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2. The cold water tap fills the bathtub in 19 minutes, while the hot water tap fills it in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it? | Answer: in 2 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 9.5 minutes. Therefore, the hot water tap should be open for 2 minutes longer.
Evaluation. 12 points for the correct solution. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A certain mechanism consists of 25 parts, some of which are large, and some are small. It is known that among any 12 parts taken, there will be at least one small part, and among any 15 parts, there will be at least one large part. How many of each type of part does the mechanism contain? | Answer: 11 large parts and 14 small parts.
Solution. Since among any 12 parts there is a small one, there are no more than 11 large parts. Since among any 15 parts there is a large one, there are no more than 14 small parts. If there were fewer than 11 large parts or fewer than 14 small parts, the total number of part... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) Masha's tablet, which she needed for a presentation at school, was completely drained. Using additional equipment, the tablet can be fully charged in 2 hours and 40 minutes, without it in 8 hours. Masha first put the discharged tablet on regular charging, and when she found the equipment, she switched to... | Answer: 288.
Solution. The tablet charges in 160 minutes on fast charging, and in 480 minutes on regular charging. Therefore, on fast charging, $\frac{1}{160}$ of the full charge is completed in 1 minute, and on regular charging, $\frac{1}{480}$ of the full charge is completed in 1 minute. Let $t-$ be the total chargi... | 288 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) There are ten weights of different weights, each weighing an integer number of grams. It is known that the weight of the lightest and heaviest weight differs by 9 grams. One weight is lost. Find its weight if the total weight of the remaining weights is 2022 grams. | Answer: 223.
Solution. Let $x$ be the weight of the lightest weight. Denote the weight of the lost weight as $(x+y)$ $(0<y<9)$. Then $x+(x+1)+(x+2)+\cdots+(x+$ $9)-(x+y)=2022$. Combine like terms: $10 x+45-x-y=$ 2022 or $9 x=1977+y$. From this, $1977+y$ is divisible by 9. Considering the condition $0<y<9$, we get that... | 223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12 \%$. However, this was not enough, so it was decided to red... | Answer: 18.
Solution. Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $2(0.9 x + 0.8 y) = 0.88 \cdot 2(x + y)$ or $x = 4 y$. The original perimet... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) A one-and-a-half kilogram model of a sports car body was made from carbon fiber for aerodynamic studies at a scale of 1:10. What is the mass of the actual body if it is also entirely made of carbon fiber? | Answer: 1500 kg.
Solution. All dimensions of the body are 10 times larger compared to the model. Therefore, the volume of the body is larger by $10 \cdot 10 \cdot 10=1000$ times. Mass is directly proportional to volume, therefore, the mass of the body:
$$
m_{\text {body }}=1000 \cdot m_{\text {model }}=1500 \text { k... | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A car traveled half of the distance at a speed 20 km/h faster than the average speed, and the second half of the distance at a speed 20% lower than the average. Determine the average speed of the car. | Answer: 60 km/h.
Solution. The average speed $v=\frac{s+s}{t_{1}+t_{2}}=\frac{s+s}{\frac{s}{v+20}+\frac{s}{0.8 v}}=\frac{2}{\frac{1}{v+20}+\frac{1}{0.8 v}}$. Solving this equation, we get $v=60$ km/h. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (15 points) A pedestrian is moving towards a crosswalk along a straight path at a constant speed of 3.6 km/h. At the initial moment, the distance from the pedestrian to the crosswalk is 20 m. The length of the crosswalk is $5 \mathrm{~m}$. At what distance from the crosswalk will the pedestrian be after half a minut... | Answer: 5 m.
Solution. The pedestrian's speed is 3.6 km/h = 1 m/s. In half a minute, he walked $s = v t = 1 \cdot 30 s = 30$ m. From the crossing, he is at a distance of $l = s - 20 - 5 = 5$ m.
## Tasks, answers, and evaluation criteria | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) Masha's tablet, which she needed for a presentation at school, was completely drained. Using additional equipment, the tablet can be fully charged in 3 hours, without it in 9 hours. Masha first put the discharged tablet on regular charging, and when she found the equipment, she switched to fast charging ... | Answer: 324.
Solution. The tablet charges in 180 minutes on fast charging, and in 540 minutes on regular charging. Therefore, on fast charging, $\frac{1}{180}$ of the full charge is completed in 1 minute, and on regular charging, $\frac{1}{540}$ of the full charge is completed in 1 minute. Let $t$ be the total chargin... | 324 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) There are ten weights of different weights, each weighing an integer number of grams. It is known that the weight of the lightest weight and the heaviest differs by 9 grams. One weight is lost. Find the weight of the lightest weight if the total weight of the remaining weights is 2022 grams.
# | # Answer: 220.
Solution. Let $x$ be the weight of the lightest weight. Denote the weight of the lost weight as $(x+y)$ $(0<y<9)$. Then $x+(x+1)+(x+2)+\cdots+(x+$ 9) - $(x+y)=2022$. Combine like terms: $10 x+45-x-y=$ 2022 or $9 x=1977+y$. From this, $1977+y$ is divisible by 9. Considering the condition $0<y<9$, we get ... | 220 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12.5 \%$. However, this was not enough, so it was decided to r... | Answer: 14.
Solution. Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $\quad 2(0.9 x + 0.8 y) = 0.875 \cdot 2(x + y) \quad$ or $\quad x = 3 y$. T... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) A one-kilogram model of a sports car body was made from carbon fiber for aerodynamic studies at a scale of 1:11. What is the mass of the actual body if it is also entirely made of carbon fiber? | Answer: 1331 kg.
Solution. All dimensions of the body are 11 times larger compared to the model. Therefore, the volume of the body is larger by $11 \cdot 11 \cdot 11=1331$ times. The mass is directly proportional to the volume, therefore, the mass of the body:
$$
m_{\text {body }}=1331 \cdot m_{\text {model }}=1331 \... | 1331 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A car traveled half of the distance at a speed 30 km/h faster than the average speed, and the second half of the distance at a speed 30% lower than the average. Determine the average speed of the car. | Answer: 40 km/h.
