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1. The head of the fish weighs as much as the tail and half of the body, the body weighs as much as the head and the tail together. The tail weighs 1 kg. How much does the fish weigh? | Answer: 8 kg.
Solution 1. The body weighs as much as the head and tail, i.e., two tails and half the body. This means that half the body weighs as much as two tails, i.e., the body weighs 4 kg. Then the head weighs $1+2=3$ kg, and the whole fish weighs $4+3+1=8$ kg.
Solution 2. Let $\Gamma, T, X$ be the weight of the... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The sum of the minuend, subtrahend, and difference is 555. Can the minuend be an integer? If yes, provide an example; if no, explain why. | # Answer. Hem.
Solution. Since the sum of the subtrahend and the difference is equal to the minuend, then the sum of the minuend, subtrahend, and difference is equal to twice the minuend, i.e., the minuend is 555/2 - a non-integer. | 555 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a psychiatric hospital, there is a chief doctor and many lunatics. During the week, each lunatic bit someone (possibly even themselves) once a day. At the end of the week, it turned out that each of the patients had two bites, and the chief doctor had a hundred bites. How many lunatics are there in the hospital | Answer: 20 madmen.
Solution: Let there be n madmen in the hospital. Then, by the end of the week, on the one hand, 7n bites were made, and on the other hand, $2 n+100$. i.e.
$7 n=2 n+100$, from which $n=20$. | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Real numbers $x, y, z$ (non-zero) satisfy the equation: $x+y=z$. Find all possible values that the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}$ can take. | 1. Answer: 3.
Solution. Transform the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}=$
$=\frac{z}{x}-\frac{y}{x}+\frac{z}{y}-\frac{x}{y}+\frac{y}{z}+\frac{x}{z}=\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}$.
Using the condition $x+y=z$ and the derived formulas $y=z-x$ and $x=$ $z-y$... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. For different numbers $a$ and $b$, it is known that $\frac{a}{b}+a=\frac{b}{a}+b$. Find $\frac{1}{a}+\frac{1}{b}$. | Answer: -1.
Solution. The given equality can be written as $\frac{a}{b}-\frac{b}{a}=b-a$, from which $\frac{a^{2}-b^{2}}{a b}=b-a$ or $\frac{(a-b)(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the requi... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked respectively such that $\angle B A M = \angle C K M = 30^{\circ}$. Find $\angle A K D$. | Answer: $75^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see Fig. 8.5a). Further reasoning can be done in different ways.
First method. Let $A H$ be the perpendicular from vertex $A$ to $M K$. Then the right triangle... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.6. Sasha drew a square of size $6 \times 6$ cells and alternately colors one cell at a time. After coloring the next cell, he writes down the number of colored cells adjacent to it. After coloring the entire square, Sasha adds up the numbers written in all the cells. Prove that no matter in what order Sasha colors th... | Solution. Consider all unit segments that are common sides for two cells. There are exactly sixty such segments - 30 vertical and 30 horizontal. If a segment separates two shaded cells, we will say that it is "painted." Note that when Sasha writes a number in a cell, he indicates the number of segments that were not pa... | 60 | Combinatorics | proof | Yes | Yes | olympiads | false |
Task 1. In a confectionery store, three types of candies are sold: caramels for 3 rubles, toffees for 5 rubles, and chocolates for 10 rubles. Varya wanted to buy exactly 8 candies of each type and took 200 rubles with her. In the morning, she saw announcements in the store: “When paying for three chocolates, get a free... | Answer: 72 rubles.
Solution. Since Varya will buy more than six but less than nine chocolates, she will get two free caramels. Then she will need to buy another 6 caramels, for which she will get two free toffees, after which she will need to buy another 6 toffees. Then she will spend a total of $8 \cdot 10 + 6 \cdot ... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. During a math test, Oleg was supposed to divide a given number by 2, and then add 6 to the result. But he hurried and instead multiplied the given number by 2, and then subtracted 6 from the result. Nevertheless, he got the correct answer. What number was given to Oleg | Answer: 8.
Solution. Since Oleg multiplied the number by 2 instead of dividing it, he got a result that is four times greater than the required one. This means that the difference between these two results is three times greater than the required result. However, according to the condition, this difference is equal to... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. In an album, a grid rectangle $3 \times 7$ is drawn. Robot Igor was asked to trace all the lines with a marker, and it took him 26 minutes (the robot draws lines at a constant speed). How many minutes will it take him to trace all the lines of a grid square $5 \times 5 ?$ | Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = \frac{1}{2}$ minutes on the side of one cell.
The rec... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. A messenger was riding a horse to deliver a message to Ilya Muromets. At some point, he noticed that Ilya Muromets had passed by him (and continued walking in the opposite direction). After 10 seconds (when the horse stopped), the messenger dismounted and ran to catch up with Ilya. How many seconds will it t... | Answer: 110.
Solution. Let Ilya Muromets walk at a speed of $x$ meters per second. Then the messenger's speed is $2x$, and the horse's speed is $10x$. Therefore, 10 seconds after the meeting, the distance between the messenger and the horse will be $10 \cdot x$ (Ilya Muromets has walked) $+10 \cdot 10 x$ (the horse ha... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. In the kingdom, there live counts, dukes, and marquises. One day, each count dueled with three dukes and several marquises. Each duke dueled with two counts and six marquises. Each marquise dueled with three dukes and two counts. It is known that all counts dueled with an equal number of marquises. How many ... | Answer: With 6 marquises.
