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3. How many ways are there to rearrange the letters of the word ГЕОМЕТРИЯ so that no two consonants stand next to each other?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Answer: 21600.
Solution. First, arrange the vowels, among which there are two letters E. They can be permuted in $5!/2=60$ ways. Now, 6 positions are formed where consonants can be placed (no more than one consonant per position) - these are the four gaps between the vowels, as well as the positions at the beginning a... | 21600 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 17,484 |
4. There are 15 students in the class. For homework, they were assigned 6 geometry problems. More than half of the students solved each problem.
Prove that there will be two students who, together, solved all 6 problems. | Solution. Each problem was solved by no fewer than 8 students, and a total of no fewer than $8 \cdot 6=48$ problems were solved. On average, each student solved no fewer than $\frac{48}{15}=3 \frac{1}{3}$ problems. Therefore, there is a student $A$ who solved no fewer than 4 problems. If he solved all 6 problems, there... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,485 |
5. Among five coins, one is counterfeit. Genuine coins have the same mass, while the counterfeit coin has a different mass. There are balance scales that work as follows: when the masses of the weights being weighed are equal, either of the pans can go down; when the masses are different, the scales work correctly. Can... | Answer: It is possible.
Solution. We will present an algorithm (not optimal in terms of the number of weighings) that allows solving the problem.
Let the total weight on the first scale pan be $a$, and on the second $b$. If the first pan has sunk, then $a \geqslant b$. In this case, we will say that the first pan is ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,486 |
1. Find the number of integer points $(x, y)$ satisfying the equation $\frac{1}{|x|}+\frac{1}{|y|}=\frac{1}{2017}$. | Solution. Transform the original equation $\frac{1}{|x|}+\frac{1}{|y|}=\frac{1}{2017} \Rightarrow$
$2017(|x|+|y|)-|x||y|=0 \Rightarrow 2017 \cdot|x|+2017 \cdot|y|-|x||y|-2017^{2}=-2017^{2}$, from which it follows that $(|x|-2017)(|y|-2017)=2017^{2}$. Since 2017 is a prime number and $|x|$ and $|y|$ are natural numbers... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,487 |
2. On a chessboard, four rooks are placed in such a way that they attack all the white squares.
a) Provide an example of such an arrangement.
b) Determine the number of such arrangements
Zhenodarov R.G. | Solution. There are 32 white squares on a chessboard. If a rook is placed on a white square, it attacks 7 white squares; if it is placed on a black square, it attacks 8 squares. 1) Since there are only 4 rooks and 32 white squares, the rooks must be placed only on black squares. Consider the division of the chessboard ... | 2\times4! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,488 |
4. Let $A_{1} A_{2} A_{3} \ldots A_{2016} A_{2017}$ be a convex 2017-gon, $B_{k}$ the midpoints of the edges $A_{k} A_{k+1}$, $k=1, \ldots, 2016$, and $B_{2017}$ the midpoint of the edge $A_{1} A_{2017}$. Prove that the area of the polygon $B_{1} B_{2} B_{3} \ldots B_{2016} B_{2017}$ is not less than half the area of $... | Solution. Let $M$ be an arbitrary internal point of the polygon $A_{1} A_{2} A_{3} \ldots A_{2016} A_{2017}$. Suppose that $M$ belongs to the triangle $\Delta A_{k-1} A_{k} A_{k+1}$. The triangle $\Delta A_{k-1} A_{k} A_{k+1}$ intersects only with two triangles $\Delta A_{k-2} A_{k-1} A_{k}$ and $\Delta A_{k} A_{k+1} A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,489 |
5. The base of the pyramid $A B C D P$ with vertex $P$ is a cyclic quadrilateral inscribed in a circle. It is known that the lines $A D$ and $B C$ intersect at point $K$, outside this circle. The angles $\angle A D P = \angle B C P = 90^{\circ}$, and the angles $\angle A P K$ and $\angle B P K$ are acute. Prove that th... | Solution. Drop a perpendicular $A T$ to the line $K P$. Points $A, D, T, P$ lie on the same circle with diameter $A P$. Therefore, considering that points $A, B, C, D$ also lie on the same circle, we get the equalities $K D \cdot K A = K T \cdot K P = K C \cdot K B$. From the last equality, it follows that points $C, B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,490 |
7.1. Halfway between Vasya's house and school stands a traffic light. On Monday, Vasya caught the green light. On Tuesday, he walked at the same speed but waited at the traffic light for 5 minutes, and then doubled his speed. On both Monday and Tuesday, he spent the same amount of time on the journey from home to schoo... | # 7.1. Answer: 20.
By doubling his speed, Vasya covered half the distance in half the time. According to the problem, this took 5 minutes less than usual. This means that normally he covers half the distance in 10 minutes. And the entire journey from home to school takes 20 minutes. By doubling his speed, Vasya covere... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,491 |
7.4. Petya knows only the digit 1. Prove that he can write a number divisible by 2019. | 7.4. Let's write down 2019 numbers: $1 ; 11 ; \ldots ; 1422 \cdot 443$. Suppose none of them is divisible by 2019. Then they are divided by 2019 with a remainder. The remainder can be one of the numbers $1, 2, 3, \ldots, 2018$, i.e., there are 2018 different remainders. By the Pigeonhole Principle, among the 2019 remai... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,494 |
1. The brother left the house 6 minutes after his sister, following her, and caught up with her after 12 minutes. How many minutes would it have taken him to catch up if he had walked twice as fast? Both the brother and the sister walk at a constant speed. | # Answer: 3 minutes.
Solution. Since the brother walked for 12 minutes before meeting his sister, and the sister walked for 18 minutes, the brother's speed was $3 / 2$ times the sister's speed. If the brother's speed is 3 times the sister's speed, which is 2 times faster than before, then the difference of 6 minutes w... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,496 |
3. In the neighboring houses on Happy Street in Sunny Village, two families of 10 people each lived, with the average weight of the members of one family being 1 kg more than the average weight of the members of the other family. After the eldest sons from both families left to study in the city, it turned out that the... | Answer: by 1 kg or by 19 kg.
Solution. The total weight of the members of one of the families before the departure of the elder sons was 10 kg more than the total weight of the members of the other family. If the heavier family remained heavier, then its total weight became 9 kg more than the total weight of the other... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,498 |
4. Four friends - Alexei, Boris, Viktor, and Gennady went fishing. Boris caught more fish than Alexei and Viktor together. Boris and Alexei together caught as much as Viktor and Gennady, and Boris and Viktor together caught less than Alexei and Gennady. Arrange the friends in the order of the amount of fish caught, fro... | Answer: In descending order of catch: Gennady, Boris, Alexei, Viktor.
