problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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|---|---|---|---|---|---|---|---|---|
5. In triangle $\mathrm{ABC}$, the sides $A B=4, B C=6$. Point $M$ lies on the perpendicular bisector of segment $A B$, and lines $A M$ and $A C$ are perpendicular. Find $M A$, if the radius of the circumscribed circle around triangle $A B C$ is 9. | Solution. Introduce a coordinate system with the origin at point A such that point C lies on the x-axis. From the problem statement, point M lies on the y-axis. Let's introduce notations for the unknown coordinates: $\mathrm{A}(0,0), \mathrm{B}\left(\mathrm{x}_{\mathrm{B}}, \mathrm{y}_{\mathrm{B}}\right), \mathrm{C}\le... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find all prime numbers whose representation in base 14 has the form 101010 ... 101 (ones and zeros alternate). | Solution: Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$. Let $q=$ 14 - the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac... | 197 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. How many solutions of the equation $x^{2}-2 x \cdot \sin (x \cdot y)+1=0$ fall within the circle $x^{2}+y^{2} \leq 100$? | Solution. We will interpret the left side of the equation as a quadratic trinomial in terms of $x$. For the roots to exist, the discriminant must be non-negative, i.e., $D=4 \sin ^{2}(x \cdot y)-4 \geq 0 \Leftrightarrow \sin ^{2}(x \cdot y)=1 \Leftrightarrow \cos 2 x y=-1 \Leftrightarrow x y=\frac{\pi}{2}+\pi n, n \in ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$... | Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}
$$
Then, according to the problem, we have
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x... | -1216 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A robot is located in one of the cells of an infinite grid and can be given the following commands:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the ... | Solution. For brevity, let's denote the command to the left as L, to the right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Find all prime numbers whose decimal representation has the form 101010 ... 01. | Solution. Let $2 n+1$ be the number of digits in the number $A=101010 \ldots 101$ under investigation. Let $q=$ 10 be the base of the number system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25$ (712) is 3.) | Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.). As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$. We obtain the non-repeating part 123 and the repeating part 63. Let's discuss why the non-repeating ... | 48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$... | Solution. The vertices of Anya's triangle divide the circle into three arcs (see figure). Let $x, y$, and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one o... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (4 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks, none of which attack each other, exist such that the numbers on the squares occupied by the rooks include all numbers from 0 to 7? $\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7... | # Solution:
Let's number the rows of the board from 1 to 8 from top to bottom. On the first row, the position of the rook can be chosen in 8 ways. For example, we choose the first cell with the number 0. On the second row, there are 7 options left. Six of them (cells from the second to the seventh) have the property t... | 3456 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) From point $A$, lying on a circle of radius 3, chords $A B, A C$ and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{4}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer area of triangle $A B C$.
# | # Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) From point $A$, lying on a circle, chords $A B$, $A C$, and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{6}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer part of the radius of the circle if th... | # Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. There is an unlimited number of test tubes of three types - A, B, and C. Each test tube contains one gram of a solution of the same substance. Test tubes of type A contain a $10\%$ solution of this substance, type B $-20\%$ solution, and type C $-90\%$ solution. Sequentially, one after another, the contents of the t... | Solution: Let the number of test tubes of types A, B, and C be $a$, $b$, and $c$ respectively. According to the problem, $0.1a + 0.2b + 0.9c = 0.2017 \cdot (a + b + c) \Leftrightarrow 1000 \cdot (a + 2b + 9c) = 2017 \cdot (a + b + c)$. The left side of the last equation is divisible by 1000, so the right side must also... | 73 | Other | math-word-problem | Yes | Yes | olympiads | false |
9. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ). | Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Co... | 5035485 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. In a triangle with sides $a, b, c$ and angles $\alpha, \beta, \gamma$, the equality $3 \alpha + 2 \beta = 180^{0}$ is satisfied. The sides $a, b, c$ lie opposite the angles $\alpha, \beta, \gamma$ respectively. Find the length of side $c$ when $a=2, b=3$.
