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11.7. Given a non-constant polynomial $P(x)$ with integer coefficients and a natural number $n$. Let $a_{0}=n, a_{k}=P\left(a_{k-1}\right)$ for all natural $k$. It turns out that for any natural number $b$, the sequence $a_{0}, a_{1}, a_{2}, \ldots$ contains a number that is a $b$-th power of a natural number greater t... | Solution. Note immediately that for each natural number $b$, the sequence $a_{0}, a_{1}, a_{2}, \ldots$ will contain infinitely many $b$-th powers of natural numbers greater than one. Indeed, if their number is finite, and the largest one is $N=x^{b}$, then the sequence will not contain any $N b$-th power, which is imp... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,148 |
11.8. Given a natural number $n$. From 26 unit white cubes and one black cube, a $3 \times 3 \times 3$ cube is assembled such that the black cube is at its center. From $n^{3}$ such cubes with edge 3, a cube with edge $3 n$ is formed. What is the minimum number of white cubes that can be repainted red so that each whit... | Answer. $(n+1) n^{2}$.
Solution. Introduce a coordinate system so that the centers of the cubes have coordinates from 1 to $3 n$ along each axis. Assign each cube the coordinates of its center. Thus, a cube is black if and only if all its coordinates give a remainder of 2 when divided by 3.
Paint red all white cubes ... | (n+1)n^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,149 |
10.1. Let these numbers be A and B. Let B be the smallest of these numbers. Then the larger number $\mathrm{A}=10 \mathrm{~B}+\mathrm{n}$, where $\mathrm{n}$ is the crossed-out digit, and
$0 \leq \mathrm{n} \leq 9$. According to the condition, $\mathrm{A}+\mathrm{B}=11 \mathrm{~B}+\mathrm{n}=2022$. Therefore, $\mathrm... | Answer: $\mathrm{B}=183, \mathrm{~A}=1839$. | B=183,A=1839 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,150 |
10.5. Let's consider the cut lines formed in the original 100 x 100 square. It is easy to see that one cut of the folded square corresponds to cutting the original square along several parallel lines, with the distances between these lines being no less than 2 (since at least one fold line must pass between any two cut... | Answer: 2601
 | 2601 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,151 |
Problem 11.1. Twelve different natural numbers are written in a circle, one of which is 1. Any two adjacent numbers differ by either 10 or 7. What is the greatest value that the largest written number can take? | Answer: 58.
Solution. For convenience, let's number all the numbers in a circle clockwise, starting from the number 1: $a_{1}=1, a_{2}, \ldots, a_{12}$. Notice that for all $1 \leqslant i \leqslant 6$ we have
$$
\begin{gathered}
a_{i+1}-a_{1}=\left(a_{i+1}-a_{i}\right)+\left(a_{i}-a_{i-1}\right)+\ldots+\left(a_{2}-a_... | 58 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,152 |
Problem 11.2. Let $\alpha$ and $\beta$ be the real roots of the equation $x^{2}-x-2021=0$, with $\alpha>\beta$. Denote
$$
A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7
$$
Find the greatest integer not exceeding $A$. | Answer: -6055.
Solution. By Vieta's theorem, we have $\alpha+\beta=1$ and $\alpha \beta=-2021$. Also, $\beta^{2}-\beta-2021=0$, since $\beta$ is a root of the equation. Therefore,
$A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7=(\alpha+\beta)^{2}-3\left(\beta^{2}-\beta-2021\right)-6063+7=1^{2}-3 \cdot 0-6063+7=-60... | -6055 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,153 |
Problem 11.3. Let $k_{1}$ be the smallest natural number that is a root of the equation
$$
\sin k^{\circ}=\sin 334 k^{\circ}
$$
(a) (2 points) Find $k_{1}$.
(b) (2 points) Find the smallest root of this equation that is a natural number greater than $k_{1}$. | # Answer:
(a) (2 points) 36.
(b) (2 points) 40.
Solution.
$$
0=\sin 334 k^{\circ}-\sin k^{\circ}=2 \sin \frac{333 k^{\circ}}{2} \cos \frac{335 k^{\circ}}{2}
$$
- Let $\sin \frac{333 k^{\circ}}{2}=0$. This is equivalent to $\frac{333 k}{2}=180 \mathrm{~m}$ for some integer $m$, that is, $333 k=360 m$ and $37 k=40 m... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,154 |
Problem 11.4. In a sports school, 55 people are training, each of whom is either a tennis player or a chess player. It is known that there are no four chess players who would have the same number of friends among the tennis players. What is the maximum number of chess players that can train in this school? | Answer: 42.
Solution. Let there be $a$ tennis players and $55-a$ chess players in the school. Each of the chess players has a number of tennis player friends that is no less than 0 and no more than $a$, meaning it can take on $a+1$ values. If there were more than $3(a+1)$ chess players, by the pigeonhole principle, th... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,155 |
Problem 11.5. Points $A, B, C, D, E, F$ are arranged clockwise on a circle, as shown in the figure. Chords $A D$ and $C E$ intersect at point $X$ at a right angle, and chords $A D$ and $B F$ intersect at point $Y$.
