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1. Solve the problem: octopuses with an even number of legs always lie, while octopuses with an odd number of legs always tell the truth. Five octopuses met, each having between 7 and 9 legs.
The first said: "We have 36 legs in total";
The second: "We have 37 legs in total";
The third: "We have 38 legs in total";
T... | Solution. All answers are different, so one is telling the truth or all are lying. If all were lying, they would have 8 legs, i.e., a total of 40, which matches the last answer, a contradiction.
Therefore, four are lying, one is telling the truth, and the one telling the truth has an odd number of legs since $4 \cdot ... | 39 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,681 |
2. Two cars started from the same point on a circular track 150 km long, heading in opposite directions. After some time, they met and continued moving in the same directions. Two hours after the start, the cars met for the second time. Find the speed of the second car if the speed of the first car is 60 km/h. | Solution. Let the speed of the first car be $x$ km/h, then their closing speed will be $(x+60)$ km/h. Obviously, in two hours they traveled two laps, i.e., 300 km, hence $2 \cdot(x+60)=300$, from which we get that $x=90($ km/h).
Answer: 90 km/h | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,682 |
3. Prove that there do not exist integers $x$ and $y$ such that the equation $(x+7)(x+6)=8y+3$ holds. | Solution. Let $x$ and $y$ be integers. Notice that $x+6$ and $x+7$ are consecutive integers, so one of them will be even, i.e., divisible by 2. Therefore, the product $(x+7)(x+6)$ will be divisible by 2. Next, we see that $8y$ is an even number for any integer value of $y$, and thus $8y+3$ is an odd number (as the sum ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,683 |
4. In triangle $\mathrm{ABC}$, $\angle A=3 \angle C$. Point $D$ on side $B C$ has the property that $\angle D A C=2 \angle C$. Prove that $A B+A D=B C$. | Solution. Let $\angle C=x$, then $\angle B A C=3 x$. Extend the segment $B A$ beyond point $A$ and mark a segment $A E=A D$ on it (see the figure). Clearly, $\angle E A C=180^{\circ}-\angle B A C=180^{\circ}-3 x$.
Notice that $\triangle E A C=\triangle A D C$ by the first criterion ( $A C$ is common, $A D=A E$ (by con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,684 |
5. At the railway platform, many people gathered early in the morning waiting for the train. On the first train, one-tenth of all those waiting left, on the second train, one-seventh of the remaining left, and on the third train, one-fifth of the remaining left. How many passengers were initially on the platform if 216... | Solution. Solving from the end, we get that before the departure of the third train, there were $216: 6 \cdot 5=270$ passengers on the platform, before the departure of the second - $270: 6 \cdot 7=315$ passengers, before the departure of the first - $315: 9 \cdot 10=350$ passengers.
Answer: 350 passengers. | 350 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,685 |
11.1. On the board, there are $n$ different integers. The product of the two largest is 77. The product of the two smallest is also 77. For what largest $n$ is this possible?
(R. Zhenodarov, jury) | Answer. For $n=17$.
Solution. The numbers $-11, -7, -6, -5, \ldots, 6, 7, 11$ provide an example for $n=17$.
Assume that there are at least 18 such numbers. Then, at least 9 of them will have the same sign (all positive or all negative). Among these 9 numbers, the absolute values of the two largest will be at least 8... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,686 |
11.2. Set $A$ consists of $n$ distinct natural numbers, the sum of which is equal to $n^{2}$. Set $B$ also consists of $n$ distinct natural numbers, the sum of which is equal to $n^{2}$. Prove that there will be a number that belongs to both set $A$ and set $B$.
(D. Khramov) | Solution. Suppose the opposite: sets $A$ and $B$ do not intersect. Then their union contains $2n$ distinct natural numbers. Consequently, the sum $S$ of all elements in the union of sets $A$ and $B$ will be no less than the sum $1+2+\ldots+2n=n(2n+1)$. On the other hand, by the condition $S=2n^2$, which is less than $n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,687 |
11.3. In a right triangle $ABC$ with hypotenuse $AC$, a height $BH$ is dropped. On side $BC$, point $D$ is marked, on segment $BH$ - point $E$, and on segment $CH$ - point $F$ such that $\angle BAD = \angle CAE$ and $\angle AFE = \angle CFD$. Prove that $\angle AEF = 90^{\circ}$. | Solution. Construct point $E^{\prime}$, symmetric to point $E$ with respect to side $A C$ (see Fig. 3). Note that point $F$ lies on line $D E^{\prime}$, because $\angle D F C = \angle E F A = \angle E^{\prime} F A$ due to symmetry. From the right triangles $A B C$ and $B C H$, we get $\angle E^{\prime} B C = 90^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,688 |
11.4. Let $p$ be a prime number greater than 3. Prove that there exists a natural number $y$, less than $p / 2$, such that the number $p y+1$ cannot be represented as the product of two integers, each greater than $y$. | Solution. Let $p=2k+1$. Assume the opposite: for each of the numbers $y=1,2, \ldots, k$ there exists a factorization $p y+1=a_{y} b_{y}$, where $a_{y}>y, b_{y}>y$. Note that each of the numbers $a_{y}$ and $b_{y}$ is strictly greater than 1, and that $a_{y} > p y + 1$. Therefore, each of the $p-1$ numbers in the set $a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,689 |
11.5. In an $N \times N$ table, all natural numbers from 1 to $N^{2}$ are arranged. A number is called large if it is the largest in its row, and small if it is the smallest in its column (thus, a number can be both large and small at the same time, or it can be neither). Find the smallest possible difference between t... | Answer: $\frac{N(N-1)(2 N+5)}{6}$.
Solution: Note that the number $N^{2}$ is large, and 1 is small, their difference is $N^{2}-1$. Let's delete the row containing $N^{2}$ and the column containing 1. Then, if $A$ and $B$ are the largest and smallest of the remaining $(N-1)^{2}$ numbers, $A-B \geqslant(N-1)^{2}-1$. At ... | \frac{N(N-1)(2N+5)}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,690 |
1. Find all pairs of real numbers ( $\mathrm{x}, \mathrm{y}$ ), for which the equality $\sqrt{x^{2}+y^{2}-1}=1-x-y$ holds. | Answer: All pairs of numbers $(1, t),(t, 1)$, where $\mathrm{t}$ is any non-positive number.
Hint. By squaring and factoring, we get that $\mathrm{x}=1$ or $\mathrm{y}=1$. The square root is non-negative, so the sum $1-x-y \geq 0$. Substituting the possible value of the variable into this inequality, we get that the o... | (1,),(,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,691 |
4. Find all functions $\mathrm{f}$ defined on the set of real numbers and taking real values, such that $2 \mathrm{f}(\mathrm{x})+2 \mathrm{f}(\mathrm{y})-\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \geq 4$ for all real $\mathrm{x}, \mathrm{y}$. | Answer: $\mathrm{f}(\mathrm{x})=2$ (a constant function, all values of which are equal to 2).
