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3. Consider the function $f(x)=\sqrt[3]{1+x}+\sqrt[3]{3-x}$, defined on the real number line. For which $x$ is the value of the function an integer? | 3. Answer: for $x \in\left\{1 \pm \sqrt{5}, 1 \pm \frac{10}{9} \sqrt{3}\right\}$.
Let $y=\sqrt[3]{1+x}, z=\sqrt[3]{3-x}$. Then $y^{3}+z^{3}=4$. Let $y+z=k \in \mathbb{Z}$. Clearly, $k \neq 0$. Then $\frac{4}{k}=\frac{y^{3}+z^{3}}{y+z}=y^{2}-y z+z^{2}=(y+z)^{2}-3 y z=k^{2}-3 y z$, from which $y z=\frac{k^{3}-4}{3 k}$.
... | x\in{1\\sqrt{5},1\\frac{10}{9}\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,865 |
4. Let $A K$ be the bisector of triangle $A B C$, and $L$ be the point of intersection of the bisectors of this triangle. Prove that $A L > L K$. | 4. We use a known property of the angle bisector of a triangle: it divides the opposite side in the ratio equal to the ratio of the lengths of the other two sides. Let's recall that this fact can be derived either using the similarity of triangles or by comparing the ratios of the areas of two triangles that share a si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,866 |
5. Petya and Vasya are playing a game, taking turns naming non-zero decimal digits. One cannot name a digit that is a divisor of any of the digits already named by anyone. The player who has no move loses. Petya starts. Who wins with correct play from both sides?
For a complete solution to each problem, 7 points are a... | 5. Answer: Petya wins.
Consider the same game but with digits from 2 to 9. It lasts a finite number of moves, and there are no draws. Therefore, one of the players has a winning strategy.
Consider the case when in the new game, the one who starts wins. Then he makes the first move by naming some digit \( a \neq 1 \),... | Petyawins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,867 |
10.1. The student was given an assignment consisting of 20 problems. For each correctly solved problem, they receive 8 points, for each incorrectly solved problem - minus 5 points, and for a problem they did not attempt - 0 points. The student scored a total of 13 points. How many problems did the student attempt? | # Solution.
Let $x$ be the number of problems solved correctly, $y$ be the number of problems solved incorrectly. Then we get the equation $8x - 5y = 13$. This equation can be solved in two ways.
1st method. Rewrite the equation as $8(x + y) = 13(1 + y)$. We see that the number $x + y$ is divisible by 13. On the othe... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,868 |
10.2. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$. | # Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,869 |
10.3. When manufacturing a batch of $N \geq 5$ coins, a worker mistakenly made two coins from a different material (all coins look the same). The supervisor knows that there are exactly two such coins, that they weigh the same, but differ in weight from the others. The worker knows which coins are fake and that they ar... | # Solution
Let $\mathrm{m} 1$ and $\mathrm{m} 2$ be the fake coins, and $\mathrm{m} 3, \mathrm{~m} 4$ and $\mathrm{m} 5$ be any three of the real coins. The worker performs two weighings: $\mathrm{m} 1 \mathrm{~V} 3, \mathrm{~m} 4+\mathrm{m} 5 \mathrm{V} \mathrm{m} 2+\mathrm{m} 3$.
As a result, the supervisor is conv... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,870 |
10.5. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out of the plane? | Solution. Consider 100 nodes - the intersection points of lines from the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second directions. The second sector - nodes lying on the second lines (excluding points lying in the first sector) and so... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,871 |
1. Denis housed chameleons that can change color only to two colors: red and brown. Initially, the number of red chameleons was five times the number of brown chameleons. After two brown chameleons turned red, the number of red chameleons became eight times the number of brown chameleons. Find out how many chameleons D... | Solution. Let $t$ be the number of brown chameleons Denis had. Then the number of red chameleons was $5t$. From the problem statement, we get the equation $5 t+2=8(t-2)$. Solving this, we find $t=6$. Therefore, the total number of chameleons is $6 t$, which is 36.
