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3. On the coordinate plane, lines of the form $y=a x+b$ are drawn, where $a$ and $b$ are natural numbers from 1 to 7. Among all the intersection points of these lines, select the point with the greatest sum of coordinates. Write the answer in the form $(x ; y)$. | Answer: $(6 ; 43)\|(6,43)\| 6 ; 43 \| 6,43$
## Examples of answer notation:
$(-4 ; 5)$
# | (6,43) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,469 |
3. Given a regular 11-gon. In how many ways can 3 diagonals be drawn in it so that every two of them intersect (inside the 11-gon)? | Answer: 462
## Examples of answer recording:
45
## Problem 6 (3 points). | 462 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,471 |
1. (2 points) Anya and Kolya were collecting apples. It turned out that Anya collected as many apples as the percentage of the total apples collected by Kolya, and vice versa, Kolya collected as many apples as the percentage of the total apples collected by Anya. How many apples did Anya collect and how many did Kolya ... | Answer: 50 apples each.
Solution: If we add the two given conditions, it turns out that the total number of apples equals the total percentage. Therefore, Anya and Kolya collected a total of 100 apples. If one of them had collected more apples, they would have also collected more percentages. On the other hand, the co... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,472 |
2. (2 points) Boy Vasya had two numbers written in his notebook. He decreased each of them by 1 and found that the product of the numbers remained the same. Find the sum of the original numbers. | Answer: 1.
## Solution:
Let the numbers be $x$ and $y$. We can set up the equation: $(x-1)(y-1)=x y$. Expanding the brackets and removing $x y$ from both sides, we get $-x-y+1=0$. Therefore, $x+y=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,473 |
3. (2 points) Among the numbers from 1 to 56000 - which are there more of - those that are divisible by 7 but not by 8, or those that are divisible by 8? | Answer: The same.
Solution: Since 56000 is divisible by 8, among the numbers of interest, every eighth number is divisible by 8.
Since 56000 is divisible by 7, among the numbers of interest, every seventh number is divisible by 7. The number of these numbers is divisible by 8, meaning exactly $\frac{7}{8}$ of them ar... | The\same | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,474 |
4. (3 points) Point $C$ is located on the segment $A E$. On one side of the line $A E$, points $B$ and $D$ are marked such that $A B C$ is an equilateral triangle, and $C D E$ is an isosceles right triangle with a right angle at $D$. It turns out that triangle $B C D$ is isosceles with base $B C$. Find the angle $A D E... | # Answer: $105^{\circ}$
Solution: The acute angles of triangle $C D E$ are each $45^{\circ}$, and the angles of triangle $A B C$ are each $60^{\circ}$. Therefore, angle $B C D$ is $75^{\circ}$. Consequently, $\angle C B D=75^{\circ}, \angle C D B=30^{\circ}$.
The altitudes of triangles $A B C$ and $B C D$, dropped to... | 105 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,475 |
5. (3 points) Petya and Vasya are playing a game. There are a total of three moves in the game. On the first move, Petya breaks a stick that is 10 cm long into two parts. Then Vasya breaks one of the resulting sticks into two parts. On the last move, Petya breaks one of the three resulting sticks into two parts. Vasya ... | Answer: Petya.
Solution: Petya's first move is to break the original stick into two 5 cm sticks. Vasya then breaks one of these sticks into two parts, with lengths $a$ and $b$. Assume $a \geqslant b$. Petya then breaks the stick of length $b$ into two arbitrary parts $c$ and $d$.
The 5 cm segment is equal to the sum ... | Petya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,476 |
6. (3 points) Given the puzzle: AB + BC + DE = FGH. Different letters represent different digits, no digit is equal to 9, and a number cannot start with 0. Find the smallest possible value of FGH. | Answer: 108.
## Solution:
The sum of all the digits at our disposal is 36. Therefore, AB $+\mathrm{B} \Gamma+\mathrm{DE}+$ ZHI is divisible by 9. But AB + BG + DE + ZHI $=2$ ZHI, so ZHI is divisible by 9.
The smallest three-digit number divisible by 9 is 108. Since, for example, $108=25+36+$ 47, this number fits our... | 108 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,477 |
8. (5 points) In the cells of a $5 \times 7$ table, the numbers 1, 2, and 3 are arranged such that in any $2 \times 2$ square, there are all three different numbers. What is the smallest value that the sum of the numbers in the entire table can take? | Answer: 55
Solution:
Example:
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 3 | 2 | 3 | 2 | 3 | 2 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 3 | 2 | 3 | 2 | 3 | 2 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Evaluation
Consider the left upper $2 \times 2$ square of the $5 \times 7$ rectangle... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,478 |
# Problem 1. (1 point)
Albert thought of a natural number and said about it: “If you divide 1 by my number and add $1 / 2$ to it, the result will be the same as if you add 2, divided by the number one greater than my guessed number, to $1 / 3$.” Little Bobby, thinking for a moment, replied: “I can't guess, as there ar... | # Problem 1. (1 point)
Albert thought of a natural number and said about it: “If you divide 1 by my number and add $1 / 2$ to it, the result will be the same as if you add 2, divided by the number one greater than my guessed number, to $1 / 3$.” Little Bobby, thinking for a moment, replied: “I can't guess, because the... | 2,3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,479 |
# Problem 5. (3 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | # Problem 5. (3 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | 1040 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,483 |
# Problem 1. (2 points)
Quadratic equations $x^{2}+p x+q$ and $x^{2}+a x+b$ have one root each. Among the numbers $p, q, a, b$ are 16, 64, and 1024. What can the fourth number be?
If there are multiple possible answers, enter the larger one in the system, and indicate all in the written solution. | Answer: $\pm 8, \pm 64,262144$
## Solution:
The discriminant of the equation $x^{2}+p x+q$ must be equal to zero, so $p^{2}=4 q$. Similarly, $a^{2}=4 b$.
Therefore, if $p$ is 16, 64, or 1024, then $q=\frac{p^{2}}{4}$, which means 64, 1024, or $2^{18}=262144$. Similarly for $a$ and $b$.
If $q$ is 16, 64, or 1024, th... | \8,\64,262144 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,489 |
# Problem 2. (3 points)
Natural numbers $a$ and $b$ are such that $2a + 3b = \operatorname{LCM}(a, b)$. What values can the number $\frac{\operatorname{LCM}(a, b)}{a}$ take? List all possible options in ascending or descending order, separated by commas. If there are no solutions, write the number 0. | Answer: 0 (no solutions)
## Solution:
Let $d=\operatorname{GCD}(a, b)$. Then $a=x d, b=y d, \operatorname{LCM}(a, b)=x y d$, where $x$ and $y$ are coprime. We get $2 x d+3 y d=x y d$. Dividing by $d$ yields $2 x+3 y=x y$.
