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# Task 1. (2 points)
Find the smallest three-digit number ABC that is divisible by the numbers AB and BC (the digit A cannot be 0, and the digit B can be; different letters do not necessarily represent different digits) | Answer: 110
Solution:
The number 110 clearly fits. A smaller answer cannot be, as all three-digit numbers less than 110 start with 10. At the same time, none of them are divisible by 10, except for 100, and 100 is not divisible by 0. | 110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,603 |
# Task 2. (2 points)
Does there exist a right triangle whose area is numerically equal to three times its 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 $a$:
$$
\frac{a^{2}}{2}=3 \sqrt{2} a
$$
Obviously, this equation has a solution $a=6 \sqrt{2}$. | 6\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,604 |
# 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 five have a root common to all. | Answer: 4
## Solution:
Consider 4 such equations. There are two cases:
1) These equations have a common root $a$. Then they can be represented in the form $(x-a)(x-b)=0,(x-a)(x-c)=0,(x-a)(x-d)=0$ and $(x-a)(x-e)=0$, where $b, c, d$ and $e$ are their remaining roots (they are not equal to each other, otherwise the eq... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,605 |
# Problem 4. (3 points)
In each cell of a $2020 \times 2020$ 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:
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 cell, so we... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 20,606 |
# Problem 5. (3 points)
In trapezoid $ABCD$, a point $X$ is taken on the base $BC$ such that segments $XA$ and $XD$ 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 5. Find $XC \cdot BX$.
Answer: 25 | Solution:
The triangles are scalene, meaning all their angles are different.
$\alpha=\angle B X A=\angle X A D \neq \angle A X D=\beta$. The angle $\angle C X D$ together with these angles sums up to $180^{\circ}$, so it is the third angle of the triangle and cannot be equal to either $\alpha$ or $\beta$.
If $\angle... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,607 |
# Problem 6. (3 points)
Solve the equation $a b c d e f g=a+b+c+d+e+f+g$ in natural numbers. | Answer: $(1,1,1,1,1,2,7)$ and $(1,1,1,1,1,3,4)$. In any order.
Solution:
Let the largest number be $a$. The total sum of the numbers does not exceed $7a$.
Suppose there are fewer than four ones in our set. Then the total product is at least $2 \cdot 2 \cdot 2 \cdot a = 8a > 7a$. This leads to a contradiction.
There... | (1,1,1,1,1,2,7)(1,1,1,1,1,3,4) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,608 |
# 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 $(9 ; 8)$. 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 with respect 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 $C(9 ; 0)$).
Then $S_{O B X}=S_{O A B C}-S_{O A B}-S_{O X Y}-S_{B ... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,609 |
# Problem 8. (4 points)
Given a checkerboard $8 \times 8$, 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, Pythagoras only attacks cells of the opposite color to the cell he is on.
Example: we color the board in a checkerboard pattern and place Pythagoruses on the cells of the color that is more prevalent.
Estimate: we mark the field as follows:
| 1 | 2 | 3 | 4 | 5 | 6... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,610 |
1. Find the value of the expression
$$
\frac{\cos 0.000001^{\circ}+\cos 0.000003^{\circ}+\cos 0.000005^{\circ}+\ldots+\cos 29.999999^{\circ}}{\sin 0.000001^{\circ}+\sin 0.000003^{\circ}+\sin 0.000005^{\circ}+\ldots+\sin 29.999999^{\circ}}
$$ | Do not round the answer. For writing the square root, use the sign «?».
Leave the source text's line breaks and format, and output the translation result directly. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,612 |
2. Find the value of the expression
$$
\frac{\cos 0.000001^{\circ}+\cos 0.000003^{\circ}+\cos 0.000005^{\circ}+\ldots+\cos 44.999999^{\circ}}{\sin 0.000001^{\circ}+\sin 0.000003^{\circ}+\sin 0.000005^{\circ}+\ldots+\sin 44.999999^{\circ}}
$$ | Do not round the answer. For writing the square root, use the sign «?».
Leave the source text's line breaks and format, and output the translation result directly. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,613 |
3. Find the value of the expression
$$
\frac{\sin 30.000001^{\circ}+\sin 30.000003^{\circ}+\sin 30.000005^{\circ}+\ldots+\sin 59.999999^{\circ}}{\sin 0.000001^{\circ}+\sin 0.000003^{\circ}+\sin 0.000005^{\circ}+\ldots+\sin 29.999999^{\circ}}
$$ | Do not round the answer. To write a square root, use the symbol «?».
```
#examples of writing answers:
17
-1.7
1/7
1+?7
``` | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,614 |
1. Given a parallelogram $A B C D$ with angle $A=30^{\circ}$. The circumcircles of triangles $A B D$ and $C B D$ intersect the diagonal $A C$ at two points, which divide it into three equal parts. Find the ratio of the longer side of the parallelogram to the shorter. | Do not round the answer. For writing the square root, use the sign «?».
Leave the source text's line breaks and format, and output the translation result directly. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,615 |
2. Given a parallelogram $A B C D$ with an angle $\arccos \frac{\sqrt{3}}{3}$. The circumcircles of triangles $A B D$ and $C B D$ intersect the diagonal $A C$ at two points, which divide it into three equal parts. Find the ratio of the longer side of the parallelogram to the shorter. | Do not round the answer. For writing the square root, use the sign «?».
Leave the source text's line breaks and format, and output the translation result directly. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,616 |
3. Given a parallelogram $A B C D$ with angle $A=45^{\circ}$. The circumcircles of triangles $A B D$ and $C B D$ intersect diagonal $A C$ at two points, which divide it into three equal parts. Find the ratio of the longer side of the parallelogram to the shorter. | Do not round the answer. To write a square root, use the symbol «?».
## Examples of writing answers:
17
$-1.7$
$1 / 7$
$1+? 7$
# | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,617 | |
1. (2 points) Solve the system:
$$
\left\{\begin{array}{l}
\left|\frac{\ln x}{\ln 2}\right|+\frac{4 x^{2}}{15}-\frac{16}{15}=0 \\
\frac{\ln \left(x+\frac{2}{3}\right)}{\ln 7}+12 x-5=0
\end{array}\right.
$$ | Answer: No solutions.