Answer. 36
Recommendations for checking. Only the cor... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,872 |
2. For various positive numbers $a$ and $b$, it is known that
$$
a^{3}-b^{3}=3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)
$$
By how many times is the larger number greater than the smaller one? | Solution. Let's consider and transform the difference:
$$
\begin{aligned}
& 0=a^{3}-b^{3}-3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}\right)-3\left(2 a b(a-b)-b^{2}(a-b)\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}-6 a b+3 b^{2}\right)= \\
& (a-b)\left(a^{2}-5 a b+4 b^{2}\right)=(a-b)(a-4 b)(a-... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,873 |
6. Next to the number 2022 on the board, an unknown positive number less than 2022 was written. Then one of the numbers on the board was replaced by their arithmetic mean. This replacement was performed 9 more times, and the arithmetic mean was always an integer. Find the smaller of the numbers originally written on th... | Solution. Let at some point the numbers $a$ and $b$ be written on the board, with $a > b$. Then notice that after the specified operation, the difference between the numbers will become twice as small, regardless of which number we erase, since
$$
a - b = 2\left(a - \frac{a + b}{2}\right) = 2\left(\frac{a + b}{2} - b\... | 998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,875 |
7. Airline "Vpered" served 50 airports, and there was at least one route from each. Due to a reduction in the number of aircraft, the airline has to cancel some routes. To avoid passenger dissatisfaction, it is required that at least one route remains from each airport. Prove that regardless of the initial route scheme... | Solution. If there is a route connecting 2 cities, from which there are more than one route, - then it can be shortened. We will perform this procedure as long as possible, after which no more than one route will remain from each city. From a city where there is more than one route, all routes lead to cities with only ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,876 |
Problem 1. Find the maximum value of the expression $a^{3} b-b^{3} a$, if $a^{2}+b^{2}=1$.
Answer. $\frac{1}{4}$
# | # Solution.
1st method.
Since $\mathrm{a}^{2}+\mathrm{b}^{2}=1$, there exists a number $\alpha$ such that $\mathrm{a}=\cos \alpha, \mathrm{b}=\sin \alpha$. Then $\mathrm{a}^{3} \mathrm{~b}-\mathrm{b}^{3} \mathrm{a}=\mathrm{ab}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)$ $=\cos \alpha \sin \alpha\left(\cos ^{2} \alpha-... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,877 |
Problem 2. Petya marked a point $X$ on the edge $\mathrm{AB}$ of the cube $\mathrm{ABCDA}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$, dividing the edge $\mathrm{AB}$ in the ratio $1: 2$, starting from vertex $\mathrm{A}$. Provide an example of how Petya can mark points $\mathrm{Y}$ and $\mathrm{Z}$ on the edges... | # Solution.
Let's mark points $\mathrm{Y}$ and $\mathrm{Z}$ such that $\mathrm{A}_{1} \mathrm{Z}: \mathrm{ZD}_{1}=2: 1, \mathrm{C}_{1} \mathrm{Y}: \mathrm{YC}=2: 1$. The equality of the sides of triangle XYZ follows, for example, from the equality of the broken lines $\mathrm{XAA}_{1} \mathrm{Z}, \mathrm{ZD}_{1} \math... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,878 |
Problem 3. The maze is represented by an $8 \times 8$ square, with each $1 \times 1$ cell containing one of four arrows (up, down, right, left). The top side of the top-right cell is the exit of the maze. In the bottom-left cell, there is a token that moves one cell in the direction indicated by the arrow on each move.... | # Solution
Assume that the chip never leaves the maze. Then the chip will land on the cell numbered 1 (see fig.) a finite number of times (less than four), because otherwise, when the arrow points to the exit, the chip would leave the maze.
| 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
| :---: | :---: | :---: | :---: | :---: | :... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 15,879 |
Problem 5. On the circle circumscribed around rectangle $\mathrm{ABCD}$, a point K is chosen. It turns out that the line $\mathrm{CK}$ intersects the segment $\mathrm{AD}$ at a point $\mathrm{M}$ such that $\mathrm{AM}: \mathrm{MD}=2$. Let $\mathrm{O}$ be the center of the rectangle. Prove that the point of intersectio... | # Solution
Let's mark a point $\mathrm{T}$ on the extension of segment AD such that $\mathrm{AT}=\mathrm{DM}$. Then $\mathrm{BCMT}$ is a parallelogram. Since DT $=\mathrm{DA}+\mathrm{AT}=3 \mathrm{DM}+\mathrm{DM}=4 \mathrm{DM}$, by Thales' theorem, the line $\mathrm{CM}$ intersects segment $\mathrm{BD}$ at a point $\m... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,881 |
1. A square of size $2019 \times 2019$ is divided by two horizontal and two vertical lines into 9 rectangles. The sides of the central rectangle are 1511 and 1115. Find the sum of the areas of the corner rectangles. | 1. Answer: 508*904. Solution. Notice that the corner rectangles form a rectangle $508 * 904$.