 the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1
2) the rows of the matrix do not repeat. | Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the sam... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A motorist set off from point A to point B, the distance between which is 100 km. At the moment when the navigator showed that 30 minutes of travel remained, the motorist first reduced the speed by 10 km/h, and at the moment when the navigator showed that 20 km remained to travel, the motorist reduced the speed by t... | Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be denoted as
$x$ (km), then the distance from B to C will be $(80 - x)$ km. Let $v \frac... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find all roots of the equation $\frac{1}{\cos ^{3} x}-\frac{1}{\sin ^{3} x}=4 \sqrt{2}$, lying in the interval $\left(-\frac{\pi}{2}, 0\right)$. Write the answer in degrees. | # Solution:
$\sin ^{3} x-\cos ^{3} x=4 \sqrt{2} \sin ^{3} x \cos ^{3} x \Leftrightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x+\cos ^{2} x\right)=4 \sqrt{2} \sin ^{3} x \cos ^{3} x$. Substitution: $\sin x-\cos x=t, \sin x \cos x=\frac{1-t^{2}}{2}$. Then $t\left(3-t^{2}\right)=\sqrt{2}\left(1-t^{2}\right)^{3}$. ... | -45 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $\left.1^{2}+2^{2}+4^{2}=21\right)$.
| Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Co... | 5035485 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A circle touches the sides of an angle at points $A$ and $B$. A point $M$ is chosen on the circle. The distances from $M$ to the sides of the angle are 24 and 6. Find the distance from $M$ to the line $A B$.
=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$... | Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$: $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$.
Then, according to the problem statement, we have:
$f(\quad)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\rig... | -1216 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Find the number of matrices that satisfy two conditions:
3) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1 and the rows of the matrix do not repeat. | Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the sam... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Non-zero numbers $a$ and $b$ are roots of the quadratic equation $x^{2}-5 p x+2 p^{3}=0$. The equation $x^{2}-a x+b=0$ has a unique root. Find $p$. | Solution. Since the equation $x^{2}-a x+b=0$ has a unique root, then $b=\frac{a^{2}}{4}$. By Vieta's theorem, we have the equalities: $a+b=5 p ; a b=2 p^{3}$. Substituting $b=\frac{a^{2}}{4}$ into the last equality, we get: $a=2 p$. Considering that $a$ and $b$ are non-zero, we find $p=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$. | Solution. If point $D$ is reflected across line $A F$, and then across line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\circ}$ rotati... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (5 points) The sides of a parallelogram are 5 and 13, and the angle between them is $\arccos \frac{6}{13}$. Two mutually perpendicular lines divide this parallelogram into four equal-area quadrilaterals. Find the lengths of the segments into which these lines divide the sides of the parallelogram.
# | # Solution:
Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, $\angle BAC = \alpha = \arccos c$. Clearly, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.
Denote $|DN| =... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) For a natural number $x$, five statements are made:
$$
3 x>91
$$
$$
\begin{aligned}
& x37 \\
& 2 x \geq 21 \\
& x>7
\end{aligned}
$$
It is known that only three of them are true, and two are false. Find $x$. | Solution. Let's transform the original system of inequalities
$$
\begin{aligned}
& x>\frac{91}{3} \\
& x \geq \frac{21}{2} \\
& x>\frac{37}{4} \\
& x>7 \\
& x<120
\end{aligned}
$$
The first inequality does not hold, since otherwise, four inequalities would be satisfied immediately. Therefore, $x \leq \frac{91}{3}$, a... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Fina... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent ... | Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be t... | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent ... | Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be t... | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate). | Solution: Let $2n+1$ be the number of digits in the number $A=101010$...101 under investigation. Let $q=$ 10 be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} ... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$... | Solution. The vertices of Anya's triangle divide the circle into three arcs. Let $x, y$ and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one of the arcs. Cl... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A square table consists of 2014 rows and 2014 columns. In each cell at the intersection of the row with number $i$ and the column with number $j$, the number $a_{i, j}=(-1)^{i}(2015-i-j)^{2}$ is written. Find the sum of all the numbers in the table | Problem 3. Answer: $\mathbf{0 .}$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.
... | Solution. We will prove that ${ }^{1}$
Multiply the equation (a) of the original system
$$
a_{2} b_{3}=a_{3} b_{2}
$$
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cas... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. It is known that there exists a natural number $N$ such that
$$
(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}
$$
Find $N$.
 | Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (here $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with integer coefficients that are not important to us now) of the form $a^{N-n}(b \sqrt{3})^{n}$. Th... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) It is known about the numbers $x_{1}$ and $x_{2}$ that $x_{1}+x_{2}=2 \sqrt{1703}$ and $\left|x_{1}-x_{2}\right|=90$. Find $x_{1} \cdot x_{2}$. | Solution: Let $A=x_{1}+x_{2}, B=x_{1} \cdot x_{2}, C=\left|x_{1}-x_{2}\right|$. Since $C^{2}=A^{2}-4 \cdot B$, we find $B=-322$.
