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10.5. What is the maximum number of 4-element subsets that can be selected from a set of 8 elements such that the intersection of any three of the selected subsets contains no more than one element? | Answer. Eight.
Solution. We will provide two different examples of choosing eight 4-element subsets in a set $\mathrm{X}$ of eight elements, satisfying the condition of the problem. Both examples are constructed as geometric objects.
Example 1. We will consider the elements of $\mathrm{X}$ as the vertices of a unit c... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. At the Olympiad, students from gymnasiums, lyceums, and regular schools met. Some of them stood in a circle. Gymnasium students always lie to regular school students, lyceum students lie to gymnasium students, and regular school students lie to lyceum students. In all other cases, the students tell the truth. Each... | Answer: There were no ordinary school students in the circle.
Solution: Let's assume there was an ordinary school student in the circle. Consider his left neighbor. This neighbor could not be another ordinary school student or a lyceum student, because they would tell the truth. But he also cannot be a gymnasium stude... | 0 | Logic and Puzzles | 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 |
8.5. Given a convex quadrilateral $A B C D$ with side $A D$ equal to 3. Diagonals $A C$ and $B D$ intersect at point $E$, and it is known that the areas of triangles $A B E$ and $D C E$ are both 1. Find the side $B C$, given that the area of $A B C D$ does not exceed 4. | # Answer: 3.
Solution: Triangles $A B D$ and $A C D$ have the same area, as they are composed of the common $A E D$ and equal-area $A B E$ and $D C E$. Since $A B D$ and $A C D$ have the same base $A D$, their heights to this base are equal. Therefore, $B C$ is parallel to $A D$, meaning our quadrilateral is a trapezo... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find all triples of distinct natural numbers, the least common multiple of which is equal to their sum. The least common multiple of several numbers is the smallest natural number that is divisible by each of these numbers. | Answer. All triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n$.
Solution. Let the required numbers be $a<b<c$. By the condition, their least common multiple, which is equal to $a+b+c$, is divisible by $c$, so $c$ divides the sum $a+b<2 c$. Therefore, $a+b=c$. Next, $a+b+c=2 a+2 b$ is divisible by $b$, so... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. For non-negative numbers $a, b, c, d$, the following equalities are satisfied:
$\sqrt{a+b}+\sqrt{c+d}=\sqrt{a+c}+\sqrt{b+d}=\sqrt{a+d}+\sqrt{b+c}$. What is the maximum number of distinct values that can be among the numbers $a, b, c, d$? | Answer. Two.
Solution. Square the first equality, combine like terms, cancel by 2, square again, combine like terms once more, ultimately obtaining the equality $a c+b d=a b+c d$, equivalent to $(a-d)(c-b)=0$, from which either $a=d$ or $b=c$. Performing similar manipulations with the second equality, we get $a=b$ or ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. What is the maximum number of colors needed to color all cells of a 4 by 4 square so that for every pair of different colors, there are two cells of these colors that are either in the same row or in the same column of the square? | Answer: In 8 colors.
Solution. If the cells were painted in 9 or more colors, there would be a color in which only one cell is painted. There are only 6 cells located in the same row or column with it, so there are no more than 6 other colors forming a pair with it, as required by the condition - a contradiction. Ther... | 8 | 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 |
8.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed into several bags (each bag contains more than one barrel). We will call a bag good if the number of one of the barrels in it equals the product of the numbers of the other barrels in the sam... | # Answer: 8.
Solution: In each good bag, there are no fewer than three barrels. The smallest number in each good bag must be unique, otherwise the largest number in this bag would be no less than $10 \times 11=110$, which is impossible. For the same reason, if a good bag contains a barrel with the number 1, it must al... | 8 | Combinatorics | 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 |
7.3. In the math test, each of the 25 students could receive one of four grades: $2, 3, 4$ or 5. It turned out that the number of students who received fours was 4 more than those who received threes. How many people received twos, if it is known that the sum of all grades for this test is 121? | # 7.3. Answer: 0.
Let's assume that the 4 students who received fours skipped school, then 21 students should have scored 105 points on the test, which is only possible if all of them received fives. Therefore, in the real situation, no one received a two.