Grading criteria. Correct solution - 7 points. It is shown that a rectangle is formed, but there is an inaccuracy in the calculations - $\mathbf{4}$ points. One of the specific cases of partitioning is discussed - 0 points. | 508\times904 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,882 |
2. The numbers from 1 to 2019 were written in a row. Which digit was written more: ones or twos, and by how many? | 2. Answer. There are 990 more ones. Solution. From 1 to 999, their quantity is the same. From 1000 to 1999, there are 1000 more ones. From 2000 to 2009, there are 10 more twos. From 2010 to 2019, it is the same again. In total, $1000-10=990$.
Grading criteria. Full solution - 7 points. In other cases $-\mathbf{0}$ poi... | 990 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,883 |
3. The length of the road from the Capital to city O is 2020 kilometers. Along the road are kilometer markers. On the first marker, on the front side, 1 is written, and on the back - 2019; on the second - 2 and 2018 ..., on the last - 2019 and 1. A marker is called good if the two written numbers have a common divisor ... | 3. Answer: 800. Solution. Note that the sum of two numbers on a pillar is 2020. If both numbers are divisible by some common divisor, then $2020=4 * 5 * 101$ is also divisible by this divisor. All even pillars are good, all divisible by 5 are good, all divisible by 101 are good. In total, there are odd pillars $2020 / ... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,884 |
4. In quadrilateral $\mathrm{ABCD}$, the diagonals intersect at point $\mathrm{O}$. Given that $\mathrm{AB}=\mathrm{OD}, \mathrm{AD}=\mathrm{OC}$, and angles $\mathrm{BAC}$ and $\mathrm{BDA}$ are equal. Prove that this quadrilateral is a trapezoid. | 4. Solution. We will show that triangles ABD and DOC are equal (by two sides and the angle between them). Two sides are equal by condition. Angle COD is external for triangle AOD, therefore it equals $\mathrm{CAD}+\mathrm{BDA}=\mathrm{CAD}+\mathrm{BAC}=\mathrm{BAD}$. Then angle $\mathrm{ACD}$ equals angle BDA, and line... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,885 |
5. In one class, there are blondes and brunettes. Light-haired students lie in response to any question, while those with dark hair tell the truth. But some children love to dye their hair and dye it every day to the opposite color! (While others never dye their hair). On Monday, it was Petya's birthday, and all the ch... | 5. Answer. There are more brunettes. Solution. On Friday, the children have the same hair color as on Monday, so the answers could only change if the month changed. Therefore, the month definitely did not change from Friday to Monday. Only the children who dye their hair could have changed their answer from Friday to M... | There\\\brunettes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,886 |
6. On November 15, a tournament of the game of dodgeball was held. In each game, two teams competed. A win was worth 15 points, a draw 11, and a loss earned no points. Each team played against each other once. At the end of the tournament, the total number of points scored was 1151. How many teams were there? | 6. Answer: 12 teams. Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are earned. Therefore, the number of games was no less than $53(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 te... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,887 |
1. Prove that the number $2014 \cdot 2015 \cdot 2016 \cdot 2017+1$ is composite. | 1. Let the number 2015 be denoted by $a$. Then $2014 \cdot 2015 \cdot 2016 \cdot 2017+1=$ $=(a-1) a(a+1)(a+2)+1=\left(a^{2}+a\right) \cdot\left(a^{2}+a-2\right)+1=\left(a^{2}+a\right)^{2}-2\left(a^{2}+a\right)+1=$ $=\left(a^{2}+a-1\right)^{2}$.
Therefore, 2014$\cdot$2015$\cdot$2016 $\cdot$2017+1$=\left(2015^{2}+2015-1... | 4062239\cdot4062239 | Number Theory | proof | Yes | Yes | olympiads | false | 15,888 |
2. How many positive numbers are there among the first 100 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$ | 2. Note that all members of the sequence, starting from $\sin 1000^{\circ}$, are equal to each other, since the difference between the numbers $10^{k+1}$ and $10^{k}$ for natural $k>2$ is a multiple of 360. Indeed, $10^{k+1}-10^{k}=10^{k}(10-1)=9 \cdot 10^{k}=$ $9 \cdot 4 \cdot 10 \cdot 25 \cdot 10^{k-3} \vdots 360$. B... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,889 |
3. In triangle $A B C \quad A C=18 \text{ cm}, B C=21$ cm. Point $K$ is the midpoint of side $B C$, and point $M$ is the midpoint of side $A B$, point $N$ lies on side $A C$ and $A N=6 \text{ cm}$. Given that $M N=K N$. Find the length of side $A B$. | 3. Since $M$ and $K$ are midpoints of the sides, we will extend the segments $NM$ and $NK$ beyond the specified points by the same distance and connect the points $L, B, A, N$; as well as $F, B, N, C$. Then the quadrilaterals $ALBN$ and $NBFC$ become parallelograms. Since in a parallelogram the sum of the squares of th... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,890 |
4. Prove that the equation $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1=0$ has no real roots. | 4. First method. Consider different cases:
1) If $x \leq 0$, then the left side of the equation will be positive, so the equation has no solutions.