## Solution:
Consider the second equation of the system. For $x \geqslant \frac{5}{12}$, the logarithm is positive, and $12 x-5$ is non-negative. For $x \leqslant \frac{1}{3}$, the logarithm is non-positive, and $12 x-5$ is negative. Therefore, all roots of this equation, however many there are,... | Nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,618 |
2. (2 points) In space, there are a hundred unit vectors, the sum of which is equal to the zero vector. Can one choose several of them so that their sum has a length of at least 60? | Answer: No.
Solution: Suppose we have chosen such a set of vectors. The length of the sum of vectors does not exceed the sum of their lengths - therefore, if we have chosen no more than 59 vectors, the length of their sum cannot exceed 59. Consequently, we must have chosen at least 60 vectors. But then there are no mo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,619 |
3. (2 points) It is known that $\operatorname{tg} a$ and $\operatorname{tg} b$ are natural numbers, and $\operatorname{tg}(a+b)$ is an integer. Find $\operatorname{tg} a$ and $\operatorname{tg} b$. | Answer:
$$
\left[\begin{array}{l}
\operatorname{tg} a=1, \operatorname{tg} b=2 \\
\operatorname{tg} a=1, \operatorname{tg} b=3 \\
\operatorname{tg} a=2, \operatorname{tg} b=1 \\
\operatorname{tg} a=3, \operatorname{tg} b=1 \\
\operatorname{tg} a=2, \operatorname{tg} b=3 \\
\operatorname{tg} a=3, \operatorname{tg} b=2
... | [\begin{pmatrix}\operatorname{tg}=1,\operatorname{tg}b=2\\\operatorname{tg}=1,\operatorname{tg}b=3\\\operatorname{tg}=2,\operatorname{tg}b=1\\\operatorname{tg}=3,\operatorname{tg}b=1\\\operatorname{tg}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,620 |
4. (3 points) The program "Cheerful Polynomial" can perform the following operations on a polynomial $P(x)$:
1) transform $P(x)$ into $x P^{\prime}(x)$, where $P^{\prime}(x)$ is the derivative of the polynomial $P(x)$;
2) divide the coefficient of $x^{k}$ by $k$ for any natural number $k$;
3) add any constant to the ex... | Answer: No.
Solution: Operations 1), 2), or 3) do not change the coefficient of $x$. Operation 4) can either leave it unchanged or turn it into 0. Thus, the coefficient 3 cannot be obtained from the coefficient 2. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,621 |
5. (4 points) In a regular triangular pyramid (not a regular tetrahedron), the area of the base is four times smaller than the area of a lateral face. The height of the pyramid is 130 cm.
Let's construct the following (infinite) sequence of spheres. Let \( S_{1} \) be the inscribed sphere of this pyramid. Then \( S_{2... | Answer: $\frac{8788}{2598} \pi$ dm $^{3}$.
Solution: Let the area of the base be $S$, the height of the pyramid be $h$, and the radius of the sphere $S_{1}$ be $r$. Then, expressing the volume of the tetrahedron in two ways, we get the equality $\frac{1}{3} \cdot 13 S \cdot r=\frac{1}{3} S h$, from which $r=\frac{h}{1... | \frac{8788}{2598}\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,622 |
6. (4 points) $A B C D E$ is a cyclic pentagon. Circle $\omega$ is tangent to its circumcircle at point $E$. Line $B C$ intersects circle $\omega$ at points $K$ and $L$, with $K$ lying on ray $E D$ and $L$ on ray $E A$. $P$ and $Q$ are the intersection points of the circumcircles of triangles $L C D$ and $A C K$ with $... | # Solution:
Points $K$ and $L$ lie at the intersections of the extensions of the sides of the pentagon, hence they are outside its circumscribed circle, i.e., the circumscribed circle of the pentagon lies inside circle $\omega$.
Since the circumscribed circle of pentagon $A B C D E$ is tangent to circle $\omega$ at v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,623 |
7. (5 points) Positive numbers $a, b, c$ are related by the equation $1+a+b+c=2 a b c$. Prove the inequality
$$
\frac{a b}{1+a+b}+\frac{b c}{1+b+c}+\frac{c a}{1+c+a} \geqslant \frac{3}{2}
$$ | Solution:
Let's make a substitution of variables: $x=1+\frac{1}{a}, y=1+\frac{1}{b}, z=1+\frac{1}{c}$. Rewrite the original relation as: $\frac{1}{a b c}+\frac{1}{a b}+\frac{1}{a c}+\frac{1}{b c}=2$.
Then $x y z=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a b}+\frac{1}{a c}+\frac{1}{b c}+\frac{1}{a b c}=1+(x+y+z-3... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 20,624 |
8. (5 points) In the cells of a $(2 k+1) \times(2 n+1)$ table, where $k \leqslant n$, 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 maximum value that the sum of the numbers in the entire table can take? | Answer: $9 k n+6 n+5 k+3$.
## Solution:
Throughout the solution, the first of the specified sides of the rectangle will be horizontal, and the second will be vertical.
We will prove this estimate by induction on $k+n$. As the base case, consider a strip $(2 n+1) \times 1$ (for $k=0$). For this strip, the sum of the ... | 9kn+6n+5k+3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,625 |
5. (3 points) In a regular triangular pyramid (not a regular tetrahedron), the area of the base is three times smaller than the area of a lateral face. The height of the pyramid is 10 cm.
Let's construct the following (infinite) sequence of spheres. Let \( S_{1} \) be the inscribed sphere of this pyramid. Then \( S_{2... | Answer: $\frac{500}{183} \pi \mathrm{cm}^{3}$. | \frac{500}{183}\pi\mathrm{}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,626 |
5. (3 points) In a regular triangular pyramid (not a regular tetrahedron), the area of the base is twice the area of a lateral face. The height of the pyramid is 50 cm.
Consider the following (infinite) sequence of spheres. Let \( S_{1} \) be the inscribed sphere of this pyramid. Then \( S_{2} \) is the sphere that to... | Answer: $\frac{1000}{93} \pi$ dm $^{3}$. | \frac{1000}{93}\pi^3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,627 |
5. (3 points) In a regular triangular pyramid (not a regular tetrahedron), the area of the base is half the area of a lateral face. The height of the pyramid is 70 cm.