Answer: -322 . | -322 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) Find the number of integers from 1 to 1000 inclusive that give the same remainder when divided by 11 and by 12. | Solution. Let $r_{n}(a)$ be the remainder of the division of the number $a$ by the number $n$. Suppose $a \in [1 ; 1000]$ and $r_{11}(a)=r_{12}(a)=t$. Then $t \in \{0, \ldots, 10\}$ and the following equality holds:
$$
t+11 k=t+12 m=a, \quad k, m \in N_{0}
$$
From the last equality, it follows that $k$ is divisible b... | 87 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of all natural numbers $n$ that are multiples of three and for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{3}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.) | Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have the... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (4 points) Find the area of a triangle if two of its medians are equal to $\frac{15}{7}$ and $\sqrt{21}$, and the cosine of the angle between them is $\frac{2}{5}$.
# | # Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \\... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (5 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks are there such that no rook attacks another and the numbers on the squares occupied by the rooks include all numbers from 0 to 7?
\(\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7... | Answer: 3456.
## Variant 2 | 3456 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. (4 points) Find the area of a triangle if two of its medians are equal to 3 and $2 \sqrt{7}$, and the cosine of the angle between them is $-\frac{3}{4}$.
# | # Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \c... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$. | Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Multiply equation (a) of the original system
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cases}
... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. [6] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 residents were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The residents answered in turn, and everyone could hear their answers. Knights answered tr... | Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadyr) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values ... | 1009 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. [8] A $7 \times 7$ board is either empty or has an invisible $2 \times 2$ ship placed on it "by cells." It is allowed to place detectors in some cells of the board and then turn them on simultaneously. An activated detector signals if its cell is occupied by the ship. What is the smallest number of detectors needed ... | Answer: 16 detectors. Solution: Evaluation. In each $2 \times 3$ rectangle, there should be at least two detectors: the rectangle consists of three $1 \times 2$ dominoes, and if a detector is in the outer domino, we cannot determine whether there is a ship on the other two dominoes, and if a detector is in the middle d... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. [5] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly 4 chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev) | Answer: For 61 rubles. Solution: Let's number the chips and cells in order from 0 to 99. The free operation does not change the remainder of the cell number when divided by 5.
Estimate. Let's mentally arrange the piles of chips in a circle. First, the pile of chips with a remainder of 0, then with 1, and so on up to 4... | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a triangular pyramid $S A B C$, the base of which is an equilateral triangle $A B C$, and all plane angles at vertex $S$ are equal to $\alpha$. For what least $\alpha$ can we assert that this pyramid is regular?
M. Malkin | Answer: $60^{\circ}$.
We will prove that when $\alpha=60^{\circ}$, the pyramid is regular. Let the sides of the triangle $ABC$ be equal to 1. Note that in any triangle with an angle of $60^{\circ}$, the side opposite this angle is the middle-length side (and if it is strictly less than one of the sides, it is strictly... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles required to rearrange the chips in reverse order?
(Egor Bakaev) | Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells i... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. [9] Inside an isosceles triangle $A B C$, a point $K$ is marked such that $C K=A B=B C$ and $\angle K A C=30^{\circ}$. Find the angle $A K B$.
E. Bakayev | Answer: $150^{\circ}$. Solution 1. Construct an equilateral triangle $B C L$ (see the figure; points $A$ and $L$ are on the same side of the line $B C$). Points $A, C$, and $L$ lie on a circle with radius $B A$ and center at point $B$. Since $K$ lies inside triangle $A B C$, angle $A B C$ is greater than $60^{\circ}$, ... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. [12] There are 100 piles, each with 400 stones. In one move, Petya selects two piles, removes one stone from each, and earns as many points as the absolute difference in the number of stones in these two piles. Petya must remove all the stones. What is the maximum total number of points he can earn?
M. Diden | Answer: 3920000. Solution. Estimation. We will assume that the stones in the piles are stacked on top of each other, and Petya takes the top (at the moment) stones from the selected piles. We will number the stones in each pile from bottom to top with numbers from 1 to 400. Then the number of points Petya gets on each ... | 3920000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. [5] Several natural numbers are written in a row with a sum of 2019. No number and no sum of several consecutive numbers equals 40. What is the maximum number of numbers that could have been written?