It is known that $C X=12, X E=27, X Y=15, B Y=9, Y F=11$.
 (2 points) 36.
(b) (2 points) 19.5.
Solution. Let $A Y=a, X D=b$ (Fig. 8). We use the fact that the products of the segments of intersecting chords through a given point inside a circle are equal. This means that $324=12 \cdot 27=a(15+b)$ and $99=9 \cdot 11=b(a+15)$. Subtracting the second equation fro... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,156 |
Problem 11.6. Given the set of numbers $\{-1,-2,-3, \ldots,-26\}$. On the board, all possible subsets of this set were written, each containing at least 2 numbers. For each written subset, the product of all numbers belonging to that subset was calculated. What is the sum of all these products? | Answer: 350.
Solution. Consider the expression $(1-1)(1-2)(1-3) \ldots(1-26)$, the value of which is 0. Expand this expression without combining like terms, resulting in the sum $1^{26}-1^{25}(1+2+3+\ldots+26)+S$. Notice that the sum $S$ coincides with the sum from the problem's condition. Therefore, it equals $0-1+(1... | 350 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,157 |
Problem 11.7. All vertices of a regular tetrahedron $ABCD$ are on the same side of the plane $\alpha$. It turns out that the projections of the vertices of the tetrahedron onto the plane $\alpha$ form the vertices of a certain square. Find the value of $AB^2$ if it is known that the distances from points $A$ and $B$ to... | Answer: 32.
Fig. 9: to the solution of problem 11.7
Solution. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the projections of points $A, B, C, D$ onto the plane $\alpha$. Let $M$ be the midpoint of s... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,158 |
Problem 11.8. In each cell of a strip $1 \times N$ there is either a plus or a minus. Vanya can perform the following operation: choose any three cells (not necessarily consecutive), one of which is exactly in the middle between the other two cells, and change the three signs in these cells to their opposites. A number... | Answer: 1396.
Solution. We will prove that all the numbers under consideration are positive, except for 4 and 5. Then the answer to the problem will be $1398-2=1396$.
For each $N$, we will number the cells of the strip $1 \times N$ from left to right with numbers from 1 to $N$.
- Let $N=3$. By applying the operation... | 1396 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,159 |
Problem 4.1. Four arithmetic examples were written on the board. Vera erased one "plus" sign, one "minus" sign, one "multiply" sign, one "divide" sign, and four "equals" signs.
Instead of the same signs, she wrote the same letters, and instead of different signs, different letters. Restore the examples.
![](https://c... | Answer: $\mathrm{a} 4 \mathrm{~b} 5 \mathrm{c} 2 \mathrm{~d} 1 \mathrm{e} 3$.
Solution. Notice that the letter $B$ appears in all examples. It turns out that the letter $B$ must be replaced by the equals sign:
$$
\begin{aligned}
& 4 A 2=2 \\
& 8=4 C 2 \\
& 2 D 3=5 \\
& 4=5 E 1
\end{aligned}
$$
Now we will look at th... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,160 | |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,162 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,163 |
Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "P", as shown in the image, with the number of points on any two touching faces being the same.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,167 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,169 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,173 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,174 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,176 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,177 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
This results in 8 rectan... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,178 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,179 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,180 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,181 |
8.3. Troublesome eighth-graders Vova and Dima tore up the school wall newspaper that criticized their behavior, diligence, and speech culture. Moreover, each piece of the newspaper that fell into Vova's hands he tore into 5 pieces, while Dima only tore into 3 (he was criticized less). Later, the school cleaner collecte... | Solution. In all operations, the number of newspaper scraps increases by 4 or by 2, so the remainder of the total number of scraps divided by 2 remains unchanged at any time. At the beginning of the process (when the newspaper was whole), this remainder was 1. But the number 2018 has a zero remainder when divided by 2,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,184 |
8.4. All vertices of the parallelogram $A_{1} B_{1} C_{1} D_{1}$ lie on the sides of the parallelogram $A B C D$, with $A_{1} \in A B, B_{1} \in B C, C_{1} \in C D, D_{1} \in D A$. Prove that the centers of both parallelograms coincide. | Solution. The center of a parallelogram can be defined, on the one hand, as the point of intersection of the diagonals, and on the other hand, as the point of intersection of the midlines. The midpoint of the diagonal $A_{1} C_{1}$ of the smaller parallelogram lies on the midline of the larger one, parallel to the side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,185 |
8.5. In the parliament of the island state of Promenade-and-Tornado, only the indigenous inhabitants of the island can be elected, who are divided into knights and liars: knights always tell the truth, liars always lie. A secret ballot on 8.09.19 re-elected 2019 deputies. At the first meeting, all deputies were present... | Solution. A rectangle $42 \times 48$ can be tiled with $3 \times 3$ squares (224 are required). In each such square, there must be at least one knight (otherwise - if the square contains only liars, the liar in the central cell would have told the truth, which is impossible). Thus, the minimum number of knights in the ... | 225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,186 |
1. Can 4 natural numbers be written on 4 cards such that when each is increased by 1, their product increases by 3 times | Answer: It is possible.
Solution. The numbers were 2, 3, 4, 5. They became 3, 4, 5, 6.