Consider the following (infinite) sequence of spheres. Let \( S_{1} \) be the inscribed sphere of this pyramid. Then \( S_{2} \) is the sphere that is ... | Answer: $\frac{686}{327} \pi$ dm $^{3}$. | \frac{686}{327}\pi^3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,628 |
8. (5 points) In the cells of the table $(2 m-1) \times(2 n-1)$, where $m \geqslant n$, 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: $7 m n-5 m-4 n+3$. | 7n-5-4n+3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,629 |
1. In parallelograms $A B C D$ and $A E F D$, heights $B P=7$ and $F Q=10$ are drawn to the side $A D$. Find the length of the perpendicular from point $C$ to $E F$. If there are multiple possible values, list them in any order separated by a semicolon. | Answer: $3 ; 17 \| 17 ; 3$ | 3;17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,630 |
3. In parallelograms $A B C D$ and $A E F D$, heights $B P=5$ and $F Q=9$ are drawn to side $A D$. Find the length of the perpendicular from point $C$ to $E F$. If there are multiple possible values, list them in any order separated by a semicolon. | Answer: $4 ; 14 \| 14 ; 4$ | 4;14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,632 |
1. In triangle $A B C$, a point $D$ is taken on side $A C$. It turns out that $B C>A B=5$, $B D=4$, and triangles $A B D$ and $B C D$ are isosceles. Find $A C$. If there are multiple answers, list them in any order separated by a semicolon. | Answer: $9 ; 10 \| 10 ; 9$ | 9;10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,633 |
2. In triangle $A B C$, a point $D$ is taken on side $A C$. It turns out that $A B>B C=6$, $B D=7$, and triangles $A B D$ and $B C D$ are isosceles. Find $A C$. If there are multiple answers, list them in any order separated by a semicolon. | Answer: $12 ; 13 \| 13 ; 12$ | 12;13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,634 |
3. In triangle $A B C$, a point $D$ is taken on side $B C$. It turns out that $A B>A C=4$, $A D=3$, and triangles $A B D$ and $A C D$ are isosceles. Find $B C$. If there are multiple answers, list them in any order separated by a semicolon. | Answer: $7 ; 8 \| 8 ; 7$
Allowed for input: digits, a dot or comma as a decimal separator, division sign, semicolon
# | 7;8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,635 |
3. How many pairs of natural numbers exist for which the number 189 is the LCM? (The numbers in the pair can be the same, the order of the numbers in the pair does not matter) | Answer: 11
Only digits are allowed as input
# | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,636 |
1. Find the largest three-digit number that gives a remainder of 2 when divided by 13, and a remainder of 6 when divided by 15. | Answer: 951
2nd option. Find the largest three-digit number that gives a remainder of 3 when divided by 13, and a remainder of 2 when divided by 14.
Answer: 926
3rd option. Find the largest three-digit number that gives a remainder of 5 when divided by 14, and a remainder of 3 when divided by 15.
Answer: 873
Only ... | 951 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,638 |
2. Find a polynomial of the third degree with the leading coefficient 1, given that its three roots $x_{1}, x_{2}, x_{3}$ satisfy the equations: $\alpha=x_{1}+x_{2}, \beta=x_{1}+x_{3}, \gamma=x_{2}+x_{3}$, where $\alpha, \beta, \gamma-$ are all the roots of the polynomial $x^{3}-12 x^{2}+44 x-46$.
Example of answer fo... | Answer: $x^{3}-6 x^{2}+8 x-2$ | x^{3}-6x^{2}+8x-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,641 |
3. Let $f(x)$ be an increasing continuous function defined on the interval $[0 ; 3]$, and $g(x)$ be its inverse function, such that $g(x) > f(x)$ for all positive $x$ for which both functions are defined. Additionally, $f(0)=0, f(3)=2$.
The area under the graph of $f(x)$ on the interval $[0 ; 3]$ is 2. Find the area o... | Answer: $4.5 \mid 9 / 2$
# | 4.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 20,642 |
1. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. What is the smaller part of the volume separated from it by a plane passing through the midpoint of edge $B C$, the center of face $C D D_{1} C_{1}$, and point $X$ on edge $A A_{1}$ such that $A X: A_{1} X=1: 3 ?$ | Provide the answer in the form of a proper fraction, do not round.
Example of answer format: $13 / 81$
Answer: $25 / 96$ | \frac{25}{96} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,643 |
3. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. What is the smaller part of the volume separated from it by a plane passing through the midpoint of edge $A D$ and points $X$ and $Y$ on edges $A A_{1}$ and $C C_{1}$ such that $A X: A_{1} X=C Y: C_{1} Y=1: 7 ?$ | Provide the answer in the form of a proper fraction, do not round.
Example of answer format: $13 / 81$
Answer: $25 / 192$
# | \frac{25}{192} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,645 |
# Problem 1. (2 points)
$A B C$ is an equilateral triangle with a side length of 10. On side $A B$, a point $D$ is taken; on side $A C$, a point $E$ is taken; on side $B C$, points $F$ and $G$ are taken such that triangles $A D E$, $B D G$, and $C E F$ are also equilateral. $A D=3$. Find $F G$. | # Answer: 4
## Solution:
$B D=A B-A D=10-3=7$, so all sides of triangle $B D G$ are equal to 7.
$C E=A C-A E=A C-A D=10-3=7$, so all sides of triangle $C E F$ are also equal to 7.
Thus, $B G+C F=14$. This is greater than the length of segment $B C=10$, so $G F=14-10=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,649 |
Problem 2. (3 points)
Alice the Fox thought of a two-digit number and told Pinocchio that this number is divisible by $2, 3, 4, 5$, and $6$. However, Pinocchio found out that exactly two of these five statements are actually false. What numbers could Alice the Fox have thought of? In your answer, indicate the number o... | # Answer: 8
## Solution:
If a number is not divisible by 2, then it is not divisible by 4 or 6 either, and we already have three false statements. Therefore, the number must be divisible by 2. If an even number is divisible by 3, it is also divisible by 6, which means the statements about divisibility by 4 and 5 are ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,650 |
# Problem 3. (3 points)
All residents of the island are either blondes or brunettes with green or blue eyes. The proportion of brunettes among blue-eyed residents is $65 \%$. The proportion of blue-eyed residents among blondes is $70 \%$. Finally, the proportion of blondes among green-eyed residents is 10\%. What perc... | Answer: 54
## Solution:
Let the number of blue-eyed brunettes be $a$, blue-eyed blondes be $b$, green-eyed blondes be $c$, and green-eyed brunettes be $d$.