Solution. Let there be $x$ counts, $y$ dukes, and $z$ marquises in the kingdom. Each count fought with three dukes, so there were $3 x$ duels between counts and dukes. But each duke fought with two counts, meaning there were $2 y$ such duels. Therefore, $3 x=2 y$.
Each duke fought with six m... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. On an island, there live knights who always tell the truth, and liars who always lie. One day, 15 natives, among whom were both knights and liars, stood in a circle, and each said: "Of the two people standing opposite me, one is a knight, and the other is a liar." How many of them are knights? | Answer: 10 knights.
Solution. Consider any knight. He tells the truth, which means that opposite him stand a knight and a liar. One of the people opposite the found liar is the initial knight, so next to him stands another knight. Opposite this new knight stands the previously found liar and another person, who must b... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Some squares of the table are mined. Each number written in a square shows the number of mines in the squares adjacent to the given square. (See the figure. Adjacent are squares that have at least one common point; a square with a number is not mined). In how many ways can the mines be placed in the table? Justify y... | # Answer: 14.
Solution. The mine adjacent to the number 1 is in the second, third, or fourth column.
1 case. If the mine is in the second column, it has three possible positions, and it is adjacent to the number two in the first column, which means the second mine adjacent to this two is in the first column, and it h... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A family of four octopuses came to a shoe store (each octopus has 8 legs). The father octopus already had half of his legs booted, the mother octopus had only 3 legs booted, and their two sons had 6 legs booted each. How many boots did they buy if they left the store fully booted? | Answer: 13.
Solution. The daddy octopus had half of his legs booted, that is, 4 legs. Thus, 4 legs were not booted.
The mommy octopus had 3 legs booted, meaning 5 legs were not booted.
Each of the two sons had 6 legs booted, meaning 2 legs were not booted.
Thus, a total of $4+5+2+2=13$ boots were bought.
## Criter... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task No. 1.1
## Condition:
On the potion-making exam, each student at Hogwarts School had to brew 4 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same s... | # Answer: 36
## Exact match of the answer -1 point
## Solution.
Let's calculate the potion-making speed of each of the three students, measured in potions per hour. Hermione made 4 potions in half an hour, so in one hour she would make twice as many potions, which is 8. Therefore, her speed is 8 potions per hour. Ha... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.3
## Condition:
At the Potions exam, each student of Hogwarts School had to brew 8 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron needed 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speed... | Answer: 72
Exact match of the answer -1 point
Solution by analogy with task №1.1
# | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.4
## Condition:
At the potion-making exam, each student from Hogwarts School had to brew 10 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the sam... | Answer: 90
Exact match of the answer -1 point
Solution by analogy with task №1.1
# | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.1
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 5 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables i... | Answer: 43750000
Exact match of the answer -1 point
## Solution.
Note that the number of vowels cannot be more than half of the total number of letters in the word, that is, more than 4, since in each syllable with a vowel there is at least one consonant. Also, the number of vowels cannot be less than $1 / 3$ of the... | 43750000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.2
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 8 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables i... | Answer: 194560000
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 194560000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.4
## Condition:
In a certain language $\mathrm{N}$, letters denote a total of 20 consonants and 3 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into sylla... | Answer: 272160000
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 272160000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.1
## Condition:
Given a triangle $\mathrm{ABC}$, in which $\mathrm{AB}=5$. The median $\mathrm{BM}$ is perpendicular to the bisector $\mathrm{AL}$. Find $\mathrm{AC}$. | Answer: 10
Exact match of the answer -1 point
Solution.
Consider triangle ABM: since the bisector drawn from vertex A is perpendicular to side $\mathrm{BM}$, triangle ABM is isosceles.
Therefore, $\mathrm{AB}=\mathrm{AM}=\mathrm{MC}$. Hence, $\mathrm{AC}=10$.
# | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.1
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid ... | Answer: 6
Exact match of the answer -1 point
## Solution.
It is claimed that each painting by Ivan Konstantinovich was sold for $1000 \times 2^{\text {x }}$ rubles, where $\mathrm{x}$ is some natural number. Indeed, initially, the painting's price is 1000 rubles, and each participant doubles its price, meaning the p... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.2
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid ... | Answer: 4
Exact match of the answer - 1 point
Solution by analogy with task №7.1
# | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 7.3
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid on... | Answer: 6
Exact match of the answer - 1 point
Solution by analogy with task №7.1
# | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 8.1
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=74^{\circ}$. On ray $\mathrm{BC}$, segment $\mathrm{CD}=\mathrm{CB}$ is laid out. Then, from point $\mathrm{D}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, dr... | Answer: 37
## Exact match of the answer -1 point
## Solution.
Let $\mathrm{F}^{\prime}$ be the midpoint of segment $\mathrm{AC}$. Then $\mathrm{F}^{\prime} \mathrm{C}=0.5 \mathrm{AC}=\mathrm{BC}=\mathrm{CD}=0.5 \mathrm{BD}$. From this, it follows that triangle BDF' is a right triangle, since its median is equal to h... | 37 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 8.2
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=106^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CX}=$ CB is laid out. Then, from point $\mathrm{X}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, draw... | Answer: 53
Exact match of the answer -1 point
Solution by analogy with task №8.1
# | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8.3
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=46^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CM}=$ CB is marked. Then, from point M, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\... | Answer: 23
Exact match of the answer - 1 point
Solution by analogy with task №8.1
# | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2. In triangle $ABC$, the median $BM$ was drawn. It turned out that $\angle ABM=40^{\circ}$, $\angle MBC=70^{\circ}$. Find the ratio $AB: BM$. Justify your answer.