The product increased by 3 times.
## Grading Criteria.
Correct answer without justification: 0 points. Correct example: 7 points. | 2,3,4,5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,187 |
2. On the Island of Logic, there live 40 knights (who always tell the truth), 25 liars (who always lie), and several sophists. A sophist can only utter phrases that a knight or a liar could not say in their place. For example, standing next to a liar, a sophist can say the phrase "We are both liars" (because if he were... | Answer: 1) "There are exactly 40 knights on the island";
2) "There are exactly 26 liars on the island."
Solution. 1) If the sophist were a liar, he would have to tell the truth and then he would say that there are 40 knights. But if he were a knight, there would be 41 knights, and he would have said a lie.
2) If the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,188 |
3. Cut the figure composed of equal equilateral triangles into two equal parts (see the figure on the right).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution
is shown in
the figure.
## Obtained
figures
are equal, so
as they consist
of triangles with side 3 and trapezoids, consisting of three small triangles. And they can be superimp... | Combinatorics | MCQ | Yes | Yes | olympiads | false | 17,189 | |
4. Vasya must plant 7 trees in a row. Each planted tree is either a birch or an oak. The planting must satisfy the condition: next to each tree of one type, there must be a tree of the other type. In how many different ways can Vasya perform the planting? Two ways are different if there is a spot where different types ... | Solution. Note that by changing the type of each planted tree to another, we get another way. Two such ways will be called complementary. Thus, it is enough to count the number of such pairs of complementary ways. Which tree to plant at the edge is completely determined by its neighbor. Therefore, it is enough to deter... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,190 |
11.1. Can a cube $40 \times 40 \times 40$ be cut into rectangular parallelepipeds such that the length, width, and height of each of them, in some order, are consecutive odd numbers? | # Answer: No.
Solution. Suppose such a cutting is possible. Since the length, width, and height of the parallelepiped are three consecutive odd numbers, this means that one of these dimensions is a number divisible by 3. Therefore, the volume of each parallelepiped resulting from the cutting will be divisible by 3. Th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,191 |
11.2. It is known that $\frac{1}{\cos (2022 x)}+\operatorname{tg}(2022 x)=\frac{1}{2022}$.
Find $\frac{1}{\cos (2022 x)}-\operatorname{tg}(2022 x)$. | Answer. 2022.
## Solution.
Consider the product $\left(\frac{1}{\cos \alpha}+\operatorname{tg} \alpha\right)\left(\frac{1}{\cos \alpha}-\operatorname{tg} \alpha\right)=\frac{1}{\cos ^{2} \alpha}-\operatorname{tg}^{2} \alpha=\frac{1-\sin ^{2} \alpha}{\cos ^{2} \alpha}=1$. Therefore, the desired expression is 2022.
Co... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,192 |
11.3. 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$. What is the smallest integer value that the sum of the four written numbers can have? | Answer: 5.
Solution: Let the roots of the polynomial $P(x)$ be $x_{1}, x_{2} > 0$. Then 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{1}{x_{1}} + \frac{1}{x_{2}}$. For a positive $k$, the inequality $k + \frac{1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,193 |
11.4. One hundred and one non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of negative numbers that can be among these 101 written numbers? | Answer: 67.
Solution: Consider any 3 consecutive numbers. All of them cannot be negative. Among them, there is a positive one. Fix this positive number and its neighbor (this number can be negative), and divide the remaining 99 numbers into 33 groups of 3 consecutive numbers. In each such group, there will be no more ... | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,194 |
11.5. The lateral faces of the pentagonal pyramid $S A B C D E$ are acute-angled triangles. We will call a lateral edge of the pyramid good if it is equal to the height of the opposite lateral face, drawn from the vertex of the pyramid (for example, the edge $S A$ is good if it is equal to the height of the triangle $S... | # Answer. 2.
Solution. We will show that two non-adjacent lateral edges of the pyramid cannot both be good. Suppose this is not the case, and, for example, edges $S A$ and $S C$ are good, meaning $S A = S P$ is the height of face $S C D$, and $S C = S Q$ is the height of face $S A E$. However, by the property of the h... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,195 |
1. Martians love to dance dances where they have to hold hands. In the dance "Pyramid," no more than 7 Martians can participate, each of whom has no more than three hands. What is the maximum number of hands that the dancers can have if any hand of one Martian holds exactly one hand of another Martian? | Solution. In each handshake, two hands are involved, so the total number of hands shaken will be even. The maximum variant (7 Martians with three hands) gives 21 hands. Therefore, the greatest number of hands will be achieved if 6 Martians have three hands each and one has two hands. The dancers have 20 hands.
$, and therefore there will be two numbers with equal remainders. The difference between these numbers (if subtracti... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,198 |
5. In a right triangle $ABC\left(\angle C=90^{\circ}\right)$, point $D$ is located inside segment $BC$, $CP$ and $CQ$ are the altitudes of triangles $ABC$ and $ADC$ respectively. Prove that triangles $APQ$ and $ADB$ are similar. | Solution. Since $\angle A P C=\angle A Q C=90^{\circ}$, the circle constructed on side $A C$ as a diameter will pass through points $Q$ and $P$. The inscribed angles in this circle, $A C P$ and $A Q P$, are equal because they subtend the same arc. Triangles $A P C$ and $A C B$ are similar by two angles, so $\angle A C ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,200 |
9.5. Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the sum of the numbers written next to its r... | Answer: 1250 sums.
