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5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
We will call suc... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment ... | Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$. | 219 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Winnie-the-Pooh stocked up on chocolate bars for the winter: $60\%$ of the total number were "Snickers", $30\%$ were "Mars", and $10\%$ were "Bounty". In the spring, it turned out that the number of "Bounty" bars eaten by Winnie-the-Pooh was $120\%$ of the number of "Mars" bars eaten and $30\%$ of the number of "Sni... | # Solution.
Let there be $3 k$ "Bounty" chocolate bars in total. Then there were $9 k$ "Mars" bars and $18 k$ "Snickers" bars. Since $k$ "Bounty" bars were eaten, $\frac{k}{1.2}=\frac{5 k}{6}$ "Mars" bars were eaten. Therefore, $k$ is divisible by 6. "Snickers" bars eaten were $\frac{k}{0.3}=\frac{10 k}{3}$, and the r... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. In a bag, there were cards with numbers from 1 to 20. Vlad drew 6 cards and said that all these cards can be divided into pairs so that the sums of the numbers in each pair are the same. Lena managed to peek at 5 of Vlad's cards: the numbers on them were $2, 4, 9, 17, 19$. What number was on the card that Lena ... | Answer: 12.
Solution. To calculate the answer, one needs to select four numbers out of the given five such that the sum of two of them equals the sum of the other two. By enumeration, it is not difficult to verify that these numbers are $2,4,17,19(2+19=4+17)$.
Thus, the number on the remaining card is $12(2+19=4+17=9... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Anya, Borya, Vika, and Gena are on duty at school for 20 days. It is known that exactly three of them are on duty each day. Anya was on duty 15 times, Borya - 14 times, Vika - 18 times. How many times was Gena on duty | Answer: 13.
Solution. Since 3 people are on duty in school every day, a total of $3 \cdot 20=60$ people are needed for the duty. Therefore, Gena was on duty $60-15$ (Anya's duties) -14 (Borya's duties) -18 (Vika's duties) $=13$ times.
| Anya | $\checkmark$ | $\checkmark$ | $\checkmark$ | | $\checkmark$ | $\checkmark... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. During the physical education class, the entire class lined up by height (all children have different heights). Dima noticed that the number of people taller than him is four times the number of people shorter than him. And Lёnya noticed that the number of people taller than him is three times less than the ... | Answer: 21.
Solution. Let $x$ be the number of people who are shorter than Dima. Then, the total number of students in the class is $x$ (people who are shorter than Dima) $+4 x$ (people who are taller than Dima) +1 (Dima) $=5 x+1$ (total number of people in the class).
Let $y$ be the number of people who are taller t... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A row of 11 numbers is written such that the sum of any three consecutive numbers is 18. Additionally, the sum of all the numbers is 64. Find the central number. | Answer: 8.
Solution. Number the numbers from left to right from 1 to 11.
Notice that the sum of the five central numbers (from the fourth to the eighth) is 64 (the sum of all numbers) $-2 \cdot 18$ (the sum of the numbers in the first and last triplets) $=28$.
Then the sixth (central) number is 18 (the sum of the fo... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Koschei the Deathless has 11 large chests. In some of them lie 8 medium chests. And in some of the medium chests lie 8 small chests. The chests contain nothing else. In total, Koschei has 102 empty chests. How many chests does Koschei have in total? | Answer: 115.
Solution. Let $x$ be the number of non-empty chests.
Consider the process when Koschei just started placing chests inside each other. Initially, he had 11 empty large chests. Each time he placed 8 smaller chests into a single empty chest, the total number of empty chests increased by 7 ( -1 old empty che... | 115 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. The dragon has 40 piles of gold coins, and the number of coins in any two of them differs. After the dragon plundered a neighboring city and brought back more gold, the number of coins in each pile increased by either 2, 3, or 4 times. What is the smallest number of different piles of coins that could result... | Answer: 14.
Solution. Evaluation. Suppose that no more than 13 piles increased by the same factor. Then no more than 13 piles increased by a factor of 2, no more than 13 by a factor of 3, and no more than 13 by a factor of 4. Thus, the dragon has no more than 39 piles in total. Contradiction.
Therefore, there will be... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. On one face of a die, one dot is drawn, on another face - two, on the third - three, and so on. Four identical dice are stacked as shown in the figure. How many dots in total are on the 6 faces where the dice touch?
 writing 10 digits on the board from left to right to form a ten-digit number. Moreover, it is not allowed to write two consecutive identical digits. If the resulting number is divisible by 9, then Sasha wins; otherwise, Masha wins. Who will win with correct play fro... | Answer: Sasha.
Solution. A number is divisible by 9 if the sum of its digits is divisible by 9. Therefore, one of Sasha's possible strategies is to complement each of Masha's digits to 9. That is, if Masha writes 0, then Sasha writes 9; if Masha writes 1, then Sasha writes 8, and so on. Thus, after each pair of moves,... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the numerical value of the expression
$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}+\frac{2}{x y+1}
$$
if it is known that $x$ is not equal to $y$ and the sum of the first two terms is equal to the third. | Answer: 2.
