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8.1. A number is called good if any two adjacent digits in its notation differ by at least 4. Vera wrote some good number, and then replaced identical digits with identical letters, and different ones with different letters. Could she have ended up with the word NOVOSIBIRSK? | Answer: For example, the number 82729161593 could work ( $\mathrm{H}=8, \mathrm{O}=2, \mathrm{~B}=7, \mathrm{C}=9$, I = 1, B = 6, $\mathrm{P}=5, \mathrm{~K}=3$).
Criterion: any valid example without verification - 7 points. | 82729161593 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find the angle $D A C$, given that $A B=B C$ and $A C=C D$, and the lines on which points $A, B, C, D$ lie are parallel, with the distances between adjacent lines being equal. Point $A$ is to the left of $B$, $C$ is to the left of $B$, and $D$ is to the right of $C$ (see figure).
 + 8 + 9, k \geq 5, n \geq 21 \), in the second case - \( n = 4k + 3 = 4(k-3) + 6 + 9, k \geq 4, n \geq 19 \). On the other hand, the three smallest compos... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. In the cells of an 8 by 8 board, tokens are placed such that for each token, the row or column of the board in which it lies contains no more than three tokens. What is the maximum possible number of tokens on the board? | Answer: 30.
Solution: By swapping the verticals and horizontals, we assume that only the left $x$ verticals and the bottom $y$ horizontals contain more than 3 chips. From the condition, it follows that there are no chips at all in the lower left rectangle at the intersection of these verticals and horizontals. Each ve... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. Given the number 1836549, you can take two adjacent non-zero digits and swap their places, after which you subtract 1 from each of them. What is the smallest number that can result from these operations? | Answer: 1010101
Solution: The digits in the number alternate in parity: odd, even, etc. Note that with the described operation, even and odd numbers swap places, and then 1 is subtracted from them, thereby not disrupting the order of parity. Thus, it is impossible to obtain a number less than 1010101 (in each place, t... | 1010101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. There is a steamship route between the cities of Dzerzhinsk and Lviv. Every midnight, a steamship departs from Dzerzhinsk, arriving exactly eight days later in Lviv. How many steamships will the steamship "Raritet" meet on its way to Dzerzhinsk if it departs from Lviv exactly at midnight and spends the same eight ... | Answer: 17.
Solution: As "Raritet" departs from Lviv, a steamer arrives there, which left Dzerzhinsk 8 days ago. By the time "Raritet" arrives at its final destination, 8 days have passed since the initial moment, and at this moment, the last steamer departs from Dzerzhinsk, which "Raritet" meets on its way. Thus, "Ra... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser gave up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play? | Answer: 14.
Solution: On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Among all natural numbers from 1 to 20 inclusive, some 10 numbers were painted blue, and the other 10 - red, then all possible sums of pairs of numbers, one of which is blue and the other is red, were counted. What is the maximum number of different sums that can be among the hundred obtained numbers? | 9.6. Answer. A maximum of 35 different numbers.
The sum of a blue and a red number can be a natural number from $1+2=3$ to $19+20=39$ inclusive, so there cannot be more than 37 different numbers. Moreover, note that one of the numbers $3,4, . ., 13$ must not be the sum of a blue and a red number. Otherwise, if the num... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. In the bus, there are single and double seats. In the morning, 13 people were sitting in the bus, and there were 9 completely free seats. In the evening, 10 people were sitting in the bus, and 6 seats were completely free. How many seats are there in the bus? | Answer: 16.
Solution: If in the morning passengers sat on 6 double seats (i.e., as densely as possible), then they occupied 7 seats, with 9 seats still free. In total: 16 seats. If they did not sit as densely, then they would have occupied more seats. That is, there are no fewer than 16 seats in the bus, on the one ha... | 16 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.4. In the country, there are 15 cities, some of which are connected by roads. Each city is assigned a number equal to the number of roads leading out of it. It turned out that there are no roads between cities with the same number. What is the maximum number of roads that can be in the country? | Answer: 85.
Solution: Let's order the city numbers in non-increasing order:
$$
a_{1} \geq a_{2} \geq \cdots \geq a_{15}
$$
Notice that the number of cities with number $15-i$ is no more than $i$. Indeed, if there are at least $i+1$ such cities, then they cannot be connected to each other, and thus can be connected t... | 85 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. A square with a side of $100 \mathrm{~cm}$ was drawn on the board. Alexei crossed it with two lines parallel to one pair of sides of the square. Then Danil crossed the square with two lines parallel to the other pair of sides of the square. As a result, the square was divided into 9 rectangles, and it turned out t... | Answer: 2400.
Solution 1: Without loss of generality, we will assume that the central rectangle has a width of 60 cm and a height of 40 cm. Let $x$ and $y$ be the width and height, respectively, of the lower left rectangle. Then the upper left rectangle has sides $x$ and $(60-y)$, the upper right rectangle has sides $... | 2400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser would give up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play... | Answer: 14.
Solution: On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.3. On the table, 28 coins of the same size but possibly different masses are arranged in a triangular shape (see figure). It is known that the total mass of any triplet of coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle. | Answer: 60 g.
Solution 1: Take a rhombus made of 4 coins. As can be seen from the diagram, the masses of two non-touching coins in it are equal. Considering such rhombi, we get that if we color the coins in 3 colors, as shown in the diagram, then the coins of the same color will have the same mass. Now it is easy to f... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.3. Find the smallest natural number divisible by 99, all digits of which are even. | Answer: 228888.