To solve problem 9.2 Sol... | Answer: $A B: B M=2$.
| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| An example of a triangle satisfying the conditions of the problem is provided | 0 points |
| Incorrect solution or its absence | 0 points | | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. After the bankruptcy of the company "Horns and Hooves," 17 horns, 2 hooves, and one weight remained. All this wealth was divided equally by weight between Pankovskiy and Balaganov, with the weight entirely going to Balaganov. The horns and hooves were not cut into pieces. Each horn is heavier than each hoof and li... | Solution: Let one hoof weigh $k$, and one horn weigh $k+\delta$ (all weights in the same units of measurement, for example, in puds). Then, according to the condition, the weight of the weight (the iron piece) is $k+2\delta$, and the total weight of the divided property is $20k+19\delta$. Each person received $10k+9.5\... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Masha is distributing tennis balls into identical boxes. If she uses 4 boxes, there is still room for 8 more balls in the last box, and if she uses 3 boxes, 4 balls will not fit into the boxes. How many balls is one box designed to hold? | Answer: 12
Solution.
Method 1
Notice that if the balls are in 3 boxes, there are 4 tennis balls left. Take these 4 balls and put them in the fourth box. According to the condition, there will still be room for 8 more balls in this box. This means that the box can hold a total of 12 balls.
Method 2.
Let the box hol... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1.
In the number, two digits were swapped, and as a result, it increased by more than 3 times. The resulting number is 8453719. Find the original number. | Answer: 1453789.
Solution. When the number 8453719 is reduced by at least 3 times, the result is a number less than 3000000. Therefore, 8 was swapped with a digit less than 3, and the only such digit is 1. | 1453789 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to ... | Answer: 23
Solution: In the period from April 10 to April 30, there are exactly 21 days. Dividing this period into three weeks: from April 10 to April 16, from April 17 to April 23, and from April 24 to April 30, we get exactly one weekend (Monday) in each of the three weeks. Therefore, in the second statement, Kolya ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Variant 2.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to Apri... | Answer: 11
Option 3.
The cafe "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "the cafe was open for 18 days from April 1 to April 20" and "the cafe was also open for 18 days from April 10 to April 30." It is known that he was wrong once. How many days was the cafe open from Ap... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
On a grid paper, a right triangle with sides 6 and 10 is drawn (see figure). Find the total length of the horizontal grid lines inside this triangle.
 | Answer: 27.
Solution. Extend the triangle to form a rectangle.
It is divided into two identical triangles, so the total length of the horizontal lines in the rectangle is twice the desired ... | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
Petya wrote 9 consecutive natural numbers on the board. Kolya calculated their sum and got the answer 43040102. It turned out that he made a mistake only in the first digit of the sum. What should the first digit be | Answer: 8.
Solution: Let the consecutive natural numbers be $a-4, a-3, \cdots, a+3, a+4$. The sum of nine consecutive numbers is divisible by 9, indeed, this sum is equal to $(a-4)+(a-3)+\cdots+(a+3)+(a+4)=9a$. Therefore, the desired sum must be divisible by 9. This means that the sum of the digits of the obtained ans... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1.
101 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there will be at least two even numbers. What is the minimum number of even numbers that can be among the written numbers? | Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there are at least two even numbers. Thus, the total number of even numbers is no less than $1+2 \cdot 20=41$. Such a situation is... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 3 blue points. And on any segment with endpoints at blue points, containing 2 blue points inside, there are at least 2 red point... | # Answer. 3.
Solution. Note that on a segment with endpoints at red points, not containing other red points, there cannot be 4 blue points. Indeed, in this case, between the outermost blue points there will be 2 blue points, which means there will be at least 2 more red points. Therefore, there are no more than 3 blue... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. In the castle, there are 25 identical square rooms arranged in a $5 \times 5$ square. In these rooms, 25 people—liars and knights (liars always lie, knights always tell the truth)—have settled, one person per room. Each of these 25 people said: "At least one of the rooms adjacent to mine is occupied by a liar." Wh... | Answer: 13 liars.
Solution. Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's select one corner room, and divide the remaining rooms into 12 pairs of adjacent rooms. Then, in each pair of rooms, there can be no more than one liar. Therefore, the total number of liars can... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.2. Petya runs down from the fourth floor to the first floor 2 seconds faster than his mother rides the elevator. Mother rides the elevator from the fourth floor to the first floor 2 seconds faster than Petya runs down from the fifth floor to the first floor. How many seconds does it take Petya to run down from the f... | Answer: 12 seconds.
Solution. Between the first and fourth floors, there are 3 flights, and between the fifth and first floors, there are 4. According to the problem, Petya runs 4 flights in 2 seconds longer than it takes his mother to ride the elevator, and 3 flights in 2 seconds less than his mother. Therefore, Pety... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play? | Answer: 11 games.