Solution. According to the problem, B > A + V, A + B = V + G, A + G > B + V. From this, B > A, V + 2G = G + (B + G) = G + A + B = (G + A) + B > 2B + B, which means G > B. Therefore, A + G > A + B = B + G, which means A > V. Consequently, G > B > A >... | G>B>A>V | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,499 |
5. 7 ginger and 7 black kittens are drinking water from a round basin. Is it true that there will be a kitten whose both neighbors are black kittens?
# | # Answer: correct.
Solution. Let's number the kittens in a circle from 1 to 14. Since there are 7 black kittens, at least four of them will have numbers of the same parity. Consider the chain of kittens with numbers of this parity. In this chain, there are at least 4 black kittens, so there are two adjacent black kitt... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,500 |
1. Vasya has 8 cards with the digits 1, 2, 3, and 4 - two of each digit. He wants to form an eight-digit number such that there is one digit between the two 1s, two digits between the two 2s, three digits between the two 3s, and four digits between the two 4s. Provide any number that Vasya can form. | Answer: 41312432 or 23421314.
Criteria. 7 points for indicating one of the numbers that meet the condition. 7 points for indicating both numbers that meet the condition. No need to describe the method of constructing the numbers. In all other cases (including if several numbers are indicated, among which there are tho... | 41312432or23421314 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,501 |
2. From one point on a straight highway, three cyclists start simultaneously (but possibly in different directions). Each of them rides at a constant speed without changing direction. An hour after the start, the distance between the first and second cyclist was 20 km, and the distance between the first and third - 5 k... | Answer: 25 km/h or 5 km/h.
Solution. Let's draw a numerical axis along the highway, taking the starting point of the cyclists as the origin and directing it in the direction of the second cyclist's movement. His speed is 10 km/h, so after an hour, he was at point B with a coordinate of 10. The distance from him to the... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,502 |
3. In a convex pentagon ABCDE, segment BD bisects each of the angles CBE and CDA, and segment CE bisects each of the angles ACD and BED. Diagonal BE intersects diagonals AC and AD at points K and L, respectively. Prove that CK = DL. | Solution. Triangles KEC and DEC are equal by the 2nd criterion ( $\angle \mathrm{KEC}=\angle \mathrm{DEC}, \angle \mathrm{KCE}=\angle \mathrm{DCE}, \mathrm{CE}$ is common), so $\mathrm{CK}=\mathrm{CD}$. Similarly, triangles DBL and DBC are also equal by the 2nd criterion ( $\angle \mathrm{DBL}=\angle \mathrm{DBC}, \ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,503 |
4. Three schoolchildren are sitting around a round table. In front of each of them is a notebook, on the first page of which each schoolchild wrote an arbitrary integer. Then all three simultaneously did the following: each looked at the numbers on the first pages of their neighbors' notebooks, subtracted the number of... | Answer: No.
Solution. Let the numbers written on the first pages by the students be $a, b, c$. Then the numbers on the second pages will be $a-b, b-c$, and $c-a$; on the third pages - the numbers $a+c-2b, a+b-2c$, and $b+c-2a$; on the fourth pages - the numbers $3c-3b, 3a-3c, 3b-3a$, that is, all numbers will be multi... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,504 |
5. In the room, there are 30 people, among whom there is one psychic. The psychic knows the birth date of each of the others, but none of the others know the birth date of the psychic. You can choose any two people in the room, ask one of them if they know the birth date of the other, and receive an honest answer. Can ... | Answer: Yes, it is possible.
Solution. Let's say we chose people A and B and asked A about B's birthday. If the answer is "I know," then B is definitely not a psychic, and if "I don't know," then A is definitely not a psychic. Thus, with one question, we can reliably identify a person who is not a psychic. We will com... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,505 |
1. (7 points) Compare $3^{76}$ and $5^{50}$. | Solution. $3^{76}=3^{4} \cdot 3^{72}=81 \cdot\left(3^{3}\right)^{24}=81 \cdot(27)^{24}$.
$5^{50}=5^{2} \cdot 5^{48}=25 \cdot\left(5^{2}\right)^{24}=25 \cdot(25)^{24}$.
Since $81>25$ and $27>25$, then $3^{76}>5^{50}$.
Answer. $3^{76}>5^{50}$. | 3^{76}>5^{50} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,506 |
10.1. Petya wrote ten natural numbers on the board, none of which are equal. It is known that among these ten numbers, three can be chosen that are divisible by 5. It is also known that among the ten numbers written, four can be chosen that are divisible by 4. Can the sum of all the numbers written on the board be less... | Answer. It can.
Solution. Example: $1,2,3,4,5,6,8,10,12,20$. In this set, three numbers $(5,10,20)$ are divisible by 5, four numbers $(4,8,12,20)$ are divisible by 4, and the total sum is 71.
Remark. It can be proven (but, of course, this is not required in the problem), that in any example satisfying the problem's c... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,510 |
10.2. Given a quadratic trinomial $P(x)$. Prove that there exist pairwise distinct numbers $a, b$, and $c$ such that the following equalities hold:
$$
P(b+c)=P(a), \quad P(c+a)=P(b), \quad P(a+b)=P(c).
$$
(N. Agakhanov) | Solution. Let $d$ be the abscissa of the vertex of the parabola $y=P(x)$, so that the line $x=d$ is the axis of symmetry of the parabola. Then for any numbers $t$ and $s$ with sum $2 d$ (i.e., such that the points $t$ and $s$ are symmetric with respect to $d$), it holds that $P(t)=P(s)$. Thus, any triple of pairwise di... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,511 |
10.3. Vasya has $n$ candies of several types, where $n \geqslant 145$. It is known that if any group of at least 145 candies is chosen from these $n$ candies (in particular, the group can consist of all $n$ candies), then there exists a type of candy such that the chosen group contains exactly 10 candies of this type. ... | Answer: 160.