Then $\frac{a}{a+b}=0.65$, from which $\frac{a+b}{a}=\frac{20}{13}$ and $\frac{b}{a}=\frac{20}{13}-1=\frac{7}{13}$.
Similarly, $\frac{b}{b+c}=0.7$, from which $... | 54 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,651 |
# Problem 4. (3 points)
Given three numbers: a five-digit number, a four-digit number, and a three-digit number. Each of them consists of identical digits (each from its own set). Can it be such that their sum is a five-digit number consisting of five different digits? | Answer: Yes
## Solution:
Actually, one example is enough, but let's explain how it is obtained.
If there are no carries during addition, the last three digits of the sum will be the same. Therefore, we need to ensure that there are carries from the units place to the tens place and from the tens place to the hundred... | 55555+6666+888=63109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,652 |
# Problem 5. (3 points)
In the cells of a $7 \times 7$ table, pairwise distinct non-negative integers are written. It turns out that for any two numbers in the same row or column, the integer parts of their quotients when divided by 8 are different. What is the smallest value that the largest number in the table can t... | # Answer: 54
Solution: Each row must contain seven numbers that give different incomplete quotients when divided by 8. This means that in each row, there is at least one number with an incomplete quotient of 6 or more when divided by 8 (6, not 7, since the minimum value of the incomplete quotient is 0). Therefore, the... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,653 |
# Problem 6. (3 points)
On each of the six faces of a cube, a number was written. Then, on each of the twelve edges, the sum of the numbers on the two adjacent faces was written. Can the numbers on the faces be uniquely determined from these 12 numbers on the edges? | Answer: Yes
## Solution:
Let the number on a certain face be $a$, and the numbers on the adjacent faces be $b, c, d$ and $e$ (in that order). Add the numbers on the four edges adjacent to this face (these are $a+b, a+c, a+d$ and $a+e$) and subtract the numbers on the two edges extending from the two opposite vertices... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,654 |
# Problem 8. (4 points)
In the Magic Country, there are 100 cities, some of which are connected by two-way air routes. Any two cities can be reached from each other with no more than 11 transfers, and there is a unique way to do so. If it is impossible to travel from city $A$ to city $B$ with 10 or fewer transfers, bo... | Answer: 89
## Solution:
Consider two peripheral cities and the route between them. All 11 cities where transfers are made on this route cannot be peripheral. All other cities can be peripheral if there are direct flights from them to the first or last of these 11 intermediate cities.
Thus, the maximum number of peri... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,655 |
3. Given a cube and 12 colors. Find the number of ways to paint the faces of this cube using these colors (each face in one color) such that adjacent faces are of different colors. Colorings that differ by a rotation are considered different. | Answer: 987360
## Examples of writing the answer:
## 7th grade
# | 987360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,656 |
3. In a four-digit number, the first digit was crossed out. The resulting number is 9 times smaller than the original. What is the largest value the original number could have had? | Answer: 7875
## Examples of how to write the answer:
1234 | 7875 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,657 |
3. On a plane, a certain number of triangles are drawn, the lengths of whose sides are ten-digit natural numbers, containing only threes and eights in their decimal representation. No segment belongs to two triangles, and the sides of all triangles are distinct. What is the maximum number of triangles that can be drawn... | Answer: 341
## Examples of how to write the answer:
## 17
# | 341 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,659 |
1. A natural number was added to its doubled sum of digits. The result was 2016. Find the greatest and the smallest possible value of the original number. | Answer: $1968 ; 2010$ || 2010; 1968 || $2010,1968 \| 1968,2010$ | 1968,2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,661 |
2. A natural number was added to its doubled sum of digits. The result was 3030. Find the greatest and the smallest possible value of the original number. | Answer: $2978 ; 3020$ || 3020; 2978 || 3020, 2978 || 2978, 3020 | 2978,3020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,662 |
3. A natural number was added to its doubled sum of digits. The result was 4023. Find the greatest and the smallest possible value of the original number. | Answer: $3969 ; 4011$ || 4011; 3969 || 4011, 3969 | 3969,4011
## Examples of answer notation:
$1234 ; 5678$
# | 3969,4011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,663 |
3. At a round table sat raccoons, hedgehogs, and hamsters, a total of 134 animals. When asked: "Are any of your neighbors the same type of animal as you?", everyone answered "No." What is the maximum number of hedgehogs that could have been sitting at the table, given that hamsters and hedgehogs always tell the truth, ... | Answer: 44
## Examples of how to write the answer:
17
# | 44 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,664 |
# Problem 7. (3 points)
1st variant: In a 10x10 table, 18 cells are colored, and at each vertex of a cell not lying on the border of the table, the number of colored cells of which it is a vertex is written. What is the minimum sum that could have been obtained? | Answer: 32
2nd variant: In a $9 \times 9$ table, 25 cells are colored, and at each vertex of a cell not lying on the border of the table, the number of colored cells that the vertex belongs to is written. What is the minimum sum that could be obtained?
Answer: 46
3rd variant: In an 11 x 11 table, 32 cells are colore... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,665 |
1. Solve the equation in natural numbers: $\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$. In the answer, list all possible values of the number $x$ in ascending or descending order, separated by commas or semicolons. | Answer: $4,6,12\|4 ; 6 ; 12\| 12,6,4 \| 12 ; 6 ; 4$ | 4,6,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,666 |
3. Solve the equation in natural numbers: $\frac{1}{x}+\frac{1}{y}=\frac{1}{4}$. In the answer, list all possible values of the number $x$ in ascending or descending order, separated by commas or semicolons. | Answer: $5,8,20\|5 ; 8 ; 20\| 20,8,5 \| 20 ; 8 ; 5$
## Examples of answer notation: $1 ; 2$ $1,2,3$
# | 5,8,20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,668 |
3. It is known that in a certain triangle, the radii of its excircles are $\frac{4 \sqrt{5}}{3}, \frac{12 \sqrt{5}}{7}$, and $\frac{3 \sqrt{5}}{2}$. Find the square of the radius of the inscribed circle for this triangle. | Answer: $1.25 \| 5 / 4$
Allowed for input are digits, minus, and division sign, a dot or comma as a decimal separator
# | \frac{5}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,672 |
3. Calculate the sum: $\frac{1 \cdot 2019!}{2019!}+\frac{2 \cdot 2019!}{1!\cdot 2018!}+\frac{4 \cdot 2019!}{2!\cdot 2017!}+\frac{8 \cdot 2019!}{3!\cdot 2016!}+\ldots+\frac{2^{2018} \cdot 2019!}{2018!\cdot 1!}+\frac{2^{2019} \cdot 2019!}{2019!}$ | Answer: $3\wedge2019$
To denote the power, use the caret symbol $\wedge$.