The right side of this equation is divisible by $x$, so $2 x+3 y$ is divisible by $x$, which me... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,490 |
# Problem 4. (3 points)
Inside a pentagon, 1000 points were marked and the pentagon was divided into triangles such that each of the marked points became a vertex of at least one of them. What is the smallest number of triangles that could have resulted?
# | # Answer: 1003
## Solution:
The sum of the angles of a pentagon is $540^{\circ}$. The sum of the angles at each internal point is either $180^{\circ}$, if it lies on the side of one of the triangles and is the vertex of several others, or $360^{\circ}$, if angles of triangles are located on all sides. That is, the su... | 1003 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,491 |
# Problem 5. (3 points)
In trapezoid $ABCD$ with base $AD$, the diagonals are the angle bisectors of $\angle B$ and $\angle C=110^{\circ}$. Find the degree measure of $\angle BAC$.
# | # Solution:
$\angle B D A=\angle C B D$ as alternate interior angles, and $\angle C B D=\angle A B D$, since $B D$ is the bisector. Therefore, triangle $B D A$ is isosceles, that is, $A B=A D$. Similarly, $C D=A D$, so our trapezoid is isosceles (equilateral) and $\angle B=\angle C=110^{\circ}$.
$$
\angle B A C=180^{... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,492 |
# Problem 6. (3 points)
In an $8 \times 10$ table (8 rows, 10 columns), some cells are black, and the rest are white. In each white cell, the total number of black cells in the same row is written; nothing is written in the black cells. What is the maximum value that the sum of the numbers in the entire table can take... | Answer: 200
## Solution:
If a row contains $x$ black cells and $10-x$ white cells, then the sum of the horizontal terms in this row is $x(10-x)$. This value is maximized when $x=5$ and equals 25. This can be understood as $x(10-x)-25=-x^{2}+10 x-25=-(x-5)^{2}$. Therefore, the maximum sum of all numbers is $8 \cdot 25... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,493 |
# Problem 7. (4 points)
On a plane, 13 points in general position are marked, some of which are connected by segments. In this case, the drawn segments do not form any triangles or quadrilaterals with vertices at the marked points.
Can more than 16 segments be drawn?
Points in general position - points, no three of ... | Answer: Yes
## Solution:
Since there should be no triangles or quadrilaterals, let's construct an example using pentagons:
In the diagram, there are 17 segments. In fact, self-intersecting... | Yes | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,494 |
# Problem 8. (5 points)
In triangle $ABC$, $AB=13$ and $BC=15$. A point $D$ is taken on side $AC$ such that $AD=5$ and $CD=9$. The bisector of the angle adjacent to angle $A$ intersects line $BD$ at point $E$. Find $DE$. | Answer: $7.5 \| 15 / 2$
## Solution:
Let's calculate the area of triangle $ABC$ using Heron's formula: $S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{21 \cdot 6 \cdot 7 \cdot 8}=84$ (where $p$ is the semi-perimeter of the triangle, and $a, b, c$ are its sides). Then the length of the height $BH$, dropped to side $AC$, is $\frac{2S... | 7.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,495 |
1. (2 points) What is larger: $\frac{1}{50}+\frac{1}{51}+\ldots+\frac{1}{150}$ or $\frac{2}{150}+\frac{2}{151}+\ldots \frac{2}{250}$? | Answer: The first sum is greater.
Solution: Rewrite the second sum as $\frac{1}{75}+\frac{1}{75.5}+\ldots \frac{1}{125}$. It has the same number of terms as the first sum, and the middle term $\frac{1}{100}$ is common to both.
Compare the pairs of terms $\frac{1}{100-k}+\frac{1}{100+k}=\frac{1}{100^{2}-k^{2}}$ and $\... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,496 |
3. (2 points) A poor student wrote the following incorrect formulas for the sine and cosine of a sum: $\sin (\alpha+\beta)=$ $\sin \alpha+\sin \beta$ and $\cos (\alpha+\beta)=\cos \alpha+\cos \beta$. In his defense, he said that for some $\alpha$ and $\beta$ his formulas are still correct. Find all such pairs $(\alpha,... | Answer: $\alpha= \pm \frac{\pi}{3}+2 \pi m, m \in \mathbb{Z}$ and $\beta=\mp \frac{\pi}{3}+2 \pi l, l \in \mathbb{Z}$.
Solution:
$$
\left\{\begin{array}{l}
\sin (\alpha+\beta)=\sin \alpha+\sin \beta \\
\cos (\alpha+\beta)=\cos \alpha+\cos \beta
\end{array}\right.
$$
From the first equality, it follows that $2 \sin \... | \alpha=\\frac{\pi}{3}+2\pi,\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,498 |
4. (3 points) On the side $AC$ of triangle $ABC$, a circle with a radius of 10 cm is constructed with $AC$ as its diameter. This circle intersects sides $AB$ and $BC$ at points $X$ and $Y$ respectively. Find $AX \cdot AB + CY \cdot BC$. | Answer: 400
Solution: Let point $M$ be the midpoint of side $AC$, which is also the center of the circle. Then $AB \cdot BX = BC \cdot BY = (BM + 10)(BM - 10)$. Therefore, $BX = \frac{BM^2 - 100}{AB} = \frac{2AB^2 + 2BC^2 - 400 - 400}{4AB}$. Accordingly, $AX = AB - \frac{2AB^2 + 2BC^2 - 800}{4AB} = \frac{2AB^2 - 2BC^2... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,499 |
7. (4 points) The leading coefficient of the quadratic trinomial $f(x)$ is 1. All three coefficients in some order form a geometric progression of three elements with a common ratio $q$. Find all possible values of $q$, given that it is a rational number and the difference between the roots of $f(x)$ is $q$. | Answer: There are no solutions.
Solution: The coefficients of the trinomial have the form: $a, a q, a q^{2}$ in some order. The roots can be represented as $b$ and $b+q$.
Case 1: $a=1$, the other two coefficients: $q$ and $q^{2}$.