Hint. By setting $\mathrm{y}=\mathrm{x}$ in the inequality and simplifying, we get $\mathrm{f}^{2}(\mathrm{x})-4 \mathrm{f}(\mathrm{x})+4 \leq 0$ or $(\mathrm{f}(\mathrm{x})-2)^{2} \leq 0$. From this, we have that $\mathrm{f}... | f(x)=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,692 |
Problem 6.1. Find any solution to the puzzle
$$
\overline{A B}+A \cdot \overline{C C C}=247
$$
where $A, B, C$ are three different non-zero digits; the notation $\overline{A B}$ represents a two-digit number composed of the digits $A$ and $B$; the notation $\overline{C C C}$ represents a three-digit number consisting... | Answer: 251.
Solution. The number $\overline{C C C}$ is divisible by 111 and is less than 247, so $A \cdot \overline{C C C}$ is either 111 or 222. In the first case, we get that $\overline{A B}=247-111=136$, which is impossible. In the second case, $\overline{A B}=247-222=25$, that is, $A=2, B=5$, and therefore, $C=1$... | 251 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,693 |
Problem 6.2. At a ball, princesses and knights gathered - a total of 22 people. The first princess danced with seven knights, the second with eight knights, the third with nine knights, ..., the last danced with all the knights present. How many princesses were at the ball? | Answer: 8.
Solution. Note that the number of knights the princess danced with is 6 more than her number. Let there be $x$ princesses in total, then the last one has the number $x$ and danced with all the knights, and there are $x+6$ of them in total. We get that there were $x+$ $(x+6)=2x+6=22$ people at the ball, whic... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,694 |
Problem 6.3. Dima, Misha, and Yura decided to find out who among them is the most athletic. For this, they held 10 competitions. The winner received 3 points, the second place 1 point, and the third place received nothing (in each competition, there was a first, second, and third place). In total, Dima scored 22 points... | Answer: 10.
Solution. In each competition, the boys in total received $3+1+0=4$ points. For all competitions, they scored $4 \cdot 10=40$ points. Dima and Misha in total scored $22+8=30$ points, so the remaining $40-30=10$ points were scored by Yura. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,695 |
Problem 6.4. On her birthday, Katya treated her classmates with candies. After giving out some candies, she noticed that she had 10 more candies left than Artem received. After that, she gave everyone one more candy, and it turned out that all the children in the class (including Katya) had the same number of candies. ... | Answer: 9.
Solution. Initially, the number of candies Kati and Artyom had differed by 10. When Kati gave everyone one more candy, the number of candies Artyom had increased by 1, and the number of candies Kati had decreased by the number of her classmates, and they ended up with the same amount. This means that 10 is ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,696 |
Problem 6.5. Cat Matroskin, Uncle Fyodor, Postman Pechkin, and Sharik sat down at a round table. In front of each of them was a plate with 15 sandwiches. Every minute, three of them ate a sandwich from their own plate, while the fourth ate a sandwich from their neighbor's plate. After 5 minutes of the meal, there were ... | # Answer: 7
Solution. We will call the sandwiches eaten from a neighbor's plate stolen. Note that in 5 minutes, exactly 5 sandwiches were stolen.
At the same time, 7 sandwiches disappeared from Uncle Fyodor's plate, of which he himself ate no more than 5, meaning that at least 2 were stolen. From this, it is clear th... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,697 |
Problem 6.6. Ksyusha placed a chip in one of the cells of a $4 \times 4$ square. She moved the chip from cell to cell within this square, each time moving to an adjacent cell by side, and visited each cell exactly once. In each cell, she recorded the move number on which the chip landed in that cell (the left image). S... | Answer: $\mathrm{a} 4, \mathrm{~b} 3, \mathrm{c} 1, \mathrm{~d} 2$
Solution. We can start either to the left or to the right of the cell with number 2. If we start from the cell to the right, the sequence of cells up to number 5 is uniquely determined. By enumeration, we can verify that only one route remains, and it ... | ,,, | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,698 |
Problem 6.7. A store received one box each of oranges, apples, cucumbers, and potatoes. The boxes were numbered from 1 to 4, and each had an inscription:
- First box: “Potatoes are in the second box.”
- Second box: “Oranges are not in this box.”
- Third box: “Apples are here.”
- Fourth box: “Cucumbers are in the 1st o... | Answer: $\mathrm{a} 2 \mathrm{~b} 3 \mathrm{c} 1 \mathrm{~d} 4$.
Solution. Let's look at the third box. It cannot contain vegetables (since the inscription would be false) and it cannot contain apples (since the inscription would be true); therefore, it must contain oranges.
Now let's see where the apples can be. It ... | \mathrm{~b}3\mathrm{}1\mathrm{~}4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,699 |
1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101$? | 1. Answer. It exists.
For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$. | 1234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,701 |
2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | 2. Answer: 8.
Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent to each... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,702 |
3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent? | 3. Answer. 5 cupcakes.
The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote $x, y$, and $z$ as the costs of a cake, a cupcake, and a bagel respectively, we get the equation: $(x+2 y+3 z)+(3 x+z)=12 y, \quad$ from which it follows that $\quad 4 x+4 z=10 y$, \quad t... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,703 |
4. There are 26 students in the class. They agreed that each of them would either be a liar (liars always lie) or a knight (knights always tell the truth). When they came to the class and sat down at their desks, each of them said: “I am sitting next to a liar.” Then some students moved to different desks. Could it be ... | # 4. Answer. Could not
Notice that the phrase "I sit next to a liar" could only have been said in the case where a liar and a knight sit at the same desk. This means that in the class of liars and knights, there are an equal number of each - 13 of each. The phrase "I sit next to a knight" could only have been said in ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,704 |
5. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $6 \times 6$ square of cells so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.) | 5. Answer. 6.
Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted.
Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get that at l... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,705 |
1. Write the number 1997 using 10 threes and arithmetic operations. | Solution. $1997=3 \cdot 333+3 \cdot 333-3: 3$.
## CONDITION | 1997=3\cdot333+3\cdot333-3:3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,706 |
3. Find all three-digit numbers that decrease by 5 times after the first digit is erased. | Solution. Let $n=abc$ be the desired number. According to the condition, $100a + 10b + c = 5(10b + c)$, which means $25a = 10b + c$, i.e., $c$ is divisible by 5. If $c=0$, then $5a = 2b$, so $b=5, a=2$ (if $b=0 \Rightarrow a=0$). If $c=5$, then $5a = 2b + 1$, so $b=2$ or $b=7$.