Examples of answer notation: $390625,5\wedge8$
For input, digits and the caret for denoting powers are allowed
# | 3\wedge2019 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,674 |
3. Let $m(a)$ and $M(a)$ be the minimum and maximum values of the function $f(x)=x^{2}+2 a x+ a+3$ on the interval $[-2; -1]$. Solve the equation $m(a)+M(a)=4 \frac{3}{4}$. | Answer: $1.5 \| 3 / 2$
Allowed for input are digits, minus, and division sign, a dot or comma as a decimal separator, a semicolon
# | 1.5;3/2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,675 |
3. The polynomial $P(x)=(x-4)^{4}\left(x^{2}+\frac{3}{2} x-1\right)^{5}$ is represented in the form $\sum_{k=0}^{14} a_{k}(x+1)^{k}$. Find $a_{0}+a_{2}+\ldots+a_{14}$. | Answer: -256
Allowed for input are digits, minus sign, and division sign, a dot or comma as a decimal separator
# | -256 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,676 |
3. How many pairs of natural numbers exist for which the number 392 is the LCM? (The numbers in the pair can be the same, the order of the numbers in the pair does not matter) | Answer: 18
Only digits are allowed as input
# | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,680 |
3. Given a fractional-linear (neither constant nor linear) function $f(x)=\frac{a x+b}{c x+d}$. It turns out that $f(f(x))=\frac{1}{2} f\left(\frac{1}{2} x\right)$ and $b c=-12$. Find $d$. If there are multiple possible values, list them in any order separated by a semicolon. | Answer: $-2 ; 2 | 2 ;-2$
Input allowed: digits, minus, division sign, period or comma as decimal separator, semicolon
# | -2;2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,683 |
3. The gnomes dug a system of tunnels that formed a rectangular grid 7 by 2. The main entrance is located at the bottom-left vertex $(0,0)$, and the main exit is at the top-right vertex $(7,2)$. The tunnels are numbered as follows: the tunnel connecting vertices $(x, k)$ and $(x+1, k)$ has a number equal to $7 k + x + ... | Answer: 8
Only digits are allowed as input | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,684 |
# Problem 1. (2 points)
Let $p_{1}, p_{2}, \ldots, p_{97}$ be prime numbers (not necessarily distinct). What is the largest integer value that the expression
$$
\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1}=\frac{p_{1}}{p_{1}^{2}+1}+\frac{p_{2}}{p_{2}^{2}+1}+\ldots+\frac{p_{97}}{p_{97}^{2}+1}
$$
can take? | # Answer: 38
Solution: Note that the function $\frac{x}{x^{2}+1}$ decreases when $x \geqslant 2$, which means $\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1} \leqslant 97 \cdot \frac{2}{2^{2}+1}=$ $97 \cdot 0.4=38.8$. Therefore, the answer cannot be greater than 38.
At the same time, $\frac{3}{3^{2}+1}=0.3$, which is 0.1 ... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,685 |
Problem 2. (2 points)
The sum of the sines of five angles from the interval $\left[0 ; \frac{\pi}{2}\right]$ is 3. What are the greatest and least integer values that the sum of their cosines can take? | Answer: $2 ; 4$
## Solution:
Note that $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$. The argument of the sine function ranges from $\frac{\pi}{4}$ to $\frac{3 \pi}{4}$, so $\frac{\sqrt{2}}{2} \leqslant \sin \left(x+\frac{\pi}{4}\right) \leqslant 1$. Therefore, $1 \leqslant \sin x+\cos x \leqslant \sqrt{... | 2;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,686 |
# Problem 3. (3 points)
Misha has 10 cards, each with one letter written on it. He can form $7!=$ 5040 different ten-letter words from them. How many different letters can he have on the cards? (Provide all possible variants and prove that there are no others). | # Answer: $5 ; 4$
Solution: The number of words that Misha can form is equal to $\frac{10!}{a!b!\ldots}$, where the numbers $a, b, \ldots$ are the number of times each letter repeats.
$7!=\frac{10!}{720}$, that is, our task is to represent the number 720 as a factorial or a product of several factorials, each greater... | 5;4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,687 |
# Problem 4. (3 points)
On the side $B C$ of triangle $A B C$, points $A_{1}$ and $A_{2}$ are marked such that $B A_{1}=6, A_{1} A_{2}=8$, $C A_{2}=4$. On the side $A C$, points $B_{1}$ and $B_{2}$ are marked such that $A B_{1}=9, C B_{2}=6$. Segments $A A_{1}$ and $B B_{1}$ intersect at point $K$, and $A A_{2}$ and $... | # Answer: 12
## Solution:
Let $M$ be the point of intersection of line $KL$ and side $AB$. We write two Ceva's theorems, for point $K$ and for point $L$:
$$
\frac{AM}{MB} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}=1 ; \quad \frac{AM}{MB} \cdot \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A}=1
$$
From this, we get $\... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,688 |
Problem 5. (3 points)
$P(x)$ is a polynomial of the fourth degree with integer coefficients, the leading one of which is positive. Moreover, $P(\sqrt{3})=P(\sqrt{5})$. Find the value(s) of $x$ for which $P(x)$ takes its minimum value. | # Answer: $-2 ; 2$
## Solution:
Let $P(x)=a x^{4}+b x^{3}+c x^{2}+d x+e$. Then $P(\sqrt{3})=9 a+3 \sqrt{3} b+3 c+\sqrt{3} d+e, P(\sqrt{5})=$ $25 a+5 \sqrt{5} b+5 c+\sqrt{5} d+e$. Numbers of the form $A+B \sqrt{3}$ and $C+D \sqrt{5}$ with integer $A, B, C, D$ can only be equal if $B=D=0$.