Solution. First, we show that there are at least 1250 irrational numbers in the table. Let there be $x$ irrational and $50-x$ rational numbers along the left side of the table. Then, along the top side, there are $50-x$ irrational and $x$ rational numbers. Since the sum of a rational and an irration... | 1250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,201 |
9.7. A convex polygon is cut by non-intersecting diagonals into isosceles triangles. Prove that this polygon will have two equal sides. (A. Gribalko) | Solution. We will prove the statement by induction on the number of sides $n$ of the polygon. The base case $n=3$ is obvious. Now we will derive the statement for a $k$-gon $(k \geqslant 4)$, assuming that it is true for polygons with fewer sides than $k$.
Thus, let a convex $k$-gon $P$ be divided by non-intersecting ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,202 |
9.8. Initially, 100 cards were placed on the table, each with a natural number written on it; there were exactly 43 cards with odd numbers. Then, every minute, the following procedure was carried out. For every three cards lying on the table, the product of the numbers written on them was calculated, all these products... | Solution. If at some moment there are exactly $k$ odd numbers among the numbers on the cards, then among the products of triples of numbers there are exactly $C_{k}^{3}$ odd; therefore, the number on the next added card will be odd if and only if $C_{k}^{3}$ is odd (and then $k$ will increase by 1 at that moment).
Not... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,203 |
3. Rewrite the system as
$$
\left\{\begin{array}{l}
|x-1|-y=1-a^{4}-a^{4}(x-1)^{4} \\
(x-1)^{2}+y^{2}=1
\end{array}\right.
$$
and make the substitution $\mathrm{t}=\mathrm{x}-1$.
We get
$$
\left\{\begin{array}{l}
|t|-y=1-a^{4}-a^{4} t^{4} \\
t^{2}+y^{2}=1
\end{array}\right.
$$
We see that if (t, y) is a solution, ... | Answer: $\pm \sqrt[4]{2}$. | \\sqrt[4]{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,205 |
5. Obviously, $\mathrm{b} \neq 0$. From the condition, it follows that $f(1)=a+b=11$, and $f(11)=121 a+11 b=1111$, from which $a=9, b=2$.
Now let's prove that the function $f(\mathrm{x})=9 \mathrm{x}^{2}+2 \mathrm{x}$ has this property. $f(\mathrm{x})=\mathrm{x}(9 \mathrm{x}+2)$
Let $n=\underbrace{11 \ldots 11}_{k}$,... | Answer: $\mathrm{a}=9, \mathrm{~b}=2$
Note for markers. If a and b are found, but the given property of the function is not proven, then 1 point. | =9,b=2 | Algebra | proof | Yes | Yes | olympiads | false | 17,206 |
1.1 Two couples want to sit on a bench. In how many ways can they sit so that each boy sits next to his girlfriend, if the left and right sides of the bench are distinguishable? | # Answer: 8
Solution. Let's number the seats from 1 to 4. If one of the couple sits in seat number 1, then the second must sit in seat 2. Thus, each couple either sits on the left (seats 1 and 2) or on the right (seats 3 and 4). Each couple can sit in two ways (boy on the left or right), and the couples themselves can... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,207 |
Task 4.1.2. A sports team consists of two boys and two girls. In the relay, the stages run by girls must alternate with the stages run by boys. In how many ways can the stages be distributed
# | # Answer: 8
Solution. Boys can run stages 1 and 3, or stages 2 and 4. If they run stages 1 and 3, they can be distributed in two ways, and girls can be distributed in two ways, making a total of 4 ways. In the second case, there are another 4 ways, making a total of 8 ways. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,208 |
Task 4.1.3 Two mothers with their children want to sit on a bench with 4 seats. In how many ways can they sit so that each mother sits next to her child? Each mother is walking with one child.