Solution. The average speed $v=\frac{s+s}{t_{1}+t_{2}}=\frac{s+s}{\frac{s}{v+30}+\frac{s}{0.7 v}}=\frac{2}{\frac{1}{v+30}+\frac{1}{0.7 v}}$. Solving this equation, we get $v=40$ km/h. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem No. 5 (10 points)
When constructing this structure, a homogeneous wire of constant cross-section was used. It is known that points $B, D, F$ and $H$ are located equally at the midpoints of the corresponding sides of the square ACEG. The resistance of segment $A B$ is $R_{0}=1 \Omega$. Determine the resistance ... | Answer: 2 Ohms
## Solution and grading criteria:
The resistance of a resistor is proportional to its length.
Taking this into account, the proposed circuit can be replaced by an equivalent one:
 When walking uphill, the tourist walks 2 km/h slower, and downhill 2 km/h faster, than when walking on flat ground. Climbing the mountain takes the tourist 10 hours, while descending the mountain takes 6 hours. What is the tourist's speed on flat ground? | # Answer: 8
Solution. Let $x$ km/h be the tourist's speed on flat terrain. According to the problem, we get the equation $10(x-2)=6(x+2)$. From this, we find $x=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) Mitya, Anton, Gosha, and Boris bought a lottery ticket for 20 rubles. Mitya paid $24\%$ of the ticket's cost, Anton - 3 rubles 70 kopecks, Gosha - 0.21 of the ticket's cost, and Boris contributed the remaining amount. The boys agreed to divide the winnings in proportion to their contributions. The ticket... | Answer: 365
Solution. The ticket costs 2000 kop. Mitya paid 480 kop, Anton - 370 kop, Gosha - 420 kop, therefore, Boris had to pay an additional 730 kop. Since the prize is 50 times the cost of the ticket, Boris is entitled to 365 rubles. | 365 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) When walking uphill, the tourist walks 3 km/h slower, and downhill 3 km/h faster, than when walking on flat ground. Climbing the mountain takes the tourist 8 hours, while descending the mountain takes 4 hours. What is the tourist's speed on flat ground? | # Answer: 9
Solution. Let $x$ km/h be the tourist's speed on flat terrain. According to the problem, we get the equation $8(x-3)=4(x+3)$. From this, we find $x=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) Mitya, Anton, Gosha, and Boris bought a lottery ticket for 20 rubles. Mitya paid $24\%$ of the ticket's cost, Anton - 3 rubles 70 kopecks, Gosha - $0.21$ of the ticket's cost, and Boris contributed the remaining amount. The boys agreed to divide the winnings in proportion to their contributions. The tick... | Answer: 292
Solution. The ticket costs 2000 kop. Mitya paid 480 kop, Anton - 370 kop, Gosha - 420 kop, therefore, Boris had to pay an additional 730 kop. Since the prize is 40 times the cost of the ticket, Boris is entitled to 292 rubles. | 292 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) The dividend is six times larger than the divisor, and the divisor is four times larger than the quotient. Find the dividend. | # Answer: 144
Solution. From the condition of the problem, it follows that the quotient is 6. Then the divisor is 24, and the dividend is 144. | 144 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Hooligan Vasily tore out a whole chapter from a book, the first page of which was numbered 231, and the number of the last page consisted of the same digits. How many sheets did Vasily tear out of the book?
# | # Answer: 41
Solution. The number of the last page starts with the digit 3 and must be even, so the last page has the number 312. Vasily tore out $312-231+1=82$ pages or 41 sheets. | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) Divide the number 90 into two parts such that $40\%$ of one part is 15 more than $30\%$ of the other part. Write the larger of the two parts in your answer. | Answer: 60
Solution. Let one part of the number be $x$, then the other part will be $90-x$. We get the equation $0.4 \cdot x = 0.3 \cdot (90 - x) + 15$, solving it we get $x = 60$, and the other part of the number is 30. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) A one-kilogram model of a sports car body was made from aluminum at a scale of 1:10. What is the mass of the actual body if it is also entirely made of aluminum? | Answer: 1000 kg
Solution. All dimensions of the body are 10 times larger compared to the model. Therefore, the volume of the body is larger by $10 \cdot 10 \cdot 10=1000$ times. Mass is directly proportional to volume, therefore, the mass of the body:
$$
m_{\text {body }}=1000 m_{\text {model }}=1000 \text { kg. }
$$... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) The dividend is five times larger than the divisor, and the divisor is four times larger than the quotient. Find the dividend.
# | # Answer: 100
Solution. From the condition of the problem, it follows that the quotient is 5. Then the divisor is 20, and the dividend is 100. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Hooligan Vasily tore out a whole chapter from a book, the first page of which was numbered 241, and the number of the last page consisted of the same digits. How many sheets did Vasily tear out of the book?
# | # Answer: 86
Solution. The number of the last page starts with the digit 4 and must be even, so the last page has the number 412. Vasily tore out $412-241+1=172$ pages or 86 sheets. | 86 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) Divide the number 80 into two parts such that $30\%$ of one part is 10 more than $20\%$ of the other part. Write the smaller of the two parts in your answer. | Answer: 28
Solution. Let one part of the number be $x$, then the other part will be $80-x$. We get the equation $0.3 \cdot x = 0.2 \cdot (80 - x) + 10$, solving it we get $x = 52$, and the other part of the number is 28. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) A two-kilogram model of a sports car body was made from aluminum at a scale of $1: 8$. What is the mass of the actual body if it is also entirely made of aluminum? | Answer: 1024 kg
Solution. All dimensions of the body are 8 times larger compared to the model. Therefore, the volume of the body is larger by $8 \cdot 8 \cdot 8=512$ times. The mass is directly proportional to the volume, therefore, the mass of the body:
$m_{\text {body }}=512 m_{\text {model }}=1024$ kg | 1024 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) Find the area of the triangle cut off by the line $y=3 x+1$ from the figure defined by the inequality $|x-1|+|y-2| \leq 2$. | Answer: 2.
Solution. The figure defined by the given inequality is a square.
The side of the square is $2 \sqrt{2}$ (this can be found using the Pythagorean theorem). The given line passes t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) A section of a regular triangular pyramid passes through the midline of the base and is perpendicular to the base. Find the area of the section if the side of the base is 6 and the height of the pyramid is 8. | Answer: 9.
