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|---|---|---|---|---|---|---|---|---|---|
4. Solve the equation
$$
x^{2018}+\frac{1}{x^{2018}}=1+x^{2019}
$$ | Answer: $x=1$.
Solution: $x^{2018}+\frac{1}{x^{2018}} \geq 2$ for $x \neq 0$, because $x^{4036}-2 x^{2018}+1=\left(x^{2018}-1\right)^{2} \geq 0 . x^{2019}=x^{2018}+\frac{1}{x^{2018}}-1 \geq 2-1=1 . x \geq 1$. If $\quad x>1, \quad x^{2019}+1>$ $x^{2018}+\frac{1}{x^{2018}}$, because $x^{2019}=x * x^{2018}>x^{2018}$ and ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,920 |
5. Is it possible to arrange the numbers from 1 to 30 in a circle so that the sum of any two adjacent numbers is equal to the square of some natural number? | Answer: No.
Solution: Notice that the only number that can stand next to 18 is 7. Therefore, it is impossible to arrange the numbers in a circle in this way (18 should have two neighbors). | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,921 |
Problem 9. Find the smallest number such that when divided by 2018, the remainder is 2017, when divided by 2019, the remainder is 2018, and when divided by 2020, the remainder is 1. | Answer: 4074341
Problem 10. Solve the equation
$20\{x\}-18[x]=0$, where $[x]$ denotes the integer part of the number $x$, i.e., the greatest integer not exceeding $x$; $\{x\}$ denotes the fractional part of the number $x$, i.e., $\{x\}=x-[x]$. Write the sum of the roots in the answer.
Answer: 1.9 | 1.9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,922 |
1. a) Can 10 consecutive natural numbers be found whose sum equals 2016:
b) Can 7 consecutive natural numbers be found whose sum equals 2016. | Answer: a) No; b) Yes.
Solution: a) Suppose that $a+(a+1)+(a+2)+\cdots+(a+9)=10 a+45=$ $2016 \Leftrightarrow 10 a=1971$. Since the last equation is impossible for natural $a$, the answer is "no".
b) $285+286+287+288+289+290+291=2016$.
Evaluation criteria. Correct solution - 20 points; in part a) for the proof - 11 p... | )No;b)Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,923 |
2. Real numbers $a$ and $b$ are such that $\frac{6 a+9 b}{a+b}<\frac{4 a-b}{a-b}$. Prove that $|b|<|a|<2|b|$. | Solution.
$$
\frac{6 a+9 b}{a+b}<\frac{4 a-b}{a-b} \Leftrightarrow \frac{6 a+9 b}{a+b}-\frac{4 a-b}{a-b}<0 \Leftrightarrow \frac{2 a^{2}-8 b^{2}}{a^{2}-b^{2}}<0
$$
Solving the last inequality with respect to $a^{2}$, we obtain $b^{2}<a^{2}<4 b^{2}$, which is equivalent to the inequalities $|b|<|a|<2|b|$. This complet... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 11,924 |
3. The coefficients of the quadratic trinomials: $f_{i}(x)=x^{2}+b_{i} x+c_{i}$, satisfy the equations $\frac{b_{i+1}}{b_{i}}=2, c_{i}=-32 \cdot b_{i}-1024(i=1,2, \cdots)$. It is known that the roots of the polynomial $f_{1}(x)$ are the numbers 32 and -31. a) Find the roots of the quadratic trinomial $f_{12}$; b) find ... | Answer: $2016: 32$.
Solution: By Vieta's theorem: $b_{1}=-(32+(-31))=-1, c_{1}=-32 \cdot(-1)-1024=$ $32-1024=-992$. Therefore, $b_{i}=-2^{i-1}, c_{i}=-32 \cdot\left(-2^{i-1}\right)-1024=2^{i+4}-2^{10}$.
Solving the quadratic equation
$$
x^{2}-2^{i-1} x+2^{i+4}-2^{10}=0
$$
we find:
$D=\left(2^{i-1}\right)^{2}-4\lef... | 2016,32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,925 |
4. In an isosceles trapezoid $A B C D (B C \| A D)$, angles $A B D$ and $D B C$ are $135^{\circ}$ and $15^{\circ}$ respectively, and $B D=\sqrt{6}$. Find the perimeter of the trapezoid. | Answer: $9-\sqrt{3}$.
Note that $\angle D B C=\angle A C B=\angle B D A=\angle C A D=15^{\circ}, \angle B A C=30^{\circ}-\angle D B C=15^{\circ}$, so $A C$ is the angle bisector of $\angle A$. Since $\angle B A C=\angle C A B=\angle C D B=15^{\circ}$, triangles $A B C$ and $B C D$ are isosceles and $A B=B C=C D$. By t... | 9-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,926 |
5. In the cells of a $9 \times 9$ table, odd integers are written. It is allowed to simultaneously change the sign of all numbers in some row or some column. Prove that using several such operations, one can arrive at a table where the sums of the numbers in any row and any column are positive. | Solution. Since the sum of an odd number of odd numbers is odd, the sum of the numbers in any row (any column) of the resulting table cannot be zero (i.e., it is strictly positive or strictly negative). Further, let us assume that the sum of all numbers in the table obtained by such operations does not exceed the sum o... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,927 |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be?
Solution. Let the first number be $a$, and the second $b$. Then the equation $1996 a+\frac{b}{96}=a+b$ must hold, from which we find that $2016 a=b$. Therefore, their quotient... | Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 14 points; correct answer without solution - 2 points. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,928 |
2. A black and white chocolate bar consists of individual pieces forming $n$ horizontal and $m$ vertical rows, arranged in a checkerboard pattern. Yan ate all the black pieces, and Maxim ate all the white pieces. What is $m+n$, if it is known that Yan ate $8 \frac{1}{3} \%$ more pieces than Maxim. | Solution. The number of black and white segments can only differ by 1. Therefore, Yan ate 1 segment more than Maksim. If 1 segment is 8 $\frac{1}{3} \%$, then Maksim ate 12 segments, Yan ate 13 segments, and together they ate 25 segments. This means the chocolate bar was $5 \times 5$.
Answer: 10.