Solution: According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of any two consecutive games, the first player must participate in at least one, which means there were no more than \(2 \cdot 10 + 1 = 21\) games. Therefore, a total of 21 games wer... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.4. Let x be some natural number. Among the statements:
$2 \times$ greater than 70
x less than 100
$3 x$ greater than 25
x not less than 10
x greater than 5
three are true and two are false. What is x | Answer: 9.
Solution
The first statement is equivalent to x > 35, and the third statement is equivalent to $\mathrm{x}>8$, since x is a natural number. The first statement is false, because if it were true, then the third, fourth, and fifth statements would also be true, i.e., there would be at least 4 true statements... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.2. The magpie-crow was cooking porridge and feeding her chicks. The third chick received as much porridge as the first two combined. The fourth one received as much as the second and third. The fifth one received as much as the third and fourth. The sixth one received as much as the fourth and fifth. And the seventh ... | Answer: 40 g.
Solution. Let the first chick receive $m$ g of porridge, and the second chick $-n$ g. Then the third chick received $m+n$ (g), the fourth chick $-m+2n$ (g), the fifth chick $-2m+3n$ (g), and the sixth chick $-3m+5n$ (g). Therefore, the total amount of porridge was: $m+n+(m+n)+(m+2n)+(2m+3n)+(3m+5n)=8m+12... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. $A B C D$ - a convex quadrilateral. It is known that $\angle C A D=\angle D B A=40^{\circ}, \angle C A B=60^{\circ}$, $\angle C B D=20^{\circ}$. Find the angle $C D B$. | Answer: $30^{\circ}$.
Solution. Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, triangle $A B C$ is equilateral (see Fig. 8.3a). We can reason in different ways from here.
 stands up and leaves. What is the maximum number of people that can be sitting simultaneously, if initially all the chairs were empty? | Answer: 11.
Solution. Evaluation. Note that it is impossible to occupy all chairs simultaneously, as at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.
Example. We will show how to seat 11 people. ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) In a three-digit number, the first digit (hundreds place) was increased by 3, the second digit by 2, and the third digit by 1. As a result, the number increased by 4 times. Provide an example of such an original number.
Answer: 107. | Solution. The answer can be found in the following way. Let $x$ be the desired number. Then the condition of the problem immediately leads to the equation $x+321=4x$, the only solution of which is $x=107$.
Criteria. Correct answer, even without any comments: 7 points.
Incorrect answer: 0 points. | 107 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) A movie ticket cost 300 rubles. When the price was reduced, the number of visitors increased by 50 percent, and the cinema's revenue increased by 35 percent. How many rubles does one ticket cost now? | Answer: 270.
Solution. Let the price of the new ticket be $s$ rubles. Suppose the initial number of visitors was $N$, and after increasing by $50 \%$ it became $1.5 N$. Then, according to the condition, the current revenue of the cinema $1.5 N \cdot s$ is $35 \%$ more than $N \cdot 300$, from which we have $1.5 N s = ... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Given an arithmetic progression. The sum of its first 10 terms is 60, and the sum of its first 20 terms is 320. What can the 15th term of this progression be? | Answer: 25.
Solution. Let the first term of the sequence be $a$, and the common difference be $b$. Then the sum of the first 10 terms is $a+(a+b)+\ldots+(a+9b)=$ $10a+45b$. The sum of the first 20 terms is $a+(a+b)+\ldots+(a+19b)=$ $20a+190b$. According to the problem, $10a+45b=60, 20a+190b=320$. Solving the system, w... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Consider the equation $\sin ^{3}(x)+\cos ^{3}(x)=-1$. How many solutions does it have on the interval $[0 ; 6 \pi]$?
Answer: 6. | Solution. From the main trigonometric identity, we have $\sin ^{2}(x)+$ $\cos ^{2}(x)=1$. Adding this to the given equality, we get
$$
0=\sin ^{2}(x) \cdot(1+\sin (x))+\cos ^{2}(x) \cdot(1+\cos (x))
$$
In this expression, all factors are non-negative, so both terms $\sin ^{2}(x)(1+\sin (x))$ and $\cos ^{2}(x)(1+\cos ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that this polygon has no parallel sides? | Answer. For $k=1509$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=$ $A_{2}$). We will divide them into quadruples of points $\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right)$, $\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \... | 1509 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the problem: octopuses with an even number of legs always lie, while octopuses with an odd number of legs always tell the truth. Five octopuses met, each having between 7 and 9 legs.
The first said: "We have 36 legs in total";
The second: "We have 37 legs in total";
The third: "We have 38 legs in total";
T... | Solution. All answers are different, so one is telling the truth or all are lying. If all were lying, they would have 8 legs, i.e., a total of 40, which matches the last answer, a contradiction.
Therefore, four are lying, one is telling the truth, and the one telling the truth has an odd number of legs since $4 \cdot ... | 39 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Two cars started from the same point on a circular track 150 km long, heading in opposite directions. After some time, they met and continued moving in the same directions. Two hours after the start, the cars met for the second time. Find the speed of the second car if the speed of the first car is 60 km/h. | Solution. Let the speed of the first car be $x$ km/h, then their closing speed will be $(x+60)$ km/h. Obviously, in two hours they traveled two laps, i.e., 300 km, hence $2 \cdot(x+60)=300$, from which we get that $x=90($ km/h).