Thus, $3 b+d=5 b+d=0$, from ... | 2;-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,689 |
# Problem 6. (3 points)
The sequence is defined by the initial conditions $x_{1}=1, x_{2}=1.5, x_{3}=2$ and the relation $x_{n}=\frac{x_{n-1}}{2}+\frac{x_{n-2}}{4}+\frac{x_{n-3}}{4}$. Prove that $x_{1001}$ and $x_{1000}$ differ by less than $10^{-300}$. | # Solution:
$x_{n+1}-x_{n}=\frac{x_{n}}{2}+\frac{x_{n-1}}{4}+\frac{x_{n-2}}{4}-\frac{x_{n-1}}{2}-\frac{x_{n-2}}{4}-\frac{x_{n-3}}{4}=\frac{1}{2}\left(\frac{x_{n-1}}{2}+\frac{x_{n-2}}{4}+\frac{x_{n-3}}{4}\right)-\frac{x_{n-1}}{4}-\frac{x_{n-3}}{4}=\frac{x_{n-2}}{8}-\frac{x_{n-3}}{8}$.
Thus, $x_{1001}-x_{1000}=\frac{1}... | 10^{-300} | Algebra | proof | Yes | Yes | olympiads | false | 20,690 |
# Problem 7. (4 points)
Point $I$ is the center of the inscribed circle of triangle $A B C$, point $O$ is the center of the excircle opposite to vertex $B$, touching side $A C$, segments $A C$ and $O I$ intersect at point $K$.
It turns out that $O I=50, I K=18, A K=24$. Find the length of the angle bisector of angle ... | # Answer: $576 / 7$
## Solution:
The center of the inscribed circle lies on the bisector of angle $\angle B A C$, and the center of the excircle lies on the bisector of the external angle $\angle A$. These bisectors are perpendicular, so triangle $A I O$ is a right triangle with a right angle at point $A$. Note that ... | \frac{576}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,691 |
Problem 8. (5 points)
In a certain country, there are 100 cities. Each of them is connected by two-way air routes to three other cities. Moreover, from any city, it is possible to reach any other city, possibly with transfers. Vasya wants to get from city A to city B. What is the smallest number of flights that are gu... | # Answer: 72
Solution:
Before us is a scheme of air routes in a country, where it would take more than 71 flights to travel from city 1 to city 100. The main part of the scheme consists of r... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,692 |
# Problem 1. (2 points)
The polynomial $P(x)$ is such that $P\left(x^{2}\right)$ has $2 n+1$ roots. What is the smallest number of roots that the derivative of the polynomial $P(x)$ can have?
(In both cases, distinct roots are meant, without regard to multiplicity).
# | # Answer: $n$.
## Solution:
If the polynomial $P\left(x^{2}\right)$ has a root $x_{0}$, then it also has a root $-x_{0}$, so the number of roots can only be odd if one of the roots is the number 0. For each pair of roots $x_{0},-x_{0}$ of the polynomial $P\left(x^{2}\right)$, the number $x_{0}^{2}$ is a root of the p... | n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,693 |
# Problem 2. (2 points)
Vasya came up with a new operation on the set of positive numbers: $a \star b=a^{\ln b}$. Find the logarithm of the number $\frac{(a b) \star(a b)}{(a \star a)(b \star b)}$ to the base $a \star b$. | Answer: 2
Solution: Note that $a \star b=a^{\ln b}=e^{\ln a \ln b}$. Also, $a b=e^{\ln a+\ln b}$, which means that under the operation $\star$, the logarithms of the numbers are multiplied, while under multiplication of the numbers, the logarithms are added, and under division, they are subtracted.
Let $x=\ln a$ and ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,694 |
# Problem 3. (3 points)
$4^{27000}-82$ is divisible by $3^n$. What is the greatest natural value that $n$ can take? | Answer: 5
Solution:
$4^{27000}=(1+3)^{27000}=1+27000 \cdot 3+\frac{27000 \cdot 26999 \cdot 3^{2}}{2}+\frac{27000 \cdot \ldots \cdot 26998 \cdot 3^{3}}{6}+$
$+\frac{27000 \cdot \ldots \cdot 26997 \cdot 3^{4}}{24}+\frac{27000 \cdot \ldots \cdot 26996 \cdot 3^{5}}{120} \ldots$
The last two terms listed are divisible by ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,695 |
# Problem 4. (3 points)
Circles $O_{1}$ of radius $b$ and $O_{2}$ of radius $c$ touch at point $O$ - the center of circle $O_{3}$ of radius $a$. Point $A$ is one of the intersection points of circles $O_{1}$ and $O_{3}$. Circle $O_{4}$ touches circle $O_{1}$ at point $A$ and circle $O_{2}$ at point $B$. Point $C$ is a... | Answer: $\frac{a^{2}}{2}\left(\frac{1}{b}+\frac{1}{c}\right)$.
Solution: Apply inversion with respect to circle $O_{3}$. Circle $O_{3}$ will map to itself, circles $O_{1}$ and $O_{2}$ will map to parallel lines $\ell_{1}$ and $\ell_{2}$, the first of which passes through point $A$, which maps to itself under the inver... | \frac{^{2}}{2}(\frac{1}{b}+\frac{1}{}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,696 |
# Problem 5. (3 points)
Positive numbers $x, y$, and $z$ are such that $x+y+z=5$. What is the smallest value that the quantity $x^{2}+y^{2}+2 z^{2}-x^{2} y^{2} z$ can take? | # Answer: -6
## Solution:
Rewrite the condition as $\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z=5$ and write the inequality for the arithmetic mean and the quadratic mean of these five numbers:
$$
1=\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z}{5} \leqslant \sqrt{\frac{\frac{x^{2}}{4}+\frac{x^{2}}{4... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,697 |
# Problem 6. (3 points)
Given a quadrilateral pyramid $O A B C D$, at the base of which lies a parallelogram $A B C D$. Plane $\alpha$ intersects the edges $O A, O B, O C$, and $O D$ of the pyramid at points $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$, respectively. It is known that $\frac{O A^{\prime}}{O A... | Answer: $\frac{2 a b c(a-b+c)}{a+c}$
## Solution:
Solution:
$A B C D$ is a parallelogram, so $\overrightarrow{C D}=\overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B}$, and therefore, $\overrightarrow{O D}=\overrightarrow{O C}+\overrightarrow{O A}-\overrightarrow{O B}$.