Answer. $125, 250, 375$. | 125,250,375 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,707 |
6. Ten schoolchildren, boys and girls, are sitting on a bench. Is it possible that between any two boys there is an even number of schoolchildren, and between any two girls - an odd number? | Solution. Let's look at the seats with even numbers and the seats with odd numbers. If one girl is sitting on a seat with an odd number and another on an even number, then there is an even number of students between them, which means either the even or the odd seats are free of girls. Consider the first case: then the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,709 |
8.1. The graphs of three functions $y=a x+a, y=b x+b$ and $y=c x+d$ have a common point, and $a \neq b$. Is it necessarily true that $c=d$? Justify your answer. | Answer: Yes, definitely.
Solution. The first method. The common point of the graphs of the first two functions can be found from the system:
$\left\{\begin{array}{l}y=a x+a, \\ y=b x+b\end{array} \Leftrightarrow\left\{\begin{array}{l}y=a x+a, \\ a x+a=b x+b\end{array} \Leftrightarrow\left\{\begin{array}{l}y=a x+a, \\... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,711 | |
8.2. A rectangular frame was cut out from graph paper (see figure). It was cut into nine pieces along the cell boundaries and a $6 \times 6$ square was formed from them. Could all the pieces obtained from the cutting be different? (When forming the square, the pieces can be flipped.)
Fig. 8.... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,712 |
8.3. Vertex $A$ of parallelogram $A B C D$ is connected by segments to the midpoints of sides $B C$ and $C D$. One of these segments turned out to be twice as long as the other. Determine whether angle $B A D$ is acute, right, or obtuse. | Answer: obtuse.
Solution. Let $N$ be the midpoint of $BC$, $M$ be the midpoint of $CD$, and $AN=2AM$ (see Fig. 8.3a, b).
First method. Draw a line through point $M$ parallel to $BC$. It will intersect $AB$ at point $K$, such that $AK=KB$ (see Fig. 8.3a). Then, by Thales' theorem, $AP=PN=0.5AN=AM$. In the isosceles tr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,713 |
8.4. Three pirates divided the diamonds they had obtained during the day in the evening: twelve each for Bill and Sam, and the rest went to John, who couldn't count. At night, Bill stole one diamond from Sam, Sam stole one from John, and John stole one from Bill. As a result, the average weight of Bill's diamonds decre... | Answer: 9 diamonds.
Solution. The first method (arithmetic). Note that the number of diamonds each pirate has did not change overnight. Since Bill has 12 diamonds, and their average weight decreased by 1 carat, the total weight of his diamonds decreased by 12 carats. Similarly, Sam also has 12 diamonds, and their aver... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,714 |
8.5. In triangle $A B C$, angle $B$ is equal to $120^{\circ}, A B=2 B C$. The perpendicular bisector of side $A B$ intersects $A C$ at point $D$. Find the ratio $A D: D C$. | Answer: $A D: D C=2: 3$.
Solution. First method. Let $M$ be the midpoint of side $A B$. Drop a perpendicular $C H$ to line $A B$ (see Fig. 8.5a). In the right triangle $B H C: \angle H B C=60^{\circ}$, then $\angle B C H=30^{\circ}$, so $B H=\frac{1}{2} B C=\frac{1}{4} A B=\frac{1}{2} A M$. Therefore, $H M: M A=3: 2$.... | AD:DC=2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,715 |
6. The gnomes sat around a round table and decided many issues by voting. For each issue, they could vote "for", "against", or abstain. If both neighbors of a gnome chose the same option for a certain issue, then for the next issue, the gnome would choose the same option. If the neighbors chose two different options, t... | Answer: the number of gnomes could be any multiple of 4.
Solution. If the gnomes voted unanimously on any question, they would always vote the same way thereafter. Therefore, the question about the dragon was discussed before the question about the gold.
It is possible that before the question about the gold, the gno... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,716 |
Problem 8.1. The numbers $x, y, z$ are such that $x \in[-3,7], y \in[-2,5], z \in[-5,3]$.
(a) (1 point) Find the smallest possible value of the quantity $x^{2}+y^{2}$.
(b) (3 points) Find the smallest possible value of the quantity $x y z - z^{2}$. | # Answer:
(a) (1 point) 0.
(b) (3 points) -200.
Solution. (a) Note that $x^{2} \geqslant 0$ and $y^{2} \geqslant 0$, so $x^{2}+y^{2} \geqslant 0$. The value $x^{2}+y^{2}=0$ is possible when $x=0, y=0$.
(b) Note that $|x y z|=|x||y||z| \leqslant 7 \cdot 5 \cdot 5$ and $z^{2} \leqslant 5^{2}$, so $x y z-z^{2} \geqsla... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,717 |
Problem 8.2. Given a rectangle $A B C D$. A line passing through vertex $A$ and point $K$ on side $B C$ divides the entire rectangle into two parts, the area of one of which is 5 times smaller than the area of the other. Find the length of segment $K C$, if $A D=60$.
Fig. 1: to the solution of problem 8.2
Solution. Draw a line through $K$ parallel to $AB$. Let it intersect side $AD$ at point $L$ (Fig. 1), then $ABKL$ and $DCKL$ are rectangl... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,718 |
Problem 8.3. In a row, there are 127 balls, each of which is either red, green, or blue. It is known that
- there is at least one red, at least one green, and at least one blue ball;
- to the left of each blue ball, there is a red ball;
- to the right of each green ball, there is a red ball.
(a) (1 point) What is the... | # Answer:
(a) (1 point) 125.
(b) (3 points) 43.
Solution. (a) Among 127 balls, there is at least 1 green and at least 1 blue, so there are no more than 125 red balls. Note also that there can be exactly 125 if the leftmost ball is blue, the rightmost ball is green, and all 125 balls between them are red.
(b) Suppos... | 125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,719 |
Problem 8.5. A grid rectangle of area $S$ is such that:
- it can be completely cut along the grid lines into rectangles $1 \times 13$;
- it can be completely cut along the grid lines into three-cell corners (examples of corners are shown in the figure below);
- there does not exist a grid rectangle of smaller area tha... | # Answer:
(a) (2 points) 78.
(b) ( points) 38, 58, 82.
Solution. (a) From the condition, it follows that $S$ is divisible by 39. We will prove that $S \neq 39$. Clearly, the rectangle $1 \times 39$ does not work.
Consider the rectangle $3 \times 13$ (3 rows, 13 columns). Suppose it can be cut into three-cell corner... | 38,58,82 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,721 |
Problem 8.6. The numbers $13,14,15, \ldots, 25$ are painted in five colors: one black number, three red, three blue, three yellow, and three green.