Answer: 90 km/h | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. At the railway platform, many people gathered early in the morning waiting for the train. On the first train, one-tenth of all those waiting left, on the second train, one-seventh of the remaining left, and on the third train, one-fifth of the remaining left. How many passengers were initially on the platform if 216... | Solution. Solving from the end, we get that before the departure of the third train, there were $216: 6 \cdot 5=270$ passengers on the platform, before the departure of the second - $270: 6 \cdot 7=315$ passengers, before the departure of the first - $315: 9 \cdot 10=350$ passengers.
Answer: 350 passengers. | 350 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. On the board, there are $n$ different integers. The product of the two largest is 77. The product of the two smallest is also 77. For what largest $n$ is this possible?
(R. Zhenodarov, jury) | Answer. For $n=17$.
Solution. The numbers $-11, -7, -6, -5, \ldots, 6, 7, 11$ provide an example for $n=17$.
Assume that there are at least 18 such numbers. Then, at least 9 of them will have the same sign (all positive or all negative). Among these 9 numbers, the absolute values of the two largest will be at least 8... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. Find any solution to the puzzle
$$
\overline{A B}+A \cdot \overline{C C C}=247
$$
where $A, B, C$ are three different non-zero digits; the notation $\overline{A B}$ represents a two-digit number composed of the digits $A$ and $B$; the notation $\overline{C C C}$ represents a three-digit number consisting... | Answer: 251.
Solution. The number $\overline{C C C}$ is divisible by 111 and is less than 247, so $A \cdot \overline{C C C}$ is either 111 or 222. In the first case, we get that $\overline{A B}=247-111=136$, which is impossible. In the second case, $\overline{A B}=247-222=25$, that is, $A=2, B=5$, and therefore, $C=1$... | 251 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. At a ball, princesses and knights gathered - a total of 22 people. The first princess danced with seven knights, the second with eight knights, the third with nine knights, ..., the last danced with all the knights present. How many princesses were at the ball? | Answer: 8.
Solution. Note that the number of knights the princess danced with is 6 more than her number. Let there be $x$ princesses in total, then the last one has the number $x$ and danced with all the knights, and there are $x+6$ of them in total. We get that there were $x+$ $(x+6)=2x+6=22$ people at the ball, whic... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.3. Dima, Misha, and Yura decided to find out who among them is the most athletic. For this, they held 10 competitions. The winner received 3 points, the second place 1 point, and the third place received nothing (in each competition, there was a first, second, and third place). In total, Dima scored 22 points... | Answer: 10.
Solution. In each competition, the boys in total received $3+1+0=4$ points. For all competitions, they scored $4 \cdot 10=40$ points. Dima and Misha in total scored $22+8=30$ points, so the remaining $40-30=10$ points were scored by Yura. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.4. On her birthday, Katya treated her classmates with candies. After giving out some candies, she noticed that she had 10 more candies left than Artem received. After that, she gave everyone one more candy, and it turned out that all the children in the class (including Katya) had the same number of candies. ... | Answer: 9.
Solution. Initially, the number of candies Kati and Artyom had differed by 10. When Kati gave everyone one more candy, the number of candies Artyom had increased by 1, and the number of candies Kati had decreased by the number of her classmates, and they ended up with the same amount. This means that 10 is ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. Cat Matroskin, Uncle Fyodor, Postman Pechkin, and Sharik sat down at a round table. In front of each of them was a plate with 15 sandwiches. Every minute, three of them ate a sandwich from their own plate, while the fourth ate a sandwich from their neighbor's plate. After 5 minutes of the meal, there were ... | # Answer: 7
Solution. We will call the sandwiches eaten from a neighbor's plate stolen. Note that in 5 minutes, exactly 5 sandwiches were stolen.
At the same time, 7 sandwiches disappeared from Uncle Fyodor's plate, of which he himself ate no more than 5, meaning that at least 2 were stolen. From this, it is clear th... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101$? | 1. Answer. It exists.
For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$. | 1234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | 2. Answer: 8.
Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent to each... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent? | 3. Answer. 5 cupcakes.
The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote $x, y$, and $z$ as the costs of a cake, a cupcake, and a bagel respectively, we get the equation: $(x+2 y+3 z)+(3 x+z)=12 y, \quad$ from which it follows that $\quad 4 x+4 z=10 y$, \quad t... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $6 \times 6$ square of cells so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.) | 5. Answer. 6.
Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted.
Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get that at l... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. Three pirates divided the diamonds they had obtained during the day in the evening: twelve each for Bill and Sam, and the rest went to John, who couldn't count. At night, Bill stole one diamond from Sam, Sam stole one from John, and John stole one from Bill. As a result, the average weight of Bill's diamonds decre... | Answer: 9 diamonds.
Solution. The first method (arithmetic). Note that the number of diamonds each pirate has did not change overnight. Since Bill has 12 diamonds, and their average weight decreased by 1 carat, the total weight of his diamonds decreased by 12 carats. Similarly, Sam also has 12 diamonds, and their aver... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. The numbers $x, y, z$ are such that $x \in[-3,7], y \in[-2,5], z \in[-5,3]$.
