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3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $50^{\circ}$. Find the angle $A B C$. | Answer: 115.
Solution. Complete triangle $A B C$ to parallelogram $A B C D$.
Then, $D B=2 N B=A B$. Therefore, triangle $A B D$ is isosceles and $\angle A D B=\frac{180^{\circ}-50^{\circ}}{... | 115 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Determine the angle between the hour and minute hands at the moment when they show 13 hours and 20 minutes. | Answer: $80^{\circ}$.
Solution. The minute hand has moved away from twelve by $\frac{20}{60} \cdot 360=120^{0}$. The hour hand has moved away from twelve by $\frac{1}{12} \cdot 360+\frac{20}{60} \cdot \frac{1}{12} \cdot 360=40^{\circ}$. The angle between the hands $120^{\circ}-40^{\circ}=80^{\circ}$. | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, six points on $A B$ and seven - on $B C$. How many different triangles exist with vertices at the chosen points? | Answer: 231.
Solution. To form a triangle, one needs to choose two points on one side and one point on another. There are 6 ways to choose the first point on $AB$, 5 ways to choose the second, and since the triangle does not change with the permutation of its vertices, we divide $6 \cdot 5$ by 2. Thus, $\frac{6 \cdot ... | 231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $40^{\circ}$. Find the angle $A B C$. | Answer: 110.
## Solution.
Extend triangle $ABC$ to parallelogram $ABCD$. Then, $DB=2NB=AB$. Therefore, triangle $ABD$ is isosceles and $\angle ADB=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Determine the angle between the hour and minute hands at the moment when they show 15 hours and 40 minutes. | Answer: $130^{\circ}$.
Solution. The minute hand has moved away from twelve by $\frac{40}{60} \cdot 360=240^{\circ}$. The hour hand has moved away from twelve by $\frac{3}{12} \cdot 360+\frac{40}{60} \cdot \frac{1}{12} \cdot 360=110^{\circ}$. The angle between the hands: $240^{\circ}-110^{\circ}=130^{\circ}$. | 130 | Geometry | 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 |
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 |
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 |
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 |
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 $DO$.
 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:
$$
\eta_{2}=1-\frac{0.6 T_{X}}{1.4 T_{H}}=1-\frac{0.6 \cdot 0.6}{1.4} \appro... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) There are two circles: one with center at point $A$ and radius 5, and another with center at point $B$ and radius 15. Their common internal tangent touches the circles at points $C$ and $D$ respectively. Lines $A B$ and $C D$ intersect at point $E$. Find $C D$, if $B E=39$. | # Answer: 48
Solution. Triangles $A C E$ and $B D E$ are similar (they have vertical angles and a right angle each) with a similarity coefficient of $1 / 3$. Therefore, $A E=13$. From triangle $A C E$, using the Pythagorean theorem, we find $C E=12$. Hence, $D E=36$, and $C D=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) Find the smallest natural number that is simultaneously twice a perfect cube and three times a perfect square.
# | # Answer: 432
Solution. We have $k=2 n^{3}=3 m^{2}$. From this, the numbers $m$ and $n$ can be represented as $n=3 a, m=2 b$. After substitution, we get $9 a^{3}=2 b^{2}$. Further, we have $a=2 c, b=3 d, 4 c^{3}=d^{2}$. Here, the smallest solution is $c=1, d=2$. Then $a=2$, $b=6, n=6, m=12, k=432$. | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) The efficiency of an ideal heat engine is $50 \%$. What will it become if the temperature of the heater is increased by $50 \%$, and the temperature of the cooler is decreased by $50 \%$? | Answer: $\approx 83 \%$
Solution. The efficiency of an ideal heat engine: $\eta=1-\frac{T_{X}}{T_{H}}$. That is, at the beginning, the ratio of the temperatures of the refrigerator and the heater: $\frac{T_{X}}{T_{H}}=1-0.5=0.5$. After the changes:
$$
\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 |
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 |
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 |
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 |
# 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 |
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 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 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 |
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 |
6.3. 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 |
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 7.4. 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 |
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 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 |
Problem 5. There is a deck of 1024 cards, each with a different set of digits from 0 to 9, and all sets are distinct (including an empty card). We will call a set of cards complete if each digit from 0 to 9 appears exactly once on them.
Find all natural $k$ for which there exists a set of $k$ cards with the following ... | Solution. Answer: 512.
For each card, consider another card that complements it to a complete set (for example, for the card 3679, such a card would be 012458). It is clear that all 1024 cards can be divided into 512 non-overlapping pairs of cards that complement each other to a complete set. Next, we will prove that ... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In a certain company, there are 100 shareholders, and any 66 of them collectively own at least $50\%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? (The percentage of shares in the company owned by a shareholder can be non-integer.) | Answer: $25 \%$.