Solution. Let the sum of the digits of the desired number $X$, located in even-numbered positions (tens, thousands, etc.), be denoted by $A$, and the sum of the digits in odd-numbered positions (units, hundreds, etc.) be denoted by $B$. According to the divisibility rules for 9 and 11, $A+B$ is divisib... | 228888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. Find the smallest natural number ending in the digit 4 that quadruples when its last digit is moved to the beginning of the number. | # 7.4. Answer: 102564
If a number ending in 4 is multiplied by 4, the result is a number ending in 6, so the last digits of the desired number are 64. If such a number is multiplied by 4, the result is a number ending in 56, so the desired number ends in 564. Continuing to restore the number in this way, it turns out ... | 102564 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. Find the maximum natural number $A$ such that for any arrangement of all natural numbers from 1 to 100 inclusive in a row in some order, there will always be ten consecutively placed numbers whose sum is not less than $A$. | Answer: 505.
Solution: The sum of all numbers from 1 to 100 is 5050. Let's divide the 100 numbers in a row into 10 segments, each containing 10 numbers. Clearly, the sum of the numbers in one of these segments is not less than 505, so A is not less than 505.
We will show that among the numbers arranged in the followi... | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase? | Answer: 500 rubles.
Solution: The additional 100 rubles spent the second time brought the merchant an additional 20 rubles in profit. Therefore, the first time, to earn $5 \cdot 20=100$ rubles in profit, the merchant must have paid $5 \cdot 100=500$ rubles.
Second solution: Let the amount of the first purchase be $x$... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Ten numbers are written around a circle, the sum of which is 100. It is known that the sum of any three consecutive numbers is not less than 29. Indicate the smallest number $A$ such that in any set of numbers satisfying the condition, each number does not exceed $A$. | Answer. $A=13$
Solution. Let $X$ be the largest of the listed numbers. The remaining numbers can be divided into 3 "neighbor" triplets. The sum of the numbers in each such triplet is no less than 29, therefore, $X \leq 100 - 3 \cdot 29 = 13$. An example of a set with the maximum number 13: $13,9,10,10,9,10,10,9,10,10$... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.4. Find the smallest natural number in which each digit occurs exactly once and which is divisible by 990. | Answer: 1234758690.
Solution. The number 990 is the product of coprime numbers 2, 5, 9, and 11. Any ten-digit number composed of different digits, each used once, is divisible by 9, since their sum, which is 45, is divisible by 9. According to the divisibility rule for 10, the desired number must end in 0. It remains ... | 1234758690 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. In the list $1,2, \ldots, 2016$, two numbers $a<b$ were marked, dividing the sequence into 3 parts (some of these parts might not contain any numbers at all). After that, the list was shuffled in such a way that $a$ and $b$ remained in their places, and no other of the 2014 numbers remained in the same part where ... | Answer: $1+2+\ldots+1008=1009 * 504=508536$ ways.
## Solution:
Hypothesis. We will prove that the question is equivalent to counting the number of ways to split 2014 into three ordered non-negative addends $2014=x+y+z$, for which the non-strict triangle inequality holds, i.e., $x+y \geq z, x+z \geq y, y+z \geq x$.
N... | 508536 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2. Let $A$ be a set of ten distinct positive numbers (not necessarily integers). Determine the maximum possible number of arithmetic progressions consisting of three distinct numbers from the set $A$. | Answer: 20.
Solution: Let the elements of set $A$ be denoted as $a_{1}<a_{2}<\ldots<a_{10}$. Three numbers $a_{k}<a_{l}<a_{m}$ form a three-term arithmetic progression if and only if $a_{l}-a_{k}=a_{m}-a_{l}$. Let's see how many times each element of $A$ can be the middle term $a_{l}$ of such a progression. It is easy... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. How many cells need to be marked on an 8 by 8 grid so that each cell on the board, including the marked ones, is adjacent by side to some marked cell? Find all possible answers. Note that a cell is not considered adjacent to itself. | Answer: 20.
Solution: First, let's gather our strength and mark twenty cells on an 8 by 8 board as required by the problem. For example, as shown in the figure. In this case, the board naturally divides into 10 parts, as indicated by the bold lines in the figure. Each part consists of cells adjacent to the given pair ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Pasha and Sasha made three identical toy cars. Sasha did one-fifth of the total work. After selling the cars, they divided the proceeds in proportion to the work done. Pasha noticed that if he gave Sasha 400 rubles, and Sasha made another such car and sold it, they would have the same amount of money. How much doe... | Answer: 1000 rubles.
Solution: Sasha did one-fifth of the entire work, which means he made 0.6 of one car, while Pasha did the remaining 2.4. That is, the difference is 1.8 cars. If Sasha makes another car, the difference will be 0.8 of one car. Pasha gave 400 of his rubles, thereby reducing the amount of money he had... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. The company receives apple and grape juices in standard barrels and produces a cocktail (mixture) of these juices in standard cans. Last year, one barrel of apple juice was enough for 6 cans of cocktail, and one barrel of grape juice was enough for 10. This year, the proportion of juices in the cocktail (mixture) ... | Answer: 15 cans.
Solution. Last year, one barrel of apple juice was enough for 6 cans of cocktail, which means each can contained $1 / 6$ of a barrel of apple juice. Similarly, one barrel of grape juice was enough for 10 cans, which means each can contained $1 / 10$ of a barrel of grape juice. Therefore, the capacity ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. For what minimum $\boldsymbol{n}$ in any set of $\boldsymbol{n}$ distinct natural numbers, not exceeding 100, will there be two numbers whose sum is a prime number? | Answer. $\boldsymbol{n}=51$.
Solution. The sum of two even natural numbers is always even and greater than two, hence it cannot be a prime number. Therefore, the example of a set of all fifty even numbers not exceeding 100 shows that the minimum $\boldsymbol{n}$ is not less than 51.