# | # Answer: 8
Solution. Let's number the seats from 1 to 4. If someone from a family sits in seat number 1, then the second person must sit in seat 2. Thus, each family either sits on the left (seats 1 and 2) or on the right (seats 3 and 4). Each family can sit in two ways (child on the left or right), and the families ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,209 |
2.1 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first one said that, in his opinion, there were 20 apples, the second thought there were 22, and the third said 25. When the friends opened the bag, it turned out that one of them was off by 1, another by 3, a... | # Answer: 19
Solution. Note that the first and second friends give even predictions, which means they make an error of the same parity, while the third makes an error of a different parity. Therefore, the third friend was off by 6 apples, so there were either 19 or 31 apples in the bag. Checking both options, we get t... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,210 |
2.2 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first said that, in his opinion, there were 19 apples, the second thought there were 22, and the third said 23. When the friends opened the bag, it turned out that one of them was off by 1, another by 2, and t... | Answer: 20
Solution. Note that the first and third friends give odd predictions, which means they make errors of one parity, while the second makes errors of the other parity. Therefore, the second friend was off by 2 apples, so there are either 20 or 24 apples in the bag. Checking both options, we get the answer. | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,211 |
2.3 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first one said that, in his opinion, there were 16 apples, the second thought there were 19, and the third said 25. When the friends opened the bag, it turned out that one of them was off by 2, another by 4, a... | # Answer: 21
Solution. Notice that the second and third friends give odd predictions, which means they make an error of one parity, while the first makes an error of the other parity. Therefore, the first one was off by 5 apples, and the number of apples in the bag is either 11 or 21. Checking both options, we get the... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,212 |
2.4 In a closed bag lie apples. Three friends tried to lift the bag and guess how many fruits were inside. The first one said that, in his opinion, there were 18 apples, the second thought there were 21, and the third said 22. When the friends opened the bag, it turned out that one of them was off by 2, another by 3, a... | Answer: 24 apples
Solution. Note that the first and third give even predictions, which means they make an error of one parity, while the third makes an error of the other parity. Therefore, the second one was off by 3 apples, so there are either 18 or 24 apples in the bag. Checking both options, we get the answer. | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,213 |
3.1 Let's say that number A hides number B if you can erase several digits from A to get B (for example, the number 123 hides the numbers 1, 2, 3, 12, 13, and 23). Find the smallest natural number that hides all numbers from 2000 to 2021.
Natural numbers are numbers used for counting objects.
Answer: 20012013456789 | Solution. The number must definitely contain three zeros (for the number 2000), two ones (for the number 2011), and two twos (for the number 2020). Other digits must appear at least once, so the result should have a minimum of 13 digits. Before the third zero, there must be both twos (for the number 2020) and at least ... | 20012013456789 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,214 |
3.2. Let's say that number A hides number B if you can erase several digits from A to get B (for example, the number 123 hides the numbers 1, 2, 3, 12, 13, and 23). Find the smallest natural number that hides the numbers 121, 221, 321, ..., 1921, 2021, 2121.
Natural numbers are numbers used for counting objects.
Answ... | Solution: The number must contain one zero, three ones (for the number 1121), two twos (for the number 1221), and at least one of each of the other digits, meaning it must have a minimum of 13 digits. After two ones and the other digits, one two and one one stand consecutively, so they will be at the end. The smallest ... | 1201345678921 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,215 |
4.1 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared three times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a hurry, stumbled, and some of the balloons burst. Eeyore received a ... | # Answer: 1
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 6 times more balloons than Piglet prepared. The next number after 31 that is a multiple of 6 is $36, 36: 6=6$, which means Piglet prepared no less than 6 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,216 |
4.2 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared three times as many balloons as Piglet, and Owl prepared four times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a great hurry, stumbled, and some of the balloons burst. Eeyore ... | # Answer: 4
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 8 times more balloons than Piglet prepared. The next number after 60 that is a multiple of 8 is $64, 64: 8=8$, which means Piglet prepared no less than 8 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,217 |
4.3 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared four times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a great hurry, stumbled, and some of the balloons burst. Eeyore receiv... | # Answer: 2
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 7 times more balloons than Piglet prepared. The next number after 44 that is a multiple of 7 is 49, $49: 7=7$, which means Piglet prepared no less than 6 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,218 |
5.1 Six children stood in a circle and whispered their favorite number to the neighbors on their right and left. Each child added the two numbers they heard and said them out loud. The picture shows what three of the children said. What number did the girl think of, next to whom there is a question mark?
 for a game. Each had to say one of the following phrases: "There is a liar below me" or "There is a k... | Answer: 17
Solution. The formation cannot consist of only liars (in such a case, the 3-7th person would be telling the truth). Consider the tallest knight, let's call him P. He could only say the first phrase, so the 1st and 2nd places must be occupied by liars, and P stands from the 3rd to the 7th place. Moreover, th... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,224 |