Solution: Estimation. We will prove that $n>160$ "does not work". Let us have a set of $n$ candies. We will call a flavor critical if there are exactly 10 candies of this flavor (among all the given $n$ candies). Let us have $k$ critical flavors, then the total number of candies is not less than $10k: n \... | 160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,512 |
10.4. Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$, where $n-$ is a natural number. It is known that the numbers $a_{0}, a_{1}, \ldots, a_{n}$ are integers, with $a_{n} \neq 0, a_{n-k}=a_{k}$ for all $k=0,1, \ldots, n$, and $a_{n}+a_{n-1}+\ldots+$ $+a_{1}+a_{0}=0$. Prove that the number $P(2022)$ is divi... | Solution. It is sufficient to prove the statement: the polynomial $P(x)$ is divisible by $(x-1)^{2}$. Indeed, after division (for example, by long division), the quotient will be a polynomial $Q(x)$ with integer coefficients, and then the polynomial equality $P(x)=(x-1)^{2} Q(x)$ implies the equality $P(2022)=2021^{2} ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,513 |
10.5. A hexagon $A E C D B F$ is inscribed in a circle $\Omega$. It is known that point $D$ bisects the arc $B C$, and triangles $A B C$ and $D E F$ have a common inscribed circle. Line $B C$ intersects segments $D F$ and $D E$ at points $X$ and $Y$, and line $E F$ intersects segments $A B$ and $A C$ at points $Z$ and ... | Solution. Let point $I$ be the center of the common inscribed circle $\omega$ of triangles $A B C$ and $D E F$. Since $D$ is the midpoint of arc $B C$, points $A, I, D$ lie on the same line. The circle $\omega$ is inscribed in angle $\angle F D E$, so $D I$ is the angle bisector of $\angle F D E$, and point $A$ is the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,514 |
9.1 In a $3 \times 3$ table, non-repeating natural numbers not greater than 20 were placed. Then the numbers in each row and each column were multiplied. Could it happen that all 6 products are perfect squares? | Solution: Yes, it could, for example, like this:
| 5 | 10 | 8 |
| :---: | :---: | :---: |
| 15 | 20 | 12 |
| 3 | 2 | 6 |
## Criteria:
- A correct example is provided and it is shown that the products are perfect squares, -7 points;
- The example is correct, but it is not shown that the products are perfect squares, ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,516 |
9.2 Find all natural numbers that, when added to the sum of their digits, give 2021. | Solution: Since the remainders of a number and the sum of its digits when divided by 9 coincide, and the number 2021 has a remainder of 5, the sought number $n$ gives a remainder of 7. Since $n<2999$, the sum of its digits is no more than 29 and has a remainder of 7 when divided by 9. Therefore, it is equal to 7, 16, o... | 2014,1996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,517 |
9.3 Prove that for all positive $x$ the inequality
$$
\left(1+x+x^{2}\right) \cdot\left(1+x+x^{2}+x^{3}+x^{4}\right) \leqslant\left(1+x+x^{2}+x^{3}\right)^{2}
$$
holds. | Solution 1: When $x=1$, the inequality is true. For other $x$, multiply both sides by $(1-x)^{2}$ and we get $\left(1-x^{3}\right)\left(1-x^{5}\right) \leqslant\left(1-x^{4}\right)^{2}$. After expanding the brackets, this reduces to the inequality $x^{3}+x^{5} \geqslant 2 x^{4}$, which is equivalent to $x^{3}(x-1)^{2} ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,518 |
9.4 In $\triangle A B C$, the angle bisectors $A A_{1}, B B_{1}$, and $C C_{1}$ intersect at point $O$. It turns out that the areas of $\triangle O C_{1} B$ and $\triangle O B_{1} C$ are equal. Is it true that $\triangle A B C$ is isosceles? | Solution 1: Let the lengths of the sides of the triangle be $AB=c, BC=a, CA=b$. From the equality of the areas, we conclude that the areas of $\triangle BB_1C$ and $\triangle CC_1B$ are equal, having a common base, so their heights from points $B_1$ and $C_1$ are equal, hence $BC \parallel B_1C_1$. Then triangles $ABC$... | Geometry | proof | Yes | Yes | olympiads | false | 17,519 | |
9.5 The numbers $x$, $y$, and $z$ satisfy the equations
$$
x y + y z + z x = x y z, \quad x + y + z = 1
$$
What values can the sum $x^{3} + y^{3} + z^{3}$ take? | Solution 1: Let $x y z=p$. Then, from the condition $x y+y z+z x$ is also equal to $p$. Therefore, by Vieta's theorem, the numbers $x, y$, and $z$ are the roots of the polynomial $t^{3}-t^{2}+p t-p$. However, the number 1 is a root of such a polynomial, so one of the numbers is equal to 1. Then the other two numbers ar... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,520 |
9.1. It is known about three positive numbers that if you choose one of them and add to it the sum of the squares of the other two, you will get the same sum, regardless of which number you choose. Is it true that all the numbers are equal?
(L. Emelyanov) | Answer. No, incorrect.
Solution. For example, the triplet $1 / 3, 1 / 3, 2 / 3$ works.
Remark. All such triplets can be obtained by solving the corresponding system: $a+b^{2}+c^{2}=a^{2}+b+c^{2}=a^{2}+b^{2}+c$. From the first two equations, we have $a^{2}-a=b^{2}-b$; moving everything to the left side, we get $(a-b)(... | ,,1- | Algebra | proof | Yes | Yes | olympiads | false | 17,521 |
9.2. Given an isosceles triangle \(ABC (AB = AC)\). On the smaller arc \(AB\) of the circumcircle of this triangle, a point \(D\) is taken. On the extension of segment \(AD\) beyond point \(D\), a point \(E\) is chosen such that points \(A\) and \(E\) lie in the same half-plane relative to \(BC\). The circumcircle of t... | Solution. Let $\Omega$ and $\omega$ be the circumcircles of triangles $A B C$ and $B D E$. Set $\angle A C B=\angle A B C=\alpha$. Quadrilateral $B D A C$ is inscribed in $\Omega$, so $\angle A D B=180^{\circ}-\angle A C B=180^{\circ}-\alpha$. Angles $A D B$ and $E D B$ are adjacent, hence $\angle E D B=180^{\circ}-\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,522 |
9.3. Through the centers of some cells of an $8 \times 8$ chessboard, a closed non-self-intersecting broken line is drawn. Each segment of the broken line connects the centers of cells that are adjacent horizontally, vertically, or diagonally. Prove that the total area of the black parts within the polygon bounded by i... | Solution. Draw dotted vertical and horizontal lines through the centers of the cells of the board. On the resulting dotted grid, each segment of our broken line connects nodes that are adjacent vertically, horizontally, or diagonally. Therefore, the dotted lines divide the area bounded by the broken line into unit squa... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,523 |
9.4. Given positive numbers $x, y, z$. Prove the inequality
$$
\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1} \leqslant \frac{x}{y}+\frac{y}{z}+\frac{z}{x}
$$
(A. Khryabrov, B. Trushin) | Solution. Note that $\frac{a+1}{b+1}-\frac{a}{b}=\frac{b-a}{b(b+1)}$ for any positive $a$ and $b$. Therefore, after moving all terms to the left side, the required inequality takes the form
$$
\frac{y-x}{y(y+1)}+\frac{z-y}{z(z+1)}+\frac{x-z}{x(x+1)} \leqslant 0
$$
We can assume that $x$ is the largest of the three gi... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,524 |
1. (7 points) Point $O$ is the center of square $A B C D$. Find any seven pairwise distinct vectors with endpoints and starting points at points $A, B, C, D, O$, the sum of which is the zero vector. Explain your answer. | Solution. For example, the chain $\overrightarrow{O A}+\overrightarrow{A C}+\overrightarrow{C D}+\overrightarrow{D A}+\overrightarrow{A B}+\overrightarrow{B D}+\overrightarrow{D O}=\overrightarrow{0}$ is suitable. Grading criteria.