On the other hand, let's divide al... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.4. In a set $X$ of 17 elements, a family of $N$ distinct non-empty subsets is selected such that each element of the set $X$ is contained in exactly two subsets from this family. What is the maximum value of $N$? Find the number of all possible different types of such families for the maximum $N$. Two families of su... | Answer. The maximum $N$ is 25, and there exist two different types of families of 25 subsets that satisfy the condition of the problem.
Solution. Consider an arbitrary family of $N$ distinct non-empty subsets such that each element of the set $X$ is contained in exactly two subsets of this family. If there are $x$ one... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. Given a triangle $\mathrm{ABC}$ with angle $\mathrm{BAC}$ equal to $30^{\circ}$. In this triangle, the median $\mathrm{BD}$ was drawn, and it turned out that angle $\mathrm{BDC}$ is $45^{\circ}$. Find angle $\mathrm{ABC}$. | Answer: $45^{\circ}$
Solution: Draw the height $C H$. Then $H D=A D=C D$ as the median to the hypotenuse. Moreover, $\angle H C D=\angle C H A-\angle H A C=60^{\circ}$, so triangle $C H D$ is equilateral, which means $\angle H D C=60^{\circ}$ (from which it follows, in particular, that $H$ lies between $A$ and $B$). T... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. In the city of Omsk, a metro has been built, represented by a straight line. On this same line is the house where Nikita and Egor live. Every morning they leave the house for classes at the same time, after which Egor runs to the nearest metro station at a speed of 12 km/h, while Nikita walks along the metro line ... | Answer: 23 km/h
Solution: Obviously, this is only possible if the subway train first arrives at the nearest station A, where Egor runs to, and then goes to station B, where Nikita is heading.
Let $v$ be the speed of the subway, $S$ be the distance between two adjacent stations, and $R$ be the distance between this su... | 23 | Other | math-word-problem | Yes | Yes | olympiads | false |
9.1. From points A and B towards each other with constant speeds, a motorcyclist and a cyclist started simultaneously from A and B, respectively. After 20 minutes from the start, the motorcyclist was 2 km closer to B than the midpoint of AB, and after 30 minutes, the cyclist was 3 km closer to B than the midpoint of AB... | Answer: In 24 minutes.
Solution: In 10 minutes, the motorcyclist travels $1 / 4$ of the distance from A to B plus 1 km, while the cyclist travels $1 / 6$ of the distance from A to B minus 1 km. Therefore, in 10 minutes, both of them, moving towards each other, cover $1 / 4 + 1 / 6 = 5 / 12$ of the distance from A to B... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Several consecutive natural numbers are written on the board. It is known that $48 \%$ of them are even, and $36 \%$ of them are less than 30. Find the smallest of the written numbers. | Answer: 21.
Solution. $\frac{48}{100}=\frac{12}{25}, \frac{36}{100}=\frac{9}{25}$ - these are irreducible fractions, so the total number of numbers is divisible by 25. If there were 50 or more, then, by the condition, there would be at least 2 fewer even numbers than odd numbers, which is impossible for consecutive na... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a family of 4 people. If Masha's scholarship is doubled, the total income of the entire family will increase by $5 \%$, if instead the mother's salary is doubled - by $15 \%$, if the father's salary is doubled - by $25 \%$. By what percentage will the family's total income increase if the grandfather's pension ... | Answer: by $55 \%$.
Solution: When Masha's scholarship is doubled, the family's total income increases by the amount of this scholarship, so it constitutes $5 \%$ of the income. Similarly, the salaries of Masha's mother and father constitute $15 \%$ and $25 \%$. Therefore, the grandfather's pension constitutes $100-5-... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. There are 100 boxes numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all ... | Answer: 99.
Solution: To be able to definitively determine which of the 100 boxes contains the prize, it is necessary to have the possibility of receiving at least 100 different answers to one set of questions. Since the host's answers for different prize positions can only differ by the number of "yes" responses, it ... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. On a certain island, only knights, who always tell the truth, and liars, who always lie, live. One day, 1001 inhabitants of this island stood in a circle, and each of them said: "All ten people following me in a clockwise direction are liars." How many knights could there be among those standing in the circle? | Answer: 91 knights.
Solution. Note that all people cannot be liars, because then it would mean that each of them is telling the truth. Therefore, among these people, there is at least one knight. Let's number all the people so that the knight is the 1001st in the sequence. Then the 10 people with numbers from 1 to 10 ... | 91 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.4. We will call a number remarkable if it can be decomposed into the sum of 2023 addends (not necessarily distinct), each of which is a natural composite number. Find the largest integer that is not remarkable. | Answer. $4 \times 2023+3=8095$.
Solution. Replace 2023 with $n$ and solve the problem in the general case for a sum of $n \geqslant 2$ composite addends. We will prove that the answer is $4 n+3$, from which we will obtain the answer to the original problem.
Claim 1. The number $4 n+3$ is not remarkable.
Proof of Cla... | 8095 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. Let $n$ be a natural number not ending in 0, and $R(n)$ be the four-digit number obtained from $n$ by reversing the order of its digits, for example $R(3257)=7523$: Find all natural four-digit numbers $n$ such that $R(n)=4n+3$. | Answer: 1997.