7.2 On the Island of Truth and Lies, there are knights who always tell the truth, and liars who always lie. One day, 15 residents of the island lined up by height (from tallest to shortest, the tallest being the first) for a game. Each had to say one of the following phrases: "There is a liar below me" or "There is a k... | # Answer: 11
Solution. The formation cannot consist of only liars (in such a case, 4-8 would be telling the truth). Consider the tallest knight, let's call him R. He could only say the first phrase, so the positions 1, 2, and 3 must be occupied by liars, and R stands from 4 to 8. In addition, there is a liar who is sh... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,225 |
7.3 On the Island of Truth and Lies, there are knights who always tell the truth, and liars who always lie. One day, 17 residents of the island lined up by height (from tallest to shortest, the tallest being the first) for a game. Each had to say one of the phrases: "There is a liar below me" or "There is a knight abov... | Answer: 14
Solution. The formation cannot consist of only liars (in such a case, 3-6 would be telling the truth). Consider the tallest knight, let's call him P. He could only say the first phrase, so the 1st and 2nd positions must be occupied by liars, and P stands from the 3rd to the 6th position. Moreover, there is ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,226 |
8.1 On 12 cards, different natural numbers are written, their sum is 84. Masha laid out the cards in ascending order. What can the sum of the numbers on the two middle cards be? Find all possible answers. | Answer: 13,14
Solution. If the smallest possible natural numbers, i.e., the card numbers (from 1 to 12), were written on the cards, then their sum would be $1+2+...+12=78$. But the actual sum is 6 more. This means that some numbers are greater than their numbers. If the number on the sixth card is greater than 6, then... | 13,14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,227 |
8.2. On 14 cards, different natural numbers are written, their sum is 112. Masha laid out the cards in ascending order. What can the sum of the numbers on the two middle cards be? Find all possible answers. | Answer: 15,16
Solution. If the smallest possible natural numbers, i.e., the card numbers (from 1 to 14), were written on the cards, then their sum would be $1+2+\ldots+14=105$. But the actual sum is 7 more. This means that some numbers are greater than their numbers. If the number on the seventh card is greater than 7... | 15,16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,228 |
8.3. On 10 cards, different natural numbers are written, their sum is 60. Masha laid out the cards in ascending order. What can the sum of the numbers on the two middle cards be? Find all possible answers. | Answer: 11,12
Solution. If the smallest possible natural numbers, i.e., the card numbers (from 1 to 10), were written on the cards, then their sum would be equal to $1+2+. .+10=55$. But the actual sum is 5 more. This means that some numbers are greater than their numbers. If the number on the fifth card is greater tha... | 11,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,229 |
7.1. Each student in the 7th grade attends a math club or an informatics club, and two students attend both the math club and the informatics club. Two-thirds of all students attend the math club, and $40 \%$ attend the informatics club. How many students are there in this class? | Answer: 30.
Solution: Note that $40\%$ is $\frac{2}{5}$ of all students. Adding $\frac{2}{3}$ and $\frac{2}{5}$ gives $\frac{16}{15}$. The two students who are involved in both the math and informatics clubs were counted twice. Therefore, these two students represent $\frac{1}{15}$ of the entire class, meaning there a... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,230 |
7.2. Cut a $5 \times 6$ grid rectangle along the grid lines into five figures, the areas of which will be five consecutive natural numbers, and the perimeter of each figure will be twice its area. | Solution. Let's find the areas of the required polygons. Let $x$ be the area of the polygon with the smallest area. Then $x + x + 1 + x + 2 + x + 3 + x + 4 = 30$, from which $x = 4$, so the areas of the polygons are $4, 5, 6, 7, 8$. Then their perimeters are $8, 10, 12, 14$, and 16, respectively. The figure below shows... | 4,5,6,7,8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,231 |
7.3. How many seven-digit natural numbers exist, for which the product of the first three digits is 30, the product of the three digits in the center is 7, and the product of the last three digits is 15? | Answer: 4.
Solution: Let the number be $\overline{\operatorname{abcdefg}}$. According to the condition, $c d e=7$, which means one of these digits is 7, and the other two are 1. Since 30 and 15 are not divisible by $7, d=7, c=e=1$. The number $30=5 \cdot 6 \cdot 1$, so we get two three-digit numbers $\overline{a b c}$... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,232 |
7.4. In a $5 \times 5$ grid, there are 5 columns, 5 rows, and 18 diagonals (diagonals of length one are also counted). Each cell of this grid is filled with one of the numbers $1, 3, 5$, or 7, and the sum of the numbers in each column, row, and diagonal is calculated. Prove that among the obtained sums, there will be a... | Solution. We will call a row, column, or diagonal along which numbers are summed a line. Note that there are 20 lines consisting of an odd number of cells (5 lines in each direction). Since all numbers in the table are odd, the sums in these lines are also odd. At the same time, they cannot exceed \(5 \cdot 7 = 35\). T... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,233 |
7.5. At a round table, 10 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each of them was given a token. Then each of them passed their token to one of their two neighbors. After that, 5 people said: "I have one token," while the other 5 said: "I have no tok... | # Answer. 7.
Solution. Evaluation. After passing the chips, each person sitting at the table can have 0, 1, or 2 chips. The total number of chips will be 10. Note that if a person lies, they will state a number of chips that differs from the actual number by 1 or 2. Since the total number of chips according to the ans... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,234 |
1. The graphs of the functions $y=x^{2}+a x+b$ and $y=x^{2}+c x+d$ intersect at the point with coordinates $(1,1)$. Compare $a^{5}+d^{6}$ and $c^{6}-b^{5}$. | # Solution
Since the graphs pass through the point $(1,1)$, then $1=1+a+b$ and $1=1+c+d$, which means $a=-b$ and $c=-d$. Therefore, $a^{5}=-b^{5}$ and $d^{6}=c^{6}$. Adding these equations, we get $a^{5}+d^{6}=c^{6}-b^{5}$.
