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11. (3b,9-11) In the conditions of a chess match, the winner is declared as the one who outperforms the opponent by two wins. Draws do not count. The probabilities of winning for the opponents are equal. The number of decisive games in such a match is a random variable. Find its mathematical expectation. | # Solution.
Let $\mathrm{X}$ be the number of successful games. At the beginning of the match, the difference in the number of wins between the two participants is zero. Let's list the possible cases of two successful games, denoting a win by the first participant as 1 and a win by the second participant as 2: $11, 12... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
15. (4b, 8-11) In Anchuria, a checkers championship is being held in several rounds. The days and cities for the rounds are determined by a draw. According to the championship rules, no two rounds can take place in the same city, and no two rounds can take place on the same day. Among the fans, a lottery is organized: ... | # Solution.
In an $8 \times 8$ tour table, you need to select $k$ cells such that no more than one cell is chosen in any row or column. The value of $k$ should be chosen to maximize the number of combinations.
The number of combinations is given by $C_{8}^{k} A_{8}^{k}=\frac{8!\cdot 8!}{(8-k)!\cdot(8-k)!\cdot k!}$, w... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. Buratino the Statistician (from 7th grade, 2 points). Every month, Buratino plays in the "6 out of 45" lottery organized by Karabas-Barabas. In the lottery, there are 45 numbered balls, and in each draw, 6 random winning balls are drawn.
Buratino noticed that in each subsequent draw, there are no balls that appeare... | Solution. The probability that no numbers from the first draw will be repeated in the second draw is $a=\frac{39 \cdot 38 \cdot 37 \cdot \ldots \cdot 34}{45 \cdot 44 \cdot 43 \cdot \ldots 40}=0.40056 \ldots$, which is slightly more than 0.4. The probability that there will be no repetitions from the previous draw in bo... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. Winter Camp. In the winter camp, Vanya and Grisha live in a room. Every evening they draw lots to decide who will turn off the light before going to bed: the switch is near the door, and the loser has to go to bed in complete darkness, bumping into chairs.
Usually, Vanya and Grisha draw lots without any complicati... | Solution. a) Let's assume that heads in a coin toss give a one, and tails give a zero in the fractional part of a binary fraction. This results in some number $x$ represented by a binary fraction. For example, if the sequence of tosses starts with HTH, then the binary fraction is 0.101.
Obviously, $0 \leq x \leq 1$, a... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11. Magic Pen (recommended for 8th grade, 1 point). Katya correctly solves a problem with a probability of $4 / 5$, while the magic pen correctly solves a problem without Katya's help with a probability of $1 / 2$. In the test, there are 20 problems, and to get a B, one needs to solve at least 13 of them correctly. How... | Solution. Let $x$ be the number of examples Katya solves herself, and $20-x$ be the number of examples solved by the pen. Then the expected number of correctly solved problems is
$$
\frac{4}{5} x+\frac{1}{2}(20-x)=0.3 x+10
$$
From the inequality $0.3 x+10 \geq 13$ we get that $x \geq 10$. Therefore, Katya needs to tr... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. New Year's Problem (recommended for 8th grade, 4 points). On the New Year's table, there are 4 glasses in a row: the first and third are filled with orange juice, while the second and fourth are empty. While waiting for guests, Vanya absent-mindedly and randomly pours the juice from one glass to another. In one mov... | Solution. We will encode full glasses with the digit 1 and empty ones with the digit 0. We will construct a graph of possible pourings (Fig. 4). This graph turns out to be the graph of an octahedron. From each state to any adjacent one, one can move with a probability of $1 / 4$, and each edge is "traversable" in both ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
15. Messengers (recommended for 9th grade, 3 points). Once, the beautiful Queen Guinevere, while staying at her parental castle, asked King Arthur to send her 20 pearls. The roads are not safe, and Arthur, just in case, decided to send 40 pearls, with different messengers, ordering them to ride on different roads. Band... | Solution. The probability of not saving at least 20 pearls if there are two messengers:
$$
\mathrm{P}_{2}=p^{2}
$$
The probability of not saving at least 20 pearls if there are three messengers:
$$
\mathrm{P}_{3}=p^{3}+2 p^{2}(1-p)=p^{2}(2-p) .
$$
The probability of not saving at least 20 pearls if there are four m... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. Lonely Cars. (From 9th grade, 4 points.) On a very long narrow highway, where overtaking is impossible, $n$ cars are driving in a random order, each with its own preferred speed. If a fast car catches up to a slower one, the fast car has to slow down and drive at the same speed as the slower one. Thus, the cars for... | Solution. Let $I_{k}$ be the indicator of the event "the $k$-th car in line is alone." For $k \leq n$, this event consists of the slowest car among the first $k+1$ cars being the last, and the second slowest being the second to last. The probability of this is $\frac{1}{(k+1) k}$. If $k=n$, then this event consists of ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The smallest set. (From 6th grade, 2 points) In a numerical set, the median is 3, the arithmetic mean is 5, and the only mode of the set is 6. What is the smallest number of numbers that can be in a set with the given properties? | Solution. It is clear that the number 6 appears at least twice in the set, and in addition, there are at least two more numbers.
If the set contains exactly four numbers $a, b, 6, 6$, then we can assume that $a \leq b \leq 3$ and, moreover, the sum of all numbers is 20, so $a+b=20-6-6=8$. Contradiction. The set cannot... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Patrick and Slippers. Every day, the dog Patrick gnaws one slipper from the existing supply in the house. With a probability of 0.5, Patrick wants to gnaw a left slipper, and with a probability of 0.5 - a right slipper. If the desired slipper is not available, Patrick gets upset. How many pairs of identical slippers... | Solution. It is clear that if 7 pairs are bought, Patrick will definitely have enough of the desired, even if he chooses only left slippers every day. The question is about the smallest number of slippers that need to be bought so that with a probability of 0.8 or higher, Patrick will not be disappointed. Probability t... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. There are fewer than 30 people in the class. The probability that a randomly chosen girl is an excellent student is $\frac{3}{13}$, and the probability that a randomly chosen boy is an excellent student is $\frac{4}{11} \cdot$ How many excellent students are there in the class? | # Solution.