- A correct set of seven vectors is provided, or it is proven that their sum equals the... | \overrightarrow{OA}+\overrightarrow{AC}+\overrightarrow{CD}+\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{DO}=\overrightarrow{0} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,526 |
2. (7 points) Can all natural numbers from 1 to 800 be paired so that the sum of any pair of numbers is divisible by 6? | Answer. No.
Solution. If what is required in the problem is possible, then the numbers divisible by six must be paired. Since $800=133 \cdot 6+2$, there are exactly 133 numbers from 1 to 800 that are divisible by six. Contradiction: 133 numbers cannot be paired.
Remark. Among the numbers from 1 to 800, 133 numbers gi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,527 |
3. (7 points) Participating in a chess tournament, Vasya played 52 games. According to the old scoring system (1 point for a win, $\frac{1}{2}$ point for a draw, and 0 points for a loss), he scored 35 points. How many points did he score according to the new scoring system (1 point for a win, 0 points for a draw, and -... | Answer: 18 points.
## Solution.
First method. Let Vasya win $a$ times, draw $b$ times, and lose $c$ times in the tournament. Then $a+b+c=52, a+\frac{b}{2}=35$. We need to find the value of $a-c$. From the second relation, it follows that $b=70-2a$. Then $a+(70-2a)+c=52$, from which $70+c-a=52, a-c=18$.
Second method... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,528 |
5. (7 points) Two vertices, the incenter of the inscribed circle, and the intersection point of the altitudes of an acute triangle lie on the same circle. Find the angle at the third vertex. | Answer: $60^{\circ}$.
Solution. Consider triangle $ABC$, in which altitudes $AA_1$ and $BB_1$ are drawn. Let point $H$ be the orthocenter, and point $I$ be the incenter. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,530 |
6. (7 points) Petya showed Vasya 37 identical-looking cards laid out in a row. He said that on the hidden sides of the cards, all numbers from 1 to 37 are written (each exactly once) such that the number on any card starting from the second is a divisor of the sum of the numbers written on all preceding cards. Then Pet... | Answer: 2.
Solution. The sum of all numbers except the last one is divisible by the last number, which means the sum of all numbers is also divisible by the last number. The sum of all numbers from 1 to 37 is $19 \cdot 37$. Therefore, the last number is 1, 19, or 37. Since 1 and 37 are in the first and second position... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,531 |
8.1 There is a set of weights, where the heaviest weight is 5 times the average weight of all weights. How many weights can be in the set? Provide all possible answers and prove that there are no others. | Solution. Let the average weight of $n$ weights be $x$. Then the heaviest weight is $5x$, and the weight of the remaining $n-1$ weights is $nx - 5x$. This number must be positive, so $n > 5$. For any such $n$, the situation is possible, for example, if all the other weights are equal, each weighing $\frac{x(n-5)}{n-1}$... | Anygreaterthan5 | Other | math-word-problem | Yes | Yes | olympiads | false | 17,532 |
8.2 Inside an equilateral triangle $ABC$, a point $P$ is marked, and on the sides $AB$, $BC$, and $CA$ - points $K$, $L$, and $M$ respectively, such that $PK \parallel BC$, $PL \parallel AC$, and $PM \parallel BA$. Prove that the sum of the segments $PK$, $PL$, and $PM$ is equal to the side of the triangle. | Solution. Extend the segment $L P$ beyond point $P$ to intersect side $A B$ at point $N$ (see the figure). Then 1) Quadrilateral $A N P M$ is a parallelogram, so $A N = P M$. 2) Triangle $N P K$ is equilateral because its sides are parallel to the sides of the original triangle, so $N K = P K$. 3) Quadrilateral $L P K ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,533 |
8.3 The number $\sqrt{1+2019^{2}+\frac{2019^{2}}{2020^{2}}}+\frac{2019}{2020}$ is an integer. Find it. | Solution. Let $a=2020$. Then the desired number is $\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+$ $\frac{a-1}{a}$. We will perform equivalent transformations:
$$
\begin{gathered}
\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+\frac{a-1}{a}=\frac{\sqrt{a^{2}(a-1)^{2}+(a-1)^{2}+a^{2}}}{a}+\frac{a-1}{a}= \\
=\frac{\sqrt{a^{4}... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,534 |
8.4 Workers were laying a square floor of size $n \times n$ with tiles of size $2 \times 2$ and $3 \times 1$. They managed to completely cover the floor, using the same number of each type of tile. For which $n$ is this possible? Justify your answer. | Solution. Let $a$ tiles of each type be used. Then, from the equality of areas, we have $4a + 3a = n^2$. Hence, the number $n^2$, and therefore the number $n$, must be divisible by 7. We will show that any $n$ divisible by 7 will work. Indeed, in this case, the square floor can be divided into non-overlapping $7 \times... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,535 |
8.5 On an island, there live 25 people: knights, liars, and tricksters. Knights always tell the truth, liars always lie, and tricksters answer the questions posed to them in turn, alternating between truth and lies. All the islanders were asked three questions: "Are you a knight?", "Are you a trickster?", "Are you a li... | Solution. Each knight will answer "yes" to the first question and "no" to the other two. Each liar will answer "yes" to the first two questions and "no" to the last one. The tricksters can be divided into those who answered the first question truthfully (tricksters of the first type) and those who answered the first qu... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,536 |
8.6 Each student in the eighth grade is friends with exactly two students in the seventh grade, and each student in the seventh grade is friends with exactly three students in the eighth grade. There are no more than 29 students in the eighth grade, and no fewer than 17 in the seventh grade. How many students are there... | Solution. Let there be $s$ students in the 8th grade and $-t$ in the 7th grade. Then the number of pairs of friends from different grades is $2s$ and it is also equal to $3t$. From the equality $2s = 3t$, we see that $s$ is a multiple of 3, and $t$ is a multiple of 2. Therefore, $2s = 3t$ is a multiple of 6. Additional... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,537 |
5.1. Insert parentheses and operation signs in the record 22222 so that the result is 24.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Solution. One of the possible options: $(2+2+2) \times(2+2)$.