(a) (1 point) Find the smallest possible value of the quantity $x^{2}+y^{2}$.
(b) (3 points) Find the smallest possible value of the quantity $x y z - z^{2}$. | # Answer:
(a) (1 point) 0.
(b) (3 points) -200.
Solution. (a) Note that $x^{2} \geqslant 0$ and $y^{2} \geqslant 0$, so $x^{2}+y^{2} \geqslant 0$. The value $x^{2}+y^{2}=0$ is possible when $x=0, y=0$.
(b) Note that $|x y z|=|x||y||z| \leqslant 7 \cdot 5 \cdot 5$ and $z^{2} \leqslant 5^{2}$, so $x y z-z^{2} \geqsla... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. Given a rectangle $A B C D$. A line passing through vertex $A$ and point $K$ on side $B C$ divides the entire rectangle into two parts, the area of one of which is 5 times smaller than the area of the other. Find the length of segment $K C$, if $A D=60$.
Fig. 1: to the solution of problem 8.2
Solution. Draw a line through $K$ parallel to $AB$. Let it intersect side $AD$ at point $L$ (Fig. 1), then $ABKL$ and $DCKL$ are rectangl... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In a row, there are 127 balls, each of which is either red, green, or blue. It is known that
- there is at least one red, at least one green, and at least one blue ball;
- to the left of each blue ball, there is a red ball;
- to the right of each green ball, there is a red ball.
(a) (1 point) What is the... | # Answer:
(a) (1 point) 125.
(b) (3 points) 43.
Solution. (a) Among 127 balls, there is at least 1 green and at least 1 blue, so there are no more than 125 red balls. Note also that there can be exactly 125 if the leftmost ball is blue, the rightmost ball is green, and all 125 balls between them are red.
(b) Suppos... | 125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given a right isosceles triangle $ABC$ with a right angle at $A$. A square $KLMN$ is positioned as shown in the figure: points $K, L, N$ lie on sides $AB, BC, AC$ respectively, and point $M$ is located inside triangle $ABC$.
Find the length of segment $AC$, if it is known that $AK=7, AN=3$.
Fig. 2: to the solution of problem 8.7
Solution. Mark a point $H$ on the segment $B K$ such that $L H \perp B K$ (Fig. 2). Triangle $B H L$ is a right isosceles triangle, $H B=... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar? | Answer: 16.
Solution. If the first jar contains no less than 11 beetles, then the second jar contains no less than 12, the third jar no less than 13, ..., and the tenth jar no less than 20. And in all jars, there are no less than $11+12+\ldots+20=155$ beetles. Contradiction. Therefore, the first jar contains no more t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.2 (7 points)
The shares of the company "Nu-i-Nu" increase in price by 10 percent every day. Businessman Borya bought shares of the company for 1000 rubles every day for three days in a row, and on the fourth day, he sold them all. How much money did he make from this operation? | Solution: $1000 \cdot 1.1^{3}+1000 \cdot 1.1^{2}+1000 \cdot 1.1-3 \cdot 1000=1331+1210+1100-3000=641$.
| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Correct approach with arithmetic error | 4 |
| Correct answer without justification | 0 |
## Answer: 641 rubles
# | 641 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8.5 (7 points)
Find the smallest natural number that is divisible by $48^{2}$ and contains only the digits 0 and 1. | Solution:
$48^{2}=2^{8} \cdot 3^{2}$. For a number to be divisible by $2^{8}$, it must end with at least 8 zeros according to the divisibility rule, because otherwise, with fewer zeros ($n \leq 7$), it would have the form $1 \ldots .1 \cdot 10^{\text {n }}$ and would only be divisible by the n-th power of two, but we ... | 11111111100000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. In a store, there are 20 items, the costs of which are different natural numbers from 1 to 20 rubles. The store has decided to have a promotion: when buying any 5 items, one of them is given as a gift, and the customer chooses which item to receive for free. Vlad wants to buy all 20 items in this store, pa... | Answer: 136.
Solution. Vlad can take advantage of the offer no more than 4 times, so he will get no more than 4 items for free. The total cost of these 4 items does not exceed $17+18+$ $19+20$ rubles. Therefore, the rubles Vlad needs are not less than
$$
(1+2+3+\ldots+20)-(17+18+19+20)=1+2+3+\ldots+16=\frac{16 \cdot ... | 136 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. Vanya thought of two natural numbers, the product of which equals 7200. What is the greatest value that the GCD of these numbers can take? | Answer: 60.
Solution. Since each of these numbers is divisible by their GCD, their product is divisible by the square of this GCD. The greatest exact square that divides the number $7200=2^{5} \cdot 3^{2} \cdot 5^{2}$ is $3600=\left(2^{2} \cdot 3 \cdot 5\right)^{2}$, so the GCD of the two numbers in question does not ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are whole numbers of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
. Each of them said one of two phrases:
- "Among those gathered, at least 5 liar... | Answer: 70.
Solution. Suppose there are at least 11 liars. Arrange the numbers on their T-shirts in ascending order and select the liar with the 6th number. Then he must be telling the truth, as there are at least 5 liars with a smaller number and at least 5 liars with a larger number. Thus, there are no more than 10 ... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25, CP = 20, \angle APC = \angle A + \angle B$.