It is known that:
- all four sums of three same-colored numbers are equal;
- the number 13 is red, 15 is yellow, 23 is blue.
(a) (1 point) Find the black number.
(b) (3... | # Answer:
(a) (1 point) 19.
(b) (3 points) 14, 21, 22.
Solution. (a) Let's calculate the sum of the numbers from 13 to 25:
$$
13+14+\ldots+25=\frac{(13+25) \cdot 13}{2}=247
$$
The sum of all numbers except the black one should be divisible by 4. Since 247 gives a remainder of 3 when divided by 4, there are exactly... | 19,14,21,22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,722 |
Problem 8.7. Given a right isosceles triangle $ABC$ with a right angle at $A$. A square $KLMN$ is positioned as shown in the figure: points $K, L, N$ lie on sides $AB, BC, AC$ respectively, and point $M$ is located inside triangle $ABC$.
Find the length of segment $AC$, if it is known that $AK=7, AN=3$.
Fig. 2: to the solution of problem 8.7
Solution. Mark a point $H$ on the segment $B K$ such that $L H \perp B K$ (Fig. 2). Triangle $B H L$ is a right isosceles triangle, $H B=... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,723 |
4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar? | Answer: 16.
Solution. If the first jar contains no less than 11 beetles, then the second jar contains no less than 12, the third jar no less than 13, ..., and the tenth jar no less than 20. And in all jars, there are no less than $11+12+\ldots+20=155$ beetles. Contradiction. Therefore, the first jar contains no more t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,725 |
5. In each of the three schools in the microdistrict, 100 first-graders enrolled. On September 1, exactly 100 first-graders came to each school. However, some children got confused about which school they were supposed to go to, and exactly 40 children came to the wrong school. Prove that it is possible to choose two l... | Solution. Let x students from other schools come to the first school (and, accordingly, x students from the first school went to the wrong place), y to the second school, and z to the third school, with $x \geq y \geq z$. Since 40 is not divisible by 3, then x > z and in the first school, there cannot be only "foreign"... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 16,726 |
# Task 8.2 (7 points)
The shares of the company "Nu-i-Nu" increase in price by 10 percent every day. Businessman Borya bought shares of the company for 1000 rubles every day for three days in a row, and on the fourth day, he sold them all. How much money did he make from this operation? | Solution: $1000 \cdot 1.1^{3}+1000 \cdot 1.1^{2}+1000 \cdot 1.1-3 \cdot 1000=1331+1210+1100-3000=641$.
| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Correct approach with arithmetic error | 4 |
| Correct answer without justification | 0 |
## Answer: 641 rubles
# | 641 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,727 |
# Problem 8.4 (7 points)
In triangle $\mathrm{ABC}$, the bisector $\mathrm{AD}$ intersects the median $\mathrm{CE}$ at a right angle. Prove that one of the sides of this triangle is twice as small as another. | # Solution:
Proof.
Let in triangle $\mathrm{ABC}$, the bisector $\mathrm{AD}$ and median $\mathrm{CE}$ intersect at point $\mathrm{F}$. Then $\mathrm{AF}$ is the bisector and altitude in triangle $\mathrm{ACE}$, which means this triangle is isosceles ($\mathrm{AC}=\mathrm{AE}$), and since $\mathrm{CE}$ is a median, $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,729 |
# Problem 8.5 (7 points)
Find the smallest natural number that is divisible by $48^{2}$ and contains only the digits 0 and 1. | Solution:
$48^{2}=2^{8} \cdot 3^{2}$. For a number to be divisible by $2^{8}$, it must end with at least 8 zeros according to the divisibility rule, because otherwise, with fewer zeros ($n \leq 7$), it would have the form $1 \ldots .1 \cdot 10^{\text {n }}$ and would only be divisible by the n-th power of two, but we ... | 11111111100000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,730 |
Problem 9.1. In a store, there are 20 items, the costs of which are different natural numbers from 1 to 20 rubles. The store has decided to have a promotion: when buying any 5 items, one of them is given as a gift, and the customer chooses which item to receive for free. Vlad wants to buy all 20 items in this store, pa... | Answer: 136.
Solution. Vlad can take advantage of the offer no more than 4 times, so he will get no more than 4 items for free. The total cost of these 4 items does not exceed $17+18+$ $19+20$ rubles. Therefore, the rubles Vlad needs are not less than
$$
(1+2+3+\ldots+20)-(17+18+19+20)=1+2+3+\ldots+16=\frac{16 \cdot ... | 136 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,731 |
Problem 9.2. Vanya thought of two natural numbers, the product of which equals 7200. What is the greatest value that the GCD of these numbers can take? | Answer: 60.
Solution. Since each of these numbers is divisible by their GCD, their product is divisible by the square of this GCD. The greatest exact square that divides the number $7200=2^{5} \cdot 3^{2} \cdot 5^{2}$ is $3600=\left(2^{2} \cdot 3 \cdot 5\right)^{2}$, so the GCD of the two numbers in question does not ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,732 |
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are whole numbers of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
. Each of them said one of two phrases:
- "Among those gathered, at least 5 liar... | Answer: 70.
Solution. Suppose there are at least 11 liars. Arrange the numbers on their T-shirts in ascending order and select the liar with the 6th number. Then he must be telling the truth, as there are at least 5 liars with a smaller number and at least 5 liars with a larger number. Thus, there are no more than 10 ... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,735 |
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25, CP = 20, \angle APC = \angle A + \angle B$.
$, whose leading coefficient is 1. On the graph of $y=P(x)$, two points with abscissas 10 and 30 are marked. It turns out that the bisector of the first quadrant of the coordinate plane intersects the segment between them at its midpoint. Find $P(20)$. | Answer: -80.
Solution. The midpoint of this segment has coordinates $\left(\frac{10+30}{2}, \frac{P(10)+P(30)}{2}\right)$. Since it lies on the bisector of the first quadrant, i.e., on the line $y=x$, these coordinates are equal. From this, we get $P(10)+P(30)=40$.
Since $P(x)$ is a monic polynomial, it can be writte... | -80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,737 |
Problem 9.8. In an $8 \times 12$ table, some $N$ cells are black, and the rest are white. In one operation, it is allowed to paint three cells forming a three-cell corner to white (some of them could already be white before repainting). It turned out that it is impossible to make the entire table white in fewer than 25... | Answer: 27.
Solution. Divide the $8 \times 12$ table into 24 squares of $2 \times 2$ (Fig. 9a).
(a)
 must be equal to $\pm 2$, and the other 2011 are $\pm 1$. But the sum of 2011 odd numbers and one even number is odd and therefore not equal to zero. Contradiction.