## Answer these numbers are equal | ^{5}+^{6}=^{6}-b^{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,235 |
2. Having walked $4 / 9$ of the length of the bridge, the traveler noticed that a car was catching up to him, but it had not yet entered the bridge. Then he turned back and met the car at the beginning of the bridge. If he had continued his movement, the car would have caught up with him at the end of the bridge. Find ... | # Solution
From the condition, it follows that the time it takes for the car to approach the bridge is equal to the time it takes for the traveler to walk $4 / 9$ of the bridge. Therefore, if the traveler continues moving, by the time the car reaches the bridge, he will have walked ${ }^{8} / 9$ of the bridge. This me... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,236 |
3. $M$ - the point of intersection of the diagonals of trapezoid $A B C D$. A point $P$ is chosen on the base $B C$ such that $\angle A P M=\angle D P M$. Prove that the distance from point $C$ to line $A P$ is equal to the distance from point $B$ to line $D P$.
# | # Solution
Let's denote the distance from point $X$ to line $l$ as $d(X, l)$.
Notice that $d(C, A P)=d(M, A P) \cdot \frac{A C}{A M}, \quad d(B, D P)=d(M, D P) \cdot \frac{D B}{D M}$.
But $d(M, A P)=d(M, D P)$, since $P M$ is the angle bisector of $\angle A P D$.
On the other hand, from the similarity of triangles ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,237 |
4. Vovochka is playing a computer game. If he scores less than 1000 points, the computer will add $20 \%$ of his score. If he scores from 1000 to 2000 points, the computer will add $20 \%$ of the first thousand points and $30 \%$ of the remaining points. If Petya scores more than 2000 points, the computer will add $20 ... | # Solution
It is clear that Petya scored more than 1000 points (otherwise his result would not exceed 1200) and less than 2000 (otherwise the result would not be less than 2500). Let's discard the 1200 points (the first thousand plus the prize for it). The remaining 1170 points constitute $130 \%$ of the points Petya ... | 470 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,238 |
5. The bisector, median, and altitude of a certain triangle, drawn from three different vertices, intersect at one point and divide this triangle into six smaller triangles (see figure). The areas of the three shaded triangles are equal. Is it true that the original triangle is equilateral?
 and a common height. $S_{COK} = 1/2 \cdot C K \cdot C O \cdot \sin \angle O C K$, $S_{COM} = 1/2 \cdot C K \cdo... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,239 |
1. Given various real numbers $p$ and $q$. It is known that there exists a number $x$ such that the equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ are satisfied. What values can the sum $p + q$ take? | Solution. Let $x$ be such that both equalities from the problem statement are satisfied. Then $x^{2}+p x+q=x^{2}+q x+p$, from which $(p-q) x=p-q$. Since $p \neq q$, it follows that $x=1$. Substituting this value into any of the equations, we find that $p+q+1=0$, that is, $p+q=-1$. Answer: -1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,240 |
2. Find four such numbers that all their pairwise sums are consecutive natural numbers, the smallest of which is 2023. | Solution. Note that the pairwise sums of the numbers $1, 3, 5$, and $9$ give six consecutive even numbers from $4$ to $14$. Then the pairwise sums of the numbers $n+1, n+3, n+5, n+9$ also form six consecutive even numbers from $2n+4$ to $2n+14$. Dividing them by 2, we get six consecutive natural numbers. Therefore, the... | 1011,1012,1013,1015 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,241 |
3. In a football tournament, seven teams played: each team played once with each other. Teams that scored thirteen or more points advance to the next round. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round? | Solution. Evaluation. Note that the total number of games will be $\frac{7 \cdot 6}{2}=21$. Then the total number of points does not exceed $21 \cdot 3=63$. This means that no more than $\left[\frac{63}{13}\right]=4$ teams can advance to the next round.
Example. We will show that 4 teams can advance to the next round.... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,242 |
4. Quadrilateral $ABCD$ is inscribed, point $K$ is the midpoint of the arc $AD$ that does not contain any other vertices of the quadrilateral. Let $X$ and $Y$ be the points of intersection of lines $BK$ and $CK$ with the diagonals. Prove that line $XY$ is parallel to $AD$. | Solution. Since $K$ is the midpoint of arc $A D$, arcs $A K$ and $K D$ are equal. Since arcs $A K$ and $K D$ are equal, the inscribed angles $K B D$ and $A C K$ are equal, which means angles $X B Y$ and $X C Y$ are equal. Since segment $X Y$ is seen from the same angle at points $B$ and $C$, quadrilateral $B C Y X$ is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,243 |
5. Positive real numbers $a, b, x, y$ satisfy the conditions $a x \leqslant 5, a y \leqslant 10, b x \leqslant 10, b y \leqslant 10$. Does it follow that
$$
a x + a y + b x + b y \leqslant 30 ?
$$ | Solution. Yes, it should. Let's generalize the condition. Suppose $a x \leqslant p, a y \leqslant q, b x \leqslant q$, $b y \leqslant q$. We will prove that then $a x+a y+b x+b y \leqslant 2(p+q)$. From the condition, it follows that $a x+b y \leqslant p+q$. Consider the sum of the remaining terms $a y+b x=a x+b x+a(y-... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 17,244 |
# 5. CONDITION
For which $n$ does there exist a convex $n$-gon, one of whose sides has length 1, and the lengths of all diagonals are integers? | Solution. For $n=4$ and $n=5$, it is possible to construct polygons where one side is equal to 1, and the diagonals "bounding" this side are equal to 2. If, however, $n \geq 6$, then if $A B=1$, the diagonals extending from one of the vertices and connecting it to $A$ and $B$ are equal (as integer values, the differenc... | n=4n=5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,246 |
Problem 9.1. Students from 9A, 9B, and 9C gathered for a line-up. Maria Ivanovna decided to count the number of attendees from each class. She found that there were 27 students from 9A, 29 students from 9B, and 30 students from 9C. Ilya Grigoryevich decided to count the total number of attendees from all three classes,... | Answer: 29.