Subcase 1.1: $2 b+q=-q \Rightarrow b=q$. Then, $b(b+q)=q^{2} \Rightarrow q \cdot 2 q=q... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,500 |
8. (5 points) A $10 \times 10$ table is filled with zeros. In one operation, the smallest number in the table (if there are several, any one is chosen) is found, and one is added to it, as well as to all numbers in the cells adjacent to it by side or corner. What is the largest number that can appear in one of the cell... | Answer: 20.
Solution: Let's call the $n$-th phase a sequence of consecutive operations applied to numbers equal to $n$. If such operations did not occur, we will say that this phase consists of zero operations. Everything starts with the zero phase.
During one phase, no number can increase by more than 4. Indeed, if ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,501 |
1. (2 points) What is larger: $\frac{1}{25}+\frac{1}{26}+\ldots+\frac{1}{75}$ or $\frac{2}{75}+\frac{2}{76}+\ldots \frac{2}{125}$? | Answer: The first sum is greater.
Solution: Rewrite the second sum as $\frac{1}{37.5}+\frac{1}{38}+\ldots \frac{1}{62.5}$. It has the same number of terms as the first sum, and the middle term $\frac{1}{50}$ is common to both.
Compare pairs of terms $\frac{1}{50-k}+\frac{1}{50+k}=\frac{1}{50^{2}-k^{2}}$ and $\frac{1}... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 20,502 |
3. (2 points) A poor student wrote the following incorrect formulas for the sine and cosine of the difference: $\sin (\alpha-\beta)=$ $\sin \alpha-\sin \beta$ and $\cos (\alpha-\beta)=\cos \alpha-\cos \beta$. In his defense, he said that for some $\alpha$ and $\beta$ his formulas are still correct. Find all such pairs ... | Answer: $\alpha=2 \pi n, n \in \mathbb{Z}$ and $\beta=\mp \frac{\pi}{3}+2 \pi l, l \in \mathbb{Z}$.
Solution:
$$
\left\{\begin{array}{l}
\sin (\alpha-\beta)=\sin \alpha-\sin \beta \\
\cos (\alpha-\beta)=\cos \alpha-\cos \beta
\end{array}\right.
$$
From the first equality, it follows that $2 \sin \left(\frac{\alpha-\... | \alpha=2\pin,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,504 |
4. (3 points) On the side $B C$ of triangle $A B C$ as a diameter, a circle with a radius of 20 cm is constructed. This circle intersects sides $A B$ and $A C$ at points $X$ and $Y$ respectively. Find $B X \cdot A B + C Y \cdot A C$. | Answer: 1600
Solution: Let point $M$ be the midpoint of side $B C$, which is also the center of the circle. Then $A B \cdot A X = A C \cdot A Y = (A M + 20)(A M - 20)$. Therefore, $A X = \frac{A M^{2} - 400}{A B} = \frac{2 A B^{2} + 2 A C^{2} - 1600 - 1600}{4 A B}$. Accordingly, $B X = A B - \frac{2 A B^{2} + 2 A C^{2... | 1600 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,505 |
7. (4 points) The leading coefficient of the quadratic trinomial $f(x)$ is 1. All three coefficients in some order form an arithmetic progression of three elements with a common difference $d$. Find all possible values of $d$, given that it is a rational number and the roots of $f(x)$ differ from each other by a factor... | Answer: $-1,-\frac{1}{2}$.
Solution: The coefficients of the trinomial have the following form: $a, a+d, a+2 d$ in some order. The roots of the trinomial can be represented as $b$ and $b d$.
Case 1: $a=1$, the other two coefficients: $1+d$ and $1+2 d$.
Subcase 1.1: $b+b d=-1-d$, so $b=-1$ or $d=-1$. If $b=-1$, we al... | -1,-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,506 |
8. (5 points) A $7 \times 7$ table is filled with zeros. In one operation, the smallest number in the table (if there are several, any one is chosen) is found, and one is added to it, as well as to all numbers in the cells adjacent to it by side or corner. What is the largest number that can appear in one of the cells ... | Solution: Let's call the $n$-th phase a sequence of consecutive operations applied to numbers equal to $n$. If such operations did not occur, we will say that this phase consists of zero operations. Everything starts with the zero phase.
During one phase, no number can increase by more than 4. Indeed, if we increased ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,507 |
1. Find all possible integer values of the expression $\frac{2}{a}+5+\sqrt{2}$, given that $a-\sqrt{2}$ is also an integer. Write the answers in any order, separated by commas or semicolons.
Example of answer format: $1 ; 2$ | Answer: $3,7\|7,3\| 3 ; 7 \| 7: 3$ | 3,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,508 |
1. (2 points) Solve the equation $p^{3}-q^{3}=11 r$, where $p, q, r-$ are prime numbers. | Solution: If all three unknown numbers are odd or exactly one of them is odd, then the right and left sides have different parity. The case where all three numbers are equal to two also does not fit. Therefore, among our numbers, there is exactly one two. Moreover, $p>q$, so the two is either $r$ or $q$.
Let $r=2$. Th... | p=13,r=199 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,515 |
2. (2 points) Find the minimum value of the expression $x^{2}+4 x \sin y-4 \cos ^{2} y$. | Answer: -4.
Solution: Add and subtract $4 \sin ^{2} y$. We get the expression $x^{2}+4 x \sin y+4 \sin ^{2} y-4=(x+2 \sin y)^{2}-4$. It is clear that this expression has a minimum value of -4.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,516 |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{100}$ be natural numbers greater than 1 (not necessarily distinct). In a $100 \times 100$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{4}$ is written. Find the smallest possible... | Answer: -10000
Solution: $\log _{x_{k}} \frac{x_{i}}{4}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 4. Add the minuend to the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this loga... | -10000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,517 |
8. (5 points) The founder of a noble family received a plot of land. Each man in the family, upon dying, divided the land he inherited equally among his sons. If he had no sons, the land went to the state. No other members of the family gained or lost any land in any other way. In total, there were 180 people in the fa... | Solution: Consider the person with the smallest share. Let his father have $a_{1}$ sons, his grandfather $a_{2}$, and so on up to the founder of the lineage, who had $a_{n}$. Then the share of this person is $\frac{1}{a_{1} a_{2} \ldots a_{n}}$, and we know that $a_{1}+a_{2}+\ldots+a_{n}$ does not exceed 179 (since the... | \frac{1}{2\cdot3^{59}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,518 |
1. (2 points) Solve the equation $p^{3}-q^{3}=5 r$, where $p, q, r$ are prime numbers.