Solution. Consider the decimal representation of the original four-digit number $n=\overline{a b c d}$, then $R(n)=\overline{d c b a}=4 n+3$ is also a four-digit number. Therefore, $4 a \leq 9$, so $a=1$ or $a=2$. Moreover, the number $R(n)=\overline{d c b a}=4 n+3$ is odd and ends in $a$, hence $a=1$. I... | 1997 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya? | Solution: If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adj... | 18 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find the largest four-digit number, all digits of which are different, and which is divisible by each of its digits. Of course, zero cannot be used. | Solution: Since the number of digits is fixed, the number will be larger the larger the digits in its higher places are. We will look for the number in the form $\overline{98 a}$. It must be divisible by 9. Therefore, the sum $a+b$ must give a remainder of 1 when divided by 9. At the same time, this sum does not exceed... | 9864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.4. Point $A$ is located midway between points $B$ and $C$. The square $A B D E$ and the equilateral triangle $C F A$ are in the same half-plane relative to the line $B C$. Find the angle between the lines $C E$ and $B F$.
. What is the minimum number of operations required to make all the signs in the table pluses? | Answer. In 100 operations.
Solution. There are 19 cells in total in the row and column passing through a given cell, so if we perform operations on all pairs of rows and columns of the table (a total of $10 \times 10=100$ operations), each sign in the table will change 19 times, turning from minus to plus, so 100 oper... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone h... | Answer: 120
Solution: Consider any pair of friends. Their "friendship" was counted exactly once, as it was included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Si... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. In a school, there are 1000 students and 35 classes. On the forehead of each student, the number of students in their class is written. What can the sum of the reciprocals of these numbers be? List all the options and prove that there are no others. Recall that the reciprocal of a number $a$ is the number $1 / a$. | Answer: 35.
Solution: Let there be a people in the class, then the sum of the fractions corresponding to the numbers from this class is 1 (a fractions, each equal to 1/a). There are 35 classes in total. Therefore, the total sum is 35.
Criteria:
Answer - 0 points.
Answer with examples - 0 points.
The idea of partit... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a competition, there are 2018 Dota teams, all of different strengths. In a match between two teams, the stronger one always wins. All teams paired up and played one game. Then they paired up differently and played another game. It turned out that exactly one team won both games. How could this be | Solution: Let's number the teams in ascending order of strength from 1 to 2018. In the first round, we will have the matches 1 - 2, $3-4, \ldots, 2017$ - 2018, and in the second round - $2018-1, 2-3, 4$ - 5, ..., 2016 - 2017. It is obvious that only the team with the number 2018 will win in both rounds.
## Criteria:
... | 2018 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. Eleven of the best football teams each played one match against each other. It turned out that each team scored 1 goal in their first match, 2 goals in their second match, ..., 10 goals in their tenth match. What is the maximum number of matches that could have ended in a draw? | Solution: According to the condition, each team scored 1 goal in the first game. In the case of a draw, it also conceded 1 goal. Then for the other team, this match was also the first. Since the number of teams is odd, they cannot be paired. Therefore, at least one of the teams played a non-draw in their first match. T... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. In triangle $\mathrm{ABC}$, the bisectors of angles $\mathrm{BAC}$ and $\mathrm{BCA}$ intersect sides ВС and АВ at points К and Р, respectively. It is known that the length of side АС is equal to the sum of the lengths of segments АР and СК. Find the measure of angle $\mathrm{ABC}$. | Answer: $60^{\circ}$.
Solution 1. Let the angles of triangle ABC be denoted by the corresponding letters $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$, and the intersection point of the angle bisectors by I. Reflect points P and K with respect to the angle bisectors AK and CP, respectively; their images will be points $\... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.5. On some cells of a rectangular board of size 101 by 99, there is one turtle each. Every minute, each of them simultaneously crawls to one of the cells adjacent to the one they are in, by side. In doing so, each subsequent move is made in a direction perpendicular to the previous one: if the previous move was hori... | Answer: 9800.
Solution. Examples of unlimited movement on the board of 9800 turtles.
Example 1. Place the reptiles in the cells of a rectangle consisting of cells at the intersection of the 98 lower horizontals and 100 leftmost verticals. They will all move in the same way: first all to the right, then all up, then a... | 9800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. In each cell of a $5 \times 5$ table, one of the numbers $1,2,3,4,5$ is written in such a way that each row, each column, and each of the two diagonals of the table contains each of the numbers from 1 to 5. What is the maximum value that the sum of the five numbers written in the cells marked with dots on the diag... | Answer: 22.
Solution. If all 4 numbers marked with a dot and not in the top right corner are different, then the sum of all numbers marked with a dot does not exceed $5+5+4+3+2=19$. Let's f... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. A bus with programmers left Novosibirsk for Pavlodar. When it had traveled 70 km, Pavel Viktorovich set off from Novosibirsk in a car along the same route, and caught up with the programmers in Karasuk. After that, Pavel drove another 40 km, while the bus traveled only 20 km in the same time. Find the distance fro... | Solution. Since by the time the car has traveled 40 km, the bus has traveled half that distance, its speed is exactly half the speed of the car. However, when the bus has traveled 70 km after the car's departure, the car will have traveled 140 km and will just catch up with the bus. According to the problem, this happe... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. In triangle $A B C$, a point $D$ is marked on side $A C$ such that $B C = C D$. Find $A D$, given that $B D = 13$, and angle $C A B$ is three times smaller than angle $C B A$.
 | Solution. Let $\angle C A B=x$. Then $\angle C B A=3 x$ and $\angle A C B=180^{\circ}-4 x$. According to the problem, triangle $B C D$ is isosceles, so $\angle C D B=\angle C B D=\left(180^{\circ}-\angle B C D\right) / 2=2 x$. Therefore, $\angle D B A=\angle A B C-\angle D B C=3 x-2 x=x=\angle D A B$. Hence, triangle $... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. There are 5 pieces of transparent glass of the same square shape and the same size. Each piece of glass is conditionally divided into 4 equal parts (right triangles) by its diagonals, and one of these triangles is painted with an opaque paint of its individual color, different from the colors of the painted parts of... | Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then... | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find a natural number that has six natural divisors (including one and the number itself), two of which are prime, and the sum of all its natural divisors is 78. | Solution: The desired natural number $n$ can be represented as $n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}, 12,1+p_{2}+p_{2}^{2}>7$, then no factorization of the number 78 into a product of two natural factors fits $78=1 \cdot 78,78=2 \cdot 39,78=3 \cdot 26,78=6 \cdot 13$ up to the order).