## Comments on Evaluation.
Answer only: 1 point. Answer with... | 0 | Algebra | 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 |
9.5. Let $M$ - be a finite set of numbers (distinct). It is known that among any three of its elements, there will be two whose sum belongs to $M$. What is the maximum number of elements that can be in $M$? | Answer: 7.
Solution: An example of a set with 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that a set $M=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ with $n>7$ numbers does not have the required property.
We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a_{4}>0$ (changing the signs of all elements does ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the wagon, several kilograms of apple jam were loaded, of which $20 \%$ was good and $80 \%$ was bad. Every day, half of the existing bad jam rotted, and it was thrown away. After several days, it turned out that $20 \%$ of the jam in the wagon was bad and $80 \%$ was good. How many days have passed since the l... | Answer: 4 days.
Solution 1: Let the initial total amount be $x$, and the final amount be $y$ kilograms of jam. Then, since the amount of good jam did not change, $0.2 x = 0.8 y$, which means $x = 4 y$. Therefore, initially, the amount of bad jam was $0.8 x = 3.2 y$, and it became $0.2 y$, meaning the mass of bad jam d... | 4 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.2. Once Alexei and Daniil were playing such a game. If a number \( x \) is written on the board, it can be erased and replaced with \( 2x \) or \( x - 1000 \). The player who gets a number not greater than 1000 or not less than 4000 loses. Both players aim to win. At some point, the boys stopped playing. Who lost if ... | Answer: no one lost.
Solution: note that if a number is less than 2000 but greater than 1000, then by multiplying by 2, you can get a number that is less than 4000. If a number is less than 4000 but greater than 2000, then by subtracting 1000 (possibly twice), you can get a number between 1000 and 2000. Thus, the only... | 1 | Logic and Puzzles | 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 |
9.2. Around a round table, 15 boys and 20 girls sat down. It turned out that the number of pairs of boys sitting next to each other is one and a half times less than the number of pairs of girls sitting next to each other. Find the number of boy - girl pairs sitting next to each other. | Answer: 10.
Solution. Let's call a group several children of the same gender sitting in a row, with children of the opposite gender sitting to the left and right of the outermost ones. Let $X$ be the number of groups of boys, which is equal to the number of groups of girls sitting in a row. It is easy to see that the ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. What is the maximum number of triangles with vertices at the vertices of a regular 18-gon that can be marked so that no two different sides of these triangles are parallel? The triangles can intersect and have common vertices, coinciding segments are considered parallel. | Answer: 5.
Solution. Estimating the number of triangles. Let's number the vertices of the 18-gon from 1 to 18 clockwise. The sides of the triangles are the sides and diagonals of the regular 18-gon. We will call a diagonal even if an even number of sides lies between its ends, and odd otherwise. The parity of a diagon... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.3. Find the maximum length of a horizontal segment with endpoints on the graph of the function $y=x^{3}-x$ | Answer: 2.
Solution 1. A horizontal segment of length $a>0$ with endpoints on the graph of the function $y=x^{3}-x$ exists if and only if the equation $(x+a)^{3}-(x+a)=x^{3}-x$ has at least one solution for the given value of the parameter $a$. Expanding the brackets, combining like terms, and dividing by $a>0$, we ob... | 2 | Calculus | 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 |
8.4. From identical isosceles triangles, where the angle opposite the base is $45^{\circ}$ and the lateral side is 1, a figure was formed as shown in the diagram. Find the distance between points $A$ and $B$. | Answer: 2.
Solution: Let's denote the points $K, L, M$, as shown in the figure. We will construct an isosceles triangle $A K C$ equal to the original one. Connect vertex $C$ to other points as shown in the figure.
In the original triangles, the angle at the vertex is $45^{\circ}$. Therefore, the other two angles are ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. On a plane, there are points $A, B, C, D, X$. Some segment lengths are known: $A C=2$, $A X=5$, $A D=11$, $C D=9$, $C B=10$, $D B=1$, $X B=7$. Find the length of the segment $C X$. | Answer: 3.