Answer: $p=7, q=2, r=67$. | Solution: If all three unknown numbers are odd or exactly one of them is odd, then the right and left sides have different parity. The case where all three numbers are equal to two also does not fit. Therefore, among our numbers, there is exactly one two. Moreover, $p>q$, so the two is either $r$ or $q$.
Let $r=2$. Th... | p=7,q=2,r=67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,519 |
2. (2 points) Find the minimum value of the expression $x^{2}-6 x \sin y-9 \cos ^{2} y$. | Answer: -9.
Solution: Add and subtract $9 \sin ^{2} y$. We get the expression $x^{2}-6 x \sin y+9 \sin ^{2} y-9=(x-3 \sin y)^{2}-9$. It is clear that this expression has a minimum value of -9.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,520 |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{60}$ be natural numbers greater than 1 (not necessarily distinct). In a $60 \times 60$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{8}$ is written. Find the smallest possible va... | Solution: $\log _{x_{k}} \frac{x_{i}}{8}=\log _{x_{k}} x_{i}-\log _{x_{k}} 8$. Add the minuend with the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this logarithm
equals o... | -7200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,521 |
8. (5 points) The founder of a noble family received a plot of land. Each man in the family, upon dying, divided the land he inherited equally among his sons. If he had no sons, the land went to the state. No other members of the family gained or lost any land in any other way. In total, there were 120 people in the fa... | Solution: Consider the person with the smallest share. Let his father have $a_{1}$ sons, his grandfather $a_{2}$, and so on up to the founder of the lineage, who had $a_{n}$. Then the share of this person is $\frac{1}{a_{1} a_{2} \ldots a_{n}}$, and we know that $a_{1}+a_{2}+\ldots+a_{n}$ does not exceed 119 (since the... | \frac{1}{2\cdot3^{39}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,522 |
1. (2 points) Solve the equation $11 p=q^{3}-r^{3}$, where $p, q, r-$ are prime numbers. | Answer: $q=13, r=2, p=199$.
Solution: If all three sought numbers are odd or exactly one of them is odd, then the right and left sides have different parity. The case where all three numbers are equal to two also does not fit. Therefore, among our numbers, there is exactly one two. In addition, $q>r$, so the two is ei... | q=13,r=2,p=199 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,523 |
2. (2 points) Find the minimum value of the expression $4 x^{2}+4 x \sin y-\cos ^{2} y$. | Answer: -1.
Solution: Add and subtract $\sin ^{2} y$. We get the expression $4 x^{2}+4 x \sin y+\sin ^{2} y-1=(x+2 \sin y)^{2}-1$. It is clear that this expression has a minimum value of -1.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and f... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,524 |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{200}$ be natural numbers greater than 2 (not necessarily distinct). In a $200 \times 200$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{9}$ is written. Find the smallest possible... | Answer: -40000
Solution: $\log _{x_{k}} \frac{x_{i}}{4}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 9. Add the minuend with the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this lo... | -40000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,525 |
8. (5 points) The founder of a noble family received a plot of land. Each man in the family, upon dying, divided the land he inherited equally among his sons. If he had no sons, the land went to the state. No other members of the family gained or lost any land in any other way. In total, there were 150 people in the fa... | Answer: $\frac{1}{2 \cdot 3^{49}}$
Solution: Consider the person with the smallest share. Let his father have $a_{1}$ sons, his grandfather $a_{2}$, and so on up to the founder of the lineage, who had $a_{n}$. Then the share of this person is $\frac{1}{a_{1} a_{2} \ldots a_{n}}$, and we know that $a_{1}+a_{2}+\ldots+a... | \frac{1}{2\cdot3^{49}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,526 |
1. (2 points) Solve the equation $5 p=q^{3}-r^{3}$, where $p, q, r$ are prime numbers. | Answer: $q=7, r=2, p=67$.
Solution: If all three sought numbers are odd or exactly one of them is odd, then the right and left sides have different parity. The case where all three numbers are equal to two also does not fit. Therefore, among our numbers, there is exactly one two. In addition, $q>r$, so the two is eith... | q=7,r=2,p=67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,527 |
3. (3 points) Let $x_{1}, x_{2}, \ldots, x_{100}$ be natural numbers greater than 1 (not necessarily distinct). In an $80 \times 80$ table, the numbers are arranged as follows: at the intersection of the $i$-th row and the $k$-th column, the number $\log _{x_{k}} \frac{x_{i}}{16}$ is written. Find the smallest possible... | Answer: -19200
Solution: $\log _{x_{k}} \frac{x_{i}}{16}=\log _{x_{k}} x_{i}-\log _{x_{k}}$ 16. Add the minuend with the similar minuend in the symmetric cell: $\log _{x_{k}} x_{i}+\log _{x_{i}} x_{k}=\log _{x_{k}} x_{i}+\frac{1}{\log _{x_{k}} x_{i}} \geqslant 2$, since this logarithm is positive. If $i=k$, then this ... | -19200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,529 |
8. (5 points) The founder of a noble family received a plot of land. Each man in the family, upon dying, divided the land he inherited equally among his sons. If he had no sons, the land went to the state. No other members of the family gained or lost any land in any other way. In total, there were 200 people in the fa... | Answer: $\frac{1}{4 \cdot 3^{65}}$
Solution: Consider the person with the smallest share. Let his father have $a_{1}$ sons, his grandfather $a_{2}$, and so on up to the founder of the lineage, who had $a_{n}$. Then the share of this person is $\frac{1}{a_{1} a_{2} \ldots a_{n}}$, and we know that $a_{1}+a_{2}+\ldots+a... | \frac{1}{4\cdot3^{65}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,530 |
# Problem 1. (2 points)
The sum of the first six terms of a geometric progression with a negative common ratio is exactly 702 times greater than the sum of the next three terms. Find this common ratio. | Indicate the answer in the form of a proper fraction.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,531 |
# Problem 8. (3 points)
Quadrilateral $ABCD$ is inscribed in a circle with diameter $AC=65$. It is also known that $BC=$ $39, AD=25$. Find the distance between the feet of the perpendiculars dropped from $B$ and $D$ to $AC$. | Write the answer in the form of an improper fraction, do not round.
# | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,532 |
1. (2 points) Among six different quadratic trinomials that differ by the permutation of coefficients, what is the maximum number that can have no roots? | Answer: 6
Solution:
Consider, for example, the coefficients 1000, 1001, 1002. The square of any of them is obviously less than the quadruple product of the other two, so all discriminants will be negative. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,533 |
2. (3 points) On a $10 \times 10$ grid, there are 400 chips. We will call two cells neighbors if they share a side, or if they are the two end cells of the same row or column. Thus, each cell will have exactly 4 neighbors.