According to classical probability theory, the probability that a randomly chosen girl is an excellent student is equal to the ratio of the number of excellent girl students to the total number of girls in the class. Accordingly,
$$
\frac{3}{13}=\frac{\text { number of girls-excellent students }}{\text { ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Disks (from 9th grade. 3 points). At a familiar factory, metal disks with a diameter of 1 m are cut out. It is known that a disk with a diameter of exactly 1 m weighs exactly 100 kg. During manufacturing, there is a measurement error, and therefore the standard deviation of the radius is 10 mm. Engineer Sidorov bel... | Solution. Given $\mathrm{E} R=0.5 \mathrm{m}, \mathrm{D} R=10^{-4}$ (sq.m). Let's find the expected value of the area of one disk:
$$
\mathrm{ES}=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(D R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi
$$
Thus, the expected value of the mass o... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Stem-and-leaf plot. (From 6th grade, 2 points). To represent whole numbers or decimal fractions, a special type of diagram called a "stem-and-leaf plot" is often used. Such diagrams are convenient for representing people's ages. Suppose that in the studied group, there are 5 people aged 19, 34, 37, 42, and 48. For t... | Solution. The digits from 0 to 5, representing decades of years, can be placed immediately (Fig. 4a). It is clear that less than 10 years have passed, otherwise there would be no digits in line "1".
If 7 or more years had passed, then the person who is 13 years old would have moved to line "2", and there would be fewe... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Solution. Suppose for clarity of reasoning that when a bite occurs, the Absent-Minded Scholar immediately pulls out and re-casts the fishing rod, and does so instantly. After this, he waits again. Consider a 6-minute time interval. During this time, on average, there are 3 bites on the first fishing rod and 2 bites ... | Answer: 1 minute 12 seconds.
Evaluation Criteria
| Correct and justified solution | 3 points |
| :--- | :---: |
| It is shown that on average there are 5 bites in 6 minutes, or an equivalent statement is proven | 1 point |
| The solution is incorrect or missing (in particular, only the answer is given) | 0 points | | 1 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 8. Solution.
a) Suppose there are 9 numbers in the set. Then five of them do not exceed the median, which is the number 2. Another four numbers do not exceed the number 13. Therefore, the sum of all numbers in the set does not exceed
$$
5 \cdot 2 + 4 \cdot 13 = 62
$$
Since the arithmetic mean is 7, the sum of the ... | Answer: a) no; b) 11.
Scoring criteria
| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| The correct estimate of the number of numbers in part (b) is found, but no example is given | 2 points |
| Part (a) is solved correctly | 1 point |
| The solution is incorrect or missing (in particula... | 11 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 8. Solution.
a) Suppose the set contains 7 numbers. Then four of them are not less than the median, which is the number 10. Another three numbers are not less than one. Then the sum of all numbers in the set is not less than
$$
3+4 \cdot 10=43
$$
Since the arithmetic mean is 6, the sum of the numbers in the set is... | Answer: a) no; b) 9.
## Grading Criteria
| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| Correct estimate of the number of numbers in part (b), but no example | 2 points |
| Part (a) solved correctly | 1 point |
| Solution is incorrect or missing (including only the answer) | 0 points | | 9 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. All students in the class scored different numbers of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did... | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. One mole of an ideal gas undergoes a closed cycle, in which:
$1-2$ - isobaric process, during which the volume increases by 4 times;
$2-3$ - isothermal process, during which the pressure increases;
$3-1$ - a process in which the gas is compressed according to the law $T=\gamma V^{2}$.
Find how many times... | Solution. Let the initial volume and pressure be denoted as $\left(V_{0} ; P_{0}\right)$. Then $V_{2}=4 V_{0}$.
From the Mendeleev-Clapeyron law, we have three relationships:
$$
P_{0} V_{0}=R T_{1}, P_{0} V_{2}=R T, P_{3} V_{3}=R T
$$
Dividing the third relationship by the second, we get: $\frac{P_{3}}{P_{0}}=\frac{... | 2 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. All students in the class scored a different number of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points di... | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A ten-liter bucket was filled to the brim with currants. Gavrila immediately said that there were 10 kg of currants in the bucket. Glafira thought about it and estimated the weight of the berries in the bucket more accurately. How can this be done if the density of the currant can be approximately considered equal t... | Solution. In approximate calculations, the sizes of the berries can be considered the same and much smaller than the size of the bucket. If the berries are laid out in one layer, then in the densest packing, each berry will have 6 neighbors: the centers of the berries will be at the vertices of equilateral triangles wi... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Hot oil at a temperature of $100^{\circ} \mathrm{C}$ in a volume of two liters is mixed with one liter of cold oil at a temperature of $20^{\circ} \mathrm{C}$. What volume will the mixture have when thermal equilibrium is established in the mixture? Heat losses to the external environment can be neglected. The coeff... | Answer: 3
Let $V_{1}=2$ L be the volume of hot oil, and $V_{2}=1$ L be the volume of cold oil. Then we can write $V_{1}=U_{1}\left(1+\beta t_{1}\right), V_{2}=U_{2}\left(1+\beta t_{2}\right)$, where $U_{1}, U_{2}$ are the volumes of the respective portions of oil at zero temperature; $t_{1}=100^{\circ} \mathrm{C}, t_{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Grandma baked 19 pancakes. The grandchildren came from school and started eating them. While the younger grandson eats 1 pancake, the older grandson eats 3 pancakes, and during this time, grandma manages to cook 2 more pancakes. When they finished, there were 11 pancakes left on the plate. How many pancakes ... | Solution. In one "cycle", the grandsons eat $1+3=4$ pancakes, and the grandmother bakes 2 pancakes, which means the number of pancakes decreases by 2. There will be ( $19-11$ ) $/ 2=4$ such cycles. This means, in these 4 cycles, the younger grandson ate 4 pancakes, the older grandson ate 12 pancakes, and the grandmothe... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. All students in the class scored different numbers of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did... | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3.1. The friends who came to visit Gavrila occupied all the three-legged stools and four-legged chairs in the room, but there was no place left for Gavrila himself. Gavrila counted that there were 45 legs in the room, including the "legs" of the stools and chairs, the legs of the visiting guests (two for each!), and Ga... | Answer: 9. Solution. If there were $n$ stools and $m$ chairs, then the number of legs in the room is $3 n+4 m+2 \cdot(n+m)+2$, from which we get $5 n+6 m=43$. This equation in integers has the solution $n=5-6 p, m=3+5 p$. The values of $n$ and $m$ are positive only when $p=0$. Therefore, there were 5 stools and 3 chair... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. A grenade lying on the ground explodes into a multitude of small identical fragments, which scatter in a radius of $L=90$ m. Determine the time interval (in seconds) between the moments of impact on the ground of the first and the last fragment, if such a grenade explodes in the air at a height of $H=10 \mathrm{m}... | Answer: 6. Solution. From the motion law for a body thrown from ground level at an angle $\alpha$ to the horizontal, the range of flight is determined by the relation $L=\frac{V_{0}^{2}}{g} \sin 2 \alpha$. Therefore, the maximum range of flight is achieved at $\alpha=45^{\circ}$ and is equal to $L=\frac{V_{0}^{2}}{g}$.... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Gavriil and Glafira took a glass filled to the brim with water and poured a little water into three ice cube trays, then placed them in the freezer. When the ice froze, they put the three resulting ice cubes back into the glass. Gavriil predicted that some water would spill out of the glass because ice expands in vo... | Solution. Let $V$ be the volume of water in the molds. Then the volume $W$ of ice in the molds can be determined from the law of conservation of mass $V \cdot \rho_{\text {water }}=W \cdot \rho_{\text {ice }}$. When ice of volume $W$ is floating, the submerged part of this volume $U$ can be determined from the conditio... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. What is the greatest whole number of liters of water that can be heated to boiling temperature using the amount of heat obtained from the combustion of solid fuel, if in the first 5 minutes of combustion, 480 kJ is obtained from the fuel, and for each subsequent five-minute period, 25% less than the previous one. Th... | Answer: 5 liters
Solution: The amount of heat required to heat a mass $m$ of water under the conditions of the problem is determined by the relation $Q=4200(100-20) m=336 m$ kJ. On the other hand, if the amount of heat received in the first 5 minutes is $Q_{0}=480$ kJ. Then the total (indeed over an infinite time) amo... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A new model car travels $4 \frac{1}{6}$ kilometers more on one liter of gasoline compared to an old model car. At the same time, its fuel consumption per 100 km is 2 liters less. How many liters of gasoline does the new car consume per 100 km? | Answer: 6 liters.
Instructions. The fuel consumption of the new car is $x$ liters, and the consumption of the old car is $x+2$
liters. Equation: $\frac{100}{x}-\frac{100}{x+2}=\frac{25}{6} \Leftrightarrow \frac{4(x+2-x)}{x(x+2)}=\frac{1}{6} \Leftrightarrow x^{2}+2 x-48=0 \Leftrightarrow x=-8 ; x=6$. Therefore, $x=6$... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Solve the equation $\sqrt{15 x^{2}-52 x+45} \cdot(3-\sqrt{5 x-9}-\sqrt{3 x-5})=1$. | Solution. Rewrite our equation in the form
$$
\sqrt{3 x-5} \cdot \sqrt{5 x-9} \cdot(3-\sqrt{5 x-9}-\sqrt{3 x-5})=1
$$
Such a transformation is possible because the solution to the original equation exists only for $x>\frac{9}{5}$. Let $\sqrt{3 x-5}=a>0, \sqrt{5 x-9}=b>0$. We have
$$
a+b+\frac{1}{a b}=3
$$
Apply the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Solve the equation $\log _{5}(3 x-4) \cdot \log _{5}(7 x-16) \cdot\left(3-\log _{5}\left(21 x^{2}-76 x+64\right)\right)=1$. | Solution. Rewrite our equation in the form
$$
\log _{5}(3 x-4) \cdot \log _{5}(7 x-16) \cdot\left(3-\log _{5}(3 x-4)-\log _{5}(7 x-16)\right)=1
$$
Such a transformation is possible because the solution to the original equation exists only for $x>\frac{16}{7}$. Let $\log _{5}(3 x-4)=a>0, \log _{5}(7 x-16)=b>0$. Note t... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (4 points)
The price of a new 3D printer is 625000 rubles. Under normal operating conditions, its resale value decreases by $20 \%$ in the first year, and then by $8 \%$ each subsequent year. After how many years will the resale value of the printer be less than 400000 rubles? | Solution:
Let's calculate the cost of the printer year by year:
1 year $=625000 * 0.8=500000$ rubles
2 year $=500000 * 0.92=460000$ rubles (1 point)
3 year $=460000 * 0.92=423200$ rubles (1 point)
4 year $=423200 * 0.92=397694$ rubles. (1 point)
Answer: in 4 years. ( $\mathbf{1}$ point) | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. What was NOT used as money?
1) gold
2) stones
3) horses
4) dried fish
5) mollusk scales
6) all of the above were used | Answer: 6. All of the above were used as money. | 6 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
7. What is a sign of a financial pyramid?
1) an offer of income significantly above average
2) incomplete information about the company
3) aggressive advertising
4) all of the above | Answer: 4. All of the above are signs of a financial pyramid. | 4 | Other | MCQ | Yes | Yes | olympiads | false |
Problem 10. (4 points)
To buy new headphones costing 275 rubles, Katya decided to save money on sports activities. Until now, she has been buying a single-visit ticket to the swimming pool, including a visit to the sauna for 250 rubles, to warm up. However, summer has arrived, and the need to visit the sauna has disap... | Solution: one visit to the sauna costs 25 rubles, the price of one visit to the swimming pool is 225 rubles. Katya needs to visit the swimming pool 11 times without going to the sauna in order to save up for buying headphones. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Leshа has 10 million rubles. Into what minimum number of banks should he deposit them to receive the full amount through ACB insurance payouts in case the banks cease operations? | Answer: 8. The maximum insurance payout is 1,400,000, which means no more than this amount should be deposited in each bank. | 8 | Other | math-word-problem | Yes | Yes | olympiads | false |
1. How much did the euro exchange rate change over the 2012 year (from January 1, 2012 to December 31, 2012)? Provide the answer in rubles, rounded to the nearest whole number. | Answer: 1 or -1. On January 1, 2012, the euro was worth 41.6714, and on December 31, it was 40.2286. $40.2286-41.6714=-1.4428 \approx-1$.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/23.html | -1 | Other | math-word-problem | Yes | Yes | olympiads | false |
4. The Petrovs family has decided to renovate their apartment. They can hire a company for a "turnkey renovation" for 50,000 or buy materials for 20,000 and do the renovation themselves, but for that, they will have to take unpaid leave. The husband earns 2000 per day, and the wife earns 1500. How many working days can... | Answer: 8. The combined daily salary of the husband and wife is $2000+1500=3500$ rubles. The difference between the cost of a turnkey repair and buying materials is $50000-20000=30000$.