Solution. Let's bring the condition to a common denominator
$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}=\frac{2}{x y+1}
$$
we get
$$
\frac{\left(x^{2}+y^{2}+2\right)(x y+1)-2\left(x^{2}+1\right)\left(y^{2}+1\right)}{\left(x^{2}+1\right)\left(y^{2}+1\right)(x y+1)}=0
$$
expand all brackets in the numerator, c... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1 Karlson eats three jars of jam and one jar of honey in 25 minutes, while Little Man takes 55 minutes. One jar of jam and three jars of honey Karlson eats in 35 minutes, while Little Man takes 1 hour 25 minutes. How long will it take them to eat six jars of jam together? | 9.1 20 minutes.
From the condition, it follows that if Karlson eats three jars of jam and one jar of honey, and then immediately eats one jar of jam and three jars of honey, he will spend $25+35=60$ minutes. For Little Man, this time will be 140 minutes. Therefore, Karlson will spend 15 minutes on one jar of jam and o... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. A football team coach loves to experiment with the lineup. During training sessions, he divides 20 available field players into two teams of 10 players each, adds goalkeepers, and arranges a game between the teams. He wants any two field players to end up on different teams at some training session. What is the mi... | # 9.5. 5 Training Sessions.
Note that four training sessions are insufficient. Let's take 10 football players who played on the same team during the first training session. During the second training session, at least five of them will be on the same team again. During the third training session, at least three of the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. At the first stop, 18 passengers entered an empty bus. Then at each stop, 4 people got off and 6 people got on. How many passengers were in the bus between the fourth and fifth stops? | Answer: 24 people.
## Solution.
## Method 1.
After each stop, except the first one, the number of passengers in the bus increases by 2 people. Therefore, from the second to the fourth stop, the number of people increased by 6 people. That is, it became $18+6=24$ people.
Method 2.
From the second to the fourth stop... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. A sign engraver makes signs with letters. He engraves identical letters in the same amount of time, and different letters possibly in different times. For two signs “ДОМ МОДЫ” and “ВХОД” together, he spent 50 minutes, and one sign “В ДЫМОХОД” he made in 35 minutes. How long will it take him to make the sign “ВЫХОД”? | Answer: 20 minutes.
## Solution.
In the signs FASHION HOUSE ENTRANCE and IN CHIMNEY, we separate the letters that form the word EXIT, then from the first sign, D, O, M, M, O, D will remain, and from the second - D, M, O. Note that FASHION HOUSE ENTRANCE differs from IN CHIMNEY by the letters D, O, M, and in time - by... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. The younger brother takes 25 minutes to reach school, while the older brother takes 15 minutes to walk the same route. How many minutes after the younger brother leaves home will the older brother catch up to him if he leaves 8 minutes later? | Answer: in 17 minutes. Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 25 minutes, in 8 minutes he will cover the distance $\frac{8 S}{25}$. After the older brother leaves the house, the rate at which they are closing the distance between them will be $\frac{S}{1... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Petya tells his neighbor Vova: «In our class, there are 30 people, and there is an interesting situation: any two boys have a different number of girlfriends in the class, and any two girls have a different number of friends among the boys in the class. Can you determine how many boys and how many girls we have in... | Answer. a) Vova is wrong; b) 15 boys and 15 girls. Solution. a) To show that Vova is wrong, we will provide an example of a class that meets the conditions of the problem. For a class with 15 girls and 15 boys, we will number them and present a "friendship" table. In the cells of the table, there is a "+", if the corre... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the USA, the date is typically written as the month number, followed by the day number, and then the year. In Europe, however, the day comes first, followed by the month and the year. How many days in a year cannot be read unambiguously without knowing which format it is written in? | 2. It is clear that these are the days whose date can be the number of the month, i.e., takes values from 1 to 12. There are such days $12 \cdot 12=144$. However, the days where the number matches the month are unambiguous. There are 12 such days. Therefore, the number of days sought is $144-12=132$. | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A traveler arrived on an island inhabited by liars (L) and truth-tellers (P). Each L, when asked a question "How many..?", gives a number that is 2 more or 2 less than the correct answer, while each P answers correctly. The traveler met two residents of the island and asked each how many L and P live on the island. ... | 5. I - L, II - P. On the island, there are 1000 L and 1000 P. The answers of the first and second are different, so the option P and P is impossible. The option L and L is also impossible, because the numbers 1001 and 1000 differ by 1, while the answers of the liars regarding the number of L should differ by 4, or coin... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. Fishermen caught several carp and pike. Each caught as many carp as all the others caught pike. How many fishermen were there if the total number of carp caught was 10 times the number of pike? Justify your answer. | # Solution:
Method 1. Each fisherman caught as many carp and pike together as the total number of pike caught. Summing the catches of all fishermen, we get that the total catch of all fishermen (in terms of the number of fish) is equal to the total number of pike caught, multiplied by the number of fishermen. On the o... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. We consider all possible pairs of quadratic equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ such that each equation has two distinct roots. Is it true that the expression $\frac{1}{x_{1} x_{3}} + \frac{1}{x_{1} x_{4}} + \frac{1}{x_{2} x_{3}} + \frac{1}{x_{2} x_{4}}$, where the numbers $x_{1}, x_{2}$ are... | Solution: According to Vieta's theorem
$$
x_{1} x_{2}=-\left(x_{3}+x_{4}\right)=q \text { and } x_{3} x_{4}=-\left(x_{1}+x_{2}\right)=p