Criteria for checking. From the equality of the product to two, it ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,739 |
4. 4 variables $x_{1}, x_{2}, x_{3}, \ldots, x_{2012}$ belong to the interval $[0 ; 1]$. Prove the inequality
$$
\mathrm{x}_{1} \mathrm{x}_{2} \mathrm{x}_{3} \ldots \mathrm{x}_{2012}+\left(1-\mathrm{x}_{1}\right)\left(1-\mathrm{x}_{2}\right)\left(1-\mathrm{x}_{3}\right) \ldots\left(1-\mathrm{x}_{2012}\right) \leq 1
$$... | Solution If a number $m$ lies in the interval [0;1], then the number $1-\mathrm{m}$ also lies in this interval, and the product of numbers from this interval also belongs to it, and since for any two numbers $a$ and $b$ from the interval $[0 ; 1]$ the inequality $a b \leq a$ holds, then
$\mathrm{x}_{1} \mathrm{x}_{2} ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,740 |
10.1. The graph of the quadratic function $y=a x^{2}+c$ intersects the coordinate axes at the vertices of an equilateral triangle. What is the value of ac? | Answer: -3.
Solution. Since the graph intersects the OX axis at two points, the numbers a and c have different signs. The points of intersection are: $A\left(\sqrt{-\frac{c}{a}} ; 0\right) ; B\left(-\sqrt{-\frac{c}{a}} ; 0\right)$. The graph intersects the OY axis at point $C(0 ; c)$. Then the side $AB$ of the equilat... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,741 |
10.2. The number 2018 is written on the board. Igor appends such a digit to the end of this number so that the resulting number is divisible by 11, and then divides it by 11. Then he appends a suitable digit to the end of the obtained result and divides it by 11, and so on. Can this process continue indefinitely? | Answer: It cannot.
Solution. First method. We will prove that after each step, the multi-digit number on the board will decrease. Indeed, consider a number $n \geq 10$ and append the digit $a$ to its end. We get the number $10n + a$. Dividing it by 11, we get a number that is less than $n$.
Now, let's prove that as a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,742 |
10.3. Inside triangle $A B C$, a point $P$ is marked. The bisectors of angles $B A C$ and $A C P$ intersect at point $M$, and the bisector of angle $P B A$ and the line containing the bisector of angle $B P C$ intersect at point $N$. Prove that the point of intersection of lines $C P$ and $A B$ lies on line $M N$. | Solution: Let the lines $C P$ and $A B$ intersect at point $K$ (see Fig. 10.3). Then $M-$ is the point of intersection of the angle bisectors of triangle $A K C$, hence, $K M$ is the bisector of angle $A K C$.
Now consider triangle $K B P$. $N-$ is the point of intersection of the bisector of the internal angle at ver... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,743 |
10.5. Find all pairs $(x ; y)$ of real numbers that satisfy the conditions: $x^{3}+y^{3}=1$ and $x^{4}+y^{4}=1$. | Answer: $(0 ; 1) ;(1 ; 0)$.
Solution. First method. From the second equality, it follows that $|x| \leq 1$ and $|y| \leq 1$. For the first equality to hold, at least one of the numbers $x$ or $y$ must be positive. Since both equalities are symmetric with respect to the variables, let $0 < x \leq 1$. Consider the funct... | (0;1);(1;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,745 |
10.6. The sides of the base of a brick are 28 cm and 9 cm, and the height is 6 cm. A snail crawls in a straight line along the faces of the brick from one vertex of the lower base to the opposite vertex of the upper base. The horizontal and vertical components of its velocity, $v_{x}$ and $v_{y}$, are related by the eq... | Answer: 35 min.
Fig. $10.6 \mathrm{a}$
Solution. Let the given brick be a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D$, where $A B-=9 \mathrm{~cm}, B C=28 \mathrm{~cm}, A A^{\prime}=6 \mathrm{~cm}$.
First method. Increase the height of the brick and the vertical component of the snail's sp... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,746 |
1. Given $n>2$ natural numbers, among which there are no three equal, and the sum of any two of them is a prime number. What is the largest possible value of $n$? | 1. Answer: 3. Note that the triplet $1,1,2$ satisfies the condition. Suppose there are more than 3 numbers. Consider any 4 of them a, b, c, d. Among these four numbers, there cannot be two even numbers, otherwise their sum would be an even prime greater than two. Therefore, at least 3 of them must be odd. Their pairwis... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,747 |
2. The product of four different numbers, which are roots of the equations
$$
x^{2}+6 b x+c=0 \text{ and } x^{2}+6 c x+b=0
$$
is equal to 1. At least one of the numbers $\mathrm{b}$ and c is an integer. Find $\mathrm{b}+\mathrm{c}$ given that b and c are positive. | 2. Answer: 2.5. From the condition, it follows that $\mathrm{b}^{*} \mathrm{c}=1$. Express c, substitute into the equations and find the discriminants, which must be positive.
$36 b^{2}-\frac{4}{b}>0, \frac{36}{b^{2}}-4 b>0$, from which $b>\frac{1}{\sqrt[3]{9}}, b<\sqrt[3]{9}, \quad$ and the same constraints will be s... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,748 |
3. Find the smallest natural number $\mathrm{n}$ such that the number of zeros at the end of the number ( $\mathrm{n}+20$ )! is exactly 2020 more than the number of zeros at the end of the number n!. | 3. Answer: $5^{2017}-20$. From the condition, it follows that in the product $(\mathrm{n}+1)^{*} \ldots *(\mathrm{n}+20)$, the power of the prime factor 5 in the prime factorization is 2020. Among 20 consecutive numbers, exactly 4 are multiples of 5, of which only one can be a multiple of 25 or higher powers of 5. That... | 5^{2017}-20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,749 |
4. Can a regular 2021-gon be divided into isosceles triangles by drawing non-intersecting diagonals? | 4. Answer: No, it cannot. Since the number of sides of the polygon is odd, there will be a triangle that includes exactly one side of the original polygon, and the other two sides are equal diagonals (such diagonals, drawn from one vertex, exist because the number of vertices of the polygon is 2021, which is odd). Remo... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,750 |
5. A circle is circumscribed around a right triangle $\mathrm{ABC}$ with hypotenuse $\mathrm{AB}$. On the larger leg $\mathrm{AC}$, a point $\mathrm{P}$ is marked such that $\mathrm{AP}=\mathrm{BC}$. On the arc $\mathrm{ACB}$, its midpoint $\mathrm{M}$ is marked. What can the angle $\mathrm{PMC}$ be equal to? | 5. Answer: 90 degrees
Consider triangles АРМ and
ВСМ. In them, ВС=АР by condition,
АМ=ВМ - chords subtending equal
arcs, angles МВС and МАС are equal,
since they subtend the same arc.