$30000: 3500 \approx 8.57$, so the family can spend no more than 8 days on the repair. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive, which means the largest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend
$$
\text { 4800+1500-4800*0.2=5340 rubles. }
$$
This is the most cost-effective way to m... | # Solution:
The root mean square value of the last purchases is $\sqrt{(300 * 300+300 * 300+300 * 300) / 3}=300$ rubles. Therefore, the permissible first purchase is no more than $300 * 3=900$ rubles, with which 18 chocolates can be bought. It remains to buy 22 chocolates for a total of $22 * 50=1100$ rubles.
For the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. The park is a $10 \times 10$ grid of cells. A lamp can be placed in any cell (but no more than one lamp per cell).
a) The park is called illuminated if, no matter which cell a visitor is in, there is a $3 \times 3$ square of 9 cells that contains both the visitor and at least one lamp. What is the minimum number of... | Solution. a) 4. Divide the park into 4 quarters (squares $5 \times 5$), then there must be at least one lamp in each quarter (to illuminate, for example, the corner cells). By placing one lamp in the center of each quarter, we get an example.
b) 10.
Estimate. In each corner square $3 \times 3$ there must be at least ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Once Valera left home, walked to the cottage, painted 11 fence boards there, and returned home 2 hours after leaving. Another time, Valera and Olga went to the cottage together, painted 9 fence boards (without helping or hindering each other), and returned home together 3 hours after leaving. How many boards will Ol... | Solution. The strange result (working together for a longer time, the characters managed to do less work) is explained by the different times spent walking, since the speed of "joint" walking is equal to the lower of the two walkers' speeds. The second time, Valery's working time decreased, which means the travel time ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A natural number $n>5$ is called new if there exists a number that is not divisible by $n$, but is divisible by all natural numbers less than $n$. What is the maximum number of consecutive numbers that can be new? | Solution. Answer: 3.
Example: the number 7 is new (60 is divisible by the numbers from 1 to 6, but not by 7);
the number 8 is new (420 is divisible by the numbers from 1 to 7, but not by 8);
the number 9 is new (840 is divisible by the numbers from 1 to 8, but not by 9).
Evaluation: every fourth number has the form... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The pond has a square shape. On the first frosty day, the part of the pond within 10 meters of the nearest shore froze. On the second day, the part within 20 meters froze, on the third day, the part within 30 meters, and so on. On the first day, the area of open water decreased by $19 \%$. How long will it take for ... | Solution. It is not hard to understand that a pond of $200 \times 200$ fits, for which the answer is - in 10 days (since each day the side decreases by 20 meters). There are no other options, as the larger the side of the pond, the smaller the percentage that will freeze on the first day.
More rigorously: let the side... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In several packages, there are 20 candies, and there are no two packages with the same number of candies and no empty packages. Some packages may be inside other packages (then it is considered that a candy lying in the inner package also lies in the outer package). However, it is forbidden to place a package inside... | Solution. 8. Example: ((6)(2)) ((3)(4)) ((1)4) (there are other examples).
There cannot be more than 8 packages. Indeed, then the sum of the number of candies in the packages (or rather, the number of incidences of candies to packages) is not less than $1+2+\ldots+9=45$. But there are 20 candies, so at least one of th... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The pond has a rectangular shape. On the first frosty day, the part of the pond within 10 meters of the nearest shore froze. On the second day, the part within 20 meters froze, on the third day, the part within 30 meters, and so on. On the first day, the area of open water decreased by 20.2%, and on the second day, ... | Solution. First method. Let the sides of the pond be $a$ and $b$ meters, then
$(a-20)(b-20)=(1-0.202) a b, (a-40)(b-40)=(1-0.388) a b$,
from which $20(a+b)-400=0.202 a b, 40(a+b)-1600=0.388 a b$, that is, $800=0.016 a b, a b=5000$ and further $a+b=525$. It turns out that the sides are 400 and 125 meters.
Answer: on ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. The park is a $10 \times 10$ grid of cells. A lamp can be placed in any cell (but no more than one lamp per cell).
a) The park is called illuminated if, no matter which cell a visitor is in, there is a $3 \times 3$ square of 9 cells that contains both the visitor and at least one lamp. What is the minimum number of... | Solution. a) 4. Divide the park into 4 quarters (squares $5 \times 5$), then there must be at least one lamp in each quarter (to illuminate, for example, the corner cells). By placing one lamp in the center of each quarter, we get an example.
b) 10.
Estimate. In each corner square $3 \times 3$ there must be at least t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Once Valera left home, walked to the cottage, painted 11 fence boards there, and returned home 2 hours after leaving. Another time, Valera and Olga went to the cottage together, painted 8 fence boards (without helping or hindering each other), and returned home together 3 hours after leaving. How many boards will Ol... | Solution. This is a more complex version of problem 5 for 5th grade.