In one move, it is allowed to take 4 chips from one cell and place them on 4 neighboring cells. ... | Answer: No, not for any.
Solution: Let's color the board in a checkerboard pattern. The number of chips on black cells changes by 4 each time, meaning, in particular, that its parity does not change. At the end, the number of chips on black cells must be even, so if it is odd at the beginning, nothing will work.
Ther... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,534 |
3. (3 points) Equilateral triangles $A B C$ and $A_{1} B_{1} C_{1}$ with side length 10 are inscribed in the same circle such that point $A_{1}$ lies on the arc $B C$, and point $B_{1}$ lies on the arc $A C$. Find $A A_{1}^{2}+B C_{1}^{2}+C B_{1}^{2}$. | Answer: 200
Solution:
Notice that the arcs $A B_{1}, B A_{1}$, and $C C_{1}$ are equal. Let their degree measure be $2 \alpha$. Then the lengths of the arcs $A C_{1}=B B_{1}=C A_{1}=120^{\circ}-2 \alpha$.
By the Law of Sines in triangle $A C A_{1}$, we get $\frac{A A_{1}}{\sin A C A_{1}}=$ $\frac{A C}{A A_{1} C}$, f... | 200 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,535 |
4. (3 points) Prove that in a right triangle, the area does not exceed the square of the perimeter divided by 23.
# | # Solution:
Let the sides of the triangle have lengths $x$ and $y$. Then the perimeter $P=x+y+\sqrt{x^{2}+y^{2}}$. By the inequalities of means, we get $x+y+$ $\sqrt{x^{2}+y^{2}} \geqslant 2 \sqrt{x y}+\sqrt{2} \sqrt{x y}$. Since the area $S=\frac{x y}{2}$, we obtain $P \geqslant(2+\sqrt{2}) x y$, from which $P^{2} \g... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,536 |
5. (3 points) Solve the equation $x^{2}+3 y^{2}=2^{z}$ in natural numbers.
Answer: $x=y=2^{n}, z=2 n+2$, where $n-$ is a non-negative integer.
# | # Solution:
Notice that $2^{z}$ is always even for natural numbers, so $x$ and $y$ must have the same parity.
First, consider the case when both numbers are odd. In this case, the sum of their squares gives a remainder of 4 when divided by 8. This means that $2^{z}=4$, so $x=y=1, z=2$.
If $x$ and $y$ are even, then ... | 2^{n},2n+2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,537 |
6. (4 points) Given three circles with radii 3, 4, and 5, each pair of which touch each other at points $A, B$, and $C$. Find the sum of the distances from the center of the circumscribed circle of triangle $A B C$ to its sides.
Answer: $\frac{5}{\sqrt{14}}+\frac{5}{\sqrt{21}}+\frac{5}{\sqrt{30}}$. | Solution:
Let's introduce the following notations: for the first circle, the center is point $O_{1}$, and the radius is $a$; for the second circle, the center is point $O_{2}$, and the radius is $b$; for the third circle, the center is point $O_{3}$, and the radius is $c$. Since points $A, B, C$ are the points of tang... | \frac{5}{\sqrt{14}}+\frac{5}{\sqrt{21}}+\frac{5}{\sqrt{30}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,538 |
7. (4 points) On the plane, there is a set of points, it is known that any three can be moved by a parallel translation into a square with vertices $(0,2),(2,0),(0,-2)$ and $(-2,0)$. Then all of them can be moved there at once by one parallel translation.
# | # Solution:
Let's introduce a coordinate system where the x-axis and y-axis are at a $45^{\circ}$ angle to the current axes, and the origin remains the same. The vertices of the square in this system will have coordinates $( \pm \sqrt{2} ; \pm \sqrt{2})$.
Consider the leftmost point (i.e., the point with the smallest... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,539 |
8. (4 points) For February 23, the boy Zhenya was given a chocolate bar of size $3 \times 3$, on each piece of which a unique picture is drawn, with each picture appearing only once. In each move, Zhenya can bite off one piece that has no more than three common sides with other pieces that have not yet been eaten. In h... | Solution:
If Zhenya had no restrictions, he could eat the pieces in any order, and he would have $9!=362880$ ways.
Not suitable are the ways in which the central cell is eaten before any of the four adjacent to it. Since for each of these five cells, there is an equal number of options where it is earlier than the ot... | 290304 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,540 |
1. (2 points) Among six different quadratic trinomials that differ by the permutation of coefficients, what is the maximum number that can have two distinct roots | Answer: 6
Solution: Let's take, for example, the coefficients, $-5,1,2$.
If the number -5 is the leading coefficient or the constant term, the equation obviously has two roots of different signs.
For the case when -5 is the second coefficient, let's calculate the discriminant: $5^{2}-1 \cdot 2=23>0$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,541 |
3. (3 points) Equilateral triangles $A B C$ and $A_{1} B_{1} C_{1}$ with side length 12 are inscribed in circle $S$ such that point $A$ lies on arc $B_{1} C_{1}$, and point $B$ lies on arc $A_{1} B_{1}$. Find $A A_{1}^{2}+B B_{1}^{2}+C C_{1}^{2}$. | Answer: 288
Solution:
Notice that the arcs $A B_{1}, B A_{1}$, and $C C_{1}$ are equal. Let their degree measure be $2 \alpha$. Then the lengths of the arcs $A C_{1}=B B_{1}=C A_{1}=120^{\circ}-2 \alpha$.
By the Law of Sines in triangle $A C A_{1}$, we get $\frac{A A_{1}}{\sin A C A_{1}}=$ $\frac{A C}{A A_{1} C}$, f... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,543 |
4. (3 points) Prove that in a right triangle, the area does not exceed the square of the semiperimeter divided by 5 and a half.
# | # Solution:
Let the sides of the triangle have lengths $x$ and $y$. Then the semiperimeter $p=\frac{x+y+\sqrt{x^{2}+y^{2}}}{2}$. By the inequalities of means, we get $\frac{x+y+\sqrt{x^{2}+y^{2}}}{2} \geqslant \sqrt{x y}+\frac{\sqrt{x y}}{\sqrt{2}}$. Since the area $S=\frac{x y}{2}$, we have $p \geqslant\left(1+\frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,544 |
5. (3 points) Solve the equation $x^{2}+y^{2}=2^{z}$ in natural numbers.