2) Transform the equation $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1=0$ to the form: $\left(x^{4}+x^{2}+1\right)(x-1)=x^{6}$. This equation has no solutions for $0<x<1$.
3) Transfo... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,891 |
5. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive semi-prime numbers that can be semi-prime? | 5. Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assume the opposite. Then we get that the numbers $2n-1, 2n+1, 2n+3$ are prime, and all of them are greate... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,892 |
4. 10 people collected a total of 46 mushrooms, and it is known that no two people collected the same number of mushrooms.
How many mushrooms did each person collect? Justify your answer and explain why there are no other possibilities. | 4. Consideration of particular cases - 1 point. Use of inequalities in estimating the quantity without the correct solution - 3 points. Full solution - 7 points. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,896 |
1. Out of four classes, $28\%$ of the students received a "5" for the final math test, "4" - $35\%$, "3" - $25\%$, and "2" - $12\%$. How many students took the test if there are no more than 30 students in each class? | 1. Answer: 100.
From the condition, it follows that the number of schoolchildren must be divisible by 25, 20, and 4. The smallest suitable number is 100, the next one is 200, but it would not work for us since, according to the condition, there cannot be more than 120 people in 4 classes. | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,898 |
2. Rita, Luba, and Varya were solving problems. To speed things up, they bought candies and agreed that the girl who solved a problem first would get four candies, the second to solve it would get two, and the last to solve it would get one. Could it be that each of them solved all the problems and received 20 candies,... | 2. Answer: could not.
For each task, the girls should receive a total of 7 candies. This means that the total number of candies they receive should be divisible by 7. But according to the problem, this number 7 should be equal to $20 \cdot 3=60$, and 60 is not divisible by 7. Therefore, the situation described in the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,899 |
3. The store received 206 liters of milk in barrels of 10 and 17 liters. How many barrels of each type were there? | # 3. Answer: 7 and 8.
The total volume will be $10 x+17 y$ and equals 206 liters. Let's look at the last digit of the sum $10 x+17 y$. Clearly, it depends only on y, because the number $10 x$ always ends in zero. Therefore, we need to find such a y that $17 y=$ ...6 and $17 y \leq 206$. Only y $=8$ fits, as $17 \times... | 78 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,900 |
4. In the nodes of a grid plane, 5 points are marked. Prove that there are two of them, the midpoint of the segment between which also falls on a node. | 4. Out of 5 points, at least 3 will have x-coordinates of the same parity. From these $3 \mathrm{x}$, at least 2 will have y-coordinates of the same parity. This means that the distances along the x-axis and y-axis between these points will be even. This implies that the midpoint of the segment will fall exactly on a g... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,901 |
5. In the office, each computer was connected by wires to exactly 5 other computers. After some computers were infected by a virus, all wires from the infected computers were disconnected (a total of 26 wires were disconnected). Now, each of the uninfected computers is connected by wires to only 3 others. How many comp... | # 5. Answer: 8.
Let $\mathrm{m}$ be the number of infected computers, and $\mathrm{n}$ be the number of uninfected computers. Then, before the infection, there were $5(\mathrm{~m}+\mathrm{n}) / 2$ cables, and after the disconnection, there were $3 \mathrm{n} / 2$ cables (from which it follows that $\mathrm{n}$ is even... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,902 |
8-1. Kolya made a figure from four identical blocks as shown in the picture. What is the surface area of this figure? Express your answer in square centimeters.
 | Answer: 64.
Solution. The surface area of one block is 18 cm². Out of this area, 2 cm² is "lost" at the joints with other blocks, leaving a total area of $18-2=16$ cm². Since there are 4 blocks, the answer is $4 \cdot 16=64$ cm². | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,903 |
8-3. Olya bought three gifts and packed them in three rectangular boxes: blue, red, and green. She tried to place these gifts in different ways: one on the table, and two on top of each other on the floor. Some distances are given in the diagram. Find the height of the table $h$. Express your answer in centimeters.