$, whose leading coefficient is 1. On the graph of $y=P(x)$, two points with abscissas 10 and 30 are marked. It turns out that the bisector of the first quadrant of the coordinate plane intersects the segment between them at its midpoint. Find $P(20)$. | Answer: -80.
Solution. The midpoint of this segment has coordinates $\left(\frac{10+30}{2}, \frac{P(10)+P(30)}{2}\right)$. Since it lies on the bisector of the first quadrant, i.e., on the line $y=x$, these coordinates are equal. From this, we get $P(10)+P(30)=40$.
Since $P(x)$ is a monic polynomial, it can be writte... | -80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. In an $8 \times 12$ table, some $N$ cells are black, and the rest are white. In one operation, it is allowed to paint three cells forming a three-cell corner to white (some of them could already be white before repainting). It turned out that it is impossible to make the entire table white in fewer than 25... | Answer: 27.
Solution. Divide the $8 \times 12$ table into 24 squares of $2 \times 2$ (Fig. 9a).
(a)
 ; B\left(-\sqrt{-\frac{c}{a}} ; 0\right)$. The graph intersects the OY axis at point $C(0 ; c)$. Then the side $AB$ of the equilat... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.6. The sides of the base of a brick are 28 cm and 9 cm, and the height is 6 cm. A snail crawls in a straight line along the faces of the brick from one vertex of the lower base to the opposite vertex of the upper base. The horizontal and vertical components of its velocity, $v_{x}$ and $v_{y}$, are related by the eq... | Answer: 35 min.
Fig. $10.6 \mathrm{a}$
Solution. Let the given brick be a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D$, where $A B-=9 \mathrm{~cm}, B C=28 \mathrm{~cm}, A A^{\prime}=6 \mathrm{~cm}$.
First method. Increase the height of the brick and the vertical component of the snail's sp... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given $n>2$ natural numbers, among which there are no three equal, and the sum of any two of them is a prime number. What is the largest possible value of $n$? | 1. Answer: 3. Note that the triplet $1,1,2$ satisfies the condition. Suppose there are more than 3 numbers. Consider any 4 of them a, b, c, d. Among these four numbers, there cannot be two even numbers, otherwise their sum would be an even prime greater than two. Therefore, at least 3 of them must be odd. Their pairwis... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A circle is circumscribed around a right triangle $\mathrm{ABC}$ with hypotenuse $\mathrm{AB}$. On the larger leg $\mathrm{AC}$, a point $\mathrm{P}$ is marked such that $\mathrm{AP}=\mathrm{BC}$. On the arc $\mathrm{ACB}$, its midpoint $\mathrm{M}$ is marked. What can the angle $\mathrm{PMC}$ be equal to? | 5. Answer: 90 degrees
Consider triangles АРМ and
ВСМ. In them, ВС=АР by condition,
АМ=ВМ - chords subtending equal
arcs, angles МВС and МАС are equal,
since they subtend the same arc.
Therefore, the triangles are equal.
Thus, angles ВМС and АМР are equal. Since angle АМВ is 90 degrees, then angle $\mathrm{PMC}=\mathr... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) The kids were given the task to convert the turtle's speed from centimeters per second to meters per minute. Masha got an answer of 25 m/min, but she thought there were 60 cm in a meter and 100 seconds in a minute. Help Masha find the correct answer. | Answer: 9 m/min.
Solution. The turtle covers a distance of 25 Machine "meters" in one Machine "minute," meaning it crawls $25 \cdot 60$ centimeters in 100 seconds. Therefore, the speed of the turtle is $\frac{25 \cdot 60}{100}=15$ cm/sec. Thus, in 60 seconds, the turtle will crawl $15 \cdot 60$ centimeters, which is $... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) At a certain moment, Anya measured the angle between the hour and minute hands of her clock. Exactly one hour later, she measured the angle between the hands again. The angle turned out to be the same. What could this angle be? (Consider all cases.) | Answer: $15^{\circ}$ or $165^{\circ}$.
Solution. After 1 hour, the minute hand remains in its original position. During this time, the hour hand has turned $30^{\circ}$. Since the angle has not changed, the minute hand must bisect one of the angles between the positions of the hour hand (either the $30^{\circ}$ angle ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon (i.e., exactly at 12 o'clock) and, without stopping, arrived: one at $B$ at 4 PM, and the other at $A$ at 9 PM. At what time was dawn that day? | Answer: at 6 AM.
Solution. Let's denote the meeting point as $C$. Let $x$ be the number of hours from dawn to noon.
The speed of the first pedestrian on segment $A C$ is $A C / x$, and on segment $B C$ it is $B C / 4$. Since his speed is constant, we have $\frac{A C}{x}=\frac{B C}{4}$, which can be rewritten as $\fra... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Determine the number of points at which 10 lines intersect, given that only two of them are parallel and exactly three of these lines intersect at one point. | Answer: 42.
Solution. Let's number the lines so that lines 1, 2, and 3 intersect at one point (denote this point as $X$). List all possible pairs of lines (1 and 2, 1 and 3, 1 and $4, \ldots, 8$ and 9, 8 and 10, 9 and 10) and their points of intersection. There are a total of 45 pairs of lines (there are 9 pairs of th... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. There are three acrobat brothers. Their average height is 1 meter 74 centimeters. The average height of two of these brothers: the tallest and the shortest - 1 meter 75 centimeters. What is the height of the middle brother? Justify your answer. | Answer: 1 meter 72 centimeters.