Answer: $x=y=2^{n}, z=2 n+1$, where $n-$ is a non-negative integer.
# | # Solution:
Notice that $2^{z}$ is always even for natural numbers, so $x$ and $y$ must have the same parity.
First, consider the case when both numbers are odd. In this case, the sum of their squares gives a remainder of 2 when divided by 4. This means that $2^{z}=2$, so $x=y=z=1$.
If $x$ and $y$ are even, then $x^... | 2^{n},2n+1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,545 |
6. (4 points) Given three circles with radii 3, 5, and 7, each pair of which touch each other at points $A, B$, and $C$. Find the sum of the distances from the center of the circumscribed circle of triangle $A B C$ to its sides.
Answer: $\frac{7}{4}+\frac{7}{3 \sqrt{6}}+\frac{7}{\sqrt{30}}$. | # Solution:
Let's introduce the following notations: for the first circle, the center is point $O_{1}$, the radius is $a$; for the second circle, the center is point $O_{2}$, the radius is $b$; for the third circle, the center is point $O_{3}$, the radius is $c$. Since points $A, B, C$ are the points of tangency of tw... | \frac{7}{4}+\frac{7}{3\sqrt{6}}+\frac{7}{\sqrt{30}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,546 |
7. (4 points) On the plane, there is a set of points, it is known that any three can be moved by a parallel translation into a rectangle with vertices $(0,0),(1,0),(0,2)$ and $(1,2)$. Then all of them can be moved there at once by one parallel translation.
# | # Solution:
Consider the leftmost point (i.e., the point with the smallest x-coordinate) and the rightmost point. Draw vertical lines through them. The distance between these lines is no more than 2, otherwise, the two outermost points cannot be moved into the specified rectangle. Then draw horizontal lines through th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,547 |
8. (4 points) For Valentine's Day, the boy Zhenya was given a chocolate bar of size $2 \times 4$, on each piece of which a unique picture is drawn. In each move, Zhenya can bite off one piece that has no more than two common sides with other pieces that have not yet been eaten. In how many ways can Zhenya eat his choco... | Answer: 6720.
## Solution:
The first cell to be eaten is necessarily a corner cell (4 ways). After this, there are 4 options:
1) The second cell to be eaten is the adjacent corner cell (1 way). A $2 \times 3$ rectangle remains. From this, the first cell to be eaten is again a corner cell (4 ways). After this, one ce... | 6720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,548 |
1. Initially, the numbers 1, 2, and 4 are written on the board. Every minute, Anton erases the numbers written on the board and writes down their pairwise sums instead. After an hour, three enormous numbers are written on the board. What are their last digits? List them in any order, separated by semicolons.
Example o... | Answer: $6 ; 7 ; 9\|6 ; 9 ; 7\| 7 ; 9 ; 6\|7 ; 6 ; 9\| 9 ; 6 ; 7 \| 9 ; 7 ; 6$ | 6;7;9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,549 |
2. Initially, the numbers 2, 3, and 4 are written on the board. Every minute, Anton erases the numbers written on the board and writes down their pairwise sums instead. After an hour, three enormous numbers are written on the board. What are their last digits? List them in any order, separated by semicolons.
Example o... | Answer: $7 ; 8 ; 9\|7 ; 9 ; 8\| 7 ; 9 ; 8\|7 ; 8 ; 9\| 9 ; 7 ; 8 \| 9 ; 7 ; 3$ | 7;8;9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,550 |
3. Initially, the numbers 2, 3, and 6 are written on the board. Every minute, Anton erases the numbers written on the board and writes down their pairwise sums instead. After an hour, three enormous numbers are written on the board. What are their last digits? List them in any order, separated by semicolons.
Examples ... | Answer: $8 ; 7 ; 1\|8 ; 1 ; 7\| 7 ; 1 ; 8\|7 ; 8 ; 1\| 1 ; 8 ; 7 \| 1 ; 7 ; 8$ | 8;7;1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,551 |
3. What is the greatest value that the sum $\sin ^{2} a+\sin ^{2}\left(a+60^{\circ}\right)+\sin ^{2}\left(a+120^{\circ}\right)+\ldots+\sin ^{2}\left(a+300^{\circ}\right)$ can take? | Answer: 3
## Examples of how to write the answer:
1,7
$1 / 7$
17
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,555 |
3. All seven-digit numbers consisting of different digits from 1 to 7 were listed in ascending order. What is the position of the number 3241765? | Answer: 1590
## Examples of how to write the answer:
1,7
# | 1590 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,556 |
3. What integer can be written in the form $\sqrt{12-\sqrt{12-\sqrt{12-\ldots}}}$ (the number of roots is infinite). | Answer: 3
## Examples of how to write the answer:
1,7
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,557 |
3. The function $f(x)$ takes each value no more than twice. Moreover, for any non-zero $x$, the equality $f(x)+f(1-x)+f\left(\frac{1}{x}\right)=6$ holds. Find such an $x$, not equal to $\frac{5}{7}$, that $f(x)=f\left(\frac{5}{7}\right)$. If such an $x$ does not exist, write "no". | Answer: $-2.5 \|-5 / 2$
## Examples of how to write the answer:
# | -\frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,558 |
3. On a unit cube, a vector was drawn along each edge in one of the two possible directions. On each face, one diagonal was chosen and a vector was drawn along it in one of the two possible directions, and the 6 drawn diagonals turned out to be non-parallel. In total, 18 vectors were obtained. What is the maximum value... | Answer: 116
## Examples of how to write the answer:
## 1,7
$1 / 7$
17
# | 116 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,559 |
1. In a $13 \times 13$ grid, numbers are arranged such that the numbers in each row and each column form an arithmetic progression in the order they are written. The grid is colored in two colors in a checkerboard pattern. The numbers on the corner white cells of the grid are $1, 2, 3,$ and $6$. Find the sum of the num... | Answer: 252
## Examples of how to write the answer:
1,7
# | 252 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,560 |
3. A circle of radius 60 touches three sides of quadrilateral $A B C D$: side $A B$ at point A, side $C D$ at point D, and side $B C$ at point $\mathrm{X}$. It turns out that $\mathrm{BX}=20, \mathrm{CX}=45$. Find the area of the quadrilateral. | Answer: 2215.2
## Examples of how to write the answer:
1.7
$1 / 7$
17
# | 2215.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,561 |
3. At a rock festival, vocalists, guitarists, and drummers met, a total of 121 people. Each vocalist gave a slap on the back of each guitarist, each guitarist gave a slap on the back of each drummer, and each drummer gave a slap on the back of each vocalist. What is the maximum number of slaps on the back that the part... | Answer: 4880.