Let $B C=2 a, O B=b$, then $A D=3 a, O A=3 b / 2$. Drop a perpendicular $O M$ from point $O$ to line $B C$ (from the isosceles property of the trapezoid, it foll... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,943 |
# 5. Option 1.
In the district, there are three villages A, B, and C connected by dirt roads, with any two villages being connected by several (more than one) roads. Traffic on the roads is two-way. We will call a path from one village to another either a road connecting them or a chain of two roads passing through a ... | Answer: 26.
Solution.
Let there be $k$ roads between cities A and B, $m$ roads between cities B and V, and $n$ roads between cities A and V. Then the number of paths from A to B is $k + m n$, and the number of paths from B to V is $m + k n$. We have the system of equations $k + m n = 34$, $m + k n = 29$, where the un... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,944 |
# 6. Variant 1.
The vertices of the triangle have coordinates $A(1 ; 3.5), B(13.5 ; 3.5), C(11 ; 16)$. Consider horizontal lines given by the equations $y=n$, where $n$ is an integer. Find the sum of the lengths of the segments cut off on these lines by the sides of the triangle. | Answer: 78.
Solution.
Draw the line $y=4$, and let it intersect the triangle $A B C$ at points $F$ and $G$. Construct a rectangle $F K M G$ such that $K M$ passes through point $C$ parallel to the $O x$ axis. Let $L_{1}, L_{2}, L_{3}$ be the sums of the lengths of the segments cut by the lines $y=n$ in the triangles ... | 78 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,945 |
7. Variant 1.
Numbers $x$ and $y$ satisfy the equation $\frac{x}{x+y}+\frac{y}{2(x-y)}=1$. Find all possible values of the expression $\frac{5 x+y}{x-2 y}$, and in the answer, write their sum. | Answer: 21.
Solution.
By bringing to a common denominator and combining like terms, we get the equality $3 y^{2}=x y$. If $y=0$, then $x$ is any non-zero number. In this case, the value of the expression is 5. If $x=3 y \neq 0$, then in this case, the value of the expression is 16. The final answer is $5+16=21$. | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,946 |
# 8. Variant 1.
Consider the quadratic trinomial $P(x)=a x^{2}+b x+c$, which has distinct positive roots. Vasya wrote four numbers on the board: the roots of $P(x)$, as well as the roots of the trinomial $Q(x)=c x^{2}+b x+a$ multiplied by 4. What is the smallest integer value that the sum of the written numbers can ha... | Answer: 9.
Solution.
Let the roots of the quadratic polynomial $P(x)$ be $x_{1}, x_{2}>0$. Then, by Vieta's formulas, the roots of the polynomial $Q(x)$ will be $\frac{1}{x_{1}}, \frac{1}{x_{2}}$.
The sum of the numbers written on the board is $S=x_{1}+x_{2}+\frac{4}{x_{1}}+\frac{4}{x_{2}}$. For a positive $k$, the ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,947 |
7.2. The pages of a book are numbered consecutively from the first to the last. Vasya, a troublemaker, tore out 25 pages from different parts of the book and added up the numbers of all fifty torn-out pages. He got the number 2020. When Kolya, an excellent student, found out about this, he claimed that Vasya must have ... | Solution. On each of the torn-out leaves - there are two pages. The number of one of the pages is an even number, and the other is an odd number. Then, in the sum of all the numbers of the torn-out pages, there will be 25 even and 25 odd addends. Therefore, the sum will be odd, and thus it cannot be equal to 2020. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,948 |
7.4. On the board, the number 321321321321 is written. Which digits need to be erased to obtain the largest possible number divisible by 9? | Answer: erase the last two threes.
Solution: From the divisibility rule for 9, it follows that the sum of the erased digits must be equal to 6. Since the larger the number, the more digits it has, we need to erase two threes. This will leave a number with 10 digits. To make this number as large as possible, we need to... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,950 |
7.5. The children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will have as many mushrooms as all the others combined. How many children went to pick mushrooms | Answer: 6 children.