A. Shapovalov | Answer: 1019 numbers.
Lemma. The sum of any forty consecutive numbers is not less than 80.
Proof. Let the numbers $a_{1}, \ldots, a_{40}$ be written consecutively. Among the numbers $b_{0}=0, b_{1}=a_{1}$, $b_{2}=a_{1}+a_{2}, \ldots, b_{40}=a_{1}+a_{2}+\ldots+a_{40}$, there will be two $b_{i}$ and $b_{j}(i1019$ numbe... | 1019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On the first day, $2^{n}$ schoolchildren played ping-pong in a "knockout" format: first, two played, then the winner played with the third, the winner of this pair played with the fourth, and so on until the last schoolchild (there are no draws in ping-pong). On the second day, the same schoolchildren played for the... | Answer: 3.
Solution. Let's construct the following graph: vertices are players, edges are played matches. According to the condition, for both tournaments, this graph is the same. Consider the first tournament and select the matches where the winners had not won before (for example, the first match). Then the correspo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. There is a 1001-digit natural number $A$. The 1001-digit number $Z$ is the same number $A$, written from end to beginning (for example, for four-digit numbers, these could be 7432 and 2347). It is known that $A > Z$. For which $A$ will the quotient $A / Z$ be the smallest (but strictly greater than 1)? | Answer. For $A$, whose notation (from left to right) is: 501 nines, an eight, and 499 nines.
Solution 1. Let $A=\overline{a_{1000} a_{999 \ldots a_{0}}}$. Since $A>Z$, among the digits $a_{0}, a_{1}, \ldots, a_{499}$ there is at least one non-nine. Therefore, $Z \leqslant Z_{0}=\underbrace{99 \ldots 9}_{499} 8 \underb... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev) | Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells i... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. [5] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 inhabitants were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The inhabitants answered in turn, and everyone could hear their answers. Knights answere... | Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadir) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values ... | 1009 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. [8] The three medians of a triangle divide its angle into six angles, of which exactly $k$ are greater than $30^{\circ}$. What is the greatest possible value of $k$?
(N. Sedrakyan) | Answer. $k=3$.
Evaluation. First method. Let $h_{1} \leq h_{2} \leq h_{3}$ be the heights of the triangle, and $m_{1}, m_{2}, m_{3}$ be the medians from the corresponding vertices (in fact, $m_{1} \leq m_{2} \leq m_{3}$, but this will not be used). Then we have $m_{3} \geq h_{i}$ for $i=1,2$, 3. Dropping perpendicular... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. [5] Pentagon $A B C D E$ is circumscribed around a circle. The angles at its vertices $A, C$ and $E$ are $100^{\circ}$. Find the angle $A C E$. | Answer: $40^{\circ}$. It is not hard to understand that such a pentagon exists.
Solution 1. The lines connecting the vertices with the center $O$ of the inscribed circle $\omega$ are the angle bisectors of the pentagon. Therefore, $\angle O A E=\angle O E A=50^{\circ}, \angle A O E=80^{\circ}$. Let $\omega$ touch the ... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.
It is known that three numbers $a_{1}, a_{2}, a_{3}$ were obtained as follows. First, a natural number $A$ was chosen and the numbers $A_{1}=[A]_{16}, A_{2}=$ $[A / 2]_{16}, A_{3}=[A / 4]_{16}$ were found, where $[X]_{16}$ is the remainder of the integer part of the number $X$ divided by 16 (for example, $[... | # Answer: 16.
## Conditions and answers to the problems of the final stage of the 2012-13 academic year
# | 16 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Task 2.
In a table consisting of $n$ rows and $m$ columns, numbers (not necessarily integers) were written such that the sum of the elements in each row is 408, and the sum of the elements in each column is 340. After that, $k$ columns were added to the table, the sum of the elements in each of which is 476, and a col... | Answer: 4 when $n=65, m=78, k=18$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $ik$. The user... | 5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The layout of the digits on the keyboard changes after entering the code based on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user ent... | 3212 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., ... | 3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., ... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The use... | 6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The use... | 3212 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Workers at an aluminum plant played the following game during their break. They drew cards from a deck (36 cards) until they had cards of all 4 suits among the drawn cards. What is the minimum number of cards that must be drawn to ensure that all 4 suits are among them?
a) 4
b) 9
c) 18
d) 28
e) 36 | 8. answer: g. There are 9 cards of each suit. If you draw 27 cards, they could all be from only three suits. Suppose you draw 28 cards and among the first 27 cards, they are only from three suits, then in the deck, only cards of the fourth suit will remain. And the 28th card will already be of the 4th suit. | 28 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
12. The director of an aluminum plant wants the advertisement of his plant, which lasts 1 min. 30 sec., to be shown exactly every 18 min. 30 sec. after the end of the previous advertising break. How much time will the advertisement of the aluminum plant be shown in a week?
a) 12 h.
b) 12 h. 12 min.
c) 12 h. 24 min.