Solution: Note that $A D=11=2+9=A C+C D$. Therefore, points $A, C, D$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Note that $C B=10=9+1=C D+D B$. Therefore, points $C, D, B$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Therefore, all fou... | 3 | Geometry | 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 |
10.5. In each cell of a 5 by 5 table, a letter is written such that in any row and in any column there are no more than three different letters. What is the maximum number of different letters that can be in such a table | Answer: 11.
Solution: If each row contains no more than two different letters, then the total number of letters does not exceed $10=5 * 2$. Further, we can assume that the first row contains exactly three different letters. If each of the remaining rows has at least one letter in common with the first, then the total ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Let $a^{2}+b^{2}=c^{2}+d^{2}=1$ and $a c+b d=0$ for some real numbers $a, b, c, d$. Find all possible values of the expression $a b+c d$. | Answer: 0.
Solution: Let's first assume $b \neq 0$. From the second equation, express $d=\frac{-a c}{b}$ and substitute it into the equation $c^{2}+d^{2}=1$. Eliminating the denominator, we get $c^{2}\left(a^{2}+b^{2}\right)=b^{2}$, from which, given $a^{2}+b^{2}=1$, we obtain $b= \pm c$. Substituting this into the eq... | 0 | Algebra | 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 |
8.1. Large sandglasses measure an hour, and small ones measure 11 minutes. How can you use these sandglasses to measure a minute? | Solution: We will run the large hourglass twice in a row and the small one eleven times in a row. A minute will be measured between the second time the large hourglass finishes (120 minutes) and the 11th time the small one finishes (121 minutes).
Criteria: Any correct example - 7 points. | 1 | Logic and Puzzles | 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 |
10.1. Vikentiy walked from the city to the village, and at the same time Afanasiy walked from the village to the city. Find the distance between the village and the city, given that the distance between the pedestrians was 2 km twice: first, when Vikentiy had walked half the way to the village, and then, when Afanasiy ... | Answer: 6 km.
Solution. Let the distance between the village and the city be denoted as $S$ km, the speeds of Vikentiy and Afanasy as $x$ and $y$, and calculate the time spent by the travelers in the first and second cases. In the first case, we get: $\frac{S / 2}{x}=\frac{S / 2-2}{y}$, in the second case $\frac{2 S /... | 6 | Algebra | 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 |
9.1. From two cities, the distance between which is 105 km, two pedestrians set out simultaneously towards each other at constant speeds and met after 7.5 hours. Determine the speed of each of them, knowing that if the first walked 1.5 times faster, and the second 2 times slower, they would have met after $8 \frac{1}{1... | Answer: 6 and 8 km per hour.
Solution: Let their speeds be $x$ and $y$ km per hour, respectively. From the condition, we get: $\frac{15}{2}(x+y)=105, \frac{105}{13}\left(\frac{3}{2} x+\frac{1}{2} y\right)=105$, from which $x=6, y=8$. | 6 | 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 |
11.3. Two circles intersect at points A and B. A tangent to the first circle is drawn through point A, intersecting the second circle at point C. A tangent to the second circle is drawn through point B, intersecting the first circle at point D. Find the angle between the lines $\mathrm{AD}$ and $\mathrm{BC}$. | Answer. The lines are parallel, the angle is $0^{\circ}$.
Solution. Mark point $\mathrm{P}$ on the extension of $\mathrm{CA}$ beyond point A and point M on the extension of DB beyond point B. The inscribed angle ABD in the first circle, which subtends the chord AD, is equal to the angle PAD between the chord AD and th... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. There are two ingots of different copper and tin alloys weighing 6 and 12 kg respectively. From each of them, a piece of the same weight was cut off and the first piece was alloyed with the remainder of the second ingot, and the second piece - with the remainder of the first ingot, after which the ratio of copper ... | Answer: 4 kilograms.