Criteria: 14 points ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,929 |
6. In the inscribed quadrilateral $ABCD$, the circumcircle of triangle $\operatorname{COD}$ (where $O$ is the intersection point of the diagonals) passes through the center of the circumcircle of quadrilateral $ABCD$. Prove that quadrilateral $ABCD$ is a trapezoid. | Solution. Let $O_{1}$ be the center of the circumscribed circle of quadrilateral $ABCD$. From the condition, it follows that $\angle COD = \angle CO_{1}D = \smile CD$. On the other hand, $\angle COD = \frac{\smile AB + \smile CD}{2}$. From these two equalities, we get that $\smile AB = \smile CD$. Therefore, $BC \| AD$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,931 |
7. Do there exist 2016 consecutive natural numbers among which there are exactly 16 prime numbers? | Solution. Let's introduce the function $S(n)$, equal to the number of prime numbers from $n$ to $n+2015$. Note that $S(n)$ differs from $S(n+1)$ by no more than 1, $S(2017!+2)=0, S(1)>16$ $(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,51,53, \ldots$ - prime numbers). Therefore, there exists a number $m$ such that $S(m)=16$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,932 |
Task 2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be equal to zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$. | Task 2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be equal to zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,934 |
Given trapezoid $A B C D, A D \| B C$. A point $E$ is chosen on side $A B$. Prove that the distance between the centers of the circumcircles of triangles $A D E$ and $B C E$ does not depend on the choice of point $E$. | Given trapezoid $A B C D, A D \| B C$. A point $E$ is chosen on side $A B$. Prove that the distance between the centers of the circumcircles of triangles $A D E$ and $B C E$ does not depend on the choice of point $E$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,938 |
Task 7. Can a grid rectangle $2018 \times 2020$ be cut into grid rectangles $5 \times 8$? | Task 7. Can a grid rectangle $2018 \times 2020$ be cut into grid rectangles $5 \times 8$?
## Solutions and Evaluation Criteria
| Task Number | Solution | Criteria |
| :---: | :---: | :---: |
| 1 | Answer: 0 or 9. Solution: The sum of all 10 digits from 0 to 9 is 45. Sasha must append a digit so that the sum of the di... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,939 |
8. There are books of three colors: white, blue, and green. To make the shelf look beautiful, the boy first arranged the white books, and then placed blue books in each gap between them. Finally, he placed green books in each gap between the standing books. In the end, there were 41 books on the shelf. How many white b... | Answer: 11
9. The sum of two natural numbers is 2017. If you append 9 to the end of the first number and remove the digit 8 from the end of the second number, the numbers will be equal. Find the largest of these numbers.
Answer: 1998 | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,940 |
2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
untranslated text remains the same as requested. | 2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
Answer: 1902 | 1902 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,942 |
2. A parallelepiped is composed of white and black unit cubes in a checkerboard pattern. It is known that the number of black cubes is $1 \frac{12}{13} \%$ more than the number of white cubes. Find the surface area of the parallelepiped, given that each side of the parallelepiped is greater than 1. | Solution. The number of black and white cubes can differ by only 1. Therefore, 1 cube is $1 \frac{12}{13} \%$ of the quantity of white cubes. Thus, there are 52 white cubes, 53 black cubes, and a total of 205 cubes. That is, our parallelepiped is $3 \times 5 \times 7$. The surface area is 142.
Answer: 142.
Criteria: ... | 142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,951 |
4. In a regular quadrilateral pyramid $S A B C D$, the height is equal to the side of the base. Points $M$ and $N$ are marked on the lateral edges $S D$ and $S B$ of the pyramid, respectively, such that the lines $A M$ and $C N$ are perpendicular to each other. Prove that
$$
2 S A(S M+S N)=S A^{2}+S M \cdot S N
$$ | Solution. We will place the coordinate system such that the origin coincides with vertex $A$, and the axes are as shown in Figure 1.
Fig. 1:
We will assume that the length of the base side o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,952 |
5. Solve the equation for all values of the parameter $a$
$$
3 x^{2}+2 a x-a^{2}=\ln \frac{x-a}{2 x}
$$ | Solution: If $a=0$, then the equation will take the form $3 x^{2}=\ln \frac{1}{2}$, which has no solutions. If $a \neq 0$, then $x$ does not belong to the interval $[0, a]$ or $[a, 0]$ depending on the sign of the parameter $a$, since otherwise the natural logarithm does not exist. Transform the equation to the form;
... | - | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,953 |
6. In the inscribed quadrilateral $A B C D$, the circumcircle of triangle $C O D$ (where $O$ is the intersection point of the diagonals) passes through the center of the circumcircle of quadrilateral $A B C D$. Prove that quadrilateral $A B C D$ is a trapezoid. | Solution. Let $O_{1}$ be the center of the circumscribed circle of quadrilateral $ABCD$. From
Fig. 2:
the condition follows that $\angle COD = \angle CO_{1}D = \smile CD$. On the other hand... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,954 |
4. Solve the equation $20 x^{2}+17 y=2017$ in natural numbers. | Answer: $(10 ; 1) ;(7 ; 61)$. | (10;1);(7;61) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,955 |
2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5? | 2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5?
Answer: 80 | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,957 |
10.1. Kolya wrote a ten-digit number on the board, consisting of different digits. Sasha added one digit so that the resulting number would be divisible by 9. Which digit could Sasha have added? | Answer: 0 or 9.
Solution: The sum of all 10 digits from 0 to 9 is 45. Sasha needs to append a digit so that the sum of the digits in the resulting number is divisible by 9 (the divisibility rule for 9). The sum of the digits will be divisible by 9 if 0 or 9 is appended.
Criteria: Only 1 point for the correct answer 0... | 0or9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,966 |
10.2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$. | Answer: 620
Solution: for a single-digit number $n$, $f(n)$ will be equal to $n$ itself if it is even and 0 if it is odd. For single-digit $\mathrm{n}$, we get the sum $2+4+6+8=20$. If $\mathrm{n}$ is a two-digit number, let's consider the cases:
1) If both digits are even. Then the first digit can be $2,4,6$ or 8, a... | 620 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,967 |
10.3. Solve the system of equations in natural numbers
$$
\left\{\begin{array}{l}
a b=c+d \\
c d=a+b
\end{array}\right.
$$ | Answer: (1;5;2;3), (1;5;3;2), (5;1;2;3), (5;1;3;2), (2;2;2;2), (2;3;1;5), (2;3;5;1), (3;2;1;5), $(3 ; 2 ; 5 ; 1)$.