... | 12. answer: g. In a week, there are $7 \cdot 24=168$ hours. For every hour, advertisements are shown for $1.5 \cdot 3=4.5$ minutes. Therefore, in a week, advertisements make up $4.5 \cdot 168=756=12$ hours and 36 minutes of airtime. In a week, there are $7 \cdot 24=168$ hours. For every hour, advertisements are shown f... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. During the lunch break on the day of the "13th Element. ALchemy of the Future" competition, a cafeteria worker mixed 2 liters of one juice with a $10 \%$ sugar content and 3 liters of another juice with a $15 \%$ sugar content. What is the sugar content of the resulting mixture?
a) $5 \%$
б) $12.5 \%$
в) $12.75 \%... | 8. Answer g. In the first juice, there is $0.1 \cdot 2=0.2$ liters of sugar, and in the second juice, there is $0.15 \cdot 3=0.45$ liters of sugar. Then, in the resulting mixture, there will be $0.2+0.45=0.65$ liters of sugar. The total volume is 5 liters. Then the percentage of sugar in the mixture is $\frac{0.65}{5} ... | 13 | Algebra | MCQ | Yes | Yes | olympiads | false |
15. Hydrogen was passed over a heated powder (X1). The resulting red substance (X2) was dissolved in concentrated sulfuric acid. The resulting solution of the substance blue (X3) was neutralized with potassium hydroxide - a blue precipitate (X4) formed, which upon heating turned into a black powder (X1). What substance... | 15. 80 g/mol; $\mathrm{m}(\mathrm{CuO})=80$ g/mol; $\mathrm{X}_{1}-\mathrm{CuO} ; \mathrm{X}_{2}-\mathrm{Cu} ; \mathrm{X}_{3}-\mathrm{CuSO}_{4 ;} \mathrm{X}_{4}-\mathrm{Cu}(\mathrm{OH})_{2}$ | 80 | Other | math-word-problem | Yes | Yes | olympiads | false |
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum num... | 16. 7. Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one participant with more than three points. Since there are still 3 rounds ahead, t... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Young researcher Petya noticed that the iron at home heats up by $9^{\circ} \mathrm{C}$ every 20 seconds, and after being turned off, it cools down by $15^{\circ} \mathrm{C}$ every 30 seconds. Last evening, after turning off the iron, it cooled down for exactly 3 minutes. How long was the iron turned on?
a) 3 min 20... | 8. answer: a. 3 minutes $=180$ seconds. In 180 seconds, the iron cools down by $180 / 30 \cdot 15=90$ degrees. To heat up by 90 degrees, the iron takes $90 / 9 \cdot 20=200$ seconds. 200 seconds $=3$ minutes 20 seconds. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. The security guard of the aluminum plant works on Tuesdays, Fridays, and on odd-numbered days. What is the maximum number of consecutive days the security guard can work?
a) 3
b) 4
c) 5
d) 6
e) 7 | 12. answer: g. 6 days. An example of such a situation: 29th (odd), 30th (Tuesday), 31st (odd), 1st (odd), 2nd (Friday), 3rd (odd). | 6 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
16. A worker at an aluminum plant can produce 16 blanks or 10 parts from blanks in one shift. It is known that exactly one part is made from each blank. What is the maximum number of blanks the worker can produce in one shift to make parts from them in the same shift? | 16. 6. Let's denote by $x$ the duration of the worker's working day in hours. Then the worker makes one part from a blank in $\frac{x}{10}$ hours, one blank in $\frac{x}{16}$ hours, and a blank and a part from it in $\frac{x}{10}+\frac{x}{16}=\frac{13 x}{80}$ hours. Since $x: \frac{13 x}{80}=6 \frac{2}{13}$, the maximu... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. What is the mass fraction (%) of oxygen in aluminum oxide $\mathrm{Al}_{2} \mathrm{O}_{3}$?