## Examples of how to write the answer:
17
# | 4880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,562 |
3. A 101-gon is inscribed in a circle with diameter $\mathrm{XY}=6$ and has an axis of symmetry perpendicular to this diameter. Find the sum of the squares of the distances from the vertices of the 101-gon to the point $\mathrm{X}$. | Answer: 1818
## Examples of how to write the answer:
# | 1818 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,563 |
3. Around a circle, 300 (not necessarily integer) numbers from 20 to 400 inclusive were written. From each number, the logarithm to the base of the next number in the clockwise direction was taken, and then all the obtained logarithms were summed. What is the greatest value that the sum of these logarithms can take? | Answer: 375
## Examples of how to write the answer:
## Grade 9
# | 375 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,564 |
3. Four different numbers $a, b, c, d$, greater than one and not divisible by 5, are such that $\operatorname{GCD}(a, b)=\operatorname{GCD}(c, d)$ and $\operatorname{LCM}(a, b)=\operatorname{LCM}(c, d)$. What is the smallest possible value of $a+b+c+d$? | Answer: 24
## Examples of how to write the answer:
17 | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,566 |
3. A spider has woven a web that consists of the x-axis, the y-axis, as well as the following curves: $y=x$, $y=-x$, $x^{2}+y^{2}=9$, $x^{2}+y^{2}=25$, $x^{2}+y^{2}=49$, $x^{2}+y^{2}=81$. In one day, flies have been caught in all the nodes. The spider is sitting at the point $(0,0)$ and plans to eat all the flies. What... | Do not round the answer. For writing the number $\pi$, use the Russian letter п or the English $\mathrm{p}$. Examples of writing the answer:
$3 \mathrm{p}+10$
$1.5 \pi+7.5$
Answer: $12 \mathrm{p}+47$ || $47+12 \mathrm{p}$ || $12$ п +47 || $47+12$ п || $12^{*} \mathrm{p}+47$ || $47+12^{*} \mathrm{p}$ || $12^{*}{ }^{*... | 12\mathrm{p}+47 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,567 |
1. For two consecutive days, Petya went to the store and each time bought buns for exactly 100 rubles. On the second day, the buns were one ruble more expensive than on the first day. The buns always cost an integer number of rubles.
How many buns did Petya buy over the two days? If there are multiple possible answers... | Answer: $150 ; 45$ || $150 ; 45 ;|| 150,45$ || 45; 150 || 45, 150 || 45; 150 ; | 150;45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,568 |
2. Petya went to the store for two consecutive days and both times he bought buns for exactly 84 rubles. On the second day, a bun was one ruble more expensive than on the first day, and on the first day, it was more than two rubles. Buns always cost a whole number of rubles.
How many buns did Petya buy over the two da... | Answer: $26 ; 49$ || $26 ; 49 ;|| 26,49$ || 49; 26 || 49, 26 || 49; 26$ | 26;49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,569 |
3. Petya went to the store for two days in a row and both times he bought buns for exactly 56 rubles. At the same time, on the second day, a bun cost one ruble more than on the first day. Buns always cost a whole number of rubles.
How many buns did Petya buy over the two days? If there are several possible options, li... | Answer: $84 ; 15$ || $84 ; 15 ;\|84,15\| 15 ; 84\|15,84\| 15 ; 84 ;$ | 84;15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,570 |
1. On a very long road, a race was organized. 20 runners started at different times, each running at a constant speed. The race continued until each runner had overtaken all the slower ones. The speeds of the runners who started first and last were the same, and the speeds of the others were different from theirs and d... | Answer: $18 ; 171$ || $171 ; 18$ || 18,171 || 171,18 || $18 ; 171 ;|| 171 ; 18$
Examples of answer notation: $10 ; 20$ | 18;171 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,571 |
2. On a very long road, a race was organized. 22 runners started at different times, each running at a constant speed. The race continued until each runner had overtaken all the slower ones. The speeds of the runners who started first and last were the same, and the speeds of the others were different from theirs and d... | Answer: $20 ; 210$ || $210 ; 20$|20,210| 210,20|| 20 ; 210 ; | 210 ; 20$ | 20;210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,572 |
3. On a very long road, a race was organized. 21 runners started at different times, each running at a constant speed. The race continued until each runner had overtaken all the slower ones. The speeds of the runners who started first and last were the same, and the speeds of the others were different from theirs and d... | Answer: $19 ; 190$ || 190; 19 || 19, 190 || 190, 19 || 19; 190; | 190; 19 | 190;19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,573 |
1. On the table, there are sticks of lengths $1,2,3, \ldots, n$ centimeters. It is known that from them, 40 triangles can be formed, using each stick no more than once, but 41 cannot.
Find $n$ (the length of the longest stick). If there are multiple possible answers, list them in ascending order separated by a semicol... | Answer: $121 ; 122 ; 123 \mid 121,122,123$ | 121;122;123 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,574 |
2. On the table lie sticks of lengths $1,2,3, \ldots, n$ centimeters. It is known that from them, 50 triangles can be formed, using each stick no more than once, but 51 cannot.
Find $n$ (the length of the longest stick). If there are multiple possible answers, list them in ascending order separated by a semicolon. | Answer: $151,152,153 \mid 151 ; 152 ; 153$ | 151;152;153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,575 |
3. On the table lie sticks of lengths $1,2,3, \ldots, n$ centimeters. It is known that from them, 100 triangles can be formed, using each stick no more than once, but 101 cannot. Find $n$ (the length of the longest stick). If there are multiple possible answers, list them in ascending order separated by a semicolon. | Answer: $301,302,303 \mid 301 ; 302 ; 303$
# | 301;302;303 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,576 |
3. The fox builds "pyramids" from 5 cubes in the following way: each "pyramid" consists of one or several levels; on each level, the number of cubes is strictly less than on the previous one; each new level consists of one or several consecutive cubes. You can see an example of a "pyramid" made of ten cubes in the pict... | Answer: 7
## Examples of answer notation:
12
# | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,578 |
3. Through the point $(2,3)$ on the line $p: y=2 x-1$, a line $q$ perpendicular to the line $p$ was drawn. Find the area of the convex quadrilateral bounded by the lines $p, q$ and the coordinate axes.
| Answer: $4.75 \mid 19 / 4$
## Examples of answer notation:
## 8
8.5
$17 / 2$
# | \frac{19}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,579 |
2. $A B C D E$ - a pentagon inscribed in circle $S$. Lines $D E$ and $A B$ are parallel, $B C=C D$, $A D=D E=20, B C: B D=4: 5$. Find the radius of circle $S$. | Answer: 16
## Examples of how to write answers: $1 / 4$ 0.25
# | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,582 |
1. On the planet Sharp Teeth, animals reproduce in a special way, specifically: each hamster gives birth to four hamsters every year; each groundhog gives birth to two baby groundhogs every 3 months; and rabbits, which are mysterious creatures, reproduce faster the more time passes, specifically, if a person does not g... | Answer: $10 ; 77 ; 153 \mid 11,77,153$
2nd variant.