Solution. Since the average height of all three is 1 meter 74 centimeters, the total height of all three is 5 meters 22 centimeters. The average height of the two brothers is 1 meter 75 centimeters, so their combined height is 3 meters 50 centimeters. Therefore, the height of the middle... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Let's call a three-digit number interesting if at least one of its digits is divisible by 3. What is the maximum number of consecutive interesting numbers that can exist? (Provide an example and prove that it is impossible to have more consecutive numbers.)
| Answer: 122.
Solution. The numbers $289,290, \ldots, 299,300, \ldots, 399,400, \ldots, 409,410$ are interesting (recall that 0 is divisible by 3), and there are 122 of them in total. Let's prove that a larger number is not possible.
Suppose we managed to find a larger number of consecutive interesting numbers; choose... | 122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. In a country, there are 100 cities. Between any two cities, there is either no connection, or there is an air route, or there is a railway (both air routes and railways cannot exist simultaneously). It is known that if two cities are connected to a third city by railway, then there is an air route between th... | Solution. Note that no city is connected by railway to more than two other cities. Indeed, suppose some three cities are connected by railway to one. Then all of them are connected to each other by air routes, which is impossible by the condition of the problem. Therefore, each city is connected by railway to no more t... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false |
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are equal to an integer number of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
![](https://cdn.mathpix.com/cropped/2024_05_06_20e437e0605a8... | # Answer: 17.
Solution. We will use the triangle inequality: in any non-degenerate triangle, the sum of any two sides is strictly greater than the remaining one.
Let $x$ km be the unknown length. From the left triangle, we see that $x < 10 + 8 = 18$. But if $x \leqslant 16$, then in the right triangle, the triangle i... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25$, $CP = 20$, and $\angle APC = \angle A + \angle B$.
![](https://cdn.mathpix.com/cro... | Answer: 16.
Solution. Since $\angle P C B+\angle P B C=\angle A P C=\angle P A C+\angle P B C$, we obtain $\angle P C B=\angle P A C$. Note that the right triangles $P A C$ and $Q C P$ are similar by the acute angle, and
$$
\frac{25}{20}=\frac{A C}{C P}=\frac{P C}{P Q}=\frac{20}{P Q}
$$
from which we find $P Q=\frac... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, the number of knights and the number of liars differ by one." Player number 2 said: "In our team, the number of knights and the number of liars... | 3. Answer: the only knight plays under number 9. The solution is that the number of knights cannot be equal to the number of liars, so one of the answers must be correct. Two answers cannot be true, as they contradict each other. Therefore, there is exactly 1 knight and 10 liars in the team, which is stated by the play... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A three-digit number is called cool if one of its digits is half the product of the other two digits. A three-digit number is called supercool if such digits are in two or three of its positions. How many different supercool numbers exist? (Zero cannot be part of the representation of cool or supercool numbers.) | 4. Answer: 25 numbers. Solution: let the record of a superclass number include digits a, b, c in some order. Then, two equalities are satisfied: $2 \mathrm{a}=\mathrm{bc} 2 \mathrm{~b}=\mathrm{ac}$. If the numbers a and b are different, then by swapping them, we will violate the existing equality. Therefore, $\mathrm{a... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
 | Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides som... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cine... | Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats ... | 1035 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4. In triangle $A B C$, points $D$ and $F$ are marked on sides $A B$ and $A C$ respectively such that lines $D C$ and $B F$ are perpendicular to each other and intersect at point $E$ inside triangle $A B C$. It turns out that $A D=D C$ and $D E \cdot E C=F E \cdot E B$. What degree measure can angle $B A C$ have? (7... | # Solution:
$D E \cdot E C=F E \cdot E B \Rightarrow \frac{D E}{E F}=\frac{E B}{E C}, \angle D E B=\angle F E C=90^{\circ}$, so triangles $\triangle D E B$ and $\triangle F E C$ are similar.... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. On an $8 \times 8$ battleship game field, a "piglet" figure is placed. What is the minimum number of shots needed to definitely hit one "piglet"? (7 points)
# | # Solution:
## Evaluation:
Divide the board into rectangles $4 \times 2$. To ensure that a piglet cannot be placed inside one such figure, at least two shots are required. There are 8 such ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $O$ be the center of the circumcircle of triangle $ABC$, points $O$ and $B$ lie on opposite sides of line $AC$, $\angle AOC = 60^\circ$. Find the angle $AMC$, where $M$ is the center of the incircle of triangle $ABC$. | # Solution.
Since points $O$ and $B$ lie on opposite sides of line $A C$, the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$. Therefore,
$\angle A B C=(1 / 2) \cdot 300^{\circ}=150^{\circ}$. The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{\circ... | 165 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On a chessboard of size $2015 \times 2015$, dominoes are placed. It is known that in every row and every column there is a cell covered by a domino. What is the minimum number of dominoes required for this? | Solution Let's start filling the 2015 × 2015 square with dominoes starting from the lower left corner (see figure). The first domino will be placed horizontally, covering two vertical lines and one horizontal line. The next domino will be placed vertically, covering two horizontal lines and one vertical line. In total,... | 1344 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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