Therefore, the triangles are equal.
Thus, angles ВМС and АМР are equal. Since angle АМВ is 90 degrees, then angle $\mathrm{PMC}=\mathr... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,751 |
1. (7 points) The numerator and denominator of the fraction are positive numbers. The numerator was increased by 1, and the denominator by 100. Can the resulting fraction be greater than the original? | Answer: Yes.
Solution. For example, $\frac{1}{200}<\frac{2}{300}$. There are many other examples.
Criteria. Any correct example: 7 points.
Answer without an example or incorrect answer: 0 points. | \frac{1}{200}<\frac{2}{300} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,752 |
2. (7 points) The kids were given the task to convert the turtle's speed from centimeters per second to meters per minute. Masha got an answer of 25 m/min, but she thought there were 60 cm in a meter and 100 seconds in a minute. Help Masha find the correct answer. | Answer: 9 m/min.
Solution. The turtle covers a distance of 25 Machine "meters" in one Machine "minute," meaning it crawls $25 \cdot 60$ centimeters in 100 seconds. Therefore, the speed of the turtle is $\frac{25 \cdot 60}{100}=15$ cm/sec. Thus, in 60 seconds, the turtle will crawl $15 \cdot 60$ centimeters, which is $... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,753 |
3. (7 points) At a certain moment, Anya measured the angle between the hour and minute hands of her clock. Exactly one hour later, she measured the angle between the hands again. The angle turned out to be the same. What could this angle be? (Consider all cases.) | Answer: $15^{\circ}$ or $165^{\circ}$.
Solution. After 1 hour, the minute hand remains in its original position. During this time, the hour hand has turned $30^{\circ}$. Since the angle has not changed, the minute hand must bisect one of the angles between the positions of the hour hand (either the $30^{\circ}$ angle ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,754 |
4. (7 points) Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon (i.e., exactly at 12 o'clock) and, without stopping, arrived: one at $B$ at 4 PM, and the other at $A$ at 9 PM. At what time was dawn that day? | Answer: at 6 AM.
Solution. Let's denote the meeting point as $C$. Let $x$ be the number of hours from dawn to noon.
The speed of the first pedestrian on segment $A C$ is $A C / x$, and on segment $B C$ it is $B C / 4$. Since his speed is constant, we have $\frac{A C}{x}=\frac{B C}{4}$, which can be rewritten as $\fra... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,755 |
5. (7 points) Determine the number of points at which 10 lines intersect, given that only two of them are parallel and exactly three of these lines intersect at one point. | Answer: 42.
Solution. Let's number the lines so that lines 1, 2, and 3 intersect at one point (denote this point as $X$). List all possible pairs of lines (1 and 2, 1 and 3, 1 and $4, \ldots, 8$ and 9, 8 and 10, 9 and 10) and their points of intersection. There are a total of 45 pairs of lines (there are 9 pairs of th... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,756 |
Problem 1. There are three acrobat brothers. Their average height is 1 meter 74 centimeters. The average height of two of these brothers: the tallest and the shortest - 1 meter 75 centimeters. What is the height of the middle brother? Justify your answer. | Answer: 1 meter 72 centimeters.
Solution. Since the average height of all three is 1 meter 74 centimeters, the total height of all three is 5 meters 22 centimeters. The average height of the two brothers is 1 meter 75 centimeters, so their combined height is 3 meters 50 centimeters. Therefore, the height of the middle... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,757 |
Problem 2. The radius of the circumscribed circle of isosceles triangle $ABC$ ($AB = BC$) is equal to the base $AC$. A square $AKLC$ is constructed on the base $AC$ such that the segment $KL$ intersects the lateral sides of the triangle. Prove that triangle $BKL$ is equilateral. | Solution. Let point $O$ be the center of the circumscribed circle of triangle $ABC$. From the condition, we get that $OA = OC = AC$, which means triangle $AOC$ is equilateral. Since $AKLC$ is a square, we have $AK = KL = LC = AC$. Note that $BO \parallel LC$, since both lines are perpendicular to $AC$, and also $BO = L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,758 |
Problem 3. Let's call a three-digit number interesting if at least one of its digits is divisible by 3. What is the maximum number of consecutive interesting numbers that can exist? (Provide an example and prove that it is impossible to have more consecutive numbers.)
| Answer: 122.
Solution. The numbers $289,290, \ldots, 299,300, \ldots, 399,400, \ldots, 409,410$ are interesting (recall that 0 is divisible by 3), and there are 122 of them in total. Let's prove that a larger number is not possible.
Suppose we managed to find a larger number of consecutive interesting numbers; choose... | 122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,759 |
Problem 4. The difference of the roots of the quadratic equation with real coefficients $2018 x^{2} + a x + b = 0$ is an integer (while the roots themselves are not necessarily integers). Prove that the discriminant of this equation is divisible by $2018^{2}$. | Solution. Let $D$ be the discriminant of this equation. Denote the roots of the equation by $x_{1}=\frac{-a+\sqrt{D}}{4036}$ and $x_{2}=\frac{-a-\sqrt{D}}{4036}$. Then $x_{1}-x_{2}=\frac{\sqrt{D}}{2018}=n-$ an integer. Thus, $\sqrt{D}=2018 \cdot n$ and $D=2018^{2} n^{2}$, which is divisible by $2018^{2}$.
## Criteria
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,760 |
Problem 5. Find all pairs of natural numbers $a$ and $b$ such that
$$
\operatorname{LCM}(a, b)=\text { GCD }(a, b)+19
$$
(and prove that there are no others).
GCD $(a, b)$ is the greatest common divisor, i.e., the largest natural number that divides both $a$ and $b$. $\operatorname{LCM}(a, b)$ is the least common mu... | Solution. Let $d=\gcd(a, b)$. Note that both the LCM and the GCD are divisible by $d$, which means that 19 is also divisible by $d$. Since 19 is a prime number, we get that $d=1$ or $d=19$.
- If $d=1$, then the numbers $a$ and $b$ are coprime, and $\operatorname{lcm}(a, b)=a \cdot b=1+19=20$. This gives the options $(... | (,b)=(1,20),(20,1),(4,5),(5,4),(19,38),(38,19) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,761 |
Problem 6. In a country, there are 100 cities. Between any two cities, there is either no connection, or there is an air route, or there is a railway (both air routes and railways cannot exist simultaneously). It is known that if two cities are connected to a third city by railway, then there is an air route between th... | Solution. Note that no city is connected by railway to more than two other cities. Indeed, suppose some three cities are connected by railway to one. Then all of them are connected to each other by air routes, which is impossible by the condition of the problem. Therefore, each city is connected by railway to no more t... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false | 16,762 |
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are equal to an integer number of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
![](https://cdn.mathpix.com/cropped/2024_05_06_20e437e0605a8... | # Answer: 17.