Solution. The section MNP passes through the midline of the base of the pyramid $MN$ and is perpendicular to the base. Therefore, the height $PH$ of the triangle $MNP$ is parallel to the height of the pyramid $DO$.
 Find the area of the triangle cut off by the line $y=2x+2$ from the figure defined by the inequality $|x-2|+|y-3| \leq 3$. | Answer: 3.
Solution. The figure defined by the given inequality is a square.
The side of the square is $3 \sqrt{2}$ (this value can be found using the Pythagorean theorem). The given line pa... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) The cross-section of a regular triangular pyramid passes through the midline of the base and is perpendicular to the base. Find the area of the cross-section if the side of the base is 8 and the height of the pyramid is 12. | Answer: 18.
Solution. The section MNP passes through the midline of the base of the pyramid $MN$ and is perpendicular to the base. Therefore, the height $PH$ of the triangle $MNP$ is parallel to the height of the pyramid $DO$.
 There are two circles: one with center at point $A$ and radius 6, and another with center at point $B$ and radius 3. Their common internal tangent touches the circles at points $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at point $E$. Find $C D$, if $A E=10$. | Answer: 12
Solution. Triangles $A C E$ and $B D E$ are similar (they have vertical angles and a right angle each) with a similarity coefficient of 2. From triangle $A C E$, using the Pythagorean theorem, we find $C E=8$. Therefore, $D E=4$, and $C D=12$. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Find the largest root of the equation
$$
\left|\cos (\pi x)+x^{3}-3 x^{2}+3 x\right|=3-x^{2}-2 x^{3}
$$ | # Answer: 1
Solution. It is obvious that 1 is a root of the equation (when $x=1$, both sides of the equation are equal to zero). If $x>1$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) Find the smallest natural number that is simultaneously twice a perfect square and three times a perfect cube.
# | # Answer: 648
Solution. We have $k=3 n^{3}=2 m^{2}$. From this, the numbers $m$ and $n$ can be represented as $n=2 a, m=3 b$. After substitution, we get $4 a^{3}=3 b^{2}$. Further, we have $a=3 c, b=2 d, 9 c^{3}=d^{2}$. Here, the smallest solution is $c=1, d=3$. Then $a=3$, $b=6, n=6, m=18, k=648$. | 648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) The efficiency of an ideal heat engine is $40 \%$. What will it become if the temperature of the heater is increased by $40 \%$, and the temperature of the cooler is decreased by $40 \%$? | Answer: $\approx 74 \%$.
Solution. The efficiency of an ideal heat engine: $\eta=1-\frac{T_{X}}{T_{H}}$. That is, initially the ratio of the temperatures of the refrigerator and the heater: $\frac{T_{X}}{T_{H}}=1-0.4=0.6$. After the changes:
$$
\eta_{2}=1-\frac{0.6 T_{X}}{1.4 T_{H}}=1-\frac{0.6 \cdot 0.6}{1.4} \appro... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=5 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its im... | Answer: -10 Dptr $u \approx 7.1$ cm
Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain that the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{5}{10}=5 \mathrm{~cm} . \quad$ The power of the lens: $D=\frac{1}{d}-\frac{1}{f}=\... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) There are two circles: one with center at point $A$ and radius 5, and another with center at point $B$ and radius 15. Their common internal tangent touches the circles at points $C$ and $D$ respectively. Lines $A B$ and $C D$ intersect at point $E$. Find $C D$, if $B E=39$. | # Answer: 48
Solution. Triangles $A C E$ and $B D E$ are similar (they have vertical angles and a right angle each) with a similarity coefficient of $1 / 3$. Therefore, $A E=13$. From triangle $A C E$, using the Pythagorean theorem, we find $C E=12$. Hence, $D E=36$, and $C D=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Find the smallest root of the equation
$$
\sin (\pi x)+\tan x=x+x^{3}
$$ | Answer: 0
Solution. Obviously, 0 is a root of the equation (when $x=0$, both sides of the equation are equal to zero). If $x<0$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) Find the smallest natural number that is simultaneously twice a perfect cube and three times a perfect square.
# | # Answer: 432
Solution. We have $k=2 n^{3}=3 m^{2}$. From this, the numbers $m$ and $n$ can be represented as $n=3 a, m=2 b$. After substitution, we get $9 a^{3}=2 b^{2}$. Further, we have $a=2 c, b=3 d, 4 c^{3}=d^{2}$. Here, the smallest solution is $c=1, d=2$. Then $a=2$, $b=6, n=6, m=12, k=432$. | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) The efficiency of an ideal heat engine is $50 \%$. What will it become if the temperature of the heater is increased by $50 \%$, and the temperature of the cooler is decreased by $50 \%$? | Answer: $\approx 83 \%$
Solution. The efficiency of an ideal heat engine: $\eta=1-\frac{T_{X}}{T_{H}}$. That is, at the beginning, the ratio of the temperatures of the refrigerator and the heater: $\frac{T_{X}}{T_{H}}=1-0.5=0.5$. After the changes:
$$
\eta_{2}=1-\frac{0.5 T_{X}}{1.5 T_{H}}=1-\frac{0.5 \cdot 0.5}{1.5}... | 83 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=20 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its i... | Answer: 5 Dpt $i \approx 14.1$ cm
Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{20}{10}=20 \mathrm{~cm} . \quad$ The optical power of the lens: $D=\frac{1}{d}-\frac{1}{f... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. How many times in a day does the angle between the hour and minute hands equal exactly $17^{\circ}$? | Answer: 44.
Solution. Over the time interval from 0:00 to 12:00, the hour hand will make one complete revolution, while the minute hand will make 12 such revolutions. Therefore, during this time, the minute hand will catch up with the hour hand 11 times. Between two consecutive meetings of the hands, the angle between... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\sqrt{\frac{x-2}{11}}+\sqrt{\frac{x-3}{10}}=\sqrt{\frac{x-11}{2}}+\sqrt{\frac{x-10}{3}}
$$ | Answer: 13.