Solution. Consider any shareholder $A$. Divide the other 99 shareholders into three groups $B, C, D$ of 33 shareholders each. By the condition, $B$ and $C$ together have at least $50 \%$ of the company's shares, similarly for $C$ and $D$, and for $B$ and $D$. Adding all this up and dividing by two, we... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. In triangle $A B C$, the angle at vertex $B$ is $120^{\circ}$, point $M$ is the midpoint of side $A C$. Points $E$ and $F$ are chosen on sides $A B$ and $B C$ respectively such that $A E=E F=F C$. Find $\angle E M F$. | Answer: $90^{\circ}$.
Solution. Note that the sum of angles $A$ and $C$ is $60^{\circ}$. Let $A E=E F=F C=u$.
On the line $E M$, place a point $G$ such that point $M$ is the midpoint of segment $E G$. Triangles $A M E$ and $C M G$ are equal by two sides and the angle between them, so $\angle M C G=\angle M A E=\angle... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A cube with a side of 5 is made up of 125 smaller cubes with a side of 1. How many small cubes does a plane perpendicular to one of the cube's diagonals and passing through its midpoint intersect? | Answer: 55.
Let's introduce a coordinate system such that the cube is located in the first octant (the set of points with non-negative coordinates) and the mentioned diagonal extends from the origin $O$. The midpoint of the cube's diagonal has coordinates $(5 / 2, 5 / 2, 5 / 2)$, so the specified plane is given by the... | 55 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all pos... | Answer: 30.
Solution. Consider any arbitrary quartet of consecutive people. If there were at least 3 knights in it, the leftmost of them would definitely lie, which is impossible. If there were at least 3 liars, the leftmost of them would definitely tell the truth, which is also impossible. Therefore, in every quartet... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live knights who always tell the truth and liars who always lie. The population of the island is 1000 people, distributed across 10 villages (with no fewer than two people in each village). One day, every islander claimed that all their fellow villagers are liars. How many liars live on the islan... | Answer: 990.
In one village, at least two knights cannot live, because otherwise the knights would lie. Also, in the village, they cannot all be liars, since then these liars would tell the truth. Therefore, in each village there is exactly one knight, and there are 10 knights in total, and 990 liars. | 990 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Petya climbed up a moving upward escalator, counting 75 steps, and then descended the same escalator (i.e., moving against the direction of the escalator), counting 150 steps. During the descent, Petya walked three times faster than during the ascent. How many steps are there on the stopped escalator? | Answer: 120.
For convenience, let's introduce a unit of time during which Petya took one step while ascending the escalator. We will measure all speeds in steps per unit of time. Petya's speed while ascending is 1 step per unit of time, and his speed while descending is 3 steps per unit of time. Let the speed of the e... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For a convex quadrilateral $A B C D$, it is known that $A B=B C=C A=$ $C D, \angle A C D=10^{\circ}$. A circle $\omega$ is circumscribed around triangle $B C D$ with center $O$. Line $D A$ intersects circle $\omega$ at points $D$ and $E$. Find the measure of angle $E O A$, express your answer in degrees. | Answer: 65.
Fig. 2: to the solution of problem 3.
Fig. 2. Since triangle $A D C$ is isosceles, and we know the angle at its vertex, the angles at its base are $\angle D A C = \angle C D A =... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Anya writes a natural number, and Boris replaces one of its digits with a digit differing by 1. What is the smallest number Anya should write to ensure that the resulting number is guaranteed to be divisible by 11? | Answer: 909090909.
According to the divisibility rule for 11, the remainder of a number when divided by 11 is the same as the remainder of the alternating sum of its digits when divided by 11. Therefore, changing a digit by 1 also changes the remainder by 1 when divided by 11. This means the original number gives a re... | 909090909 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles. | Answer: 28.
Fig. 3: Solution to problem 5.
Example: The arrangement of 28 tiles can be seen in Fig. 3.
Estimation. Cells not covered by tiles will be called empty. First, let's prove that no... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On the parade ground, 2018 soldiers are lined up in one row. The commander can order either all soldiers standing in even positions or all soldiers standing in odd positions to leave the formation. After this order, the remaining soldiers close up into one row. In how many ways can the commander issue a series of 8 ... | Answer: 30.
Let's add 30 imaginary people to the end of the line. We will number all the people in the line from 0 to 2047. We will write all these numbers in binary using 11 digits. This will result in sequences from 00000000000 to 11111111111. The imaginary soldiers correspond to numbers from 11111100010 to 11111111... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. When a five-digit number is multiplied by 9, the result is a number composed of the same digits but in reverse order. Find the original number. | Answer: 10989.
Let $\overline{a b c d e}$ be the original number. The condition is written as the equation $9 \cdot \overline{a b c d e}=\overline{e d c b a}$. Note that $a=1$, because if $a \geqslant 2$, then $9 \cdot \overline{a b c d e} \geqslant 9 \cdot 20000>100000>\overline{e d c b a}$.
We have $9 \cdot \overli... | 10989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. We will call a pair of numbers magical if the numbers in the pair add up to a multiple of 7. What is the maximum number of magical pairs of adjacent numbers that can be obtained by writing down all the numbers from 1 to 30 in a row in some order? | Answer: 26.