2) $\alpha_{1}=2, \alpha_{... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The automatic line for processing body parts initially consisted of several identical machines. The line processed 38880 parts daily. After the production was modernized, all the machines on the line were replaced with more productive but also identical ones, and their number increased by 3. The automatic line began... | Solution: Let $x$ be the number of machines before modernization, $y$ be the productivity of each machine, i.e., the number of parts processed per day, and $z$ be the productivity of the new machines. Then we have $x y = 38880 = 2^{5} \cdot 3^{5} \cdot 5, (x+3) z = 44800 = 2^{8} \cdot 5^{2} \cdot 7, x > 1, y \frac{3888... | 1215 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The numbers $u, v, w$ are roots of the equation $x^{3}-3 x-1=0$. Find $u^{9}+v^{9}+w^{9}$. (12 points) | Solution. $\quad$ According to Vieta's theorem, $u+v+w=0, uv+vw+uw=-3, uvw=1$. Consider the sequence $S_{n}=u^{n}+v^{n}+w^{n}$. We have $S_{0}=3, S_{1}=0$. Let's find $S_{2}$: $S_{2}=u^{2}+v^{2}+w^{2}=(u+v+w)^{2}-2(uv+vw+uw)=6. \quad$ Since $\quad u^{3}=3u+1, v^{3}=$ $3v+1, w^{3}=3w+1, \quad$ then $S_{3}=u^{3}+v^{3}+w^... | 246 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In the laboratory, there are flasks of two sizes (volume $V$ and volume $V / 3$) in a total of 100 pieces, with at least 2 flasks of each size. The lab assistant randomly selects two flasks in sequence, and fills the first one with a 70% salt solution, and the second one with a 40% salt solution. Then, he pours the ... | Solution. Let $N$ be the number of large flasks in the laboratory, $N=2,3, \ldots, 98$, $n=100-N$ be the number of small flasks in the laboratory, $n=2,3, \ldots, 98$, and $\mathrm{P}(A)$ be the probability of the event $A=\{$ the salt content in the dish is between $50 \%$ and $60 \%$ inclusive\}. It is necessary to f... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The fraction $\frac{1}{5}$ is written as an infinite binary fraction. How many ones are there among the first 2022 digits after the decimal point in such a representation? (12 points) | Solution. The smallest number of the form $2^{n}-1$ divisible by 5 is 15. Then
$$
\frac{1}{5}=\frac{3}{15}=\frac{3}{16-1}=\frac{3}{2^{4}-1}=\frac{3}{16} \cdot \frac{1}{1-2^{-4}}=\left(\frac{1}{16}+\frac{1}{8}\right)\left(1+2^{-4}+2^{-8}+2^{-12}+\cdots\right)=
$$
$\left(2^{-3}+2^{-4}\right)\left(1+2^{-4}+2^{-8}+2^{-12... | 1010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. There is a cube fixed on legs, and six different paints. In how many ways can all the faces of the cube be painted (each in one color, not all paints have to be used) so that adjacent faces (having a common edge) are of different colors? (16 points) | Solution. Let's consider 4 cases of coloring a cube.
1) The top and bottom faces are the same color, and the left and right faces are the same color. We choose the color for the top and bottom faces in 6 ways, then the color for the left and right faces in 5 ways, then the color for the front face in 4 ways, and the c... | 4080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). T... | Answer: 7200 ways.
# | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). T... | Answer: 3720 ways.
# | 3720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$ | # Solution
Notice that 345 and $5 y^{2}$ are divisible by 5, so $3 x^{2}$ must also be divisible by 5. Therefore, $\quad x=5 t, t \in Z$. Similarly, $y=3 n, n \in Z$. After simplification, the equation becomes $5 t^{2}+3 n^{2}=23$. Therefore, $t^{2} \leq \frac{23}{5}$, $n^{2} \leq \frac{23}{3}$ or $|t| \leq 2,|n| \leq... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The car traveled half of the distance at a speed of 60 km/h, then one third of the remaining distance at a speed of 120 km/h, and the remaining distance at a speed of 80 km/h.
Find the average speed of the car during this trip. Give your answer in km/h. | Solution: Let x hours be the time the car traveled at a speed of 60 km/h, then $60 x=\frac{s}{2}$. Let y hours be the time the car traveled at a speed of 120 km/h, then $120 y=\frac{s}{6}$. Let z hours be the time the car traveled at a speed of 80 km/h, then $80 z=\frac{s}{3}$. By definition $v_{cp}=\frac{s}{t_{\text{t... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.
On a coordinate line, 16 points are marked and numbered from left to right. The coordinate of any point, except for the extreme points, is equal to half the sum of the coordinates of the two adjacent points. Find the coordinate of the fifth point if the first point has a coordinate of 2 and the sixteenth point has... | # Solution
Solution. Let $a, b$ and $c$ be the coordinates of three consecutive points (from left to right). Then $b=\frac{a+c}{2}$, which means the second point is the midpoint of the segment with endpoints at the neighboring points. This condition holds for any triple of consecutive points, meaning the distances bet... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
In a 6-liter vessel, 4 liters of a 70% (by volume) sulfuric acid solution are poured, and in a second vessel of the same capacity, 3 liters of a 90% solution of sulfuric acid are poured. A certain amount of the solution is transferred from the second vessel to the first so that it becomes an $r-\%$ solution of sul... | # Solution.