$$
Then
$$
\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}=\frac{x_{3}+x_{4}}{x_{1} x_{3} x_{4}}=-\frac{q}{p x_{1}}
$$
Similarly
$$
\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}=-\frac{... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
10.3. In the distant times of stagnation in the Soviet Union, 15 and 20 kopeck coins were in circulation. Schoolboy Valera had a certain amount of money only in such coins. Moreover, the number of 20 kopeck coins was greater than the number of 15 kopeck coins. Valera spent one-fifth of all his money, paying two coins f... | Solution: One fifth of Valera's capital could be either 30, 35, or 40 kopecks. Then, after buying the ticket, he should have had 120, 140, or 160 kopecks left, and the cost of the lunch was either 60, 70, or 80 kopecks. The maximum value of three coins is 60 kopecks, so the last two scenarios are impossible. Therefore,... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A mathematician left point A for point B. After some time, a physicist also left point A for point B. Catching up with the mathematician after 20 km, the physicist, without stopping, continued to point B and turned back. They met again 20 km from B. Then each continued in their respective directions. Upon reaching p... | # Answer: 45.
Solution. From the first meeting to the second meeting, the mathematician walked a total of 100 - 20 - 20 = 60 km, while the physicist walked -100 - 20 + 20 = 100 km. It is clear from this that the ratio of their speeds is 6:10 or 3:5. From the second to the third meeting, they will walk together 100 + 1... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The city center is a rectangle measuring $5 \times 8$ km, consisting of 40 blocks, each $1 \times 1$ km, with boundaries formed by streets that create 54 intersections. What is the minimum number of police officers needed to be placed at the intersections so that any intersection can be reached by at least one polic... | Solution. Evaluation. Consider the intersections on the boundary. There are 26 in total. Each police officer can control no more than 5 intersections (if he is on the boundary, then exactly 5, if he is inside the city, then no more than 3 on each side and no more than 5 in the corner). Therefore, at least 6 police offi... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. The hunter told a friend that he saw a wolf with a one-meter tail in the forest. That friend told another friend that a wolf with a two-meter tail had been seen in the forest. Passing on the news further, ordinary people doubled the length of the tail, while cowards tripled it. As a result, the 10th channel report... | Solution: Note that when information is transmitted by ordinary people, the length of the tail is multiplied by 2, and when transmitted by cowardly people, it is multiplied by 3. Therefore, the number of twos in the product equals the number of ordinary people (and the number of threes equals the number of cowardly peo... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.4. Three pirates were dividing a bag of coins. The first took 3/7 of all the coins; the second took 51 percent of the remainder. After this, the third received 8 fewer coins than the second. How many coins were in the bag? Justify your answer. | Solution: The third pirate received $49\%$ of the remainder, which is $2\%$ less than the third. These two percent of the remainder amount to 8 coins, so one percent is
4 coins, and the entire remainder is 400 coins. These 400 coins make up $1-\frac{3}{7}=\frac{4}{7}$ of the total. Therefore, the total number of coins... | 700 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.5. On cards, all two-digit numbers from 10 to 99 are written (one number per card). All these cards are lying on the table face down. What is the minimum number of cards that need to be flipped to guarantee that at least one of the revealed numbers is divisible by 7? Justify your answer. | Solution: There are exactly 13 two-digit numbers divisible by 7 (14 = 7 * 2, 21 = 7 * 3, ..., 98 = 7 * 14). If at least 13 cards are not flipped, it is impossible to exclude the situation where all these numbers remain covered. Therefore, no more than 12 cards can be left in their original position, and the rest must b... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. January first of a certain non-leap year fell on a Saturday. And how many Fridays are there in this year? | Answer: 52
Solution. In a non-leap year, there are 365 days. The first two days fell on Saturday and Sunday, followed by 51 full weeks ( $51 \cdot 7=357$ days) and 6 more days. In total, there are 52 Fridays. | 52 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. Lisa wrote a quadratic equation. Artem erased its free term, so the equation now looks like $\operatorname{\operatorname {mak}} 2 x^{2}+20 x+\ldots=0$. Lisa doesn't remember what number Artem erased, but she remembers that the equation has exactly one real root. What is this root? | # Answer: -5
Solution. A quadratic equation has one root if and only if its discriminant is 0. And if the discriminant is 0, then the root is calculated using the formula $x_{1}=-b /(2 a)=-20 / 4=-5$. | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Lisa drew graphs of all functions of the form $y=a x+b$, where $a$ and $b$ take all natural values from 1 to 100. How many of these graphs pass through the point $(3,333)$? | Answer: 23
Solution. We are looking for such $a$ and $b$ that $333=3a+b$. That is, $b=3(111-a)$. Therefore, $0<111-a \leqslant 33$, $78 \leqslant a$. There are exactly 23 such $a$ and for each of them, $b$ can be uniquely determined. | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. The sea includes a bay with more saline water. The salinity of the water in the sea is 120 per mille, in the bay 240 per mille, in the part of the sea not including the bay - 110 per mille. How many times is the volume of water in the sea larger than the volume of water in the bay? The volume of water is considere... | Answer: 13
Solution. Let the volume of salt in the bay be $s_{1}$, and the volume of water $v_{1}$; in the part of the sea not including the bay, the volume of salt is $s_{2}$, and the total volume is $v_{2}$. We have the equations
$$
\frac{s_{1}}{v_{1}}=\frac{240}{1000} ; \quad \frac{s_{2}}{v_{2}}=\frac{110}{1000} ;... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. A paper rectangle $4 \times 8$ was folded along the diagonal as shown in the figure. What is the area of the triangle that is covered twice?