Solution: By adding the equations, moving everything to one side, and adding 2 to both sides of the equation, we get
$$
\begin{aligned}
& a b-a-b+1+c d-c-d+1=2 \\
& (a-1)(b-1)+(c-1)(d-1)=2
\end{aligned}
... | (1;5;2;3),(1;5;3;2),(5;1;2;3),(5;1;3;2),(2;2;2;2),(2;3;1;5),(2;3;5;1),(3;2;1;5),(3;2;5;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,968 |
10.4. In an initially empty room, every minute either 2 people enter or 1 person leaves. Can there be exactly 2018 people in the room after 2019 minutes? | Answer: No.
Solution: Every minute, the remainder of the number of people in the room when divided by 3 changes either from 0 to 2, or from 1 to 0, or from 2 to 1. This means that after 3 times, the remainder of the number of people in the room when divided by 3 will not change. 2019 is 673 times 3, so the remainder a... | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,969 |
10.5. In a row, $n$ integers are written such that the sum of any three consecutive numbers is positive, while the sum of any five consecutive numbers is negative. For what largest $n$ is this possible | Answer: 6.
Solution: Let's provide an example for $n=6: 3,-5,3,3,-5,3$. We will prove that for $n \geq 7$ it will not be possible to write down a sequence of numbers that satisfy the condition of the problem. We will construct a table for the first 7 numbers in this sequence
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,970 |
10.6. Given a trapezoid $A B C D, A D \| B C$. A point $E$ is chosen on the side $A B$. Prove that the distance between the centers of the circumcircles of triangles $A D E$ and $B C E$ does not depend on the choice of point $E$. | Solution: The center of the circumcircle of triangle $A D E$ lies at the intersection of the perpendicular bisectors of $A D$ and $E A$. And the center of the circumcircle of triangle $B C E$ lies at the intersection of the perpendicular bisectors of $B C$ and $B E$, i.e., at the intersection of two pairs of parallel l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,971 |
10.7. Can a grid rectangle $2018 \times 2020$ be cut into grid rectangles $5 \times 8$? | Answer: No.
Solution: Note that $2018 \times 2020$ cannot be divided into rectangles of $1 \times 8$ (this can be understood from a diagonal coloring in 8 colors - each rectangle $1 \times 8$ will cover one cell of each color. If we lay out the rectangles $1 \times 8$ as rectangles $2016 \times 2016$, $2 \times 2016$,... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,972 |
Task 2. Two circles with radii 1 and 2 have a common center O. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.
 | Task 2. Two circles with radii 1 and 2 have a common center $O$. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,973 | |
Given function $f(x)$, satisfying the condition
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
What is $f(2017)$, if it is known that $f(0)=1$? | Given function $f(x)$, satisfying the condition
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
What is $f(2017)$, if it is known that $f(0)=1$? | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,976 | |
Task 6. In space, there are two equal regular tetrahedra with side $\sqrt{6}$. It is known that their centers coincide. Prove that the volume of their common part is greater than $\frac{1}{2}$. | Task 6. In space, there are two equal regular tetrahedra with side $\sqrt{6}$. It is known that their centers coincide. Prove that the volume of their common part is greater than $\frac{1}{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,977 |
Task 7. Given the infinite sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n}, \ldots$
Prove that for any natural number $k \geq 3$, from this sequence it is possible to select $k$ terms that are consecutive terms of some arithmetic progression. | Task 7. Given an infinite sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n}, \ldots$
Prove that for any natural number $k \geq 3$, from this sequence, one can select $k$ terms that are consecutive terms of some arithmetic progression.
## Solutions and Evaluation Criteria
| Problem Number | Solu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 11,978 |
Task 1. Find natural numbers $a$ and $b$, given that the number $\overline{a b b a}$ is a perfect cube. | Task 1. Find natural numbers $a$ and $b$, if it is known that the number $\overline{a b b a}$ is a perfect cube.
| Solution | Criteria |
| :---: | :---: |
| Note that the number $\overline{a b b a}$ is divisible by 11 (by the divisibility rule for 11). Since $22^{3}=10648$ is a five-digit number, $\overline{a b b a}=1... | =1,b=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,979 |
Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students? | Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students?
| Solution | Criteria |
| :---: | :---: |
| Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,980 |
Problem 3. Excellent students Alyosha and Vasya are listing four-digit numbers. Alyosha lists numbers where the first digit is equal to the product of the other three, while Vasya lists numbers where the last digit is equal to the product of the other three. Who will list more numbers and by how many? | Task 3. Excellent students Alyosha and Vasya write down four-digit numbers. Alyosha writes down numbers where the first digit is equal to the product of the other three, while Vasya writes down numbers where the last digit is equal to the product of the other three. Who will write down more numbers and by how many?
| ... | 171 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,981 |
Problem 4. On a straight line, segment $K L$ and 15 points lying outside segment $K L$ are marked. Prove that the sum of the distances from these points to point $K$ cannot be equal to the sum of the distances to point $L$. | Problem 4. On a line, a segment $K L$ and 15 points lying outside the segment $K L$ are marked. Prove that the sum of the distances from these points to point $K$ cannot be equal to the sum of the distances to point $L$.
| Solution | Criteria |
| :--- | :--- |
| Let the marked points be denoted as $A_{1}, A_{2}, \ldot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,982 |
Problem 5. In a regular 2015-gon, 807 vertices are painted. Prove that there exist three painted vertices that lie at the vertices of an isosceles triangle. | Problem 5. In a regular 2015-gon, 807 vertices are painted. Prove that there will be three painted vertices that lie at the vertices of an isosceles triangle.
| Solution | Criteria |
| :---: | :---: |
| It is known that any three vertices of a regular pentagon form an isosceles triangle. In the vertices of the 2015-go... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,983 |
Task 1. Find the value of the expression $\sqrt[3]{7+5 \sqrt{2}}-\sqrt[3]{5 \sqrt{2}-7}$ | Answer: 2
Task 2.B The bases $AD$ and $BC$ of an isosceles trapezoid $ABCD$ are $16 \sqrt{3}$ and $8 \sqrt{3}$, respectively, and the acute angle at the base is $30^{\circ}$. What is the length of the lateral side of the trapezoid?