Let $x$ liters of the solution be transferred from the second vessel to the first. Since it follows from the condition that $0 \leq x \leq 2$, to find the amount of pure acid in the new solution, we obtain the equation $2.8 + 0.9x = (4 + x) \frac{r}{100}$, from which $x = \frac{4r - 280}{90 - r}$. Now, con... | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.
In triangle $A B C$, the bisector $A L$ ( $L \in B C$ ) is drawn, and $M$ and $N$ are points on the other two bisectors (or their extensions) such that $M A=M L$ and $N A=N L, \angle B A C=50^{\circ}$.
Find the measure of $\angle M A N$ in degrees. | # Solution
We will use the auxiliary statements.
If the bisector $B K$ in triangle $A B C$ intersects the circumscribed circle at point $W$, then:
1) $A W=C W$ (since $\angle C A W=\angle C B W=\angle A B W=\angle A C W$, that is, triangle $A W C$ is isosceles and $A W=C W$).
 Pavel caught 32 crayfish and decided to sell them at the market. When part of his catch was bought, it turned out that the buyer paid 4.5 rubles less for each one than the number of crayfish that remained on the counter. At the same time, the boy earned the maximum amount of money possible. How much mone... | Solution: Let $x$ be the number of crayfish left on the counter, then (32-x) crayfish were bought, ($x-4.5$) - the cost of one crayfish.
$(32-x)(x-4.5)$ - the cost of all crayfish.
$y=(32-x)(x-4.5)$
$x_{B}=18.25$.
$x_{1}=19, y_{1}=(32-19)(19-4.5)=13 * 14.5=188.5$.
$x_{2}=18, y_{2}=(32-18)(18-4.5)=14 * 13.5=189$. T... | 189 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) In an acute-angled triangle $\mathrm{ABC}$, a point $\mathrm{D}$ is chosen on side $\mathrm{BC}$ such that $\mathrm{CD}: \mathrm{DB}=2: 1$, and a point $\mathrm{K}$ is chosen on segment $\mathrm{AD}$ such that $\mathrm{AK}=\mathrm{CD}+\mathrm{DK}$. A line is drawn through point $\mathrm{K}$ and vertex $\... | Solution: Extend SV beyond point V so that BN = BD. Draw NM || BE. NM intersects AD at point L. Draw segment $\mathrm{CH} \perp \mathrm{AD}$. Extend it to $\mathrm{P}$ such that $\mathrm{HP}=\mathrm{HC}, \mathrm{PN}|| \mathrm{AD}$.
In triangle DLN, segment $\mathrm{BK}$ is the midline, therefore, $\mathrm{DK}=\mathrm{... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A farmer initially placed his produce in boxes with a capacity of 8 kg each, but one box was not fully loaded. Then the farmer repackaged all the produce into boxes with a capacity of 6 kg each, which required 8 more boxes, but in this case, one box was also not fully loaded. When all the produce was placed in boxes... | Solution. Let $x$ kg be the weight of the farmer's produce. Then $\quad 8(n-1)<x<8 n, \quad 6(n+7)<x<6(n+8)$, $5(n+13)=x, \Rightarrow 8(n-1)<5(n+13)<8 n, \quad 6(n+7)<5(n+13)<6(n+8)$,
$\Rightarrow 21 \frac{2}{3}<n<23, \quad n=22, \quad x=35 \cdot 5=175$.
Answer: 175. | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the country of Landia, which breeds an elite breed of horses, an annual festival is held to test their speed, in which only one-year-old, two-year-old, three-year-old, and four-year-old horses can participate. For each horse that meets the speed standard, the festival organizers pay a fixed amount of money to the... | Solution. A four-year-old horse can earn a maximum of 4 landricks over its entire participation in festivals. If the horse starts participating in festivals at 1 year old, it can participate for another 3 years after that. In the case of winning every year, it will earn 1+2+3+4=10 landricks over 4 years. If the horse s... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$. | Solution. The number $N$ can be represented as
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \... | 2018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can a rectangular board of size $2 \times 18$ be covered with identical rectangular tiles of size $1 \times 2$? The tiles must be placed so that they fit entirely on the board and do not overlap.
(12 points) | Solution. Let there be a board of size $2 \times$ n. Denote the number of ways to tile it with tiles of size $1 \times 2$ by $P_{n}$. Then the following recurrence relation holds: $P_{n}=P_{n-1}+P_{n-2}$. Since $P_{1}=1, P_{2}=2$, performing sequential calculations using the recurrence relation, we arrive at the answer... | 4181 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. | Solution. Transform the equation by factoring the right-hand side
$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.
Considering that the variables are natural numbers, and 7 is a prime number, we get
$$
\left\{\begin{array} { l }
{ y - x ... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ of a circular track, a car and a motorcycle started simultaneously and in the same direction. The car drove two laps without stopping in one direction. At the moment when the car caught up with the motorcyclist, the motorcyclist turned around and increased his speed by $16 \mathrm{~km} / \mathrm{u}$, ... | Solution. Let $x$ (km/h) be the speed of the motorcyclist, $y$ (km/h) be the speed of the car, and $S$ (km) be the distance the motorcyclist travels before turning around. Then the total length of the track is $2 S + 5.25$. We have $\frac{S}{x} = \frac{3 S + 5.25}{y}$, $\frac{3 x}{8} + 6 = S$, $\frac{3 y}{8} = S + 5.25... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$. | Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiply them by 307, and consider linear combinations for integer $t$, we get values in natural numbers
$\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t \in\{... | 164 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) The creative competition at the institute consisted of four tasks. In total, there were 70 applicants. The first test was successfully passed by 35, the second by 48, the third by 64, and the fourth by 63 people, with no one failing all 4 tasks. Those who passed both the third and fourth tests were admit... | Solution. 1st and 2nd tasks were solved by at least $35+48-70=13$ people. 3rd and 4th - at least $64+63-70=57$ people. No one failed all tasks, so 1st and 2nd were solved by 13 people, 3rd and 4th - 57 people.