 | Answer: 10.
Solution. By the Pythagorean theorem, $A C=\sqrt{A B^{2}+B C^{2}}=4 \sqrt{5}$. Triangle $A E C$ is isosceles, and if we drop the height $E H$ in it, then triangle $A E H$ will be similar to triangle $A C D$. Therefore, $A H / E H=A D / D C=2 . A H=2 \sqrt{5}$, since the height in an isosceles triangle is a... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.1. In the club, there are 9 people. Every day, some three of them went to the cafe together, while the others did not go to the cafe. After 360 days, it turned out that any two people from the club had been to the cafe together the same number of times. How many times? | Answer: 30
Solution. The total number of pairs of people in the circle is $9 \cdot 8 / 2=36$. Over 360 days, the cafe was visited by $360 \cdot 3$ pairs (since three new pairs are added each day). Since all pairs visited the cafe an equal number of times, this number is $360 \cdot 3 / 36=30$. | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. Real $x, y, z$ are such that $x y + x z + y z + x + y + z = -3, x^{2} + y^{2} + z^{2} = 5$. What is $x + y + z$? | Answer: -1
Solution. Add twice the first equation to the second, we get $(x+y+z)^{2}+2(x+y+z)=-1$. Therefore, $(x+y+z+1)^{2}=0, x+y+z=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. Of all numbers with the sum of digits equal to 25, find the one whose product of digits is maximal. If there are several such numbers, write the smallest of them in the answer. | Answer: 33333334
Solution. Obviously, there is no 0 in the number. If the number contains the digit 1, then it can be removed and one of the remaining digits can be increased by 1, which does not change the sum, but increases the product. If the number contains a digit $x \geqslant 5$, then it can be replaced by the d... | 33333334 | Other | math-word-problem | Yes | Yes | olympiads | false |
11.1 The number 890 has the property that by changing any of its digits to 1 (increasing or decreasing), you can obtain a number that is a multiple of 11. Find the smallest three-digit number with the same property. | Answer: 120.
Solution. Since the last digit of the number must be changed to obtain a number divisible by 11, the required number should differ from it by 1. The smallest three-digit number divisible by 11 is 110. However, the numbers adjacent to it, 109 and 111, do not have the required property. Indeed, if the secon... | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. A $7 \times 7$ checkered board was assembled using three types of figures (see the image), not necessarily all. How many figures, composed of four cells, could have been used?
 | Answer: only one.
Solution. We will prove that only one figure consisting of four cells can be used. We will color the cells of the board as shown in Fig. 11.5a: Each of the given figures can cover no more than one shaded cell, therefore, the number of figures must be no less than 16.
Since 16 three-cell figures cove... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. All graduates of the mathematics school took the Unified State Exam (USE) in mathematics and physical education. Each student's result in mathematics turned out to be equal to the sum of the results of all other students in physical education. How many graduates are there in the school if the total number of points ... | # Answer: 51.
## Solution:
Let the total number of graduates be $n$. Denote the total score in mathematics by $M$. This sum is equal to the sum of the physical education scores of all students, counted $(n-1)$ times.
Thus, we have the equation: $M=50 \frac{M}{n-1}$.
From this, the answer is
Instructions for checki... | 51 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded? | Answer: She could.
## Solution:
Here is an example of a code table:
| A | B | K | M | P | U | Sh |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 110 | 1111 | 100 | 101 | 11100 | 11101 |
The word will look like this: 10101010100010001100110111001110111110 - a total of 38 characters. Decoding is una... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. A travel agency ran a promotion: "Buy a trip to Egypt, bring four friends who also buy a trip, and get the cost of your trip back." During the promotion, 13 customers came on their own, and the rest were brought by friends. Some of them brought exactly four new customers, while the other 100 did not bring anyone.... | Answer: 29.