Answer: 8 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,984 |
Task 4. Solve the system of equations $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=10 \\ \sqrt[4]{x}+\sqrt[4]{y}=4\end{array}\right.$ and find the value of the product $x y$. | Answer: 81
Problem 5. From the vertex of the right angle $K$ of triangle $MNK$, a perpendicular $KL$ is drawn to the plane of the triangle, equal to 280. Find the distance from point $L$ to the line $MN$, given that the height of the triangle dropped from vertex $K$ is 96.
Answer: 296
Problem 6. The bank allocated a... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,985 |
Task 9. Find the largest negative root $x_{0}$ of the equation $\frac{\sin x}{1+\cos x}=2-\operatorname{ctg} x$. In the answer, write $x_{0} \cdot \frac{3}{\pi}$. | Answer: $-3.5$
Problem 10. A group of 30 people collected cranberries in the forest. Each of them collected 2, 3, 4, or 5 buckets of cranberries, totaling 93 buckets. Moreover, the number of people who collected 3 buckets was more than those who collected 5 buckets and less than those who collected 4 buckets. Addition... | -3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,986 |
Task 12. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $3 \sqrt{2}$. Find the volume of a regular tetrahedron, one vertex of which coincides with point $A$, and the other three vertices belong to the plane $C M A_{1} N$, where $M$ and $N$ are the midpoints of edges $D D_{1}$ and $B B_{1}$. | Answer: 9
Problem 13. Solve the equations for all natural $n$:
$$
\cos ^{4} x+\sin x(\sin x+1)\left(\cos ^{2} x+\sin x-1\right)=n
$$
In the answer, write the number of roots in the interval $[0,2 \pi]$
Answer: 3
Problem 14. $M$ is the midpoint of the lateral side $A B$ of trapezoid $A B C D$, and $E$ is the inters... | 3986 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,987 |
Task 2. Prove that the graphs of the quadratic trinomials: $y=a x^{2}-b x+c, y=b x-$ $c x+a, y=c x^{2}-a x+b$ have a common point. | Task 2. Prove that the graphs of the quadratic trinomials: $y=a x^{2}-b x+c, y=b x^{2}-$ $c x+a, y=c x^{2}-a x+b$ have a common point. | proof | Algebra | proof | Yes | Yes | olympiads | false | 11,989 |
Task 3. The diagonals of a quadrilateral $A B C D$, inscribed in a circle with center at point $O$, intersect at point $P$. It is known that $O P \perp B C$. Prove that $A B=C D$. | Task 3. The diagonals of a quadrilateral $A B C D$, inscribed in a circle with center at point $O$, intersect at point $P$. It is known that $O P \perp B C$. Prove that $A B=C D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,990 |
Task 5. Is it possible to arrange the numbers from 1 to 30 in a circle so that the sum of any two adjacent numbers is equal to the square of some natural number? | Task 5. Is it possible to arrange the numbers from 1 to 30 in a circle so that the sum of any two adjacent numbers is equal to the square of some natural number?
## Solutions and Evaluation Criteria
| Task Number | Solutions | Criteria |
| :---: | :---: | :---: |
| $\mathbf{1}$ | Answer: no. Solution: The difference ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,992 |
Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
 | Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
Answer: 14 | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,996 |
Problem 5. Does there exist a value of \(a\) such that the equation
\[
x^{4}+4 x^{3} a+6 x^{2} a^{2}+8 x a^{3}+10 a^{4}=0
\]
has four distinct real roots?
| Solution | | Criteria |
| :--- | :--- | :--- |
| It is known that between | two consecutive roots | For the correct answer - 0 points. |
| of a differentiable ... | Answer: No, it does not exist. | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,003 |
Problem 5. On the table, there are 10 stacks of playing cards (the number of cards in the stacks can be different, there should be no empty stacks). The total number of cards on the table is 2015. If a stack has an even number of cards, remove half of the cards. If the number of remaining cards in the stack is still ev... | # Solution
a) Since 2014 is an odd number, in any distribution of cards into 10 piles, there will be at least one pile with an even number of cards. Therefore, the number of cards in this pile, and thus the total number of cards, will definitely decrease by at least one card (if there are 2 cards in this pile). This m... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,004 |
Task 6. In an acute-angled triangle $\mathrm{ABC}$, angle $\mathrm{B}$ is $30^{\circ}, \mathrm{BC}=12$. The altitude CD of triangle $\mathrm{ABC}$ and the altitude DE of triangle BDC are drawn. Find BE. | Answer: 9
Task 7.3a 2016 The number of books in the school library fund increased by $0.4 \%$, and in 2017 - by $0.8 \%$, remaining less than 50 thousand. By how many books did the library fund increase in 2017?
Answer: 251
Task 8. A rectangle is divided into six squares (see figure). What is the side of the larger ... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,005 |
Problem 3. The function $f$ is such that for any positive $x$ and $y$, the equality $f(x y)=f(x)+f(y)$ holds. Find $f(100)$, if $f(0.2)=2$ and $f(0.5)=5$. | Problem 3. The function $f$ is such that for any positive $x$ and $y$, the equality $f(x y)=f(x)+f(y)$ holds. Find $f(100)$, if $f(0.2)=2$ and $f(0.5)=5$.
| Solution | Criteria |
| :---: | :---: |
| . | Problem 5. Solve the equation
$x^{5}-[x]=5$, where $[x]$ is the integer part of the number $x$ (the greatest integer not exceeding $x$).