Solution: Let the weight of each of the cut pieces be $x$ kg, and the proportions of tin in the first and second ingots be $a \neq b$ respectively. Then, the proportion of tin after remelting in the first ingot will be $\frac{b x + a(6 - x)}{6}$, and in the second ingot $\frac{a x + b(12 - x)}{12}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the tournament, each of the six teams played against each other exactly once. In the end, the teams scored 12, 10, 9, 8, 7, and 6 points respectively. a) How many points were awarded for a win in a match, if 1 point was awarded for a draw, and 0 points for a loss? The answer, of course, should be a natural numb... | Answer. a) 4 points.
b) The first team had three wins, the second team had two wins and two draws, the third team had two wins and one draw, the fourth team had two wins, the fifth team had one win and three draws, the sixth team had one win and two draws. The rest of the matches were lost by the teams.
c) One exampl... | 4 | Combinatorics | 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 $B F$.
 games. Therefore, a total of 21 games were played... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. On the board, 10 natural numbers are written, some of which may be equal, and the square of each of them divides the sum of all the others. What is the maximum number of different numbers that can be among the written ones? | Answer. Four
Solution. An example for four different numbers: 1,1,1,2,2,3,5,5,5,5.
Let among the listed numbers there are exactly $n \geq 2$ different ones, the maximum of which we denote by $x$, and the sum of all numbers by $\mathrm{S}$. Then the sum of all numbers except the maximum does not exceed $(9-n) x + x - ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. The bathtub fills up from the hot water tap in 17 minutes, and from the cold water tap in 11 minutes. After how many minutes from opening the hot water tap should the cold water tap be opened so that by the time the bathtub is full, there is one third more hot water than cold water? | Answer. In 5 minutes.
Solution. Let the volume of the bathtub be $V$, then by the time it is filled, there should be $\frac{3}{7} V$ of cold water and $\frac{4}{7} V$ of hot water in it. The filling rates of cold and hot water are $\frac{V}{11}$ and $\frac{V}{17}$, respectively. Therefore, the desired time is the diff... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find the smallest natural number $n$ such that in any set of $n$ distinct natural numbers, not exceeding 1000, it is always possible to select two numbers, the larger of which does not divide the smaller one. | Answer. $n=11$.
Solution. Among the first 10 powers of two $1=2^{0}, 2=2^{1}, 4=2^{2}, \ldots, 512=2^{9}$, in each pair of numbers, the larger number is divisible by the smaller one, hence $n \geq 11$.
On the other hand, let there be some set of $n \geq 11$ numbers where the larger number of each pair is divisible by... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. In each cell of a 10 by 10 table, a minus sign is written. In one operation, it is allowed to simultaneously change the signs to their opposites in all cells of some column and some row (plus to minus and vice versa). 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 |
7.3. Anya wrote down 100 numbers in her notebook. Then Sonya wrote down in her notebook all the pairwise products of the numbers written by Anya. Artem noticed that there were exactly 2000 negative numbers in Sonya's notebook. How many zeros did Anya initially write down in her notebook? | Answer: 10 zeros.
Solution: Let Anya write down $n$ positive numbers, $m$ negative numbers, and $100-n-m$ zeros in her notebook. Then, by the condition, $n m=2000$, since a negative number can only be obtained by multiplying numbers of different signs.
Let's list all the divisors of the number $2000=2^{4} * 5^{3}$:
... | 10 | Combinatorics | 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 |
7.4. Arseny sat down at the computer between 4 and 5 PM, when the hour and minute hands were pointing in opposite directions, and got up from it on the same day between 10 and 11 PM, when the hands coincided. How long did Arseny sit at the computer? | Solution: Let's see where the hands will be 6 hours after Arseny sat down at the computer. The minute hand will go around the clock 6 times and return to its place. The hour hand will move exactly half a circle. Therefore, the angle between the hands will change by 180 degrees, i.e., the hands will coincide. It is obvi... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. Several seventh-graders were solving problems. The teacher does not remember how many children there were and how many problems each of them solved. However, he remembers that, on the one hand, each solved more than a fifth of what the others solved, and on the other hand, each solved less than a third of what the... | Answer: 5 seventh-graders.