Let each of the $x$ potential "lucky ones" bring 4 friends. Then the number of "brought" customers is $4 x$, and 13 came on their own, so the total number of tourists was $13+4 x$.
On the other hand, $x$ people brought new customers, while 100 people did not, so the total number of tourists was $x+100$.
... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Papa always picked up Misha from kindergarten at the same time and in the same car. But today, after firmly establishing that there would be pea porridge for dinner, Misha resolutely left the kindergarten and set off down the road to meet Papa. Having walked 5 km, Misha met Papa, got into the car, and they arrived h... | 3. Speed: 60 km/h (or 1 km/min). Today, Misha and his dad saved 10 minutes that were previously spent driving 10 km (from the meeting point to the kindergarten and back). This means that in 1 minute, Misha's dad drives 1 km, i.e., his speed is 60 km/h.
Criteria: Correct reasoning and correct answer - 7 points; only an... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. How many natural numbers greater than one exist, the product of which with their smallest prime divisor does not exceed 100? | Answer: 33.
Sketch of the solution. The smallest prime divisor can be 2. These are the numbers: $2, 4, \ldots, 50$ (25 numbers). The smallest prime divisor can be 3. These are the numbers: 3, 9, 15, 21, 27, 33 (6 numbers). The smallest prime divisor can be 5. This number: 5 (1 number). The smallest prime divisor can b... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Every week, listeners choose the ten most popular songs via SMS. It is known that 1) the same set of songs in the same order is never chosen twice in a row; 2) a song that has once dropped in the ranking will not rise again in the future. What is the maximum number of weeks that the same 10 songs can remain in the r... | Answer: 46 weeks.
Sketch of the solution. Suppose that for two consecutive weeks, the same set of 10 songs was on the list. Since the order of the songs is different, at least one song in the list has moved up in the ranking, and at least one has moved down. Since a song, once it has moved down, does not move up in th... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A student's income comes from three sources: scholarship, temporary part-time work, and parental support. If the government doubles the scholarship, the income will increase by $5 \%$. If the time spent on part-time work is doubled, the income will increase by $15 \%$. By what percentage will the student's income in... | Solution. Let $\mathrm{S}$ be the student's monthly income, $a, b$ and $c$ be the amounts of the scholarship, part-time job, and parental support, respectively (expressed, for example, in rubles). Clearly, $S=a+b+c$. Then, according to the conditions, $2 a+b+c=1.05 S$ and $a+2 b+c=1.15 S$. From the first equation, $a=0... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In the basket, there are fruits (no less than five). If you randomly pick three fruits, there will definitely be an apple among them. If you randomly pick four fruits, there will definitely be a pear among them. What fruits can be picked and in what quantities if you randomly pick five fruits? | Solution. From the condition of the problem, it follows that the "non-apples" in the box are no more than two fruits (otherwise, you could pull out 3 fruits, none of which would be apples). Similarly, "non-pears" are no more than three fruits (otherwise, you could pull out 4 fruits, none of which would be pears). Thus,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation: $\cos ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)+\sin ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)=\sin ^{-2}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. | # Solution.
Let's make the substitution $t=\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. Clearly, $0<t \leq \frac{\pi}{2}$.
$\cos ^{4} t+\sin ^{4} t=\sin ^{-2} t$.
$\cos ^{4} t+\sin ^{4} t=\left(\cos ^{2} t+\sin ^{2} t\right)^{2}-2 \sin ^{2} t \cdot \cos ^{2} t=1-2 \sin ^{2} t \cdot \cos ^{2} t \leq 1 . \frac{1}{\sin ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the height of the pyramid if, by cutting it only along the lateral edges and unfolding the lateral faces onto the plane of the base, outside of it, you get a square with a side of 18. If it is impossible to get such a square, explain why.
# | # Solution:
The figure shows such a net. If we flip the pyramid and place it on the face $A B C$, it is easy to calculate the volume of the pyramid $V=\frac{1}{3} \cdot \frac{1}{2} \cdot 9 \c... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. In each cell of a table consisting of 10 columns and $n$ rows, a digit is written. It is known that for any row $A$ and any two columns, there exists a row that differs from $A$ exactly in these two columns. Prove that $n \geqslant 512$.
(P. Karasev) | 10.1. Let $R_{0}$ be the first row of the table. Consider any set of an even number of columns and number them from left to right: $C_{1}, \ldots, C_{2 m}$. Then there is a row $R_{1}$ that differs from $R_{0}$ exactly in columns $C_{1}$ and $C_{2}$; further, there is a row $R_{2}$ that differs from $R_{1}$ exactly in ... | 512 | Combinatorics | proof | Yes | Yes | olympiads | false |
10.2. On the board, nine quadratic trinomials are written: $x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \ldots, x^{2}+a_{9} x+b_{9}$. It is known that the sequences $a_{1}, a_{2}, \ldots, a_{9}$ and $b_{1}, b_{2}, \ldots, b_{9}$ are arithmetic progressions. It turned out that the sum of all nine trinomials has at least o... | 10.2. Answer. 4.