Solution. We will use the triangle inequality: in any non-degenerate triangle, the sum of any two sides is strictly greater than the remaining one.
Let $x$ km be the unknown length. From the left triangle, we see that $x < 10 + 8 = 18$. But if $x \leqslant 16$, then in the right triangle, the triangle i... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,763 |
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25$, $CP = 20$, and $\angle APC = \angle A + \angle B$.
![](https://cdn.mathpix.com/cro... | Answer: 16.
Solution. Since $\angle P C B+\angle P B C=\angle A P C=\angle P A C+\angle P B C$, we obtain $\angle P C B=\angle P A C$. Note that the right triangles $P A C$ and $Q C P$ are similar by the acute angle, and
$$
\frac{25}{20}=\frac{A C}{C P}=\frac{P C}{P Q}=\frac{20}{P Q}
$$
from which we find $P Q=\frac... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,764 |
3. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, the number of knights and the number of liars differ by one." Player number 2 said: "In our team, the number of knights and the number of liars... | 3. Answer: the only knight plays under number 9. The solution is that the number of knights cannot be equal to the number of liars, so one of the answers must be correct. Two answers cannot be true, as they contradict each other. Therefore, there is exactly 1 knight and 10 liars in the team, which is stated by the play... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,765 |
4. A three-digit number is called cool if one of its digits is half the product of the other two digits. A three-digit number is called supercool if such digits are in two or three of its positions. How many different supercool numbers exist? (Zero cannot be part of the representation of cool or supercool numbers.) | 4. Answer: 25 numbers. Solution: let the record of a superclass number include digits a, b, c in some order. Then, two equalities are satisfied: $2 \mathrm{a}=\mathrm{bc} 2 \mathrm{~b}=\mathrm{ac}$. If the numbers a and b are different, then by swapping them, we will violate the existing equality. Therefore, $\mathrm{a... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,766 |
5. A square was cut into several triangles such that all angles of these triangles are different (both within the same triangle and across different triangles). Can all these angles be multiples of 15 degrees? | 5. Answer: No. Solution: three of the square's angles must be divided into parts, otherwise a right angle will repeat. Then we will have at least 6 acute angles. But there are only 5 distinct acute angles that are multiples of 15 degrees.
Criteria: Correct solution - 7 points. In all other cases - 0 points. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,767 |
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
 | Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides som... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,769 |
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cine... | Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats ... | 1035 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,770 |
2. Solve the inequality $\frac{\sin (\cos x)}{\cos (\sin x)}-1>0$.
# | # Solution
Consider the difference $\cos (\sin x)-\sin (\cos x)=\sin \left(\frac{\pi}{2}-\sin x\right)-\sin (\cos x)=$
$2 \cos \frac{\frac{\pi}{2}-\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)}{2} \sin \frac{\frac{\pi}{2}-\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)}{2}, \quad 0<\frac{\pi-2 \sqrt{2}}{4} \leq \frac{\frac{\... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 16,771 |
3. Ivan Tsarevich needs to obtain the apples of youth. Baba Yaga, Koschei, and Leshy gave him the following answers.
Baba Yaga: “Yes, Koschei has them. He took them and hasn't given them back for 100 years. And Leshy is a good guy: if he had them, he would have given them to me.”
Koschei: “Baba Yaga is a trickster, s... | # Solution
Based on Koschei's testimony, we conclude that Baba Yaga either has no apples or he does.
Let's consider the case where Baba Yaga has no apples.
Then, in one part of his testimony, the Forest Spirit did not lie about Baba Yaga. Therefore, the Forest Spirit's statement that he has no apples is a lie. Thus,... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,772 |
5. In a rectangular parallelepiped $\mathrm{ABCDA}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1} \mathrm{AB}=1$ cm, $A D=2, \mathrm{AA}_{1}=1$. Find the smallest area of triangle $\mathrm{PA}_{1} \mathrm{C}$, where vertex $\mathrm{P}$ lies on the line $\mathrm{AB}_{1}$. | # Solution
The area of triangle $\mathrm{PA}_{1} \mathrm{C}$ is $S=\frac{1}{2} A_{1} C \times P H$, where $P H$ is the distance from point $\mathrm{P}$, taken on line $\mathrm{AB}_{1}$, to line $\mathrm{A}_{1} \mathrm{C}$. The area will be the smallest when the length of segment $P H$ is the smallest, i.e., when $P H$... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,774 |
10.1. Find all possible functions of the form $f(x)=k x-b$, where $k \neq 0$, such that $f(f(x))=2 f(x)-3 f(1-x)$, or prove that they do not exist. (7 points) | # Solution:
1) $f(f(x))=f(k x-b)=k(k x-b)-b=k^{2} x-k b-b=k^{2} x+(-k b-b)$
2) $2 f(x)-3 f(1-x)=2(k x-b)-3(k(1-x)-b)=2 k x-2 b-3(k-k x-b)=2 k x-$ $2 b-3 k+3 k x+3 b=5 k x+(b-3 k)$
3) By equating the coefficients, we get: $\left\{\begin{array}{c}k^{2}=5 k \\ -k b-b=b-3 k\end{array}\right.$.
4) Using the fact that $k \n... | f(x)=5x-2\frac{1}{7} | Algebra | proof | Yes | Yes | olympiads | false | 16,775 |
10.3. On the table, there are 2021 candies. Three fat men take turns eating either one, two, or four candies until the candies run out. Can the second and third fat men conspire in such a way that the first fat man does not get to eat the last candy? (7 points)
# | # First Solution:
1) If the first fat man eats 1 candy on the first move, then the second and third will eat 4 and 1 candies. The order does not matter. If the first fat man eats 2 candies on the first move, then the second and third will eat 2 candies each. If the first fat man eats 4 candies on the first move, then ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,776 |
10.4. In triangle $A B C$, points $D$ and $F$ are marked on sides $A B$ and $A C$ respectively such that lines $D C$ and $B F$ are perpendicular to each other and intersect at point $E$ inside triangle $A B C$. It turns out that $A D=D C$ and $D E \cdot E C=F E \cdot E B$. What degree measure can angle $B A C$ have? (7... | # Solution:
$D E \cdot E C=F E \cdot E B \Rightarrow \frac{D E}{E F}=\frac{E B}{E C}, \angle D E B=\angle F E C=90^{\circ}$, so triangles $\triangle D E B$ and $\triangle F E C$ are similar.... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,777 |
10.5. On an $8 \times 8$ battleship game field, a "piglet" figure is placed. What is the minimum number of shots needed to definitely hit one "piglet"? (7 points)
# | # Solution:
## Evaluation:
Divide the board into rectangles $4 \times 2$. To ensure that a piglet cannot be placed inside one such figure, at least two shots are required. There are 8 such ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,778 |
10.1. 33 heroes go on patrol for 33 days. On the first day, one hero should go out, on the second day two, on the third day three, and so on, on the last day - all the heroes. Can Chernomor the Elder organize the patrols so that all heroes go on patrol an equal number of times? | Answer: will be able to.