| Solution | Criteria |
| :---: | :---: |
|  | 1 point ... | \sqrt[5]{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,010 |
Problem 6. In rectangle $A B C D$, diagonal $A C$ is twice as long as side $A B$. Point $E$, lying outside rectangle $A B C D$, is such that angle $\angle C E D=120^{\circ}$. Prove that the center of the inscribed circle in triangle $C E D$ lies on the line $O E$, where $O$ is the point of intersection of the diagonals... | Task 6. In rectangle $ABCD$, diagonal $AC$ is twice as long as side $AB$. Point $E$, lying outside rectangle $ABCD$, is such that angle $\angle CED=120^{\circ}$. Prove that the center of the inscribed circle in triangle $CED$ lies on the line $OE$, where $O$ is the point of intersection of the diagonals of rectangle $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,011 |
Task 7. On a circle, there are 25 non-overlapping arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the arc following it in a clockwise direction. What can the sum of all the numbers be? | Problem 7. On a circle, there are 25 non-intersecting arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the next arc clockwise. What can the sum of all the numbers be?
| Solution | Criteria |
| :---: | :---: |
| Let the... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,012 |
3. In a convex quadrilateral $A B C D: A B=A C=A D=B D$ and $\angle B A C=\angle C B D$. Find $\angle A C D$. | Answer: $70^{\circ}$.
Solution: Triangle $A B D$ is equilateral, so the angles $\angle A B D=\angle B D A=$ $\angle D A B=60^{\circ}$. Let $\angle B A C=\angle C B D=\alpha$, then $\angle A B C=60^{\circ}+\alpha . A B=A C$, thus $\angle A C B=\angle A B C=60^{\circ}+\alpha$. The sum of the angles in triangle $A B C$ i... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,015 |
4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.
---
Note: The translation maintains the original format and line breaks as requested. | Answer: $x=y=0$.
Solution: If both numbers are not equal to 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,016 |
5. In how many ways can a pile of 100 stones be divided into heaps so that the number of stones in any two heaps differs by no more than one? | Answer: 99.
Solution: We will prove that for any $k$ from 2 to 100, the pile can be divided into $\mathrm{k}$ such piles in a unique way. $100 = q k + r$, where $q$ and $r$ are the quotient and remainder of 100 when divided by $k$, respectively. Suppose there is a pile with no more than $q-1$ stones, then in the remai... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,017 |
Task 2. Determine which function's graph is depicted: a) $y=x^{3}+x^{2}+x+1$; b) $y=x^{3}+2 x^{2}+x+1$; c) $y=x^{3}+x^{2}+3 x+1$; d) $y=x^{3}+x^{2}+x-1$.
 | Answer: a)
Task 3. In triangle $ABC$, side $BC$ is equal to $\frac{5 \sqrt{6}}{3}$, and angles $BAC$ and $ABC$ are $45^{\circ}$ and $75^{\circ}$, respectively. Find side $AB$.
Answer: 5
Task 4. What is the area of the figure in the image, if the side of the cell is 1?
 | Algebra | MCQ | Yes | Yes | olympiads | false | 12,018 |
Task 6. Find the value of the expression $\left(\sqrt[3]{x^{2}} \cdot x^{-0.5}\right):\left(\left(\sqrt[6]{x^{2}}\right)^{2} \cdot \sqrt{x}\right)$ at $x=\frac{1}{2}$ | Answer: 2
Problem 7. The diagonals of trapezoid $ABCD (BC \| AD)$ are perpendicular to each other, and $CD = \sqrt{129}$. Find the length of the midline of the trapezoid, given that $BO = \sqrt{13}, CO = 2\sqrt{3}$, where $O$ is the point of intersection of the diagonals of the trapezoid.
Answer: 10
Problem 8. Four ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,019 |
7. Find the remainder when the number 20172017... 2017 (2017 written 100 times) is divided by 8. | Answer: 1
8. MN and $\mathrm{PQ}$ are two parallel chords located on opposite sides of the center of a circle with radius $10 . \mathrm{MN}=12, \mathrm{PQ}=16$. Find the distance between the chords.
Answer: 14 | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,020 |
Task 4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is not part of the translation, it is just for your guidance.)
Task 4. Solve the equation in in... | Task 4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is not part of the translation, it is just for your guidance.)
Task 4. Solve the equation in in... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,024 |
Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one? | Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one?
## Solutions and Evaluation Criteria
| Task Number | Solution | Criteria |
| :---: | :---: | :---: |
| 1 | Answer: No. Suppose this is possible. No digit is 0, because the... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,025 |
1. Find the largest natural number consisting of distinct digits such that the product of the digits of this number is 2016. | Answer: 876321.
Solution: Factorize the number 2016. $2016=2^{5} \cdot 3^{2} \cdot 7$. To make the number as large as possible, it should contain the maximum number of digits. Notice that the number must include the digit 1. Therefore, the number should consist of the digits $1,2,3,6,7,8$.
Grading Criteria. 20 points... | 876321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,026 |
2. Excellent student Vasya solves exactly 1 algebra problem and 11 geometry problems, or 3 algebra problems and 8 geometry problems, or 15 algebra problems and no geometry problems each day. Over a certain period, Vasya solved 100 algebra problems. Could he have solved 144 geometry problems during this time? | Answer: No.
Solution: Note that the number of geometry problems solved in one day differs from the number of algebra problems solved by a multiple of 5. Therefore, the total number of geometry problems solved should differ from the total number of algebra problems solved by a multiple of 5. However, $144-100=44$ is no... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,027 |
3. Ten consecutive natural numbers are written on the board. What is the maximum number of them that can have a digit sum equal to a perfect square? | Answer: 4.