Solution. If we perform the variable substitution $x=t+13$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}
$$
Now it is clear that for $t>0$ the right side of the equation is greater, and... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let in triangle $A B C$
$$
\cos (2 \angle A-\angle B)+\sin (\angle A+\angle B)=2 \text {. }
$$
Find the side $B C$, if $A B=4$. | Answer: 2.
Solution. Each term in the left part of the equation does not exceed 1. Therefore, the equality will hold only if each of them equals 1. We solve the corresponding equations, denoting $\alpha=\angle A, \beta=\angle B:$
$$
2 \alpha-\beta=2 \pi n, n \in \mathbb{Z} ; \quad \alpha+\beta=\frac{\pi}{2}+2 \pi k, ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. How many times in a day does the angle between the hour and minute hands equal exactly $19^{\circ}$? | Answer: 44.
Solution. Over the time interval from 0:00 to 12:00, the hour hand will make one complete revolution, while the minute hand will make 12 such revolutions. Therefore, during this time, the minute hand will catch up with the hour hand 11 times. Between two consecutive meetings of the hands, the angle between... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation
$$
\sqrt{\frac{x-3}{11}}+\sqrt{\frac{x-4}{10}}=\sqrt{\frac{x-11}{3}}+\sqrt{\frac{x-10}{4}}
$$ | Answer: 14.
Solution. If we perform the variable substitution $x=t+14$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}
$$
Now it is clear that for $t>0$ the right side of the equation is greater, and... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A circle is inscribed with 2019 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 2, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers. | Answer: 6060.
Solution. Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 2$ and $y+z \geqslant 6$, we get $y \geqslant 4$. Then $x \geqslant y+2 \geqslant 6$. A number not less than 6 has been found. The remaining ... | 6060 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. The infantry column stretched out over 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked 2 km $400 \mathrm{m}$ during this time. How far did the sergeant travel during this time? | Answer: 3 km $600 \mathrm{~m}$
Solution. Let the speed of the column be $x$ km/h, and the sergeant travels $k$ times faster, i.e., at a speed of $k x$ km/h. To reach the end of the column, Kim traveled $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction, $t_{2}=\frac{1}{k x+x}$ hours (meeting he... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A circle is inscribed with 1001 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 4, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers. | Answer: 3009.
Solution. Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 4$ and $y+z \geqslant 6$, we get $y \geqslant 5$. Then $x \geqslant y+4 \geqslant 9$. A number not less than 9 has been found. The remaining ... | 3009 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) A beam of light with a diameter of $d_{1}=5 \mathrm{~cm}$ falls on a thin diverging lens with an optical power of $D_{p}=-6$ Diopters. On a screen positioned parallel to the lens, a bright spot with a diameter of $d_{2}=20 \mathrm{~cm}$ is observed. After replacing the thin diverging lens with a thin con... | # Answer: 10 Dptr
Solution. The optical scheme corresponding to the condition:
(3 points)
The path of the rays after the diverging lens is shown in black, and after the converging lens in ... | 10 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 6 (10 points)
A cylinder with a mass of $M=1$ kg was placed on rails inclined at an angle $\alpha=30^{\circ}$ to the horizontal (the side view is shown in the figure). What is the minimum mass $m$ of the load that needs to be attached to the thread wound around the cylinder so that it starts rolling upward... | # Solution and evaluation criteria:
The moment of forces relative to the point of contact of the cylinder with the plane: $m g($ ( $-R \sin \alpha)=M g R \sin \alpha$ (5 points) $m\left(1-\f... | 1 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 7 (10 points)
In the electrical circuit shown in the diagram, the resistances of the resistors are $R_{1}=10$ Ohms and $R_{2}=30$ Ohms. An ammeter is connected to points A and B in the circuit. When the polarity of the current source is reversed, the ammeter readings change by one and a half times. Determi... | # Solution and Evaluation Criteria:
When the positive terminal of the power supply is connected to point $A$, the current flows only through resistor $R_{2}$, and in this case: $I_{1}=\frac{\varepsilon}{R_{2}+r}$.
## (3 points)
When the polarity is reversed, the current flows only through resistance $R_{1}$, and:
$... | 30 | Other | math-word-problem | Yes | Yes | olympiads | false |
2. The area of triangle $A B C$ is 1. On the rays $A B, B C$, $C A$, points $B^{\prime}, C^{\prime}, A^{\prime}$ are laid out respectively, such that
$$
B B^{\prime}=2 A B, \quad C C^{\{\prime}=3 B C, \quad A A^{\prime}=4 C A .
$$
Calculate the area of triangle $A^{\prime} B^{\prime} C^{\prime}$. | Answer: 36.
Solution. We will solve the problem in a general form, assuming that
$$
B B^{\prime}=q A B, \quad C C^{\{\prime}=r B C, \quad A A^{\prime}=p C A .
$$
We will calculate the ar... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation
$$
\sqrt{\frac{x-2}{11}}+\sqrt{\frac{x-3}{10}}+\sqrt{\frac{x-4}{9}}=\sqrt{\frac{x-11}{2}}+\sqrt{\frac{x-10}{3}}+\sqrt{\frac{x-9}{4}}
$$ | Solution. If we perform the variable substitution $x=t+13$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}+\sqrt{\frac{t}{9}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}+\sqrt{\frac{t}{4}+1}
$$
Now it is clear that for $t>0$ the right-hand si... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The engine of a car traveling at a speed of $v_{0}=72 \mathrm{km} / \mathbf{h}$ operates with a power of $P=50$ kW. Determine the distance from the point of engine shutdown at which the car will stop, if the resistance force is proportional to the car's speed. The mass of the car is m=1500 kg. (15
## points) | Answer: $240 m$
Solution. The power of the engine: $P=F v=F_{\text {conp }} v_{0}=\alpha v_{0}^{2}$, that is, the coefficient of resistance to the car's movement: $\alpha=\frac{P}{v_{0}^{2}}$ (3 points). The projection of the second law of Newton, for a small time interval, when moving with the engine off: $m \frac{\D... | 240 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
2. The area of triangle $A B C$ is 1. On the rays $A B, B C$, $C A$, points $B^{\prime}, C^{\prime}, A^{\prime}$ are laid out respectively, such that
$$
B B^{\prime}=A B, \quad C C^{\prime}=2 B C, \quad A A^{\prime}=3 C A
$$
Calculate the area of triangle $A^{\prime} B^{\prime} C^{\prime}$. | Answer: 18.