Comment. Any correct option is scored 7 points. | (2+2+2)\times(2+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,538 |
5.2. Anya lies on Tuesdays, Wednesdays, and Thursdays and tells the truth on all other days of the week. Vanya lies on Thursdays, Fridays, and Saturdays and tells the truth on all other days of the week. When asked: "What day of the week is it today?" Anya answered: "Friday," and Vanya - "Tuesday." On which days of the... | Answer: Tuesday, Thursday, or Friday. Both cannot be telling the truth as they name different days. If Anya is telling the truth, then it is Friday; if Vanya is telling the truth, then it is Tuesday. Another possibility is that both are lying, in which case the day is Thursday.
Comment. Only the answer without explana... | Tuesday,Thursday,orFriday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,539 |
5.4. Ivan Ivanovich's age is 48 years 48 months 48 weeks 48 days 48 hours. How many full years old is Ivan Ivanovich? Don't forget to explain your answer. | Answer: 53 years.
48 months is 4 years, 48 weeks is 336 days, 48 days and 48 hours is 50 days, in total 53 years and 21 or 20 days, hence the answer.
Comment. Correct answer only - 3 points; answer with explanations or calculations leading to the answer - 7 points. | 53 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,541 |
# Problem №1
Misha suggested that Yulia move a chip from cell $A$ to cell $B$ along the shaded cells. In one step, you can move the chip to an adjacent cell by side or corner. To make it more interesting, Misha put 30 candies in the prize fund, but said he would take 2 candies for each horizontal or vertical move and ... | Answer: 14.
Solution. From $A$ to $B$, one can travel via the top or the bottom. If traveling via the top, the first 2 moves are diagonal (a diagonal move is more advantageous than 2 horizontal moves), and the next 5 moves are horizontal. Misha will collect $2 \cdot 3 + 5 \cdot 2 = 16$ candies, and Yulia will win 14. ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,542 |
# Problem №2
Find all two-digit numbers, the sum of the digits of which does not change when the number is multiplied by $2,3,4,5,6,7,8$ and 9. | Answer: $18,45,90$ and 99.
Solution: According to the condition, the sum of the digits of the number $a$ and the number $9 a$ is the same. Therefore, according to the divisibility rule for 9, the number $a$ is divisible by 9. Two-digit numbers divisible by 9 are as follows: $18,27,36,45,54,63,72,81,90$ and 99. Among t... | 18,45,90,99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,543 |
# Problem №3
A New Year's garland hanging along the school corridor consists of red and blue bulbs. Next to each red bulb, there is definitely a blue one. What is the maximum number of red bulbs that can be in this garland if there are 50 bulbs in total? | Answer: 33 bulbs
## Solution
Let's calculate the minimum number of blue bulbs that can be in the garland. We can assume that the first bulb is red. Since there must be a blue bulb next to each red bulb, three red bulbs cannot go in a row. Therefore, among any three consecutive bulbs, at least one bulb must be blue. T... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,544 |
# Problem №4
On a grid paper, a square with a side of 5 cells is drawn. It needs to be divided into 5 parts of equal area by drawing segments inside the square only along the grid lines. Can it be such that the total length of the drawn segments does not exceed 16 cells? | Answer: yes, it can
(the total length of the segments drawn is... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,545 |
# Problem №5
In a train, there are 18 identical cars. In some of the cars, exactly half of the seats are free, in some others, exactly one third of the seats are free, and in the rest, all seats are occupied. At the same time, in the entire train, exactly one ninth of all seats are free. In how many cars are all seats... | Answer: in 13 carriages.
Solution. Let's take the number of passengers in each carriage as a unit. We can reason in different ways.
First method. Since exactly one ninth of all seats in the train are free, this is equivalent to two carriages being completely free. The number 2 can be uniquely decomposed into the sum ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,546 |
1. Variant 1.
Write the smallest number with a digit sum of 62, in the notation of which at least three different digits are used. | Answer: 17999999.
Solution: $62=9 \cdot 6+8$. That is, the number is at least a seven-digit number. But if it is a seven-digit number, then in its decimal representation there are exactly 6 nines and 1 eight. According to the problem, the number must contain at least three different digits, so it is at least an eight-... | 17999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,547 |
2. Variant 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to A... | Answer: 23
Solution: In the period from April 10 to April 30, there are exactly 21 days. Dividing this period into three weeks: from April 10 to April 16, from April 17 to April 23, and from April 24 to April 30, we get exactly one weekend (Monday) in each of the three weeks. Therefore, in the second statement, Kolya ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,548 |
# 3. Option 1.
The Ivanov family consists of three people: dad, mom, and daughter. Today, on the daughter's birthday, the mother calculated the sum of the ages of all family members and got 74 years. It is known that 10 years ago, the total age of the Ivanov family members was 47 years. How old is the mother now, if s... | Answer: 33.
Solution: If the daughter had been born no less than 10 years ago, then 10 years ago the total age would have been $74-30=44$ years. But the total age is 3 years less, which means the daughter was born 7 years ago. The mother is now $26+7=33$ years old. | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,549 |
4. Variant 1.
The figure shows a rectangle composed of twelve squares. The perimeter of this rectangle is 102 cm. What is its area? Express your answer in square centimeters.
 | Answer: 594.
Solution. We will call the squares large (one such), medium (three such), and small (eight such). Let's denote the side of the medium square as $4a$. Then the side of the large square is $12a$, and the side of the small squares is $12a: 4=3a$. Therefore, the sides of the rectangle are $12a$ and $4a+12a+3a... | 594 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,550 |
# 5. Variant 1.
The organizers of a ping-pong tournament have only one table. They call up two participants who have not yet played against each other. If, after the game, the losing participant has suffered their second defeat, they are eliminated from the tournament (there are no draws in tennis). After 29 games hav... | Answer: 16.
Solution: Each player is eliminated after exactly two losses. In the situation where two "finalists" remain, the total number of losses is 29. If $n$ people have been eliminated from the tournament, they have collectively suffered $2n$ losses, while the "finalists" could have $0 (0+0)$, $1 (0+1)$, or $2 (1... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,551 |
# 6. Variant 1.
A diagonal of a 20-gon divides it into a 14-gon and an 8-gon (see figure). How many of the remaining diagonals of the 20-gon intersect the highlighted diagonal? The vertex of the 14-gon is not considered an intersection.
, but it can consist of 2 and 3, and it equals 32. Arranging the remaining digits in descending ... | 8654232 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,553 |
# 8. Variant 1.
On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, 2023 natives, among whom $N$ are liars, stood in a circle and each said: "Both of my neighbors are liars." How many different values can $N$ take? | Answer: 337.
Solution: Both neighbors of a knight must be liars, and the neighbors of a liar are either two knights or a knight and a liar. Therefore, three liars cannot stand in a row (since in this case, the middle liar would tell the truth). We can divide the entire circle into groups of consecutive liars/knights. ... | 337 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,554 |
8.1. The students of a school went on a trip in six buses. The number of students in the buses was not necessarily equal, but on average, there were 28 students per bus. When the first bus arrived at the destination, the average number of students in the buses that continued moving became 26. How many students were in ... | Answer: 38.