Let $P_{i}(x)=x^{2}+a_{i} x+b_{i}, P(x)=P_{1}(x)+\ldots+P_{9}(x)$. Notice that $P_{i}(x)+P_{10-i}(x)=2 x^{2}+\left(a_{i}+a_{10-i}\right) x+\left(b_{i}+b_{10-i}\right)=$ $=2 P_{5}(x)$. Therefore, $P(x)=9 P_{5}(x)$, and the condition is equivalent to $P_{5}(x)$ having at least one root.
Let $x_{0}$ be ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Write the number 2021 using each of the digits from 0 to 9 exactly once, along with parentheses and arithmetic operations. (Parentheses and arithmetic operations can be used in any quantity. "Sticking" digits together to form a single number is allowed) | For example, $43 \cdot (8 \cdot 5 + 7) + 0 \cdot 1 \cdot 2 \cdot 6 \cdot 9 = 2021$. | 2021 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Anya has a chocolate bar of size 5 x 6 squares. It contains 6 squares, forming a $2 \times 3$ rectangle, in which there are nuts (the rectangle can be positioned either vertically or horizontally). Anya does not know exactly where the nuts are. She wants to eat the smallest number of squares, but in such a way that ... | 2. Answer: 5.
Solution. An example of which slices to eat.
Evaluation. Let's number the slices as shown in the figure.
| 1 | 1 | 2 | 2 | 3 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A mathematician and a physicist started running on a running track towards the finish line at the same time. After finishing, the mathematician said: "If I had run twice as fast, I would have beaten the physicist by 12 seconds." And after finishing, the physicist said: "If I had run twice as fast, I would have beate... | # Answer: 16.
Solution: Let the time it took the mathematician to run the entire distance be $2 x$ seconds, and the time it took the physicist be $2 y$ seconds. Then, under the condition that the mathematician ran twice as fast, we get $2 y-x=12$. And from the condition that the physicist ran twice as fast, we get $2 ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On the island, there live 7 natives who know mathematics and physics, 6 natives who know physics and chemistry, 3 natives who know chemistry and mathematics, and 4 natives who know physics and biology. In how many ways can a team of three people be formed who together know at least three subjects out of the four? Th... | # Answer: 1080.
## Solution.
Let's choose three people. If this trio includes representatives of at least two different groups of aborigines, then together they know no fewer than three subjects.
If this trio includes representatives of only one group, then together they know only two subjects.
Therefore, we will f... | 1080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Islandland consists of ten islands, some of which are connected by two-way air routes. If you choose any 9 islands, you can fly around them one by one and return to the starting island at the end. Find the minimum number of air routes that can exist in this country. | Answer: 15 airlines.
Solution. Evaluation. Let islands $A$ and $B$ be connected by an airline. If we choose 9 islands for the tour, excluding $A$, then a circular path passes through $B$, meaning $B$ is connected by airlines to at least two other islands. Thus, at least three airlines depart from $B$, and a similar st... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | 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?
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 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we get $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
 Mom is walking around a lake with a stroller and completes a full lap around the lake in 12 minutes. Vanya rides a scooter on the same path in the same direction and meets (overtakes) mom every 12 minutes. At what intervals will Vanya meet mom if he rides at the same speed but in the opposite direction? | Answer: Every 4 minutes.
Solution. Since Mom completes a full lap around the lake in 12 minutes and meets Vanya every 12 minutes, in 12 minutes Vanya rides around the lake exactly 2 times, while Mom completes one lap. Therefore, Vanya's speed is twice that of Mom's. From this, it follows that when Vanya was riding in ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 8.2. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander is 1 m
The third islander is 1 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then from his statement, it follows that the third one is also a knight; by the principle of exclusion, the second and fourth must be liars. The fourth said ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2.1. Given trapezoid $A B C D(A D \| B C)$. It turns out that $\angle A B D=\angle B C D$. Find the length of segment $B D$, if $B C=36$ and $A D=64$.
 | Answer: 48.
Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4.1. A rectangle was cut into six smaller rectangles, the areas of five of them are marked on the diagram. Find the area of the remaining rectangle.
 | Answer: 101.
Solution. Let's introduce the notation as shown in the figure.