Solution. We will solve the problem in a general form, assuming that
$$
B B^{\prime}=q A B, \quad C C^{\prime}=r B C, \quad A A^{\prime}=p C A .
$$
We will calculate the area... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation
$$
\sqrt{\frac{x-3}{11}}+\sqrt{\frac{x-4}{10}}+\sqrt{\frac{x-5}{9}}=\sqrt{\frac{x-11}{3}}+\sqrt{\frac{x-10}{4}}+\sqrt{\frac{x-9}{5}}
$$ | Answer: 14.
Solution. If we perform the variable substitution $x=t+14$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}+\sqrt{\frac{t}{9}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}+\sqrt{\frac{t}{4}+1} .(*)
$$
Now it is clear that for $t>0$... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. An ideal gas was expanded in such a way that during the process, the pressure of the gas turned out to be directly proportional to its volume. As a result, the gas heated up by $\Delta T=100^{\circ} \mathrm{C}$, and the work done by the gas was $A=831$ J. Determine the amount of substance that participated in this p... | Answer: 2 moles
Solution. From the condition $p=\alpha V$ (3 points). The work of the gas is equal to the area under the graph of the given process, constructed in coordinates $p-V$:
$A=\frac{p_{1}+p_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha V_{1}+\alpha V_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha}{2... | 2 | Other | math-word-problem | Yes | Yes | olympiads | false |
7. Two small balls with charges $Q=-20 \cdot 10^{-6}$ C and $q=50 \cdot 10^{-6}$ C are located at the vertices $A$ and $B$ of a mountain slope (see figure). It is known that $AB=2$ m, $AC=3$ m. The masses of the balls are the same and equal to $m=200$ g each. At the initial moment of time, the ball with charge $q$ is r... | Answer: $5 \mathrm{M} / \mathrm{c}$
Solution. The law of conservation of energy for this situation:
$k \frac{Q q}{A B}+m g \cdot A B=\frac{m v^{2}}{2}+k \frac{Q q}{B C}(5$ points $)$.
As a result, we get: $v=\sqrt{\frac{2 k Q q}{m \cdot A B}+2 \cdot g \cdot A B-\frac{2 k Q q}{m \cdot B C}}=\sqrt{\frac{2 \cdot 9 \cdo... | 5 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem No. 5 (15 points)
The system shown in the figure is in equilibrium. It is known that the uniform rod $AB$ and the load lying on it have the same mass $m=10$ kg, and the load is located at a distance of one quarter of the length of the rod from its left end. Determine the mass $m$ of the second load suspended f... | Answer: 100 kg
# Solution and grading criteria:
Diagram with forces correctly placed.
Tension force in the thread: $T=\frac{M g}{2}$
The lever rule, written relative to point v:
$m g \cdot \frac{1}{2} A B+m g \cdot \frac{3}{4} A B+T \cdot \frac{3}{4} A B=T \cdot A B$
As a result, we get: $M=10 m=100$ kg.
A pers... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem No. 5 (15 points)
The system shown in the figure is in equilibrium. It is known that the uniform rod $AB$ and the load lying on it have the same mass $m=10$ kg. The load is located exactly in the middle of the rod. The thread, passing over the pulleys, is attached to one end of the rod and at a distance of one... | # Solution and Evaluation Criteria:
Diagram with forces correctly placed.
Tension force in the string: $T=\frac{M g}{2}$
The lever rule, written relative to point v:
$m g \cdot \frac{1}{2} A B+m g \cdot \frac{1}{2} A B+T \cdot \frac{3}{4} A B=T \cdot A B$
As a result, we get: $M=8 m=80$ kg. | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In triangle $A B C$, the median $B K$ is twice as small as side $A B$ and forms an angle of $32^{\circ}$ with it. Find the angle $A B C$. | Answer: $106^{\circ}$.
Solution. Let $K$ be the midpoint of segment $B D$. Then $A B C D$ is a parallelogram. In triangle $A B D$, we have the equality of sides $A B$ and $B D$. Therefore,
$$
\angle B D A=\frac{1}{2}\left(180^{\circ}-32^{\circ}\right)=74^{\circ}
$$
Angles $A D B$ and $C B D$ are equal as alternate i... | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In triangle $A B C$, the median $B N$ is twice as short as side $A B$ and forms an angle of $20^{\circ}$ with it. Find the angle $A B C$. | Answer: $100^{\circ}$.
Solution. Let $N$ be the midpoint of segment $B D$. Then $A B C D-$ is a parallelogram. In triangle $A B D$, we have the equality of sides $A B$ and $B D$. Therefore,
$$
\angle B D A=\frac{1}{2}\left(180^{\circ}-20^{\circ}\right)=80^{\circ} \text {. }
$$
Angles $A D B$ and $C B D$ are equal as... | 100 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. What is the greatest length that a closed, non-self-intersecting broken line can have, which runs along the grid lines of a $6 \times 10$ cell field?
# | # Answer: 76.
Solution. We will color the nodes of the grid in a checkerboard pattern, with black and white colors. The length of a closed non-self-intersecting broken line is equal to the number of nodes it passes through. Each segment of the broken line connects a black and a white node. When traversing the broken l... | 76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. 100 balls of the same mass move along a trough towards a metal wall with the same speed. After colliding with the wall, a ball bounces off it with the same speed. Upon collision of two balls, they scatter with the same speed. (The balls move only along the trough). Find the total number of collisions between the bal... | Answer: 4950.
Solution. We will assume that each ball has a flag. Imagine that upon collision, the balls exchange flags. Then each flag flies to the wall at a constant speed, and after hitting the wall, it flies in the opposite direction. The number of collisions between the balls is equal to the number of flag exchan... | 4950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Let's call a number small if it is a 10-digit number and there does not exist a smaller 10-digit number with the same sum of digits. How many small numbers exist | Solution. Answer: 90.