Solution: The initial total number of schoolchildren was $28 \cdot 6=168$. After the first bus finished its trip, there were $26 \cdot 5=130$ schoolchildren left. Therefore, there were $168-130=38$ schoolchildren in the first bus.
Comment: A correct answer without justification - 0 points. | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,555 |
8.2. On weekdays (from Monday to Friday), Petya worked out in the gym five times. It is known that in total he spent 135 minutes in the gym, and the time spent in the gym on any two different days differed by at least 7 minutes. What is the maximum duration that the shortest workout could have been? | Answer: 13 minutes.
Solution. Let the minimum training time be $x$ minutes, then the second (in terms of duration) is no less than $x+7$, the third is no less than $x+14$, the fourth is no less than $x+21$, and the fifth is no less than $x+28$. Therefore, the total duration of the trainings is no less than $5 x+70$ mi... | 13 | Other | math-word-problem | Yes | Yes | olympiads | false | 17,556 |
8.3. Can a square grid $25 \times 25$ be cut into rectangular grid pieces such that the perimeter of each of them is 18? | Answer: No.
Solution: Suppose such a cutting is possible. Since the perimeter of the rectangle is 18, the sum of its length and width is 9. This means that one of these dimensions is even, and the other is odd. Therefore, the area of each rectangle in the cutting will be an even number. Thus, the sum of the areas of t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,557 |
8.4. Given a triangle $A B C$. Outside the triangle $A B C$, points $D, E, F$ are chosen such that $A D = D B, B E = E C, C F = F A$. Prove that the lines containing the bisectors of angles $A D B, B E C$, and $C F A$ intersect at one point. | The first solution. From the condition, it follows that triangle $A D B$ is isosceles, so the bisector $D M$ of angle $A D B$ is also the median and altitude of this triangle. Similarly, the bisector $E N$ of angle $B E C$ is the median and altitude of triangle $B E C$. Consider the point $O$ of intersection of the lin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,558 |
8.5. One hundred and one numbers are written in a circle. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 101 written numbers? | Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is a positive one. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there will be at least two positive numbers. Thus, the total number of positive numbers is at least $1+2 \cdot 20=41$. Such a situ... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,559 |
7.1. What is the maximum number of L-shaped pieces
| |
| :--- |
consisting of three $1 x 1$ squares, that can be placed in a 5x7 rectangle? (The L-shaped pieces can be rotated and flipped, but they cannot overlap). | Solution: The area of the corner is 3, and the area of the rectangle is 35, so 12 corners cannot fit into the rectangle. The image below shows one way to place 11 corners in the rectangle.
A... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,560 |
7.2. Angles $A O B, B O C$ and $C O D$ are equal to each other, while angle $A O D$ is three times smaller than each of them. All rays $\mathrm{OA}, \mathrm{OB}, \mathrm{OC}$, OD are distinct. Find the measure of angle $A O D$ (list all possible options). | # Solution.
Angles $A O B, B O C$ and $C O D$ follow each other in the same direction (since no rays coincide). Their sum can be less than $360^{\circ}$ (see Fig. 1) and more than $360^{\circ}$ (see Fig. 2).
Let the measure of angle AOD be denoted by x. Then each of the angles AOB, BOC, and COD is equal to 3x. In the... | 36,45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,561 |
7.3. In the notation of a natural number, there are 300 units, and the other digits are zeros. Can this number be a square of an integer?
# | # Solution.
If the natural number $n$ were the square of an integer $a$, then from the condition it would follow that $n$ is divisible by 3, since the sum of the digits of this number is 300 and is divisible by 3. But then $a^2$ is also divisible by 3, meaning that $a^2 = n$ is divisible by 9, which cannot be the case... | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,562 |
7.4. During the class chess championship, two participants, who played an equal number of games, fell ill and dropped out of the tournament, while the remaining participants finished the tournament. Did the dropped-out participants play against each other if a total of 23 games were played? (The tournament was played i... | # Solution:
A tournament with 6 participants consists of 15 matches (6 participants $\cdot 5$ matches: 2), while a tournament with 7 participants consists of 21 matches, and with 8 participants, it consists of 28 matches. Therefore, either there were 6 participants in the tournament, in addition to those who dropped o... | theydidnotplay | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,563 |
7.5. There are 11 empty boxes. In one move, you can place one coin in any 10 of them. Two players take turns. The winner is the one who, after their move, first has 21 coins in one of the boxes. Who wins with correct play? | # Solution.
Number the boxes: $1, \ldots$, 11 and denote the move by the number of the box where we did not put a coin. We can assume that the first player started the game with move 1. To win, the second player needs to, regardless of the first player's moves, make moves $2, \ldots, 11$. These ten moves, together wit... | Theplayerwins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,564 |
2. Apples of two varieties are placed in five baskets such that each basket contains apples of only one variety. It is known that the first basket contains 20 apples, the second - 30, the third - 40, the fourth - 60, and the fifth - 90. After the contents of one of the baskets were completely sold, the number of apples... | Answer: 60 or 90.
Solution: Of course, we can try to remove each basket and see if the remaining ones can be divided into two groups that meet the condition. We will offer a solution that slightly reduces the number of trials.
If the number of apples of the first type has become twice the number of apples of the seco... | 60or90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,566 |
3. On the board, three two-digit numbers are written, one of which starts with 5, the second with 6, and the third with 7. The teacher asked three students to each choose any two of these numbers and add them. The first student got 147, and the answers of the second and third students are different three-digit numbers ... | Answer: Only 78.
Solution: Let the number starting with 7 be denoted as $a$, the number starting with 6 as $b$, and the number starting with 5 as $c$. The sum $a+b \geqslant 70+60=130$, so it must be equal to 147. The maximum sum of numbers starting with 7 and 6 is $69+79=148$. The number 147 is only 1 less than 148, ... | 78 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,567 |
4. In a deck of 52 cards, each person makes one cut. A cut consists of taking the top $N$ cards and placing them at the bottom of the deck, without changing their order.
- First, Andrey cut 28 cards,
- then Boris cut 31 cards,
- then Vanya cut 2 cards,
- then Gena cut several cards,
- then Dima cut 21 cards.
The last... | Answer: 22.
Solution: Removing $N$ cards will result in the same outcome as moving $N$ cards one by one from the top to the bottom. We will consider that each of the boys moved one card several times.