On the planet Sharp Teeth, animals reproduce in a special way, specifically: every hamster gives birth to two hamsters every 3 months; every groundhog gives birth to one small groundhog every 4 months; and rabbits, which are mysterious creatures, reproduce faster th... | 13;41;130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,583 |
3. $A B D E, B C E F, C D F A$ - cyclic quadrilaterals with the intersection points of the diagonals $K, L$ and $M$ respectively. It is known that point $K$ lies on segments $B L$ and $A M$, point $M$ - on segment $C L$. Moreover, $E L=F L=K L=5, D M=4, A K=M K=6$. Find the length of segment $M C$. If there are multipl... | Answer: 4
## Examples of how to write answers:
$1 / 4$
0.25
$4 ; 10$
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,584 |
3. Given two quadratic trinomials with the leading coefficient $\frac{\sqrt{3}}{8}$. The vertices and the intersection point of their graphs form an equilateral triangle. Find the length of its side. If there are multiple possible answers, list them in any order separated by a semicolon. | Answer: 16
## Examples of answer notations:
$1 / 4$
$-0.25 ; 10$
8th grade. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,585 |
# Problem 9. (4 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | # Problem 9. (4 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | 1040 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,593 |
Task 1. (1 point)
Find the largest three-digit number ABC that is divisible by the two-digit numbers AB and BC. (Different letters do not necessarily represent different digits) | Answer: 990
Solution:
The number 990 clearly fits. A larger answer is not possible, as all three-digit numbers greater than 990 do not divide by 99: the next number divisible by 99 is 1089.
# | 990 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,595 |
# Task 2. (2 points)
Does there exist a right triangle whose area is numerically equal to twice the hypotenuse? | Answer: Yes
Solution:
There are quite a few such triangles. Let's prove that there exists an isosceles one. We will set up an equation, denoting its side as $a$:
$$
\frac{a^{2}}{2}=2 \sqrt{2} a
$$
Obviously, this equation has a solution $a=4 \sqrt{2}$. | 4\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,596 |
# Problem 3. (2 points)
What is the maximum number of different reduced quadratic equations that can be written on the board, given that any two of them have a common root, but no four have a root common to all. | Answer: 3
Solution:
Consider three such equations. There are two cases:
1) These equations have a common root $a$. Then they can be represented in the form $(x-a)(x-b)=0, (x-a)(x-c)=0$ and $(x-a)(x-d)=0$, where $b, c$ and $d$ are their remaining roots (they are not equal to each other, otherwise the equations coinci... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,597 |
# Problem 4. (3 points)
In each cell of a $2018 \times 2018$ square, both diagonals are drawn. Does there exist a closed path consisting of these diagonals, not passing through any diagonal more than once and visiting all cells of the square (i.e., containing at least one diagonal from each cell)? | Answer: Yes, it is possible.
## Solution:
Let's start the path from the top-left corner. Cross the top-left cell from its lower left to the upper right corner and continue moving diagonally to the right until the top-right cell. Due to the even length of the square, we will end up in the lower right corner of this ce... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,598 |
# Problem 5. (3 points)
In trapezoid $ABCD$, a point $X$ is taken on the base $AD$ such that segments $XB$ and $XC$ divide the trapezoid into three triangles that are similar to each other but pairwise unequal and non-isosceles. The side $AB$ has a length of 6. Find $AX \cdot DX$. | Answer: 36
## Solution:
The triangles are scalene, meaning all their angles are different.
$\alpha=\angle B X A=\angle X B C \neq \angle B X C=\beta$. The angle $\angle C X D$ together with these angles sums to $180^{\circ}$, so it is the third angle of the triangle and cannot be equal to either $\alpha$ or $\beta$.... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,599 |
# Problem 6. (4 points)
Solve the equation $a b c d e f=a+b+c+d+e+f$ in natural numbers. | Answer: $(1,1,1,1,2,6)$ in any order
Solution:
Let the largest number be $a$. The total sum of the numbers does not exceed $6a$.
Suppose there are fewer than three ones in our set. Then the total product is at least $2 \cdot 2 \cdot 2 \cdot a = 8a > 6a$. This leads to a contradiction.
Therefore, there must be at le... | (1,1,1,1,2,6) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,600 |
# Problem 7. (4 points)
$O A B C$ is a rectangle on the Cartesian plane, with sides parallel to the coordinate axes. Point $O$ is the origin, and point $B$ has coordinates $(11; 8)$. Inside the rectangle, a point $X$ with integer coordinates is taken. What is the smallest value that the area of triangle $O B X$ can ta... | Solution:
Since the picture is symmetric relative to the center of the rectangle, it is sufficient to consider the case when point $X$ lies below the diagonal $O B$. Let $Y$ and $Z$ be the projections of $X$ onto the x-axis and segment $B C$ (here $C(11 ; 0)$).
Then $S_{O B X}=S_{O A B C}-S_{O A B}-S_{O X Y}-S_{B X Z... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,601 |
# Problem 8. (4 points)
Given a $7 \times 7$ checkerboard, where the side length of each cell is one centimeter. A chess piece called Pythagoras, standing on cell $A$, attacks cell $B$ if the distance between the centers of cells $A$ and $B$ is five centimeters. What is the maximum number of non-attacking Pythagorases... | # Solution:
Notice that with a chessboard coloring of the board, Pythagoras only attacks squares of the opposite color to the square he is on.
Example: we color the board in a checkerboard pattern and place Pythagorases on the squares of the color that is more prevalent.
Estimate: we mark the field as follows:
| $A... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,602 |
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