It is clear that the sum of the digits of a 10-digit number can take any value from 1 to 90 inclusive. For each of the 90 possible sums of the digits, there is a unique smallest 10-digit number with such a sum of digits. Therefore, there are 90 small numbers.
## Criteria
One of the largest suit... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. On the board, there are $N$ natural numbers, where $N \geqslant 5$. It is known that the sum of all the numbers is 80, and the sum of any five of them is no more than 19. What is the smallest value that $N$ can take? | Solution. Answer: 26.
The condition of the problem is equivalent to the sum of the five largest numbers not exceeding 19, and the sum of all numbers being 80. Note that among the five largest numbers, there must be a number not greater than 3 (otherwise, if all of them are not less than 4, their sum is not less than $... | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1010 of them such that the sum of any two does not equal 2021 or 2022. How many
ways are there to do this | Solution. Answer: $\frac{1011 \cdot 1012}{2}=511566$.
Let's write our numbers in the following order: $2021,1,2020,2,2019,3, \ldots, 1012,1010,1011$.
Notice that if two numbers sum to 2021 or 2022, they stand next to each other in this sequence. Our task is reduced to the following: 2021 objects (for convenience, let... | 511566 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Petya was given several physics problems and several math problems for homework. All the problems solved by Petya constitute $5 \%$ of the total number of physics problems and $20 \%$ of the total number of math problems. What percentage of the total number of problems did Petya solve? | Solution. Answer: $4 \%$.
Let Pete solve $N$ problems. This constitutes $5 \%$ ( $\frac{1}{20}$ of) the physics problems, so the total number of physics problems was $20 \mathrm{~N}$. Similarly, the number of math problems was $5 N$, so the total number of problems assigned was $20 N + 5 N = 25 N$. The problems solved... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a two-digit number, each digit was increased by 2 or by 4 (different digits could be increased by different numbers), as a result of which the number increased fourfold. What could the original number have been? Find all possible options and prove that there are no others. | Solution. Answer: 14.
Let $x$ be the original number, then the resulting number equals $4 x$. In this case, after increasing two digits, the number itself was increased by 22, 24, 42, or 44. This results in four cases:
- $4 x-x=22$
- $4 x-x=24$
- $4 x-x=42$
- $4 x-x=44$.
Among them, only the third one fits, correspo... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. We will call a natural number odd-powered if all its prime divisors enter its factorization with an odd exponent. What is the maximum number of odd-powered numbers that can occur consecutively?
## Answer: 7. | Solution. Note that among any eight consecutive natural numbers, there will definitely be a number that is divisible by 4 but not by 8. Then the number 2 will appear in its factorization to the second power.
Example of seven odd-power numbers: $37,38,39,40,41,42,43$.
Problem 4/1. On side $A C$ of triangle $A B C$, a ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Grandfather Frost had 120 chocolate candies and 200 jelly candies. At the morning performance, he gave candies to the children: each child received one chocolate candy and one jelly candy. Counting the candies after the performance, Grandfather Frost found that there were three times as many jelly candies le... | Answer: 80.
Solution. Let the total number of children be $x$, then after the morning party, Grandfather Frost had $120-x$ chocolate candies and $200-x$ jelly candies left. Since there were three times as many jelly candies left as chocolate candies, we get the equation $3 \cdot(120-x)=200-x$. Solving this, we get $x=... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. On an island, there live knights who always tell the truth, and liars who always lie. One day, 30 inhabitants of this island sat around a round table. Each of them said one of two phrases: "My left neighbor is a liar" or "My right neighbor is a liar." What is the smallest number of knights that can be at the ta... | # Answer: 10.
Solution. Next to each liar, there must be at least one knight (otherwise, if liars sit on both sides of a liar, the statement made by the liar would definitely be true). Therefore, among any three consecutive residents, there is at least one knight. If we divide all the people sitting at the table into ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Five different natural numbers are written in a circle on the board. Each of them, Petya divided by the next one in the clockwise direction, and then wrote down the 5 resulting numbers (not necessarily integers) on a piece of paper. Can the sum of the 5 numbers on the paper be an integer? | Answer: Yes, it can.
Solution. For example, for the numbers $1,2,4,8,16$, after division, the sum is $1 / 2+1 / 2+1 / 2+1 / 2+16=18$.
## Criteria
One suitable criterion is used:
7 p. Any complete solution to the problem.
0 p. There is only the correct answer, but the correct example is not provided. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Carlson has three boxes, each containing 10 candies. One box is labeled with the number 4, another with 7, and the third with 10. In one operation, Carlson sequentially performs the following two actions:
- he takes from any box a number of candies equal to the number written on it;
- he eats 3 of the taken... | Answer: 27.
Solution. Since at each step, Karlson eats 3 candies, the total number of candies eaten is divisible by 3. We will prove that it does not exceed 27; for this, it is sufficient to show that it cannot equal 30, that is, Karlson cannot eat all the candies. Indeed, if this were possible, then before the last o... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The incircle of triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $I_{A}$ is the excenter of the excircle opposite side $BC$ of triangle $ABC$, and points $K$ and $L$ are the midpoints of segments $DI_{A}$ and $EI_{A}$ respectively. Lines $BK$ and $CL$ intersect at point $F$, which... | 5. The circle inscribed in triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $I_{A}$ is the center of the excircle opposite side $BC$ of triangle $ABC$, and points $K$ and $L$ are the midpoints of segments $DI_{A}$ and $EI_{A}$ respectively. Lines $BK$ and $CL$ intersect at point $F$,... | 130 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.2. Masha and the Bear ate a basket of raspberries and 40 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate both raspberries and pies 3 times faster than Masha. How many pies did Masha eat, if they ate t... | Answer: 4 pies. Solution: The bear ate his half of the raspberries three times faster than Masha. This means Masha ate pies for three times less time than the bear. Since she eats three times slower, she ate 9 times fewer pies than the bear. Dividing the pies in the ratio of $9: 1$, we see that Masha got the 10th part,... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.3. At a round table sit 10 elves, each with a basket of nuts in front of them. Each was asked, "How many nuts do your two neighbors have together?" and, going around the circle, the answers received were 110, 120, 130, 140, 150, 160, 170, 180, 190, and 200. How many nuts does the elf who answered 160 have? | Answer: 55. Solution. We will call the elves 1-m, 2-m, etc., in the order of receiving answers. The odd-numbered ones sit every other. In their answers, the number of nuts of each even number is counted twice, so the sum $110+130+150+170+190=750$ is equal to twice the number of nuts of all even numbers. Therefore, the ... | 55 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. At a party, 24 people gathered. A guest is considered an introvert if they have no more than three acquaintances among the other guests. It turned out that each guest has at least three acquaintances who are introverts. How many introverts could there have been at the party? Provide all possible answers an... | # Answer: 24.