Solution: Note that the sums of the digits of consecutive natural numbers within the same decade are consecutive natural numbers. Since there are 10 numbers, they span two decades. Also note that among ten consecutive natural numbers, there can be no more than 3 perfect squares, and three perfect squares ca... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,028 |
4. A convex polygon with an area of 7 is placed inside a circle with a radius of 2. Prove that it contains the center of the circle. | Solution: Let the polygon not contain the center of the circle. Then there exists a diameter of the circle that does not intersect with the given polygon, since the polygon is convex. That is, this polygon is completely inside a semicircle. Let's calculate the area of the semicircle $\frac{\pi \cdot 2^{2}}{2}=2 \pi<7$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,029 |
5. In a regular 2015-gon, 64 vertices are marked. Prove that among them there exist four points that are the vertices of some trapezoid. | Solution: First, we will prove that each diagonal of the polygon is parallel to one of its sides. Let $A_{1} A_{2} A_{3} \ldots A_{2015}$ be the given 2015-gon. Each diagonal $A_{i} A_{j} \parallel A_{i+1} A_{j-1} \parallel A_{i+2} A_{j-2} \ldots$ and so on until we reach a side of the polygon (we can assume that $i < ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,030 |
1. The numbers $a$ and $b$ satisfy the equation $\frac{2 a}{a+b}+\frac{b}{a-b}=2$. Find all possible values of the expression $\frac{3 a-b}{a+5 b}$. | 1. From the given equality, it follows that $2 a(a-b)+b(a+b)=2\left(a^{2}-b^{2}\right), b(3 b$ $-a)=0$, from which $b=0$ or $a=3 b$. Both cases are possible. If $b=0$, then the given equality holds for all $a \neq 0$, and the value of the expression $\frac{3 a-b}{a+5 b}$ for all such $a$ and $b$ is 3. If $a=3 b$ and $a... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,031 |
2. From the product of three consecutive natural numbers, their sum was subtracted and an odd number N was obtained. Prove that the number N is the product of some three consecutive odd numbers. | 2. Let $m-1, m, m+1-$ be the initial numbers. Then $N=(m-1) m(m+1)-$ $((m-1)+m+(m+1))=m\left(m^{2}-1\right)-3 m=m\left(m^{2}-4\right)=(m-2) m(m+2)$. The numbers $m-2, m, m+2$ are either consecutive even numbers or consecutive odd numbers. But since $N$ is odd, the numbers $m-2$, $m, m+2$ are odd, which is what we neede... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,032 |
3. Point $K$ is the midpoint of the hypotenuse $\mathrm{AB}$ of a right isosceles triangle $\mathrm{ABC}$. Points $\mathrm{L}$ and $\mathrm{M}$ are chosen on the legs $\mathrm{BC}$ and $\mathrm{AC}$ respectively such that $\mathrm{BL}=\mathrm{CM}$. Prove that triangle $\mathrm{LMK}$ is also a right isosceles triangle. | 3. The median SK of triangle $\mathrm{ABC}$ is also the altitude and bisector, since the triangle is isosceles. Therefore, $\angle KBC = \angle KCB = \angle KCA = 45^{\circ}$. Hence, $KC = KB$, and thus triangles KBL and $KCM$ are congruent by two sides ($\mathrm{KC} = \mathrm{KB}, \mathrm{BL} = \mathrm{CM}$) and the a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,033 |
4. Three brothers - Andrey, Vasily, and Sergey - have the same birthday. When the oldest of them, Andrey, turned 12 years old, it turned out that the sum of the ages of the three brothers was divisible by 12. The same thing happened when Vasily turned 12 years old. Prove that the same will happen when Sergey turns 12 y... | 4. Let on the day of Andrei's twelfth birthday, Vasily is $x$ years old. Then on the same day, Sergei is 12 - $x$ years old. Let's construct the following table:
| | A | V | S |
| :--- | :---: | :---: | :---: |
| day of Andrei's 12th birthday | 12 | $x$ | $12-x$ |
| day of Vasily's 12th birthday | $24-x$ | 12 | $24-2... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,034 |
5. There are 15 coins, among which an even (unknown to us) number are counterfeit. All genuine coins weigh the same, and all counterfeit coins also weigh the same, but they are lighter than the genuine ones. Can at least one genuine coin be found in 3 weighings using a balance scale? | 5. Let's specify the method for finding the genuine coin. For the first weighing, we will place 4 coins on each pan of the scale. There are two possible cases. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,035 |
1. Do there exist four distinct points in space such that any three of them do not have coinciding coordinate values, but any two of them have one coinciding coordinate? | Solution. Yes, they do exist, for example $A(0,0,0), C(1,1,0), B_{1}(1,0,1), D_{1}(0,1,1)$. (The notation for the points is chosen based on a simple way to construct an example - these are the vertices of a tetrahedron inscribed in the unit cube $A B C D A_{1} B_{1} C_{1} D_{1}$. The problem has a simple geometric inte... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,036 |
2. Find all real roots of the equation
$(x+1)^{5}+(x+1)^{4}(x-1)+(x+1)^{3}(x-1)^{2}+(x+1)^{2}(x-1)^{3}+(x+1)(x-1)^{4}+(x-1)^{5}=0$ | Solution. Multiply both sides of the equation by $(x+1)-(x-1)$ (this factor equals 2). Use the formula $a^{6}-b^{6}=(a-b)\left(a^{5}+a^{4} b+a^{3} b^{2}+a^{2} b^{3}+a b^{4}+b^{5}\right)$.