If point $X$ lies in the plane $A^{... | \frac{2(-b+)}{+} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,698 |
# Problem 8. (5 points)
In an $8 \times 8$ table, some 23 cells are black, and the rest are white. In each white cell, the sum of the number of black cells on the same row and the number of black cells on the same column is written; nothing is written in the black cells. What is the maximum value that the sum of the n... | Answer: 234
## Solution:
The number in the white cell consists of two addends: "horizontal" and "vertical". Consider the sum of all "horizontal" addends and the sum of all "vertical" addends separately across the entire table. If we maximize each of these two sums separately, the total sum will also be the largest.
... | 234 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,699 |
2. Six natural numbers are written on the board, such that for any two $a$ and $b$ among them, $\log _{a} b$ or $\log _{b} a$ is an integer (the second logarithm does not necessarily exist). What is the smallest value that the maximum of these numbers can take? | The answer can be written in the form of a power of a number: $m^{n}$ is denoted as $\mathrm{m}^{\wedge} \mathrm{n}$.
Answer: 65536 || $2 \wedge 16\left\|4^{\wedge} 8\right\| 16^{\wedge} 4 \| 256^{\wedge} 2$ | 65536 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,701 |
3. Five odd numbers are written on the board, such that for any two $a$ and $b$ among them (where $b>a$), $\log _{a} b$ is an integer. What is the smallest value that the maximum of these numbers can take? | The answer can be written in the form of a power of a number: $m^{n}$ is denoted as $m^{\wedge} n$.
Answer: $43046721\left\|3^{\wedge} 16\right\| 9^{\wedge} 8 \| 81^{\wedge} 4$
## Examples of writing answers:
12345678
$7^{\wedge} 14$
# | 3^{\wedge}16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,702 |
3. $A B C A_{1} B_{1} C_{1}$ - a right triangular prism with a circumscribed sphere. The perimeter of the base $A B C$ is 32 units, and the product of the sides is 896 cubic units. The surface area of the prism is 192 square units. Find the square of the radius of its circumscribed sphere. | Answer: 53
## Examples of answer notations:
7.4
$7 / 14$
714
# | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,708 |
2. For what value of the parameter $a$ does the domain of the function $f(x)=a^{2}+a-3-\ln ^{2}(x+a)$ and the set of its values not intersect and together cover the entire number line?
If there are multiple possible answers, list them in any order separated by a semicolon. | Answer: $1 ;-3\|-3 ; 1\| 1,-3$ | 1,-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,710 |
3. For what value of the parameter $a$ does the domain of the function $f(x)=a^{2}+4 a+2-\ln ^{2}(x-a)$ and the set of its values not intersect and together cover the entire number line?
If there are multiple possible answers, list them in any order separated by a semicolon. | Answer: $-1 ;-2\|-2 ;-1\|-1,-2 \|-2 ;-1$ | -1,-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,711 |
3. Circles $O_{1}$ and $O_{2}$, with radii 4 and 21, touch each other externally, and circle $O_{3}$ with radius 28 touches them internally. Find the square of the length of the segment of the common tangent to $O_{1}$ and $O_{2}$, contained inside $O_{3}$, drawn at the point of their tangency. | Answer: $3119,3536 \| 1949596 / 625$
## Examples of answer notation:
7.4
$7 / 14$
714 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,713 | |
1. (2 points) Does there exist a rectangular parallelepiped with integer sides, for which the surface area is numerically equal to the volume? | Answer: Yes, it exists.
Solution: For example, a cube $6 \times 6 \times 6$ fits. Both the surface area and the volume are numerically equal to 216. | 6\times6\times6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,714 |
2. (3 points) On the larger base $AD$ of the isosceles trapezoid $ABCD$, a point $X$ is marked such that $AB = AX$. On the ray $AB$, a point $E$ is chosen such that $DE \parallel BX$. Prove that $AD + CE \geqslant BE + ED$.
| Solution: $A B=A X, D E \| B X$, therefore, by Thales' theorem, $A D=A E$. By the triangle inequality $C E+C D \geqslant D E$, but $C D=A B$, since the trapezoid is isosceles. From all the above, $A D+C E=A E+C E=A B+B E+C E \geqslant B E+E D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,715 |
3. (3 points) The perimeter of parallelogram $A B C D$ is 20. The bisector of angle $B$ intersects lines $A D$ and $C D$ at points $K$ and $L$ respectively. Find $C L$, given that $D K=4$. | Answer: 3 or 7
Solution: $\angle A B K=\angle B L C$, as alternate interior angles. Similarly, $\angle C B L=\angle B K A$. Since points $B, K$, and $L$ lie on the bisector, all four of these angles are equal.
$\angle B K A=\angle D K L$: depending on which of the points $K$ and $L$ lies on the side of the parallelog... | 3or7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,716 |
4. (3 points) From the set of three-digit numbers that do not contain the digits $0,1,2$, $3,4,5$ in their notation, several numbers were written on paper in such a way that no two numbers can be obtained from each other by swapping two adjacent digits. What is the maximum number of such numbers that could have been wr... | Answer: 40
Solution:
Whether a number can be written or not depends only on numbers consisting of the same set of digits.
Suppose a number ABC, consisting of three different digits, is written. This means that the numbers BAC and ACB cannot be written. Among the numbers consisting of these same digits, BVA, VAB, and... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,717 |
5. (3 points) Solve the equation $a+b+\operatorname{GCD}(a, b)+\operatorname{LCM}(a, b)=143$. | Answer: No solutions
Solution:
Let $d=\text{GCD}(a, b)$. Then for some $k$ and $l$, the equations $a=k d, b=l d, \text{LCM}(a, b)=k l d$ hold. Accordingly, the equation becomes $k d+l d+d+k l d=143$ or $(k+1)(l+1) d=11 \cdot 13$. Since $k+1$ and $l+1$ are both greater than 1, one of them must be 11 and the other 13. ... | Nosolutions | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,718 |
6. (3 points) On a $3 \times 3$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 5.
Solution: The people in the corner cells are obviously liars: they simply do not have three neighbors.
If the person in the center is a knight, then all the people in the side cells are also liars. But in this case, the knight has 4 liar neighbors, not three. This leads to a contradiction.
If the person i... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,719 |
7. (4 points) There are 23 saucers arranged in a circle, and on them are 46 buns. In one move, it is allowed to take 2 buns from one saucer and place them on 2 adjacent saucers. Can it be achieved that all saucers have an equal number of buns, regardless of the initial arrangement of the buns? | Answer: No, not in any case.