Solution. Let $a$ be the number of pairs of introverts who know each other, and $b$ be the number of pairs of acquaintances where one of the pair is an introvert. Each person who came to the party is included in at least three pairs, since they know at least three other introverts, and pairs of introvert... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. It is known that the number 400000001 is the product of two prime numbers $p$ and $q$. Find the sum of the natural divisors of the number $p+q-1$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: 45864.
Solution. The number $n=400000001$ can be written in the following form
\[
\begin{aligned}
n & =4 \cdot 10^{8}+1=4 \cdot 10^{8}+4 \cdot 10^{4}+1-4 \cdot 10^{4}= \\
& =\left(2 \cdot 10^{4}+1\right)^{2}-\left(2 \cdot 10^{2}\right)^{2}=\left(2 \cdot 10^{4}+2 \cdot 10^{2}+1\right)\left(2 \cdot 10^{4}-2 \cd... | 45864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Alexei wrote several consecutive natural numbers on the board. It turned out that only two of the written numbers have a digit sum divisible by 8: the smallest and the largest. What is the maximum number of numbers that could have been written on the board? | Answer: 16.
Example: numbers from 9999991 to 10000007.
Evaluation. We will prove that there cannot be more than 16 numbers. Let's call a number $x$ good if the sum of its digits is divisible by 8. Among the listed numbers, there must be a number $x$ ending in 0, otherwise there are too few numbers. The number $x$ is ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. For positive numbers $x$ and $y$, it is known that
$$
\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}=1
$$
What values can the product $x y$ take? List all possible options and prove that there are no others. | Answer: 1.
Solution. Note that for each positive $y$ the function
$$
f_{y}(x)=\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}
$$
strictly decreases on the ray $(0 ;+\infty)$, since the denominators of all three fractions increase. Therefore, the function $f_{y}$ takes each value no more than once. Moreover, ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. Through the point with coordinates $(9,9)$, lines (including those parallel to the coordinate axes) are drawn, dividing the plane into angles of $9^{\circ}$. Find the sum of the x-coordinates of the points of intersection of these lines with the line $y=10-x$. | Answer: 190. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(9,9)$, 20 lines are drawn, the line $y=20-x$ intersects 19 of them. For each point on the line $y=20-x$, the sum of the coordinates is 20, so the total su... | 190 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Answer: 4.
Solution. Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. The miserly knight has 5 chests of gold: the first chest contains 1001 gold coins, the second - 2002, the third - 3003, the fourth - 4004, the fifth - 5005. Every day, the miserly knight chooses 4 chests, takes 1 coin from each, and places them in the remaining chest. After some time, there were no coins l... | Answer: in the fifth.
Solution. Let's look at the remainders of the number of coins in the chests when divided by 5. Each day, the number of coins in each chest either decreases by 1 or increases by 4, which means the remainder when divided by 5 always decreases by 1 (if it was 0, it will become 4). Since after some t... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. On the sides $AB$ and $AC$ of triangle $ABC$, points $X$ and $Y$ are chosen such that $\angle A Y B = \angle A X C = 134^{\circ}$. On the ray $YB$ beyond point $B$, point $M$ is marked, and on the ray $XC$ beyond point $C$, point $N$ is marked. It turns out that $MB = AC$ and $AB = CN$. Find $\angle MAN$.
... | Solution. Note that $\angle A C N=\angle C A X+\angle A X C=\angle B A Y+\angle A Y B=\angle A B M$. Adding
Fig. 1: to the solution of problem $?$ ?
## Kurchatov School Olympiad in Mathemat... | 46 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A lame rook makes moves alternating between one and two squares, with the direction of the move being freely chosen (in any of the four directions). What is the maximum number of cells on a $6 \times 6$ board it can visit, if visiting the same cell twice is prohibited, but the starting cell and the first m... | Answer: 34.
Solution. An example for 34 cells is shown in Fig. ??. The visited cells are numbered from 1 to 34.
| 5 | 2 | 4 | 3 | 13 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 6 | 1 | 7 | 8 | 14 | 9 |
| 28 | 31 | | 34 | 12 | 11 |
| 27 | 32 | | 33 | 15 | 16 |
| 29 | 30 | 20 | 19 | 21 | 18 |
| 26 | 25... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In a five-digit number, each digit was increased by 2 or by 4 (different digits could be increased by different numbers), as a result of which the number increased fourfold. What could the original number have been? Find all possible options and prove that there are no others. | Solution. Answer: 14074.
Let the original five-digit number be denoted by $N$, then the resulting number is equal to $4N$. Their difference is $3N$ and is a five-digit number consisting of 2 and 4. This difference $3N$ is not less than $3 \cdot 10000$, so it starts with a four.
By the divisibility rule for 3, the sum... | 14074 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1010 of them such that the sum of any two does not equal 2021 or 2022. How many ways are there to do this? | Solution. Answer: $\frac{1011 \cdot 1012}{2}=511566$.
Let's write our numbers in the following order: $2021,1,2020,2,2019,3, \ldots, 1012,1010,1011$. Notice that if two numbers sum to 2021 or 2022, they stand next to each other in this sequence. Our task is reduced to the following: 2021 objects (for convenience, let'... | 511566 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Petya has a $3 \times$ 3 table. He places chips in its cells according to the following rules:
- no more than one chip can be placed in each cell;
- a chip can be placed in an empty cell if the corresponding row and column already contain an even number of chips (0 is considered an even number).
What is th... | Solution. Answer: 9.