After the last move, the order of the cards returned to the initial state, meaning the total number of card moves was... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,568 |
5. Two equilateral triangles $C E F$ and $D I H$ are positioned as shown in the diagram. The diagram indicates the measures of some angles. Find the measure of angle $x$. Provide your answer in degrees.
- $\angle C E M=60^{\circ}$: this is the angle of the equilateral triangle $C E F$;
- $\angle ... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,569 |
6. In triangle $A B C$, the lengths of the sides are known: $A B=4, B C=5, C A=6$. Point $M$ is the midpoint of segment $B C$, and point $H$ is the foot of the perpendicular dropped from $B$ to the angle bisector of angle $A$. Find the length of segment $H M$. If necessary, round your answer to the hundredths.
# | # Answer. 1.
Solution. Let $D$ be the intersection point of line $B H$ with line $A C$. Triangle $A B D$ is isosceles because in it the bisector and the altitude from vertex $A$ coincide. Therefore, $H$ is the midpoint of segment $B D$. Then $H M$ is the midline of triangle $B C D$. Note that $C D = A C - A D = A C - ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,570 |
7. The numbers $a, b$, and $c$ were written on the board. They were erased, and in their place, the numbers $a-1, b+1, c^{2}$ were written. It turned out that the numbers on the board were the same as those initially written (possibly in a different order). What values can the number $a$ take, given that the sum of the... | Answer: 1003 or 1002.5.
Solution. The sum of all numbers does not change, so $a+b+c=(a-1)+(b+1)+c^{2}$. From this, $c^{2}=c$, hence $c=1$ or $c=0$. In both cases, the number $c$ equals the number $c^{2}$, then the number $a$ will be equal to the number $b+1$ (and the number $b$ will be equal to the number $a-1$).
If ... | 1003or1002.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,571 |
8. The school stage of the Magic and Wizardry Olympiad consists of 5 spells. Out of 100 young wizards who participated in the competition,
- 95 correctly performed the 1st spell
- 75 correctly performed the 2nd spell
- 97 correctly performed the 3rd spell
- 95 correctly performed the 4th spell
- 96 correctly performed... | # Answer: 8.
Solution. The number of students who correctly performed all spells is no more than 75, since only 75 students correctly performed the second spell. The number of students who made mistakes in the 1st, 3rd, 4th, or 5th spells is no more than $(100-95)+(100-97)+(100-95)+(100-96)=$ $5+3+5+4=17$. If a studen... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,572 |
7. The numbers $a, b$, and $c$ were written on the board. They were erased, and in their place, the numbers $a-2, b+2, c^{2}$ were written. It turned out that the numbers on the board were the same as those initially written (possibly in a different order). What values can the number $a$ take, given that the sum of the... | Answer: 1004 or 1003.5. | 1004or1003.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,573 |
7. The numbers $a, b$, and $c$ were written on the board. They were erased, and in their place, the numbers $a-1, b+1, c^{2}$ were written. It turned out that the numbers on the board were the same as those initially written (possibly in a different order). What values can the number $a$ take, given that the sum of the... | Answer: 1004 or 1004.5. | 1004or1004.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,574 |
7. The numbers $a, b$, and $c$ were written on the board. They were erased, and in their place, the numbers $a-2, b+2, c^{2}$ were written. It turned out that the numbers on the board were the same as those initially written (possibly in a different order). What values can the number $a$ take, given that the sum of the... | Answer: 1003 or 1003.5. | 1003or1003.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,575 |
1. Multiplying the first and fourth, and the second and third factors, we get the equation: $\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)=360$.
By making the substitution $y=x^{2}+5 x+4$, we get $y^{2}+2 y-360=0$, from which $y_{1}=-20$, $y_{2}=18$. Therefore, we have the equations:
$x^{2}+5 x+24=0, x^{2}+5 x-14=... | Answer: $x_{1}=-7, x_{2}=2$ | x_{1}=-7,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,576 |
1. Ivan and Petr are running in different directions on circular tracks with a common center, and initially, they are at the minimum distance from each other. Ivan completes one full circle every 20 seconds, while Petr completes one full circle every 28 seconds. After what least amount of time will they be at the maxim... | Answer: $35 / 6$ seconds.
Solution. Ivan and Petr will be at the minimum distance from each other at the starting points after the LCM $(20,28)=140$ seconds. During this time, Ivan will complete 7 laps, and Petr will complete 5 laps relative to the starting point. Consider this movement in a reference frame where Petr... | \frac{35}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,577 |
2. Rational numbers a, b, and c are such that $(a+b+c)(a+b-c)=4 c^{2}$. Prove that $\mathrm{a}+\mathrm{b}=0$.
---
The translation maintains the original text's line breaks and formatting. | Solution. The initial equality is equivalent to the following $(a+b)^{2}-c^{2}=4 c^{2}$, or $(a+b)^{2}=5 c^{2}$. If $c \neq 0$, we get $((a+b) / c)^{2}=5 .|(a+b) / c|={ }^{-}$. On the left, we have a rational number, since the sum, quotient, and absolute value of rational numbers are rational, while on the right, we ha... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 17,578 |
3. Let $f(x)=x^{2}-p x+q$. It turned out that $f(p+q)=0$ and $f(p-q)=0$. Find $p$ and $q$. | Answer: All pairs of the form ( $m, 0$), where $m$ is any number, and the pair $(0,-1)$.
Solution. First case. The numbers $p+q$ and $p-q$ are equal. Then $q=0$. The roots of this quadratic trinomial are $p$ and 0. Therefore, all pairs ( $m, 0$ ), where $m$ is any number, are suitable.
Second case. The numbers $p+q$ a... | (,0)whereisany,thepair(0,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,579 |
4. Find all functions $f$, defined on the set of real numbers and taking real values, such that for any real $x$ and $y$ the equality $f(x y)=f(x) f(y)-2 x y$ holds. | Answer: The solutions are the linear functions $f(x)=2x$ and $f(x)=-x$.
Solution. Substitute 1 for $x$ and $y$. Then $f(1)=f(1)^{2}-2$. Therefore, $f(1)=a-$ is a root of the quadratic equation: $a^{2}-a-2=0$. The equation has two roots 2 and -1. Substitute 1 for $y$ in the equation. We get $f(x)=f(x) f(1)-2x$. If $f(1... | f(x)=2xf(x)=-x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,580 |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find $AB$, if $CA=9, CB=4$. | Answer: 6.
Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,581 |
1. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part. | Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,582 |
2. On the coordinate plane ( $x, y$ ), plot the set of all points for which $y^{2}-y=x^{2}-x$.
# | # Solution.
Solution. $y^{2}-y=x^{2}-x \Leftrightarrow y^{2}-x^{2}=y-x \Leftrightarrow (y-x)(y+x)=y-x \Leftrightarrow y=x$ or $y+x=1$ | xory+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,583 |
4. Does there exist a natural number $n$ such that the number $n^{2012}-1$ is some power of two | Answer. No, it does not exist.