Solution: Let's number the plates in order from 0 to 22. For each bun, we take the number of the plate it is on. Note that the sum of these 46 numbers will not change its remainder when divided by 23: one of the buns always moves to a plate with a number that is 1 greater or 22 less, while... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,720 |
8. (4 points) The decimal representation of a four-digit number ABCD, consisting of different digits, was divided into three parts in three ways, after which three quadratic equations $\mathrm{A} x^{2}+B x+B \Gamma=0, \mathrm{~A} x^{2}+\mathrm{B} \mathrm{B} x+\Gamma=0$ and $\mathrm{A} x^{2}+\mathrm{B} x+\Gamma=0$ were ... | Answer: 1920 or 2910.
Solution: First, we will prove that $\Gamma=0$. From the first equation, we get that $Б^{2} \geqslant 4 \cdot \mathrm{A} \cdot \mathrm{B}$. Since $\mathrm{Б}^{2} \leqslant 81$, we get that $\mathrm{B} \Gamma \leqslant 20$, which means $\Gamma=0$ or $\mathrm{B}=1$ (since zero cannot be equal to an... | 1920or2910 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,721 |
1. (2 points) Does there exist a rectangular parallelepiped with integer sides, for which the surface area is numerically equal to the sum of the lengths of all twelve edges? | Answer: Yes, it exists.
Solution: For example, a $2 \times 2 \times 2$ cube fits. Both the surface area and the sum of the lengths of the edges are 24. | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,722 |
2. (3 points) On the larger base $AD$ of the isosceles trapezoid $ABCD$, a point $K$ is marked such that $DK = CD$. On the ray $DC$, a point $L$ is chosen such that $AL \parallel CK$. Prove that $AL + CL \leq AD + BL$.
| Solution: $D K=C D, A L \| C K$, therefore, by Thales' theorem, $A D=D L$. By the triangle inequality $A B+B L \geqslant A L$, but $C D=A B$, since the trapezoid is isosceles. From all the above, $A D+B L=D L+B L=C D+C L+B L=A B+B L+C L \geqslant A L+C L$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 20,723 |
3. (3 points) The perimeter of parallelogram $A B C D$ is 32. The bisector of angle $C$ intersects lines $A D$ and $A B$ at points $E$ and $F$ respectively. Find the value of $B F$, given that $A E=2$. | Answer: 7 or 9
Solution: $\angle B C E=\angle C E D$, as alternate interior angles. Similarly, $\angle B F C=\angle F C D$. Since points $C, E$, and $F$ lie on the bisector, all four of these angles are equal.
$\angle B F C=\angle A F E$: depending on which of the points $E$ and $F$ lies on the side of the parallelog... | 7or9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 20,724 |
4. (3 points) From the set of three-digit numbers that do not contain the digits 0, 6, 7, 8, 9 in their notation, several numbers were written on paper in such a way that no two numbers can be obtained from each other by swapping two adjacent digits. What is the maximum number of such numbers that could have been writt... | Answer: 75
Solution:
Whether a number can be written or not depends only on numbers consisting of the same set of digits.
Suppose a number ABC, consisting of three different digits, is written. This means that the numbers BAC and ACB cannot be written. Among the numbers consisting of these same digits, BVA, VAB, and... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,725 |
5. (3 points) Solve the equation $a+b+\operatorname{GCD}(a, b)+\operatorname{LCM}(a, b)=187$. | Answer: $a=10, b=16$ or vice versa.
Solution:
Let $d=$ GCD $(a, b)$. Then for some $k$ and $l$ the equations $a=k d, b=$ $l d, \operatorname{LCM}(a, b)=k l d$ hold. Accordingly, the equation takes the form $k d+l d+d+k l d=187$ or $(k+1)(l+1) d=11 \cdot 17$. Since $k+1$ and $l+1$ are greater than 1, one of them equal... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20,726 |
6. (3 points) On a $2 \times 4$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 6.
Solution: On the corner cells, liars are obviously standing: they simply do not have three neighbors.
A situation where all people on the board are liars is impossible, as it would mean that four of these liars are telling the truth.
Therefore, there is at least one knight. Then all his neighbors are liar... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 20,727 |
7. (4 points) There are 25 plates arranged in a circle, on which 50 buns are placed. In one move, it is allowed to take 2 buns from one plate and place them on 2 adjacent plates. Can it be achieved, regardless of the initial arrangement of the buns, that all plates end up with an equal number of buns? | Answer: No, not in any case.
Solution: Let's number the plates in order from 0 to 24. For each bun, we take the number of the plate it is on. Note that the sum of these fifty numbers will not change its remainder when divided by 25: one of the buns always moves to a plate with a number that is 1 greater or 24 less, wh... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 20,728 |
8. (4 points) The decimal representation of a four-digit number ABCD, which does not contain nines, was divided into three parts in three ways, after which three quadratic equations were formed: $A x^{2} + B x + B \Gamma = 0$, $A x^{2} + B B x + \Gamma = 0$, and $AB x^{2} + B x + \Gamma = 0$. It turned out that all of ... | Answer: 1710 or 1810.
Solution: From the first equation, we get that $B^2 \geq 4 \cdot A \cdot B\Gamma$. Since according to the condition $B^2 \leq 64$, we get that $B\Gamma \leq 16$, which means $B=1$ (since zero cannot be equal to any digit except $\Gamma$).
On the other hand, from the third equation, we get $1=B^2... | 1710or1810 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,729 |
3. Given the function $f(x)=\frac{\sqrt{3} x-1}{x+\sqrt{3}}$. Find $\underbrace{f(f(\ldots(1) \ldots))}_{2013 \text { times }}$. | Answer: -1
## Examples of how to write answers:
$1 / 4$
0.25
10
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,730 |
3. A number from 0 to 1 inclusive was written on the board. Each of the twenty-five students in the class approached the board, erased the number $x$ written there at the moment, and wrote $\frac{1}{2} \cos \pi x$ instead.
For which initial value of the number will the final result be the maximum?
If there are multip... | Answer: $0 ; 1 \| 1 ; 0$
## Examples of writing answers:
$1 / 4$
0, $25 ; 0.5$
# | 0;1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 20,732 |
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