The strange result (that the characters accomplished less work in more time when working together) is explained by the different times spent walking, since the speed of "joint" walking is equal to the lower of the two walkers' speeds. The second time... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Several plants and zombies (no more than 20 creatures in total) attended the "Plants VS Zombies" gathering, and it turned out that all creatures were of different heights. Plants always tell the truth to those who are shorter than them and lie to those who are taller. Zombies, on the contrary, lie to shorter creatur... | Solution. Let the total number of beings be $n$, and exactly $z$ of them are zombies. When plants greet, they say to everyone “I am taller than you”, and zombies say to everyone “I am shorter”. Each zombie said this phrase to everyone except themselves, so we get $z(n-1)=20$. Given that $n-1<20$, the possible cases are... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The magical clock, in addition to the usual pair of hands, has a second pair that is symmetrical to the first at every moment relative to the vertical axis. It is impossible to determine which hands are real from a photograph of the clock. Furthermore, just like with ordinary clocks, it is impossible to distinguish ... | Solution. See the solution to problem 6 for 10th grade.
Instead of the last paragraph of the solution to problem 6 for 10th grade, it is sufficient to provide an example with exactly three undefined photographs and prove that there are exactly three. For example, let the first three photos be taken at 5:50, 11:40, 17:... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
\frac{1}{x}=\frac{32}{y^{5}}+\frac{48}{y^{3}}+\frac{17}{y}-15 \\
\frac{1}{y}=\frac{32}{z^{5}}+\frac{48}{z^{3}}+\frac{17}{z}-15 \\
\frac{1}{z}=\frac{32}{x^{5}}+\frac{48}{x^{3}}+\frac{17}{x}-15
\end{array}\right.
$$
(A. B. Vladimirov) | Solution. Let $F(t)=32 t^{5}+48 t^{3}+17 t-15$. Then the system has the form $F\left(\frac{1}{y}\right)=\frac{1}{x}, F\left(\frac{1}{z}\right)=\frac{1}{y}$, $F\left(\frac{1}{x}\right)=\frac{1}{z}$. From this, it follows that $F\left(F\left(F\left(\frac{1}{x}\right)\right)\right)=\frac{1}{x}$. Note that $F(0.5)=0.5$ and... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Three people want to travel from city A to city B, which is 45 kilometers away from A. They have two bicycles. The speed of a cyclist is 15 km/h, and the speed of a pedestrian is 5 km/h. What is the minimum time they can reach B, if the bicycle can be left unattended on the road? | Solution. Two people ride a bicycle for 10 kilometers, then one of them leaves the bicycle by the road and walks the next 10 kilometers, while the other continues for the next 10 kilometers and also leaves the bicycle (which the first one should pick up later), the third one walks the first 10 kilometers and then rides... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the Olympionov family, it is a tradition to especially celebrate the day when a person turns as many years old as the sum of the digits of their birth year. Kolya Olympionov had such a celebration in 2013, and Tolya Olympionov had one in 2014. Who is older and by how many years? | Solution. Let's determine in which year a person could be born if adding the sum of the digits of their birth year results in 2013 or 2014. It is clear that the year of such a birth cannot be later than 2014. Since the sum of the digits of each number from 1 to 2014 does not exceed \(1+9+9+9=28\), the birth year cannot... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. A fly is sitting at vertex $A$ of a triangular room $A B C$ ( $\angle B=60^{\circ}, \angle C=45^{\circ}, A C=5$ m). At some point, it flies out in a random direction, and each time it reaches a wall, it turns $60^{\circ}$ and continues flying in a straight line (see figure). Can it happen that after some time, the f... | Solution. Let the fly take off at an angle of 60 degrees to the line $A C$. Consider the equilateral triangle $A K C$ with side $A C$. Note that its sides $A K$ and $K C$ can be divided into parts (into infinitely many parts) such that each part equals the next segment of the fly's trajectory. The sum of these parts is... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). After 12 games, Pasha is 2023 rubles richer than he was initially... | Answer: 9 (all except 4, 8, and 1024).
Solution. We need to place the signs in the equation $\pm 1 \pm 2 \pm 2^{2} \pm 2^{3} \pm \ldots \pm 2^{9} \pm 2^{10} \pm 2^{11}=2023$. If we choose all plus signs, the sum will be $2^{0}+\ldots+2^{11}=2^{12}-1=4095$, so we need to replace plus signs with minus signs before the n... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). At the beginning of the game, Pasha had a single-digit amount of ... | Solution. Let $n$ be the amount of money Pasha has become richer (and Igor poorer). Note that Pasha won the last game (otherwise, he would have lost more money than he gained in all previous stages). Therefore, the sequence of games can be divided into series, in each of which Pasha won the last game and lost all the o... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the area of the set of points on the coordinate plane that satisfy the inequality $(y+\sqrt{x})\left(y-x^{2}\right) \sqrt{1-x} \leqslant 0$. | Solution. The left side makes sense only for $0 \leqslant x \leqslant 1$. In this case, it is required that $y+\sqrt{x}$ and $y-x^{2}$ have different signs (or one of them equals zero), or that $x$ equals 1. If we exclude the case $x=1$ (which gives a zero area), we are left with a part of the plane bounded by the segm... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Flint has five sailors and 60 gold coins. He wants to distribute them into wallets, and then give the wallets to the sailors so that each gets an equal number of coins. But he doesn't know how many sailors will be alive by the time of the distribution. Therefore, he wants to distribute the coins in such a way that t... | Solution. Answer: 9 wallets. Example: $12,12,8,7,6,5,4,3,3$.
We will prove that 8 wallets are insufficient.
1) Note that each wallet should contain no more than 12 coins. Therefore, 15 coins must be made up of at least two wallets. This means that when we divide 8 wallets among four pirates, each should receive two w... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural numbers $n$ for which $2^{n}+n^{2016}$ is a prime number. | Solution. Let's consider three cases.
- If $n$ is even, then the given number is also even (and greater than two for $n>0$).
- If $n$ is odd and not divisible by 3, then $2^{n}$ gives a remainder of 2 when divided by 3, and $n^{2016}=\left(n^{504}\right)^{4}$ gives a remainder of 1 when divided by 3, so the sum is div... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit or all three numbers contain different digits.
The complexity of... | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Petya prints five digits on the computer screen, none of which are zeros. Every second, the computer removes the initial digit and appends to the end the last digit of the sum of the remaining four digits. (For example, if Petya enters 12345, after one second it will become 23454, then 34546, and so on. However, he ... | Answer: 2.