Solution. Transform: $n^{2012}-1=\left(n^{1006}\right)^{2}-1=\left(n^{1006}-1\right)\left(n^{1006}+1\right)$. Suppose that this number is a power of two, then each of the two resulting factors is also a power of two, and these factors differ by 2. This is only possible in one case, if $n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,584 |
5. Solve the system of equations:
$$
\left\{\begin{array}{l}
x y=1 \\
x+y+\cos ^{2} z=2
\end{array}\right.
$$
Answer: $x=y=1 ; z=\frac{\pi}{2}+\pi n$, where $n \in Z$. | Solution. From the first equation, $x, y$ are both positive or both negative. But from the second equation $x+y \geq 1$, so they are both positive. Then, applying the inequality of means and using the first equation of the system, we get: $x+y \geq 2 \sqrt{x y}=2$. Since $\forall z \in R \cos ^{2} z \geq 0$, it follows... | 1;\frac{\pi}{2}+\pin, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,585 |
6. Can a solid wall in the shape of a parallelepiped with dimensions $27 \times 16 \times 15$ be built a) from bricks of size $3 \times 5 \times 7$; b) from bricks of size $2 \times 5 \times 6$, if bricks cannot be broken but can be rotated? | Answer: a) no; b) no.
Solution a) Note that the volume of one brick $3 * 5 * 7$ is divisible by 7. From such bricks, only a wall with a volume divisible by 7 can be built. But in our case, the volume of the wall is not divisible by 7.
b) Suppose it is possible to build. Consider the face of the wall of size $27 \time... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,586 |
9.1. On the diagonal $B D$ of parallelogram $A B C D$, a point $K$ is taken. Line $A K$ intersects lines $C D$ and $B C$ at points $L$ and $M$ respectively. Prove that $A K^{2}=K L \cdot K M$.
... | Solution: Without loss of generality, we assume that point $L$ lies on side $A M$ (and not on its extension) - see the figure. The similarity of triangles $B K A$ and $D K M$ gives the equality $A K / B K = K M / K D$. The similarity of triangles $B K L$ and $D K A$ gives the equality $A K / D K = K L / K B$. Multiplyi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,587 |
9.2. Solve the system of equations
$$
\left\{\begin{aligned}
10 x^{2}+5 y^{2}-2 x y-38 x-6 y+41 & =0 \\
3 x^{2}-2 y^{2}+5 x y-17 x-6 y+20 & =0
\end{aligned}\right.
$$ | Solution: We will eliminate the product $xy$. For example, multiply the first equation by 5, the second by 2, and add the left and right parts of the obtained equations. We get $56x^2 + 21y^2 - 224x - 42y + 245 = 0$. Divide the equation by 7, and then complete the squares for $x$ and $y$. We have $8(x-2)^2 + 3(y-1)^2 =... | 2,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,588 |
9.3. Let for some integers $a, b, c$ the following equality holds: $|a+b+c|+2=|a|+|b|+|c|$. Prove that in this case at least one of the numbers $a^{2}, b^{2}, c^{2}$ is equal to 1. | Solution: If all numbers have the same sign, then $|a+b+c|=|a|+|b|+|c|$, and the equality is not satisfied. Without loss of generality, let $a$ and $b$ have the same sign, and $c$ have the opposite sign. Then
$$
|a+b+c|=(|a|+|b|)-|c|=|a|+|b|-|c|
$$
or
$$
|a+b+c|=-(|a|+|b|)+|c|=-|a|-|b|+|c|
$$
In the first case, tak... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,589 |
9.4. The distance between the midpoints of sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ is equal to the distance between the midpoints of its diagonals. Find the angle formed by the lines $AD$ and $BC$ at their intersection. Justify your answer.
 $E M=A D / 2=F N ; 2) F M=B C / 2=E N$ (points $M, N, E, F-$ are the midpoints of segments $A B, C D, D B, A C$, respectively). Therefore, quadrilateral $E M F N$ is a parallelogram. According to the condition, its diagonals ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,590 |
9.5. The older brother took identical uncolored cubes from Misha and used them to build a large cube. After that, he completely painted some (not all) faces of the large cube red. When the paint dried, Misha disassembled the large cube and found that exactly 343 small cubes had no red faces. How many faces of the large... | Solution: We will call a small cube that has a red face painted. The size of the large cube is greater than 7 (since only the unpainted cubes amount to $343=7^{3}$, and there are also painted ones), but less than 9 (since all "internal" cubes are unpainted - no more than $7^{3}$). Therefore, it is equal to 8. Out of $8... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,591 |
9.6. In the room, there are 30 people; each is either a knight (who always tells the truth) or a liar (who never tells the truth). Each of them made the statement: "In this room, there are as many liars as there are people with the same eye color as me." How many liars are in the room? Provide all possible answers and ... | Solution: There is at least one liar in the room, since for any knight there is at least one person with the same eye color - the knight himself. Let there be exactly $x$ liars in the room ($x$ is a natural number). Note that if two people have the same eye color, then they either both tell the truth or both lie. There... | 2,3,5,6,10,15,30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,592 |
1. The last digit in the notation of a natural number is 2016 times smaller than the number itself. Find all such numbers. Answer: $4032,8064,12096,16128$. | Solution. Let $x$ be the last digit of the number. We can reason in different ways.
The first method. The number 2016x should end with the digit $x$. Therefore, $x$ is an even digit, and $x \neq 0$. By checking, we find that the values of $x$, equal to 2, 4, 6, and 8, satisfy the condition.
We can also perform a comp... | 4032,8064,12096,16128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,593 |
3. The points of intersection of the graphs of four functions given by the formulas $y=k x+b, y=k x-b, y=m x+b$ and $y=$ $m x-b$ are the vertices of a quadrilateral. Find the coordinates of the point of intersection of its diagonals. | Answer: $(0 ; 0)$.
Solution. The graphs of the given linear functions are two pairs of parallel lines, since the slopes of the first and second lines and the slopes of the third and fourth lines are equal. Therefore, the points of intersection of the graphs are the vertices of a parallelogram. Two opposite vertices of... | (0;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,595 |
4. In the class, there are 30 students: excellent students, average students, and poor students. Excellent students always answer questions correctly, poor students always make mistakes, and average students answer the questions given to them strictly in turn, alternating between correct and incorrect answers. All stud... | Answer: 20 C-students
Solution. Let $a$ be the number of excellent students, $b$ be the number of poor students, $c$ be the number of C-students who answered the first question incorrectly, answered the second question correctly, and answered the third question incorrectly (we will call these C-students of the first t... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,596 |
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