Solution. Since $1+2+\ldots+33=\frac{1+33}{2} \cdot 33=17 \cdot 33$, each bogatyr must go on duty 17 times. Let's say, for example, Chernomor numbers the bogatyrs, and for the first 16 days, the bogatyrs go on duty according to their numbers: on the first day - the bogatyr with number one, on ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,779 |
10.2. Do there exist pairwise distinct numbers $a, b$, and $c$ such that the number $a$ is a root of the quadratic trinomial $x^{2}-2 b x+c^{2}$, the number $b$ is a root of the quadratic trinomial $x^{2}-2 c x+a^{2}$, and the number $c$ is a root of the quadratic trinomial $x^{2}-2 a x+b^{2}$? | Answer: do not exist.
Solution. Suppose such numbers were found. Then the equalities hold: $a^{2}-2 b a+$ $c^{2}=0, b^{2}-2 c b+a^{2}=0$ and $c^{2}-2 a c+b^{2}=0$. Add these equalities. By regrouping and factoring out complete squares, we get: $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$. This is only possible if $a=$ $b=c$, whi... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,780 |
10.3. Two circles intersect at points $A$ and $B$. It turns out that the radii $O A$ and $O B$ of the first circle are tangents to the second circle. A line is drawn through point $A$ and intersects the circles again at points $M$ and $N$. Prove that $M B \perp N B$. | Solution. Let $\angle B M N=\alpha, \angle B N M=\beta$ (see Fig. 10.3a, b).
First method. See Fig. 10.3a. Note that $\angle O A B=\angle O B A=\alpha$ (by the theorem on the angle between a tangent and a chord), $\angle A O B=2 \beta$ (central angle). From triangle $A O B: 2 \alpha+2 \beta=180^{\circ}$, hence $\alpha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,781 |
10.4. All divisors of a certain natural number have been listed, except for one and the number itself. Some two numbers in this list differ by six times. By how many times do the two largest numbers in this list differ? | Answer: one and a half times.
Solution. Let among the divisors of the number $N$ there be numbers $a$ and $6a$, then $N$ is divisible by $6a$. Therefore, $N$ is divisible by 2 and by 3, which means 2 and 3 are the two smallest numbers in the list. Then the two largest numbers in the list are $\frac{N}{3}$ and $\frac{N... | \frac{3}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,782 |
10.5. On the sides $AB$ and $BC$ of triangle $ABC$, points $X$ and $Y$ are marked, respectively. Segments $CX$ and $AY$ intersect at point $T$. Prove that the area of triangle $XBY$ is greater than the area of triangle $XTY$. | Solution. First method. Mark point $X_{1}$ on segment $B X$ such that $Y X_{1} \| C X$ (see Fig. 10.5a). Similarly, choose point $Y_{1}$ on segment $B Y$ such that $X Y_{1} \| A Y$. Let segments $Y X_{1}$ and $X Y_{1}$ intersect at point $S$. Then $X S Y T-$ is a parallelogram, so triangles XSY and YTX are equal, and t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,783 |
10.6. There are two boxes, one with 2017 candies, and the other with 2018. Two players take turns. On each turn, a player can eat any non-zero number of candies from any one box. The rules of the game do not allow the number of candies in one box to be divisible by the number of candies in the other after any move. The... | Answer: the first player will win.
Solution. To win, after each of his moves, the first player must create a situation where one of the boxes contains $2n$ candies, and the other contains $2n+1$ ($n$ is a natural number). In such a situation, he is guaranteed not to lose.
First, he eats two candies from the second bo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,784 |
1. For what values of $a, b$, and $c$ do the lines $y=a x+b, y=b x+c, y=c x+a$ pass through the point $(1 ; 3)$? | # Solution.
The lines $y=a x+b, y=b x+c, y=c x+a$ pass through the point $(1 ; 3)$ if and only if $a, b$ and $c$ satisfy the system of equations:
$$
\left\{\begin{array}{l}
3=a+b \\
3=b+c \\
3=c+a
\end{array}\right.
$$
By adding the equations, we get: $9=2(a+b+c) \Rightarrow a+b+c=4.5$. Subtracting the first, second... | =b==1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,785 |
2. Two entrepreneurs, Anatoly and Vladimir, built a 16 km road: Anatoly built 6 km, and Vladimir built 10 km. Their acquaintance, Boris, said he would like to use the road equally with them and is willing to contribute his share of the money - 16 million rubles. How should Anatoly and Vladimir divide this money between... | # Solution.
Each of the partners was supposed to build $5 \frac{1}{3}$ km of road. Anatoly built $6-5 \frac{1}{3}=\frac{2}{3}$ km of road instead of Boris, and Vladimir built $10-5 \frac{1}{3}=\frac{14}{3}$ km. Therefore, the money should be divided between them in the ratio $2: 14$.
Answer: Anatoly should receive 2 ... | 1000000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,786 |
3. Let $O$ be the center of the circumcircle of triangle $ABC$, points $O$ and $B$ lie on opposite sides of line $AC$, $\angle AOC = 60^\circ$. Find the angle $AMC$, where $M$ is the center of the incircle of triangle $ABC$. | # Solution.
Since points $O$ and $B$ lie on opposite sides of line $A C$, the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$. Therefore,
$\angle A B C=(1 / 2) \cdot 300^{\circ}=150^{\circ}$. The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{\circ... | 165 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,787 |
4. Is the number $\operatorname{tg} \sqrt{3 \pi}+1$ positive or negative?
# | # Solution.
$$
\text { We will prove that } \frac{3 \pi}{4}0$.
Answer: positive.
Instructions. Only the answer - 0 points; if the inequality (1) for $\pi$ uses an approximate value of 3.14 (after squaring) - 4 points; the square root is found approximately, for example, using a calculator - 0 points.
$$ | positive | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,788 |
5. In each cell of a $6 \times 6$ board, there is a grasshopper. At the whistle, each grasshopper jumps over one cell diagonally (not to the adjacent diagonal cell, but to the next one). As a result, some cells may end up with more than one grasshopper, while some cells may remain unoccupied. Prove that in this case, t... | # Solution.
Let's color the cells of the board in black and white, as shown in the figure. As a result, 24 cells will be colored black, and 12 cells will be colored white. Notice that from a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,789 |
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