Solution. The record 00000 cannot appear on the screen, as it can only result from 00000. A record with four zeros and one also cannot appear, since in that case, the last digit would not equal the remainder of the division of the sum of the first four by 10.
However, a sum of digits equal to 2 is possible... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The pond has a square shape. On the first frosty day, the part of the pond that is no more than 10 meters away from the nearest point on the shore froze. On the second day, the part no more than 20 meters away froze, on the third day, the part no more than 30 meters away, and so on. On the first day, the area of ope... | Solution. Note that the larger the side of the pond, the smaller the percentage that will freeze on the first day. If the side of the pond is 100 m, then 36% will freeze on the first day, and if the side is 120 m, then on the first day, $1-\frac{100^{2}}{120^{2}}=11 / 36<1 / 3$ of the pond's area will freeze. Therefore... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find all such numbers $k$ for which
$$
(k / 2)!(k / 4)=2016+k^{2}
$$
The symbol $n!$ denotes the factorial of the number $n$, which is the product of all integers from 1 to $n$ inclusive (defined only for non-negative integers; $0!=1$). | Solution. Note that the left side makes sense only for even values of $k$. We directly verify that $k=2,4,6,8,10$ do not work, while $k=12$ gives a correct equality.
With each further increase of $k$ by 2, the expression $(k / 2)!$ increases by at least 7 times, i.e., the left side grows by more than 7 times. At the s... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let's call a rectangular parallelepiped typical if all its dimensions (length, width, and height) are different. What is the smallest number of typical parallelepipeds into which a cube can be cut? Don't forget to prove that this is indeed the smallest number. | Solution. A cube can be cut into four typical parallelepipeds. For example, a cube $5 \times 5 \times 5$ can be cut into parallelepipeds $5 \times 3 \times 1, 5 \times 3 \times 4, 5 \times 2 \times 1, 5 \times 2 \times 4$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bfb93b3716727f813bd7g-2.jpg?height=401&width=401&... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Petya's favorite TV game is called "Sofa Lottery." During the game, viewers can send SMS messages with three-digit numbers containing only the digits $1, 2, 3$, and 4. At the end of the game, the host announces a three-digit number, also consisting only of these digits. An SMS is considered a winning one if the numb... | Answer: 8.
Solution. An example of eight suitable SMS messages: 111, 122, 212, 221, 333, 344, 434, 443. Indeed, no matter what number the host names, it contains either at least two digits from the set $\{1,2\}$, or at least two from the set $\{3,4\}$. If the third digit is from the other set, we replace it with a dig... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. What is the maximum number of numbers that can be chosen from the set $\{1,2, \ldots, 12\}$ so that the product of no three chosen numbers is a perfect cube? | Solution. 9: all except $4,9,12$. Note that to remove the cubes, we need to remove at least one element from each of the sets $\{1,2,4\},\{3,6,12\},\{2,4,8\},\{1,3,9\},\{2,9,12\}$, $\{3,8,9\},\{4,6,9\}$. Note that all numbers, except 9, are in no more than three of these seven triples. Therefore, if we remove two numbe... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest possible value of the expression
$$
\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right)
$$
where $x, y, z$ are non-zero real numbers. | Solution. Note that the signs of all six numbers $\frac{x y}{z}, \frac{z x}{y}$, etc., are the same. If all of them are negative, then replace the numbers $x, y, z$ with their absolute values. As a result, each term ($\frac{x y}{z}$, etc.) will change its sign. The modulus of each bracket will remain the same, but the ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The edge of a regular tetrahedron $A B C D$ is 1. Through a point $M$, lying on the face $A B C$ (but not on the edge), planes parallel to the other three faces are drawn. These planes divide the tetrahedron into parts. Find the sum of the lengths of the edges of the part that contains point $D$. | 5. See the figure at the end of the file.
Note that the part of the tetrahedron we are interested in is bounded by three of its faces containing point $D$, and by three planes parallel to the faces. Therefore, this is a parallelepiped.
Consider the faces $A B C$ and $A D C$. Align these faces so that vertex $B$ coinc... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. Prove that $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ is a rational number. | Solution: Let $x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$. Then $x^{3}=(\sqrt{5}+2)-(\sqrt{5}-2)-$ $3 \sqrt[3]{(\sqrt{5}+2)(\sqrt{5}-2)} \times(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2})=4-3 x ; x^{3}+3 x-4=0.0=x^{3}+$ $3 x-4=(x-1)\left(x^{2}+x+4\right), x-1=0$ or $x^{2}+x+4=0$. The second equation has no roots becau... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
4. Solve the equation
$$
x^{2018}+\frac{1}{x^{2018}}=1+x^{2019}
$$ | Answer: $x=1$.
Solution: $x^{2018}+\frac{1}{x^{2018}} \geq 2$ for $x \neq 0$, because $x^{4036}-2 x^{2018}+1=\left(x^{2018}-1\right)^{2} \geq 0 . x^{2019}=x^{2018}+\frac{1}{x^{2018}}-1 \geq 2-1=1 . x \geq 1$. If $\quad x>1, \quad x^{2019}+1>$ $x^{2018}+\frac{1}{x^{2018}}$, because $x^{2019}=x * x^{2018}>x^{2018}$ and ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A black and white chocolate bar consists of individual pieces forming $n$ horizontal and $m$ vertical rows, arranged in a checkerboard pattern. Yan ate all the black pieces, and Maxim ate all the white pieces. What is $m+n$, if it is known that Yan ate $8 \frac{1}{3} \%$ more pieces than Maxim. | Solution. The number of black and white segments can only differ by 1. Therefore, Yan ate 1 segment more than Maksim. If 1 segment is 8 $\frac{1}{3} \%$, then Maksim ate 12 segments, Yan ate 13 segments, and together they ate 25 segments. This means the chocolate bar was $5 \times 5$.
Answer: 10.
Criteria: 14 points ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. There are books of three colors: white, blue, and green. To make the shelf look beautiful, the boy first arranged the white books, and then placed blue books in each gap between them. Finally, he placed green books in each gap between the standing books. In the end, there were 41 books on the shelf. How many white b... | Answer: 11
9. The sum of two natural numbers is 2017. If you append 9 to the end of the first number and remove the digit 8 from the end of the second number, the numbers will be equal. Find the largest of these numbers.