Solution. Let $a, b, c$ be the actual number of students from 9 "A", 9 "B", 9 "C" respectively. From Maria Ivanovna's counts, it follows that $a \leqslant 29, b \leqslant 31, c \leqslant 32$, and from Ilya Grigoryevich's counts, it follows that $a+b+c \geqslant 92$. We get that
$$
92 \leqslant a+b+c \leqs... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,247 |
Problem 9.2. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=7$, $NC=39$, $AM=12$, and $MB=3$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 645. (b) 268.5.
Solution. In a rectangle, opposite sides are equal: $AB = CD$ and $BC = AD$. Then $CD = 12 + 3 = 15$. From the Pythagorean theorem for triangle $CDN$, it follows that $DN = \sqrt{CN^2 - CD^2} = \sqrt{39^2 - 15^2} = 36$. Then $BC = AD = 36 + 7 = 43$ and $S_{ABCD} = 15 \cdot 43 = 645$.
As is... | 268.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,248 |
Problem 9.3. Natural numbers $a$ and $b$ are such that $a$ is divisible by $b+1$ and 43 is divisible by $a+b$.
(a) (1 point) Indicate any possible value of $a$.
(b) (3 points) What can $b$ be? Indicate all possible options. | Answer: (a) any of the numbers $22,33,40,42$. (b) $1,3,10,21$.
Solution. Since 43 is a prime number and $a+b$ is a natural number greater than 1, we get that $a+b=43$. Therefore, $b \leqslant 42$ and $a=43-b$.
From the condition, it follows that the number $b+1$ is a divisor of the number $a+(b+1)=44$. Since $b+1$ is... | =22,33,40,42;\,b=1,3,10,21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,249 |
Problem 9.4. An empty $3 \times 51$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 153$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 152$, the num... | Answer: (a) 152. (b) 76.
Solution. Let's color the table in black and white so that the cell where the number 1 should be is black. The third rule in the condition means that consecutive numbers in the table are placed in a "snake" pattern, and the colors of their cells alternate, that is, all odd numbers, including 1... | 152 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,250 |
Problem 9.5. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees doe... | Answer: (a) $107^{\circ}$. (b) $47^{\circ}$.
Fig. 5: to the solution of problem 9.5
Solution. Given $A B=B X$ and $\angle B=34^{\circ}$, therefore $\angle A X B=\angle X A B=\frac{1}{2}\lef... | 107 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,251 |
Problem 9.6. The graph of the line $y=k x+l$ intersects the $O x$ axis at point $B$, the $O y$ axis at point $C$, and the graph of the function $y=1 / x$ at points $A$ and $D$, as shown in the figure. It turns out that $A B=B C=C D$. Find $k$, given that $O C=3$.
Fig. 6: to the solution of problem 9.6
Solution. Draw a line through point $D$ parallel to the $O y$ axis; let it intersect the $O x$ axis at point $D_{1}$ (Fig. 6). It is easy ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,252 |
Problem 9.7. In front of a sweet-tooth lie five boxes of candies: the first box contains 11 candies, the second - 22 candies, the third - 33 candies, the fourth - 44 candies, the fifth - 55 candies. In one move, the sweet-tooth can take four candies from one box and distribute them, one candy to each of the remaining f... | Answer: 159.
Solution. Consider the number of candies in the boxes. Initially, these are 5 natural numbers that give different remainders when divided by 5. We will prove that as long as the sweet-tooth does not take candies from any box, these 5 numbers will always give different remainders when divided by 5. For thi... | 159 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,253 |
Problem 9.8. For real numbers $x$ and $y$, define the operation $\star$ as follows: $x \star y=x y+4 y-3 x$.
Calculate the value of the expression
$$
((\ldots)(((2022 \star 2021) \star 2020) \star 2019) \star \ldots) \star 2) \star 1
$$ | Answer: 12.
Solution. Let $t=(\ldots(((2022 \star 2021) \star 2020) \star 2019) \star \ldots) \star 4$. Then the value of the expression from the problem condition is
$$
\begin{aligned}
((t \star 3) \star 2) \star 1 & =((3 t+12-3 t) \star 2) \star 1=12 \star 2 \star 1= \\
& =(24+8-36) \star 1=(-4) \star 1=-4+4+12=12
... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,254 |
Variant 9.2.1. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=7$, $NC=39$, $AM=12$, $MB=3$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 645. (b) 268.5. | 268.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,255 |
Variant 9.2.2. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=11$, $NC=39$, $AM=12$, $MB=3$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 705. (b) 298.5. | 705 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,256 |
Variant 9.2.3. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=3$, $NC=39$, $AM=10$, $MB=5$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 585. (b) 202.5. | 585 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,257 |
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