$$
\begin{gathered}
(x+1)^{6}-(x-1)^{6}=0 \\
(x+1)^{6}=(x-1)^{6}
\end{gathered}
$$
The equation
$$
x+1=x-1
$$
has no solutions. ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,037 |
3. Four points $A, B, C, D$ are on a plane. It is known that $A B=1, B C=$ $2, C D=\sqrt{3}, \angle A B C=60^{\circ}, \angle B C D=90^{\circ}$. Find $A D$. | Solution. Let's construct the diagram. Let the line $CD$ intersect the line $AB$ at point $O$
(according to the condition, these lines are not parallel). There are two possible positions for... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,038 |
4. The sum of several natural numbers is 972. What is the greatest possible value of their product? | Solution. Let's consider into which addends 972 should be divided to obtain the maximum possible product. Among the addends, there should be no ones - if there is a one, we can add it to any of the other addends, and the product will increase. Any addend greater than 4 can be broken down into twos and threes. That is, ... | 3^{324} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,039 |
5. The parabolas in the figure are obtained by shifting the parabola $f(x)=x^{2}$ and are arranged such that the points of their intersections with the $O X$ axis respectively coincide in pairs, and all vertices lie on the same straight line. There are a total of 2020 parabolas. The length of the segment on the $O X$ a... | Solution. Note that the arrangement of parabolas can be changed by shifting the entire set to the left or right (the lengths of the segments on the $O X$ axis, enclosed between the roots of the parabolas, do not change with the shift). The intersection point of the line drawn through the vertices of the parabolas will ... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,040 |
1. Find at least one solution to the puzzle: KO,M + K,OM = KR,EM (identical digits are denoted by identical letters, different digits by different letters). | 1. Solution. For example: $32.4 + 3.24 = 35.64$. Note. The problem has many solutions, it is sufficient to provide only one. Grading criteria: Correct example - 7 points. In other cases - 0 points. | 32.4+3.24=35.64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,041 |
3. Two escalators are moving at the same speed: one going down, the other going up. Seventh-grader Semyon stepped onto the escalator going down and tried to climb up it, but with his usual effort, he could not move at all. At the moment Semyon stepped onto the adjacent escalator, he dropped his glove on the first step.... | 3. Answer: Exactly at the midpoint of the escalator. Solution. Semyon tried to move up an escalator moving downwards and could not move from his spot. Therefore, Semyon's own speed is equal to the speed of the escalator. According to the problem, the escalators move at the same speed. Thus, when Semyon started to climb... | Exactly\at\the\midpoint\of\the\escalator | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,043 |
4. Can rectangular parallelepipeds of size $2 \times 2 \times 1$ be used to form a rectangular parallelepiped of size $3 \times 4 \times 5$? Don't forget to justify your answer. | 4. Answer: No, it is not possible. Solution. Suppose we managed to assemble a large parallelepiped from small ones, then each face of the large parallelepiped consists of some faces of the small ones. But all the faces of the small parallelepiped are rectangles with even area. And the large parallelepiped has a face $3... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,044 |
5. Three numbers are stored in a computer's memory. Every second, the following operation is performed: each number in this triplet is replaced by the sum of the other two. For example, the triplet $(1 ; 3 ; 7)$ turns into $(10 ; 8 ; 4)$. What will be the difference between the largest and smallest number in the triple... | 5. Answer: 19. Solution. Let the initial triplet be ( $a ; b ; c$). Since all numbers in the triplet are distinct, we will assume that $a<b<c$. In the next second, the triplet of numbers will look like this: $(b+c$; $a+c ; a+b)$. The largest number in this triplet is $b+c$, and the smallest is $a+b$. Their difference i... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,045 |
8.1. Given positive numbers $a, b, c, d, e$. It is known that $a b=2, b c=3, c d=4, d e=15$, $e a=10$. What is the value of $a$? | Answer: $a=\frac{4}{3}$.
Multiply the first, third, and fifth equations and divide by the second and fourth: $\frac{a b \cdot c d \cdot e a}{b c \cdot d e}=\frac{2 \cdot 4 \cdot 10}{3 \cdot 15}$. From this, $a^{2}=\frac{16}{9}$ and $a=\frac{4}{3}$. | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,046 |
8.2. Find the largest four-digit number in which all digits are different and which is divisible by any of its digits (don't forget to explain why it is the largest). | Answer: 9864.
Firstly, the desired number cannot have the form $\overline{987 a}$, because divisibility by the digit 7 would mean that $a$ is 0 or 7. This means the desired number is smaller. Secondly, consider numbers of the form $\overline{986 a}$. Divisibility by the digit 9 would mean that $9+8+6+a=a+23$ is divisi... | 9864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,047 |
8.3. On a test, the teacher gave five problems and graded the test with a score equal to the number of problems solved. All students, except for Petya, solved the same number of problems, while Petya solved one more. The first problem was solved by 9 people, the second by 7 people, the third by 5 people, the fourth by ... | Answer: None. Suppose Petya got no less than a four, then the others solved no less than 3 problems each, and the total number of problems solved by all students is no less than $3 \cdot 9=27$ (from the condition it is clear that the number of students is no less than 9). But, on the other hand, this number is equal to... | None | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,048 |
5.1. To some number $\kappa$, the sum of its digits was added and the result was 2014. Provide an example of such a number. | Answer: 1988 or 2006.
Grading Criteria:
+ a correct example is provided (one is sufficient)
$\pm$ a correct example is provided along with an incorrect one
- the problem is not solved or solved incorrectly | 1988 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,049 |
5.2. The Wolf, the Hedgehog, the Chizh, and the Beaver were dividing an orange. The Hedgehog got twice as many segments as the Chizh, the Chizh got five times fewer segments than the Beaver, and the Beaver got 8 more segments than the Chizh. Find out how many segments were in the orange, if the Wolf only got the peel. | Answer: 16 segments.
Solution. First method. Let the number of orange segments given to Chizh be $x$, then Hedgehog received $2x$ segments, and Beaver received $5x$ segments (Wolf - 0 segments). Knowing that Beaver received 8 more segments than Chizh, we set up the equation: $5x - x = 8$. The solution to this equation... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,050 |
5.3. In a seven-story building, domovoi (Russian house spirits) live. The elevator travels between the first and the last floors, stopping at every floor. On each floor, starting from the first, one domovoi entered the elevator, but no one exited. When the thousandth domovoi entered the elevator, it stopped. On which f... | Answer: on the fourth floor.
Solution. First, let's find out how many housekeepers ended up in the elevator after the first trip from the first to the seventh floor and back, until the elevator returned to the first floor. One housekeeper entered on the first and seventh floors, and on all other floors, two housekeepe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,051 |
5.5. After the hockey game, Anton said that he scored 3 goals, while Ilya only scored one. Ilya said that he scored 4 goals, and Seryozha scored as many as 5. Seryozha said that he scored 6 goals, and Anton only two. Could it be that together they scored 10 goals, given that each of them told the truth once and lied on... | Answer: No, it could not.
Solution. The first method. There are two possible cases.
1) If Anton told the truth about himself, then he scored 3 goals. Then Seryozha lied about Anton, so Seryozha told the truth about himself, meaning he really scored 6 goals. Consequently, Ilya lied about Seryozha and told the truth ab... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,053 |
10.1. Can a quadratic equation $a x^{2}+b x+c=0$ with integer coefficients have a discriminant equal to 23? | Answer: No.
Solution. Suppose that the discriminant of the given equation is equal to the number 23. Then we can write: $b^{2}-4 a c=23$, and $b^{2}-25=4 a c-2$ or $(b-5) \cdot(b+5)=2(2 a c-1)$.