Answer: 1998 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. In a row, $n$ integers are written such that the sum of any three consecutive numbers is positive, while the sum of any five consecutive numbers is negative. For what largest $n$ is this possible | Answer: 6.
Solution: Let's provide an example for $n=6: 3,-5,3,3,-5,3$. We will prove that for $n \geq 7$ it will not be possible to write down a sequence of numbers that satisfy the condition of the problem. We will construct a table for the first 7 numbers in this sequence
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Find the value of the expression $\sqrt[3]{7+5 \sqrt{2}}-\sqrt[3]{5 \sqrt{2}-7}$ | Answer: 2
Task 2.B The bases $AD$ and $BC$ of an isosceles trapezoid $ABCD$ are $16 \sqrt{3}$ and $8 \sqrt{3}$, respectively, and the acute angle at the base is $30^{\circ}$. What is the length of the lateral side of the trapezoid?
Answer: 8 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the table, there are 10 stacks of playing cards (the number of cards in the stacks can be different, there should be no empty stacks). The total number of cards on the table is 2015. If a stack has an even number of cards, remove half of the cards. If the number of remaining cards in the stack is still ev... | # Solution
a) Since 2014 is an odd number, in any distribution of cards into 10 piles, there will be at least one pile with an even number of cards. Therefore, the number of cards in this pile, and thus the total number of cards, will definitely decrease by at least one card (if there are 2 cards in this pile). This m... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.
---
Note: The translation maintains the original format and line breaks as requested. | Answer: $x=y=0$.
Solution: If both numbers are not equal to 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 6. Find the value of the expression $\left(\sqrt[3]{x^{2}} \cdot x^{-0.5}\right):\left(\left(\sqrt[6]{x^{2}}\right)^{2} \cdot \sqrt{x}\right)$ at $x=\frac{1}{2}$ | Answer: 2
Problem 7. The diagonals of trapezoid $ABCD (BC \| AD)$ are perpendicular to each other, and $CD = \sqrt{129}$. Find the length of the midline of the trapezoid, given that $BO = \sqrt{13}, CO = 2\sqrt{3}$, where $O$ is the point of intersection of the diagonals of the trapezoid.
Answer: 10
Problem 8. Four ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Ten consecutive natural numbers are written on the board. What is the maximum number of them that can have a digit sum equal to a perfect square? | Answer: 4.
Solution: Note that the sums of the digits of consecutive natural numbers within the same decade are consecutive natural numbers. Since there are 10 numbers, they span two decades. Also note that among ten consecutive natural numbers, there can be no more than 3 perfect squares, and three perfect squares ca... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find all real roots of the equation
$(x+1)^{5}+(x+1)^{4}(x-1)+(x+1)^{3}(x-1)^{2}+(x+1)^{2}(x-1)^{3}+(x+1)(x-1)^{4}+(x-1)^{5}=0$ | Solution. Multiply both sides of the equation by $(x+1)-(x-1)$ (this factor equals 2). Use the formula $a^{6}-b^{6}=(a-b)\left(a^{5}+a^{4} b+a^{3} b^{2}+a^{2} b^{3}+a b^{4}+b^{5}\right)$.
$$
\begin{gathered}
(x+1)^{6}-(x-1)^{6}=0 \\
(x+1)^{6}=(x-1)^{6}
\end{gathered}
$$
The equation
$$
x+1=x-1
$$
has no solutions. ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Four points $A, B, C, D$ are on a plane. It is known that $A B=1, B C=$ $2, C D=\sqrt{3}, \angle A B C=60^{\circ}, \angle B C D=90^{\circ}$. Find $A D$. | Solution. Let's construct the diagram. Let the line $CD$ intersect the line $AB$ at point $O$
(according to the condition, these lines are not parallel). There are two possible positions for... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.3. In a seven-story building, domovoi (Russian house spirits) live. The elevator travels between the first and the last floors, stopping at every floor. On each floor, starting from the first, one domovoi entered the elevator, but no one exited. When the thousandth domovoi entered the elevator, it stopped. On which f... | Answer: on the fourth floor.
Solution. First, let's find out how many housekeepers ended up in the elevator after the first trip from the first to the seventh floor and back, until the elevator returned to the first floor. One housekeeper entered on the first and seventh floors, and on all other floors, two housekeepe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Find the maximum possible area of a quadrilateral in which the product of any two adjacent sides is 1. | 1. Find the maximum possible area of a quadrilateral for which the product of any two adjacent sides is 1.
OTBET: 1.
SOLUTION.
Let the quadrilateral have sides $a, b, c, d$. Then $a b=b c=c d=d a=1$. From the equality $a b=b c$, it follows that $a=c$, and from the equality $b c=c d$, we get that $b=d$. Therefore, th... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The graph of the linear function $y=k x+k+1(k>0)$ intersects the $O x$ axis at point $A$, and the $O y$ axis at point $B$ (see the figure). Find the smallest possible value of the area of triangle $A B O$. | Answer: 2.
Solution. The abscissa of point $A$ of intersection with the $O x$ axis: $0=k x+k+1 ; x=$
$-\left(1+\frac{1}{k}\right)$. The ordinate of point $B$ of intersection with the $O y$ ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.7. In a magic shop, for 20 silver coins you can buy an invisibility cloak and get 4 gold coins as change. For 15 silver coins you can buy an invisibility cloak and get 1 gold coin as change. How many silver coins will you get as change if you buy an invisibility cloak for 14 gold coins? | Answer: 10.
Solution. In the first case, compared to the second, by paying 5 extra silver coins, one can receive 3 extra gold coins in change. Therefore, 5 silver coins are equivalent to 3 gold coins.
In the second case, by paying 15 silver coins (which is equivalent to $3 \cdot 3=9$ gold coins), one can get the cloa... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. Each of the 33 bogatyrs (Russian epic heroes) either always lies or always tells the truth. It is known that each bogatyr has exactly one favorite weapon: a sword, spear, axe, or bow.