Let $a, b, c$ denote the left, middle, and right columns of the table, and $1, 2, 3$ denote the bottom, middle, and top rows of the table. Petya can fill all 9 cells of the table with chips, for example, in the following order: $a 1, b 2, c 3, a 2, b 3, c 1, a 3, b 1, c 2$. Obviously, it is not po... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1011 of them such that the sum of any two does not equal 2021 or 2022. How many ways are there to do this? | Solution. Answer: 1.
Among the numbers from 1 to 2021, we can select 1010 non-overlapping pairs of numbers that sum to 2022. Specifically: $(1,2021),(2,2020), \ldots,(1009,1013),(1010,1012)$; only the number 1011 remains unpaired.
According to the condition, from each pair, we can choose no more than one number. Ther... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. In the Kurchatov School, exactly 2 people sit at each desk. It is known that exactly $70 \%$ of the boys have a boy as a desk partner, and exactly $40 \%$ of the girls have a girl as a desk partner. How many times more boys are there than girls? | Answer: 2 times.
Solution. Let the number of boys be $x$, and the number of girls be $y$. Note that $30\%$ of the boys sit at desks with girls and $60\%$ of the girls sit at desks with boys. Since exactly 2 people sit at each desk, then $0.3 x = 0.6 y$, from which $x = 2 y$. Thus, there are 2 times more boys than girl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Find the number of ways to color all natural numbers from 1 to 20 in blue and red such that both colors are used and the product of all red numbers is coprime with the product of all blue numbers. | Answer: $2^{6}-2=62$ ways.
Solution. Note that all even numbers must be of the same color. Since among them are the numbers 6, 10, and 14, the numbers divisible by 3, 5, and 7 must be of the same color. The remaining numbers are $1, 11, 13, 17$, and 19. Note that they can be distributed in any way among the two colors... | 62 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. On the board, the numbers $2,3,5, \ldots, 2003,2011,2017$ are written, i.e., all prime numbers not exceeding 2020. In one operation, two numbers $a, b$ can be replaced by the largest prime number not exceeding $\sqrt{a^{2}-a b+b^{2}}$. After several operations, only one number remains on the board. What is t... | Answer: The maximum value is 2011.
Solution. Note that if $a$ $2011^{2}$, therefore, as a result of this operation, the number 2011 will appear on the board.
Remark. The inequality $a<\sqrt{a^{2}-a b+b^{2}}<b$ for $a<b$ can be proven geometrically. The matter is that in a triangle with sides $a$ and $b$ and an angle ... | 2011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On a plane, an overlapping square and a circle are drawn. Together they occupy an area of 2018 cm². The area of intersection is 137 cm². The area of the circle is 1371 cm². What is the perimeter of the square? | Answer: 112 cm.
The area of the part of the circle outside the square is $1371-137=1234$ cm $^{2}$, therefore, the area of the square can be expressed by the formula $2018-1234=784 \mathrm{~cm}^{2}$. In conclusion, the length of the side of the square is $\sqrt{784}=28$ cm, and its perimeter is 112 cm.
## Criteria
+... | 112 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Gosha entered a natural number into the calculator. Then he performed the following operation, consisting of two actions, three times: first, he extracted the square root, and then took the integer part of the obtained number. In the end, he got the number 1. What is the largest number that Gosha could have ... | Answer: 255.
Solution. Suppose he entered a number not less than 256. Then after the first operation, he would get a number not less than $[\sqrt{256}]=16$, after the second - not less than $[\sqrt{16}]=4$, after the third - not less than $[\sqrt{4}]=2$, which is a contradiction.
Let's assume Gosha entered the number... | 255 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 3. Lёnya has cards with digits from 1 to 7. How many ways are there to glue them into two three-digit numbers (one card will not be used) so that each of them is divisible by 9? | Answer: 36.
Solution. The sum of the digits in each number is divisible by 9, which means the overall sum of the used digits is also divisible by 9. The sum of all given digits $1+2+\ldots+7$ is 28. If we remove the digit 1, the remaining sum is 27, which is divisible by 9; removing any other digit cannot result in a ... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Five numbers $2,0,1,9,0$ are written in a circle on the board in the given order clockwise (the last zero is written next to the first two). In one move, the sum of each pair of adjacent numbers is written between them. For example, such an arrangement of numbers (on the right) will be after the first move:
... | Answer: $8 \cdot 3^{5}=1944$.
Solution. Let's see how the numbers between the first and second zero (which Polina will count at the end) change. On each move, each "old" number is included as an addend in two "new" numbers. This means that if all the "new" numbers are added together, each "old" number will be summed t... | 1944 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Dodson, Williams, and their horse Bolivar want to get from city A to city B as quickly as possible. Along the road, there are 27 telegraph poles that divide the entire journey into 28 equal segments. Dodson takes 9 minutes to walk one segment, Williams takes 11 minutes, and either of them can ride Bolivar to... | Solution. Let the distance from A to B be taken as a unit, and time will be measured in minutes. Then Dodson's speed is $1 / 9$, Williams' speed is $1 / 11$, and Bolivar's speed is $-1 / 3$.
Let the desired post have the number $k$ (i.e., the distance from city A is $k / 28$). Then Dodson will arrive in time
$$
\frac... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. In the cells of an $8 \times 8$ chessboard, there are 8 white and 8 black chips such that no two chips are in the same cell. Additionally, no column or row contains chips of the same color. For each white chip, the distance to the black chip in the same column is calculated. What is the maximum value that th... | Answer: 32
Solution. An example of chip placement where the sum of distances is 32 is shown in the figure:
We will prove that the sum of distances cannot be greater than 32. For this, consid... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In each cell of a $15 \times 15$ table, a number $-1, 0$, or $+1$ is written such that the sum of the numbers in any row is non-positive, and the sum of the numbers in any column is non-negative. What is the smallest number of zeros that can be written in the cells of the table? | Answer: 15.
Evaluation. Since the sum of the numbers in all rows is non-positive, the sum of all numbers in the table is also non-positive. On the other hand, if the sum of the numbers in any column is non-negative, then the sum of all numbers in the table is non-negative. Therefore, the sum of all numbers in the tabl... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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