Note that $b-5$ and $b+5$ are numbers of the same parity, so their product, if it is even, is divisible by 4. The right sid... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,054 |
10.2. An isosceles triangle with base $a$ and angle at the base $\alpha$ is inscribed in a circle. Additionally, a second circle is constructed that is tangent to the first circle and the base of the triangle, with the point of tangency being the midpoint of the base. Find the radius of the second circle. | Answer: $\frac{a}{4} \operatorname{tg} \alpha, \frac{a}{4} \operatorname{ctg} \alpha$.
Solution. The diameter of one of the desired circles is the height of the given triangle, and
the diame... | \frac{}{4}\operatorname{tg}\alpha,\frac{}{4}\operatorname{ctg}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,055 |
10.3. Vova placed several (possibly 0) chess pieces on an $8 \times 8$ board. Lena noticed that in each $2 \times 2$ square, the same number of pieces is placed. And Vlad noticed that in each $3 \times 1$ (or $1 \times 3$) rectangle, the same number of pieces is placed. How many pieces were placed on the board? (List a... | Answer: 0 or 64.
Solution. Suppose that in each $2 \times 2$ square there are $m$ figures, and in each $1 \times 3$ rectangle there are $n$ figures. Let's select a $2 \times 6$ rectangle from the board. On one hand, this rectangle can be divided into three $2 \times 2$ squares, and thus it contains $3 m$ figures. On t... | 0or64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,056 |
10.4. Prove that if the sum $\left(x^{2}+y^{2}\right)$ is divisible by 3 and $x, y$ are integers, then $x$ and $y$ are divisible by 3. | Solution. Let $x=3 q+r_{1}, y=3 p+r_{2}$, where $r_{1}$ and $r_{2}$ are the remainders when divided by 3, that is, the numbers $0,1,2$. Then $x^{2}+y^{2}=\left(3 q+r_{1}\right)^{2}+\left(3 p+r_{2}\right)^{2}=3\left(3 q^{2}+3 p^{2}+2 q r_{1}+2 p r_{2}\right)+r_{1}{ }^{2}+r_{2}^{2}$. Since $x^{2}+y^{2}$ is divisible by 3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,057 |
10.5. Currently, there are coins of 1, 2, 5, and 10 rubles. Indicate all monetary amounts that can be paid with both an even and an odd number of coins. (You can use identical coins.) | Answer: Any amount of money greater than 1 ruble can be paid with either an even or odd number of coins.
Solution. Any amount of money greater than 10 rubles can be made up of 10-ruble and 1-ruble coins. The parity of the number of coins can be changed by exchanging 10 rubles for 2 coins of 5 rubles. An amount of mone... | Anyamountofmoneygreaterthan1rublecanbepaidwitheitheranevenoroddof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,058 |
1. Find the maximum possible area of a quadrilateral in which the product of any two adjacent sides is 1. | 1. Find the maximum possible area of a quadrilateral for which the product of any two adjacent sides is 1.
OTBET: 1.
SOLUTION.
Let the quadrilateral have sides $a, b, c, d$. Then $a b=b c=c d=d a=1$. From the equality $a b=b c$, it follows that $a=c$, and from the equality $b c=c d$, we get that $b=d$. Therefore, th... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,059 |
2. For positive numbers $a, b, c, d$, it is known that $a b c d=1$. Prove that among the numbers $\frac{a^{2}+1}{b^{2}}, \frac{b^{2}+1}{c^{2}}, \frac{c^{2}+1}{d^{2}}, \frac{d^{2}+1}{a^{2}}$ there is a number not less than 2. | 2. For positive numbers $a, b, c, d$, it is known that $a b c d=1$. Prove that among the numbers $\frac{a^{2}+1}{b^{2}}, \frac{b^{2}+1}{c^{2}}, \frac{c^{2}+1}{d^{2}}, \frac{d^{2}+1}{a^{2}}$ there is a number not less than 2.
SOLUTION.
Assume the opposite, then each number is less than 2. Multiplying all these positiv... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,060 |
3. Find all triples of pairwise distinct real numbers $x, y, z$ that are solutions to the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=-x+3 y+z \\
y^{2}+z^{2}=x+3 y-z \\
z^{2}+x^{2}=2 x+2 y-z
\end{array}\right.
$$ | 3. Find all triples of pairwise distinct real numbers $x, y, z$ that are solutions to the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=-x+3 y+z \\
y^{2}+z^{2}=x+3 y-z \\
z^{2}+x^{2}=2 x+2 y-z
\end{array}\right.
$$
ANSWER: $x=0, y=1, z=-2$ or $x=-\frac{3}{2}, y=\frac{5}{2}, z=-\frac{1}{2}$.
SOLUTION. T... | 0,1,-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,061 |
4. Find the radius of the sphere that touches all the edges of a regular tetrahedron with edge $a$.
untranslated text remains in its original form. | 4. Find the radius of the sphere that touches all the edges of a regular tetrahedron with edge length \(a\).
ANSWER. \(r = \frac{a \sqrt{2}}{4}\)
SOLUTION.
Let \(M\) be the center of the face \(ABC\) of the regular tetrahedron \(ABCD\), and \(O\) be the center of the given sphere. Since point \(O\) is equidistant fr... | \frac{\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,062 |
5. Thirteen girls and thirteen boys participated in a math competition. Each participant solved no more than four problems. For any girl and boy, there is at least one problem solved by both. Prove that there was a problem solved by no fewer than three girls and no fewer than three boys. | 5. Thirteen girls and thirteen boys participated in a math competition. Each participant solved no more than four problems. For any girl and boy, there is at least one problem solved by both. Prove that there was a problem solved by no fewer than three girls and no fewer than three boys.
SOLUTION.
By contradiction. S... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,063 |
7.2. Cards with numbers $7,8,9,4,5$, $6,1,2,3$ are laid out in a row. It is allowed to take several consecutive cards and reverse their order. Is it possible to achieve the arrangement $1,2,3,4,5,6,7,8,9$ in three such operations? | Answer: Yes.
First, rearrange the first six cards in reverse order, resulting in 654987123, then rearrange cards from 4 to 9, resulting in 654321789, and then again the first 6 cards. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,064 |
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