton ones, into 49 pairs as follows:
$$
(1,99), \quad(2,98), \quad(3,97), \quad \ldots, \qu... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on a $9 \times 9$ board such that any placement of a three-cell corner piece will touch at least two marked cells? | Answer: 56.
Solution. It is not hard to notice that in any $2 \times 2$ square, at least three cells must be marked, and in each $1 \times 2$ rectangle, at least two cells must be marked. Since from a $9 \times 9$ board, 16 $2 \times 2$ squares and 8 $1 \times 2$ rectangles can be cut out, at least $16 \cdot 3 + 8 = 5... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A cinema was visited by 50 viewers, the total age of whom is 1555 years, and among them, there are no viewers of the same age. For what largest $k$ can we guarantee to select 16 viewers whose total age is not less than $k$ years? | Answer: 776.
Solution. We will show that $k \geqslant 776$, that is, the total age of the 16 oldest viewers is always not less than 776. Arrange the viewers in order of increasing age, and let $a_{i}$ be the age of the $i$-th viewer. Since there are no viewers of the same age, we get
$$
a_{1} \leqslant a_{2}-1 \leqsl... | 776 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on a $12 \times 12$ board such that any placement of a four-cell figure $\square \square$ on the board touches at least one marked cell? (The figure can be rotated and flipped.) | Answer: 48.
Solution. It is not hard to notice that in any $2 \times 3$ rectangle, there are at least two marked cells. Since the $12 \times 12$ board is divided into 24 such rectangles, there must be no fewer than 48 marked cells. An example with 48 marked cells can be obtained by marking the cells where the sum of t... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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