Example: $1,6,8,13,15,20,22,27,29,2,5,9,12,16,19,23,26,30,3,4$, $10,11,17,18,24,25,7,14,21,28$. It is not hard to see that in this sequence, only the pairs $(29,2),(30,3),(25,7)$ are not magical.
Evaluation: Suppose it is possible to make no less than 27 pairs magical. Then there would be no more than two... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Dima has 25 identical bricks of size $5 \times 14 \times 17$. Dima wants to build a tower from all his bricks, each time adding one more brick on top (each new brick adds 5, 14, or 17 to the current height of the tower). We will call a number $n$ constructible if Dima can build a tower of height exactly $n$. How man... | Answer: 98.
In essence, we need to find the number of different towers by height that can be built from the given set of bricks.
Mentally reduce the length, width, and height of each brick by 5: from $5 \times 14 \times 17$ to $0 \times 9 \times 12$. Then the total height of a potential tower will decrease by $25 \cd... | 98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles. | Answer: 28.
Fig. 1: to the solution of problem 6.
Example: the arrangement of 28 tiles can be seen in Fig. 1.
Estimate. Cells not covered by tiles will be called empty. To begin with, let's... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. Through the point with coordinates $(10,9)$, lines (including those parallel to the coordinate axes) have been drawn, dividing the plane into angles of $10^{\circ}$. Find the sum of the x-coordinates of the points of intersection of these lines with the line $y=101-x$. | Answer: 867. Solution: Shift the entire picture to the left by 1. We get a set of lines passing through the point $(9,9)$ and intersecting the line $y=100-x$. The picture will become symmetric with respect to the line $y=x$, so the sum of the abscissas on it is equal to the sum of the ordinates. Through the point $(9,9... | 867 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 4. A pair of natural numbers is called good if one of the numbers is divisible by the other. Numbers from 1 to 30 were divided into 15 pairs. What is the maximum number of good pairs that could result? | Solution. Answer: 13.
First, let's prove that more than 13 good pairs could not have been formed. We will call a number bad if it is prime and not less than 15. There are only four bad numbers: \(17, 19, 23, 29\). None of the other numbers from 1 to 30 can be divisible by any bad number, and the bad numbers themselves... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a $3 \times 3$ table, natural numbers (not necessarily distinct) are placed such that the sums in all rows and columns are different. What is the minimum value that the sum of the numbers in the table can take? | Answer: 17.
Solution. Note that in each row and column, the sum of the numbers is no less than 3. Then, the doubled sum of all numbers in the table, which is equal to the sum of the sums of the numbers in the rows and columns, is no less than \(3+4+\ldots+8=33\), so the simple sum of the numbers in the table is no les... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, cont... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Polycarp has 2 boxes, the first of which contains $n$ coins, and the second is empty. In one move, he can either transfer one coin from the first box to the second, or remove exactly $k$ coins from the first box, where $k$ is the number of coins in the second box. For which $n$ can Polycarp make the first bo... | Solution. Let's call a move where one coin is moved from the first box to the second box a move of the first type, and a move where coins are removed from the first box a move of the second type. Suppose a total of $x$ moves of the first type and $y$ moves of the second type were made. Then, the second box contains no ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Lёnya has cards with digits from 1 to 7. How many ways are there to glue them into two three-digit numbers (one card will not be used) so that their product is divisible by 81, and their sum is divisible by 9? | Answer: 36.
Solution. If one of the numbers is not divisible by 9, then the other is not either, since their sum is divisible by 9. But then the product cannot be divisible by 81, a contradiction. Therefore, both numbers are divisible by 9.
Then the sum of the digits in each number is divisible by 9, and thus the tot... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\le... | 3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\le... | 4080401 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal? | 5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal?
Answer: 432.
Estimate. We color the cells of the table ... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all pos... | Answer: 30.
Solution. Consider any four consecutive people. If there were at least 3 knights among them, the leftmost knight would definitely lie, which is impossible. If there were at least 3 liars, the leftmost liar would definitely tell the truth, which is also impossible. Therefore, in every group of four consecut... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. Masha and the Bear ate a basket of raspberries and 60 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate raspberries 6 times faster than Masha, and pies only 3 times faster. How many pies did the Bear... | Answer: 54 pies. Solution: Divide the raspberries into 3 equal parts. The bear ate each part 6 times faster than Masha, but there are two parts, so he spent only 3 times less time on the raspberries than Masha. This means Masha ate pies in one-third of the time the bear did. Since she eats three times slower, she ate 9... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the table, there are 2018 playing cards (2018 stacks, each with one card). Petya wants to combine them into one deck of 2018 cards in 2017 operations. Each operation consists of merging two stacks. When Petya merges stacks of $a$ and $b$ cards, Vasily Ivanovich pays Petya $a \cdot b$ rubles. What is the maximum a... | Answer: $\frac{2017 \cdot 2018}{2}=2035153$
Let's mentally connect the cards in one stack with invisible threads, each with each. Then the operation of combining two stacks with $a$ and $b$ cards adds exactly $a \cdot b$ threads. In the end, each card will be connected by a thread to each other, and the number of thre... | 2035153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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