One day, Ded Chernomor asked each bogatyr four questions:
- Is your favorite weapon a sword?
- Is your favorite weapon a ... | Answer: 12.
Solution. Note that each of the truth-telling heroes answers affirmatively to only one question, while each of the lying heroes answers affirmatively to exactly three questions. Let the number of truth-telling heroes be $x$, and the number of lying heroes be $-(33-x)$. Then the total number of affirmative ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
.
By the evening of September 8, Yura realized that th... | Answer: 12.
Solution. In the first 3 days, Yura solved $91-46=45$ problems. Let's say on September 7th, he solved $z$ problems, then on September 6th, he solved $(z+1)$ problems, and on September 8th, he solved $(z-1)$ problems. We get that $45=(z+1)+z+(z-1)=3 z$, from which $z=15$.
Since $91=16+15+14+13+12+11+10$, Y... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. Petya bought himself shorts for football at the store.
- If he had bought shorts with a T-shirt, the cost of the purchase would have been twice as much.
- If he had bought shorts with cleats, the cost of the purchase would have been five times as much.
- If he had bought shorts with shin guards, the cost ... | Answer: 8.
Solution. Let the shorts cost $x$. Since the shorts with a T-shirt cost $2x$, the T-shirt also costs $x$. Since the shorts with boots cost $5x$, the boots cost $4x$. Since the shorts with shin guards cost $3x$, the shin guards cost $2x$. Then, if Petya bought shorts, a T-shirt, boots, and shin guards, his p... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. Vanya runs from home to school at a constant speed. If he had initially increased his speed by 2 m/s, he would have arrived at school 2.5 times faster. How many times faster would he have arrived at school if he had initially increased his speed by 4 m/s?
# | # Answer: 4.
Solution. Let Vasya's initial speed be $v$ m/s. If he ran at a speed of $(v+2)$ m/s, he would cover the same distance to school 2.5 times faster. This means that $\frac{v+2}{v}=2.5$, from which we find $v=\frac{4}{3}$. If he had initially run at a speed of $(v+4)$ m/s, he would have arrived at school $\fr... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. The entire surface of a cube $13 \times 13 \times 13$ was painted red, and then this cube was sawn into smaller cubes $1 \times 1 \times 1$. All the faces of the smaller cubes $1 \times 1 \times 1$ that were not painted red were painted blue. By what factor is the total area of the blue faces greater than... | Answer: 12.
Solution. Each face of the original cube consists of exactly $13^{2}$ $1 \times 1$ squares, so a total of $6 \cdot 13^{2}$ $1 \times 1$ squares were painted red. Since there are exactly $13^{3}$ $1 \times 1 \times 1$ cubes, and each has 6 faces, the number of $1 \times 1$ squares painted blue is $6 \cdot 1... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. Petya wrote down ten natural numbers in a row as follows: the first two numbers he wrote down arbitrarily, and each subsequent number, starting from the third, was equal to the sum of the two preceding ones. Find the fourth number if the seventh is 42 and the ninth is 110. | Answer: 10.
Solution. From the condition, it follows that the eighth number is equal to the difference between the ninth and the seventh, i.e., $110-42=68$. Then the sixth is $68-42=26$, the fifth is $42-26=16$, and the fourth is $26-16=10$.
Remark. In fact, the numbers on the board are the doubled Fibonacci numbers. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Oleg wrote down several composite natural numbers less than 1500 on the board. It turned out that the greatest common divisor of any two of them is 1. What is the maximum number of numbers that Oleg could have written down? | Answer: 12.
Solution. Prime numbers less than $\sqrt{1500}$ will be called small. There are exactly 12 such numbers: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37$.
Note that each of Oleg's numbers has a small divisor (otherwise it would be not less than $43^{2} > 1500$), and different numbers have different small divi... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. There are 22 batteries, 15 of which are charged and 7 are discharged. The camera works with three charged batteries. You can insert any three batteries into it and check if it works. How can you guarantee to turn on the camera in 10 such attempts? | Solution. Let's number the batteries: $1,2, \ldots, 22$. The first six tests will involve inserting batteries into the camera as follows: $1,2,3 ; 4,5,6 ; \ldots, 16,17,18$. If at least one of these groups turns on the camera, everything is fine. If not, then among the first 18 batteries, there are at least 6 discharge... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Inside triangle ABC, two points are given. The distances from one of them to the lines AB, BC, and AC are 1, 3, and 15 cm, respectively, and from the other - 4, 5, and 11 cm. Find the radius of the circle inscribed in triangle ABC. | Answer: 7 cm. First solution. Let $M_{1}$ and $M_{2}$ be the first and second given points, and let point $O$ be such that point $M_{2}$ is the midpoint of segment $O M_{1}$. Drop perpendiculars $M_{1} N_{1}, M_{2} N_{2}$, and $O N_{3}$ to line $A B$. Then segment $M_{2} N_{2}$ will be the midline of trapezoid $O N_{3}... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A teacher fills in the cells of a class journal of size $7 \times 8$ (7 rows, 8 columns). In each cell, she puts one of three grades: 3, 4, or 5. After filling in the entire journal, it turns out that in each row, the number of threes is not less than the number of fours and not less than the number of fives, and in... | Answer: 8 fives.
## Solution.
First step.
In each row, there are no fewer threes than fours, so in the entire journal, there are no fewer threes than fours. In each column, there are no fewer fours than threes, so in the entire journal, there are no fewer fours than threes. Therefore, the number of threes and fours ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A car left point A for point B, which are 10 km apart, at 7:00. After traveling $2 / 3$ of the way, the car passed point C, from which a cyclist immediately set off for point A. As soon as the car arrived in B, a bus immediately set off from there in the opposite direction and arrived in point A at 9:00. At what dis... | Solution. Let $v_{a}$ be the speed of the car, $v_{\varepsilon}$ be the speed of the cyclist, and $v_{a \varepsilon}$ be the speed of the bus. From the problem statement, we derive the following system of equations:
$$
\left\{\begin{array}{l}
\frac{20 / 3}{v_{a}}+\frac{20 / 3}{v_{s}}=3 \\
\frac{10}{v_{a}}+\frac{10}{v_... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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