Notice that
$$
2=\frac{40}{20}=\frac{S_{H I L K}}{S_{D E I H}}=\frac{H K \cdot H I}{H D \cdot H I}=\frac{H K}{H ... | 101 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. There is a set of 2021 numbers. Moreover, it is known that if each number in the set is replaced by the sum of the others, the same set will be obtained. Prove that the set contains a zero. | Solution. Let the sum of the numbers in the set be $M$, then the number $a$ in the set is replaced by the number $b=M-a$. Summing these equations for all $a$:
$$
b_{1}+\ldots+b_{2021}=2021 M-\left(a_{1}+\ldots+a_{2021}\right)
$$
from which $M=0$, since $b_{1}+\ldots+b_{2021}=a_{1}+\ldots+a_{2021}=M$. Therefore, for a... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
9.2. Toshа travels from point A to point B via point C. From A to C, Toshа travels at an average speed of 75 km/h, and from C to B, Toshа travels at an average speed of 145 km/h. The entire journey from A to B took Toshа 4 hours and 48 minutes. The next day, Toshа travels back at an average speed of 100 km/h. The journ... | Answer: 290 km.
Solution. Let $\mathrm{X}$ km be the distance between B and C, and $y$ km be the distance between A and C. From the system of equations $x / 145 + y / 75 = 24 / 5$ and $(x + y) / 100 = 2 + y / 70$, we find: $x = 290$ and $y = 210$.
. Two chips are called "connected" if they are in the same column or row, and there are no other chips between them. What is the maximum number of chips that can be placed on the board so that each has no more than two "con... | Answer: 35.
Solution. Let $x_{1}, x_{2}, \ldots, x_{20}$ be the number of chips in rows $1,2, \ldots, 20$, and $y_{1}, y_{2}, \ldots, y_{15}$ be the number of chips in columns $1,2, \ldots, 15$. Then the total number of chips $S=x_{1}+x_{2}+\ldots+x_{20}=y_{1}+y_{2}+\ldots+y_{15}$. The number of "connectivities" $k \g... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1. Petya thought of two numbers and wrote down their product. After that, he decreased the first of the thought numbers by 3, and increased the other by 3. It turned out that the product increased by 900. By how much would the product have decreased if Petya had done the opposite: increased the first number ... | Answer: 918.
Solution: Let these numbers be $a$ and $b$. Then, according to the condition, $(a-3)(b+3)-ab=600$. Expanding the brackets: $ab+3a-3b-9-ab=900$, so $a-b=303$. We need to find the difference $ab-(a+3)(b-3)=$ $ab-ab+3a-3b+9=3(a-b)+9=3 \cdot 303+9=918$. | 918 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Variant 1. At the intersection of perpendicular roads, a highway from Moscow to Kazan and a road from Vladimir to Ryazan intersect. Dima and Tolya set out with constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively. When Dima passed the intersection, Tolya had 3500 meters left to reach it. Wh... | Answer: 9100.
Solution. When Tolya has traveled 3500 meters, Dima will have traveled 4200 meters, so at the moment when Dima is 8400 meters past the intersection, Tolya will be 3500 meters past the intersection. By the Pythagorean theorem, the distance between the boys is 9100 meters. | 9100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Variant 1. Find the largest root of the equation
$(x+1)(x+2)-(x+2)(x+3)+(x+3)(x+4)-(x+4)(x+5)+\cdots-(x+1000)(x+1001)=0$. | Answer: -501.
Solution. Let's break the terms into pairs of adjacent terms and factor out the common factor, we get $(x+2)(x+1-x-3)+\cdots+(x+1000)(x+999-x-1001)=-2 \cdot(x+2+x+4+\cdots+x+1000)=$ $-2 \cdot\left(500 x+\frac{2+1000}{2} \cdot 500\right)=-1000 \cdot(x+501)=0$. From this, $x=-501-$ the only root. | -501 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. Variant 1
It is known that $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{1}{5}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{4}{5}}$. Find $\cos (\alpha-\beta)+\cos (\beta-$ $\gamma)+\cos (\gamma-\alpha)$ | Answer: -1.
Solution. Consider the expression $(\cos \alpha+\cos \beta+\cos \gamma)^{2}+(\sin \alpha+\sin \beta+\sin \gamma)^{2}$ and expand the brackets: $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2(\cos \alpha \cdot \cos \beta+\cos \alpha \cdot \cos \gamma+\cos \beta \cdot \cos \gamma)+\sin ^{2} \alpha+\sin ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 6. Option 1
Initially, there were 20 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{25}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{16}... | Answer: 4.
Solution: Let there be $a$ white balls, $b$ blue balls, and $c$ red balls in the box. We can set up the following equations:
$$
\begin{gathered}
a+b+c=20 \\
\frac{a}{20}=\frac{a}{20+b}+\frac{1}{25} \\
\frac{b}{20}+\frac{1}{16}=\frac{b}{20-a}
\end{gathered}
$$
Transform the second equation: $\frac{a b}{20(... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1
Unit cubes were used to assemble a large parallelepiped with sides greater than 4. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 836. Find the number of cubes that have no more th... | Answer: 144.