Solution: Let one seventh-grader solve $a$ problems, and the rest solve $S - a$. Then
$$
\begin{gathered}
a < (S-a) / 3 \\
3a < S - a \\
4a < S \\
a < S / 4
\end{gathered}
$$
Similarly,
$$
\begin{gathered}
(S-a) / 5 < a \\
S - a < 5a \\
S < 6a \\
S / 6 < a
\end{gathered}
$$
Thus, if ther... | 5 | Inequalities | 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 |
3. Point $M$ lies on the leg $A C$ of the right triangle $A B C$ with a right angle at $C$, such that $A M=2, M C=16$. Segment $M H$ is the altitude of triangle $A M B$. Point $D$ is located on the line $M H$ such that the angle $A D B$ is $90^{\circ}$, and points $C$ and $D$ lie on the same side of the line $A B$. Fin... | Solution. 1. A circle can be circumscribed around quadrilateral $A B C D$ with diameter $A B$ (angles $A D B$ and $A C B$ are right angles). Then $\angle A B D=\angle A C D$,
$$
\angle H A D=90^{\circ}-\angle A B D, \angle A D H=\angle A B D=\angle A C D
$$
Triangles $A C D$ and $A D M$ are similar, and $\frac{A D}{A... | 6 | Geometry | 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 |
6.
Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. # | # Solution:
Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:
$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from ... | -5 | 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 $A W=C W$).
 x+a}{x^{2}+5 x+4} \geq 0$ is the union of three non-overlapping intervals. In your answer, specify the sum of the three smallest integer values of $a$ from the obtained interval. | # Solution
Let's factorize the numerator and the denominator of the left part of the inequality, it will take the form: $\frac{(x+1)(x+a)}{(x+1)(x+4)} \geq 0$. There are five possible cases for the placement of the number ($-a$) relative to the numbers (-4) and (-1). In each case, the inequality is solved using the in... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.
Calculate the number $8^{2021}$, find the sum of the digits in this number, and write down the result. Then, in the newly written number, find the sum of the digits and write down the result again. These actions were repeated until a single-digit number was obtained. Find this number. | # Solution.
Consider the natural powers of 8. Notice that even powers of the number 8 give a remainder of 1 when divided by 9, while odd powers (including the number \(8^{2021}\)) give a remainder of 8. Indeed, let's analyze the powers of 8:
\[
\begin{gathered}
8^{2}=(9-1)^{2}=9 n+1, n \in N \\
8^{3}=(9 n+1) \cdot 8=... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. We will call a ticket with a number from 0001 to 2014 excellent if the difference between some two adjacent digits of its number is 5. Find the number of excellent tickets. | Solution. The number of excellent tickets from 0001 to 2014 is equal to the number of excellent tickets from 0000 to 2014. First, let's calculate the number of non-excellent tickets from 0000 to 2014.
The number of non-excellent tickets from 0000 to 1999 can be found as follows. Let $a_{1} a_{2} a_{3} a_{4}$ be the nu... | 543 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (15 points) 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 |
2. Solve the equation $8 \sin ^{4}(\pi x)-\sin ^{2} x=\cos ^{2} x-\cos (4 \pi x)$. In your answer, specify the sum of the roots that belong to the interval $[-1 ; 2]$.
(5 points) | Solution. Considering the basic trigonometric identity, we get
$8 \sin ^{4}(\pi x)-1+\cos (4 \pi x)=0 \quad \Rightarrow \quad 8 \sin ^{4}(\pi x)-2 \sin ^{2}(2 \pi x)=0 \quad \Rightarrow$
$\left(2 \sin ^{2}(\pi x)-2 \sin (\pi x) \cos (\pi x)\right)\left(2 \sin ^{2}(\pi x)+2 \sin (\pi x) \cos (\pi x)\right)=0 \Rightarr... | 5 | Algebra | 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
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \... | 2018 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency | Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$
$$
\begin{aligned}
& S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-\text { abscissa of the point of tangency } A, \\
& A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad \text { - ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$. | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Le... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A vessel with a capacity of 10 liters is filled with air containing $24\%$ oxygen. A certain volume of air was pumped out of the vessel and the same volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again the same volume of argon was added. In the new mixture, $11.7... | Solution. Let $x$ liters of the mixture be released each time from the vessel. Then, the first time, the amount of oxygen left in the vessel is $2.4 - 0.24x$. The percentage of oxygen in the mixture after adding argon is $(2.4 - 0.24x) \times 10$. The second time, the amount of oxygen left in the vessel is $2.4 - 0.24x... | 3 | Algebra | 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 |
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