Answer: 57 people.
Criteria.
| Points | Conditions for awarding |
| :--- | :--- |
| 10 points | Justified s... | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) In one of the regions on the planet, seismic activity was studied. 80 percent of all days were quiet. The instrument predictions promised a calm situation in 64 out of 100 cases; moreover, in 70 percent of all cases when the day was quiet, the instrument predictions came true. What percentage of days wit... | Solution. Let the total number of observed days be x. The number of actually quiet days was $0.8x$, and seismically active days were $0.2x$. The predictions of quiet days matched the actually quiet days: $0.7 \cdot 0.8x = 0.56x$. Then the number of active days that did not match the predictions was $0.64x - 0.56x = 0.0... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A car left point A for point B, and a second car left with some delay. When the first car had traveled half the distance, the second car had traveled $26 \frac{1}{4}$ km, and when the second car had traveled half the distance, the first car had traveled $31 \frac{1}{5}$ km. After overtaking the first car, the second... | Solution. S - the distance between points A and B.
$$
\frac{S-2-S / 2}{S+2-26.25}=\frac{S-2-31.2}{S+2-S / 2}, \quad 5 S^{2}-383 S+5394=0, \quad \sqrt{D}=197, \quad S=58
$$
Answer: 58. | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given six socks, all of different colors and easily stretchable. You cannot turn them inside out. In how many ways can you put on 3 socks on each foot, considering which one to put on earlier and which one later? | Solution. There is a sequence of 6 sock puttings on: $\mathrm{C}_{6}^{3}=20$ ways to choose which puttings are for the right foot. For each such choice, there are $6!=720$ ways to choose which sock to take for each putting.
Answer: 14400. | 14400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x, y, z$ be the roots of the equation $t^{3}-2 t^{2}-9 t-1=0$. Find $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}$.
(12 points) | Solution. Let's bring the desired expression to a common denominator: $\frac{y^{2} z^{2}+x^{2} z^{2}+x^{2} y^{2}}{x y z}$. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem $x+y+z=2, x y+x z+y z=-9, x y z=1$.
$$
\begin{aligned}
& x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the midline of which is equal to $5 \sqrt{3}$. The ratio of the areas of the parts of the trapezoid $\mathrm{ABCD}$, into which it is divided by the midline, is $7: 13$. All lateral faces of the pyramid $\mathrm{TABCD}$ are inclined ... | # Solution.
Let $TO$ be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, $O$ is the center of the circle inscribed in the base. Let $MP$ be the midline of the trapezoid, $\quad AD=a, BC=b. \quad$ According to the problem, we have $S_{MB CP}=7x, S_{\text{AMPD}}=13x, \quad \... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A group of schoolchildren heading to a school camp was to be seated in buses so that each bus had the same number of passengers. Initially, 22 people were to be seated in each bus, but it turned out that three schoolchildren could not be seated. When one bus left empty, all the schoolchildren were able to sit evenly... | Solution. Let $\mathrm{n}$ be the number of buses, $\mathrm{m}$ be the number of schoolchildren in each bus, and $\mathrm{S}$ be the total number of schoolchildren.
We have
$$
S=22 n+3, \quad S=(n-1) m, n \leq 10, m \leq 36, \quad 22 n+3=(n-1) m, \quad n=1+\frac{25}{m-22}
$$
Considering the constraints on $\mathrm{n... | 135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can the line $x \sin \sqrt{16-x^{2}-y^{2}}=0$ be drawn without lifting the pencil and without retracing any part of the line? (12 points) | Solution. Since $\pi^{2}<16<(2 \pi)^{2}$, the given line consists of 2 circles with radii 4 and $\sqrt{16-\pi^{2}}$ and a vertical segment.
This line is unicursal, as it has only 2 odd poin... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. For how many two-digit natural numbers n are exactly two of these three statements true: (A) n is odd; (B) n is not divisible by $3 ;$ (C) n is divisible by 5? | Solution. We can consider the first 30 two-digit numbers (from 10 to 39), and then multiply the result by 3, since the remainders when dividing by 2, 3, and 5 do not change when shifted by 30 or 60. There are three mutually exclusive cases.
1) (A) and (B) are satisfied, and (C) is not. From (A) and (B), it follows tha... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the length of the larger base $A D$ of which is $12 \sqrt{3}$. The ratio of the areas of the parts of the trapezoid $A B C D$, into which it is divided by the midline, is $5: 7$. All lateral faces of the pyramid TABCD are inclined to... | # Solution.
Let TO be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, O is the center of the circle inscribed in the base. Let MP be the midline of the trapezoid, $A D=a=12 \sqrt{3}, B C=b$.
According to the problem, we have
$S_{\text {MBCP }}=5 x, S_{\text {AMPD }}=7 x... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. If a two-digit natural number is decreased by 54, the result is a two-digit number with the same digits but in reverse order. In the answer, specify the median of the sequence of all such numbers.