Solution. Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$ and $c-2$ is greater than 2 and their product equals the product of four prime numbers $2,2,11$ and 19, we ha... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
At the base of the quadrilateral pyramid $S A B C D$ lies a square $A B C D, S A$ - the height of the pyramid. Let $M$ and $N$ be the midpoints of the edges $S C$ and $A D$. What is the maximum value that the area of triangle $B S A$ can have if $M N=3 ?$ | Answer: 9.
Solution: Let $O$ be the center of the square $ABCD$. Then $MO$ is the midline of the triangle $SAC$, so $SA = 2MO$. Similarly, $ON$ is the midline of the triangle $BDA$, so $AB = 2ON$. Therefore, $SA^2 + AB^2 = 4(MO^2 + ON^2) = MN^2 = 36$. Let $SA = x, AB = y$. From the formula $S = 0.5 \cdot SA \cdot AB =... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the village, seven people live. Some of them are liars (always lie), and the rest are knights (always tell the truth). Each of them said about each of the others whether they are a knight or a liar. Out of the 42 answers received, 24 were “He is a liar.” What is the smallest number of knights that can live in t... | Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of knight-liar, the phrase "He is a liar" will be said twice. Since this phr... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.5. Three hundred non-zero integers are written in a circle such that each number is greater than the product of the next three numbers in the clockwise direction. What is the maximum number of positive numbers that can be among these 300 written numbers? | Answer: 200.
Solution: Note that three consecutive numbers cannot all be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number preceding them (counterclockwise) is also a natural number. Since it is greater than the product of these three natural numbers, it is greater ... | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. Misha and Grisha are writing numbers on the board. Misha writes threes, and Grisha writes fives. In total, they wrote 20 numbers. How many fives were written if the sum of all the numbers is 94? | Answer: 17
Solution. If all 20 numbers were "5", their sum would be 100. We will replace "5" with "3". With each replacement, the sum decreases by 2. Since $100-94=6, 6: 2=3$, we need 3 replacements. Therefore, the number of fives is $20-3=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Five identical squares, standing in a row, were cut by two horizontal lines. The sum of the perimeters of the resulting 15 rectangles is 800 cm. Indicate in centimeters the length of the original squares.
| | | | | |
| :--- | :--- | :--- | :--- | :--- |
| | | | | | | # Answer: 20
Solution. Let's calculate how many times the side of the original square is repeated in the sum of all perimeters. The sides of the rectangle are counted once (a total of 12), and the crossbars are counted twice ($4 \cdot 2 + 10 \cdot 2 = 28$). In total, $40.800: 40=20$ cm - the side of the square. | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. In a row, 64 people are standing - each one is either a knight, who always tells the truth, or a liar, who always lies. One of the standing knights said that he stands next to a knight and a liar, and all the other 63 people repeated his phrase. Indicate how many of them were knights. | Answer: 42
Solution. The people at the ends must be liars, as they do not have a second neighbor, and by saying this phrase, those standing there lied. All others can be liars, in which case there are no knights at all, but this option contradicts the condition that at least one knight is present. If there is a knight... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. In a row, there are 27 matchboxes, each containing a certain number of matches. It is known that in any four consecutive boxes, the total is 25 matches, and in all of them, the total is 165. How many matches are in the eighth box? | Answer: 10
Solution. In the first 24 boxes, there are a total of $6 \cdot 25=150$ matches. In the last three boxes, there are 15 matches. Therefore, the 4th from the end (or 24th from the start) has 10 matches. Then, in boxes $24, 23, 22, 21$, there are 25 matches in total, meaning in boxes $21, 22, 23$, there are 15 ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A $7 \times 7$ board has a chessboard coloring. In one move, you can choose any $m \times n$ rectangle of cells and repaint all its cells to the opposite color (black cells become white, white cells become black). What is the minimum number of moves required to make the board monochromatic?
Answer: in 6 moves. | Solution. Consider segments of length equal to the side of a cell, separating pairs of cells adjacent to the side of the board. Along each side of the board, there are 6 such segments, totaling $6 \cdot 4=24$. Each of these segments separates cells that initially have different colors, so each segment must end up on th... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There is a ruler 10 cm long without divisions. What is the smallest number of intermediate divisions that need to be made on the ruler so that segments of length 1 cm, 2 cm, 3 cm, ..., 10 cm can be laid off, applying the ruler in each case only once. | Answer: 4
Solution: First, let's prove that three divisions are not enough. There are a total of 10 segments with endpoints at five points. Therefore, if three divisions are made, each length from 1 to 10 should be obtained exactly once. If any division is made at a non-integer distance from the left end, there will b... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.10. In the company, there are 100 children, some of whom are friends (friendship is always mutual). It is known that by selecting any child, the remaining 99 children can be divided into 33 groups of three such that in each group, all three are pairwise friends. Find the minimum possible number of pairs of friends.
... | Answer: 198.
Solution: Let's translate the problem into the language of graphs, associating each child with a vertex and each friendship with an edge. Then we know that in this graph with 100 vertices, after removing any vertex, the remaining vertices can be divided into 33 triples such that the vertices in each tripl... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
, adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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