Solution: Let Anya give half of her mushrooms to Vitya. Now all the children have the same number of mushrooms (this means that Vitya did not have any mushrooms of his own). For Sanya to now get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushr... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,951 |
Problem 4.1. On nine cards, the numbers from 1 to 9 are written (each one only once). These cards were laid out in a row such that there are no three consecutive cards with numbers in ascending order, and no three consecutive cards with numbers in descending order. Then, three cards were flipped over, as shown in the f... | Answer: The number 5 is written on card $A$, the number -2 on card $B$, and the number -9 on card $C$.
Solution. The missing numbers are $-2, 5$, and 9.
If the number 5 is on card $B$, then we get the consecutive numbers 3, 4, 5. If the number 5 is on card $C$, then we get the consecutive numbers 8, 7, 5. Therefore, ... | 5,2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,952 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,953 |
Problem 4.5. Hooligan Dima laid out a structure in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,954 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Let's add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
To the obtained sum, add the distance from Asya to Borya: $32+8=40$. Now all three branches (to Gala - twice) and the mai... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,955 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,956 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,957 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,958 |
Problem 6.3. In the cells of a $4 \times 4$ table, the numbers $1,2,3,4$ are arranged such that
- each number appears in each row and each column;
- in all four parts shown in the figure, the sums of the numbers are equal.
Determine in which cells the twos are located based on the two numbers in the figure.
Fig. 1: to the solution of problem 7.2
Solution. All available routes are shown in Fig. 1. One road leads out of Saint Petersburg and Yaroslavl, so ... | SaintPetersburgorYaroslavl | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,963 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,964 |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,965 |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,966 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,967 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,968 |
Problem 9.6. In triangle $ABC$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $AC$, point $K$ is marked, and on side $BC$, points $L$ and $M$ are marked such that $KL=KM$ (point $L$ lies on segment $BM$).
Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$.
... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,969 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,971 |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,972 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the largest integer value ... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,973 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
, if: 1) a pawn cannot be placed on the $e4$ square; 2) no two pawns can stand on squares that are symmetric with respect to the $e4$ square?
Answer: 39 pawns. | Solution. All fields of the board except for the vertical $a$, the horizontal 8, and the field e4 can be divided into pairs that are symmetrical relative to e4. Such pairs form 24. According to the condition, no more than one pawn can be placed on the fields of each pair. In addition, no more than one pawn can be place... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,976 |
4. Given a parallelogram ABCD. Points H and K are chosen on lines AB and BC such that triangles KAB and HCB are isosceles (KA=AB, HC=CB). Prove that triangle KDH is isosceles. | # Solution.
First case. Angle B is acute. Let $\angle K B A=\alpha$, then $\angle H B C=\alpha$ (as vertical angles). From the isosceles triangles, we get that $\angle B K A=\angle B H C=\alpha$. Then, by the sum of angles in a triangle, we get that $\angle K A B=\angle H C B=180^{\circ}-2 \alpha$.
By the properties ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,977 |
5. Borya and Vova are playing the following game on an initially white $8 \times 8$ board. Borya moves first and on each of his turns, he colors any four white cells black. After each of his moves, Vova colors an entire row (row or column) completely white. Borya aims to color as many cells black as possible, while Vov... | Solution. Let Vova make white the row with the most black cells on each of his moves. Then, as soon as Borya achieves a row of no less than four black cells (we will call such a row "rich"), Vova will remove at least four cells, meaning that Borya will not be able to increase the number of black cells compared to his p... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,978 |
1. Vasya and Petya drew a five-pointed star each. All the angles at the vertices of Petya's star are acute, while Vasya's star has an obtuse angle. Each of them claims that the sum of the angles at the vertices of his star is greater. Who is right?
\[
\begin{aligned}
& \angle 1+\angle 3=180^{\circ}-\angle 6=\angle 9 \\
& \angle 2+\angle 4=180^{\circ}-... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,979 |
2. The password consists of four different digits, the sum of which is 27. How many password options exist | Answer: 72.
Solution. Among the digits of the password, there is a 9. Otherwise, since all digits of the password are different, the maximum sum of the digits will not exceed $8+7+6+5=26$. If there are 9 and 8, then the other two digits are no more than 7, and their sum is 10. There are two possible cases: 7,3 and 6,4... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,980 |
3. A real number $x$ is called interesting if, by erasing one of the digits in its decimal representation, one can obtain the number $2x$. Find the largest interesting number. | Answer: 0.375.
Solution. When a digit from the integer part of a real number $x$ is erased, the number cannot increase, and when a digit from the fractional part is erased, it can only increase by a number less than 1. Therefore, the number $x$ must be less than 1, and if its first digit after the decimal point is not... | 0.375 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,981 |
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