# | # Solution.
Let $\overline{x y}=10 x+y$ be the original two-digit number, then $\overline{y x}=10 y+x$ is the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+54$. From the equation, it is clear that the two-digit number is greater than 54. Let's start the investigation with... | 82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The wolf saw a roe deer several meters away from him and chased after her along a straight forest path. The wolf's jump is $22\%$ shorter than the roe deer's jump. Both animals jump at a constant speed. All the roe deer's jumps are of the same length, and the wolf's jumps are also equal to each other. There is a per... | Solution: Let $x$ be the length of the roe deer's jump, then $0.78 x$ is the length of the wolf's jump; $y$ - the number of jumps the roe deer makes over the time interval specified in the condition, $y\left(1+\frac{t}{100}\right)$ - the number of jumps the wolf makes over the same time interval. The wolf will not be a... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Ilya takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 1989 applications of this rule, if the initial triplet of numbers was $\{70 ; 61; 20\}$?... | Solution. Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b$.
$B=70-20=50$.
Answer: 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given triangle $A B C$. Lines $O_{1} O_{2}, O_{1} O_{3}, O_{3} O_{2}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point $O$ is the center of the inscribed circle of triangle $A B C$. Find the angle in degrees between the lines $O_{1} O_{2}$ and $O O_{3}$.
.
1) $\triangle I G C = \triangle G F H$ - by two legs, since $I C = G H = 2, I G = H F = 1$, therefore ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} M C_{1}$. | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considering t... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{25}{3}$ times. Deter... | # Solution.
1) Let the syrup content in the initial solution be $p \%$ and let $x$ liters of the solution were poured out the first time.
2) Then after pouring out the liquid, there remained $(1000-x)$ liters of the solution, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p}{100... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A warehouse has coffee packed in bags of 15 kg and 8 kg. How many bags of coffee in total does the warehouseman need to prepare to weigh out 1998 kg of coffee, with the number of 8 kg bags being the smallest possible? | Solution. Let $x$ be the number of bags weighing 15 kg, and $y$ be the number of bags weighing 8 kg. We get the equation $15 x + 8 y = 1998$. $8(x + y) + 7 x = 1998$, let $x + y = k$,
$8 k + 7 x = 1998$,
$7(k + x) + k = 1998$, let $k + x = t$,
$7 t + k = 1998$, $k = 1998 - 7 t$. Substitute into (2), $x = 8 t - 1998$... | 136 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find the degree measure of angle $B_{1} B M$.
# | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considering... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B C A=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=60^{\circ}$. On the side $\mathrm{BC}$, a point $\mathrm{M}$ is ch... | Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.
Then: 1) $\Delta T P M = \Delta K O M$... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Right triangles $M D C$ and $A D K$ have a common right angle $D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$, counting from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of the degree measures of angles $A K D$ and $M C D$, if $A D: C D=2: 5$. | Solution. Extend triangle $A D C$ to form a square $L J C D$.
Choose point $H$ on side $L J$ such that $L H: H J=2: 3$, point $N$ on side $C J$ such that $C N: N J=3: 2$, and point $B$ on si... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).
(16 points) | Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}^{m}}$, where $l_{i} \in\left\{0,1, \l... | 25920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 21, green - 45, blue - 75, dark blue - 87, purple - 91, black -... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text{ times}}(n) \ldots)-k$-fold composition of the function $f(n)$, which is equal to $n+4$ when $n \leq 19$, and equal to $|129-2 n|$ when $n \geq 20$. Let's compute and write down the first few value... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. (Option 2).
How to build a highway? (an old problem) From a riverside city A, goods need to be transported to point B, located $a$ kilometers downstream and $d$ kilometers from the riverbank. How should the highway be built from B to the river so that the transportation of goods from A to B is as cost-effec... | Solution: Let the distance $AD$ be denoted by $x$ and the length of the highway $DB$ by $y$: by assumption, the length of $AC$ is $a$ and the length of $BC$ is $d$.
Since transportation along the highway is twice as expensive as along the river, the sum $x + 2y$ should be the smallest according to the problem's requir... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. According to the inverse theorem of Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$:... | Answer: 32
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. For a rectangle, the sum of two sides is 11, and the sum of three sides is 19.5. Find the product of all possible different values of the perimeter of such a rectangle. | Solution:
Let the sides of the rectangle be $a$ and $b$.
If the sum of adjacent sides is 11, then the system describing the condition of the problem is $\left\{\begin{array}{l}a+b=11 \\ 2 a+b=19.5\end{array}\right.$, its solution is $a=8.5, b=2.5$, the perimeter of the rectangle is $P_{1}=22$.
If, however, the numbe... | 15400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) Find the area of a convex quadrilateral with equal diagonals, if the lengths of the segments connecting the midpoints of its opposite sides are 13 and 7. | Solution. Let $MK$ and $PH$ be segments connecting the midpoints of opposite sides of a convex quadrilateral $ABCD$, with $MK = PH$, $AC = 18$, and $BD = 7$.
We have: $MP \| AC$, $MP = \frac{1}{2} AC$ (as the midline of $\triangle ABC$); $HK \| AC$, $HK = \frac{1}{2} AC$ (as the midline of $\triangle ADC$). $\Rightarr... | 63 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Four elevators of a skyscraper, differing in color (red, blue, green, and yellow), are moving in different directions and at different but constant speeds. Observing the elevators, someone started a stopwatch and, looking at its readings, began to record: 36th second - the red elevator caught up with the blue one (m... | Solution. Let's number the elevators: red - first, blue - second, green - third, yellow - fourth. The elevators move at constant speeds, so the distance traveled $S_{i}, i=1,2,3,4$, in some coordinate system depends on time according to the law $S_{i}=k_{i} t+b_{i}$. According to the problem, the red and blue elevators... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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