a) $53 \%$
b) $27 \%$
c) $16 \%$
d) $102 \%$ | 7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
The text is already in a format that is a mix... | 53 | Other | MCQ | Yes | Yes | olympiads | false |
16. Every sixth bus in the bus park of the aluminum plant is equipped with an air conditioner. After the plant director ordered to install it on 5 more buses, a quarter of the buses had an air conditioner. How many buses are there in the park of the plant, if each bus is equipped with only one air conditioner? | 16. 60. Let the number of buses in the park be denoted by $x$. Then, $\frac{1}{6} x + 5$ buses are equipped with air conditioning. This constitutes a quarter of all the buses in the park. We have the equation: $\left(\frac{1}{6} x + 5\right) \cdot 4 = x$ or $\frac{1}{3} x = 20$. From this, $x = 60$. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three is to the sum of all terms except the last three as $4: 3$. Find the number of terms in this progressio... | Answer: 20.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$. | Answer: 11000.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{d c b a}=1000 d+100 c+10 b+a$, where $a$, $b, c, d$ are digits and $a \neq 0$.
According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.
We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from w... | 11000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{31}{65} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{31}{65} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be deno... | 2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10. A cylinder of volume 9 is inscribed in a cone. The plane of the upper base of this cylinder cuts off a frustum of volume 63 from the original cone. Find the volume of the original cone. | Answer: 64.
Solution. Let the height and radius of the original cone be $H$ and $R$, and the height and radius of the cylinder be $h$ and $r$. We use the formula for the volume of a frustum of a cone: $\frac{1}{3} \pi\left(R^{2}+R r+r^{2}\right) h=63$. We also know that $\pi r^{2} h=9$. Dividing the corresponding part... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $5: 4$. Find the number of terms in this prog... | Answer: 22.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90. | Answer: 45.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.
According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.
We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so sin... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{13}{155} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{13}{155} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be de... | 2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $3: 2$. Find the number of terms in this prog... | Answer: 18.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. | Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{31}{65} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{31}{65} \overrightarrow{AO}$.
Let the v... | -2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is in the ratio $6: 5$ to the sum of all terms except the last three. Find the number of terms in... | Answer: 24.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. If the sum of the digits of a four-digit number $X$ is subtracted from $X$, the result is a natural number $N=K^{2}$, where $K$ is a natural number that gives a remainder of 5 when divided by 20 and a remainder of 3 when divided by 21. Find the number $N$. | Answer: 2025.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d$, where $a, b, c, d$ are digits and $a \neq 0$.
According to the condition, $X-a-b-c-d=999 a+99 b+9 c=K^{2}$, where $K=20 u+5, K=21 v+3$.
Notice that $999 a+99 b+9 c \vdots 9$, i.e., $K^{2} \vdots 9$, which means $K=3 M$, and $M^{2}=111 a+11 b+c \... | 2025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$. | Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{13}{155} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{13}{155} \overrightarrow{AO}$.
Let the... | -2015 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 1. What is greater: 1 or $\frac{21}{64}+\frac{51}{154}+\frac{71}{214} ?$ | Answer: One is greater.
## First solution.
$$
\frac{21}{64}+\frac{51}{154}+\frac{71}{214}<\frac{21}{63}+\frac{51}{153}+\frac{71}{213}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1
$$
## Second solution.
$$
\begin{aligned}
\frac{21}{64}+\frac{51}{154}+\frac{71}{214} & =\frac{21 \cdot 154 \cdot 214+64 \cdot 51 \cdot 214+64 \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored thirteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next rou... | Answer: 4.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 13 \leqslant 63$, f... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round... | Answer: 5.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 12 \leqslant 63$, f... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2019 \cdot 2020 \cdot 2021 \cdot 2022 + k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2020$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. What is greater: 1 or $\frac{23}{93}+\frac{41}{165}+\frac{71}{143}$? | Answer: One is greater.
## First solution.
$$
\frac{23}{93}+\frac{41}{165}+\frac{71}{143}<\frac{23}{92}+\frac{41}{164}+\frac{71}{142}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1 .
$$
## Second solution.
$$
\begin{aligned}
\frac{23}{93}+\frac{41}{165}+\frac{71}{143} & =\frac{23 \cdot 165 \cdot 143+93 \cdot 41 \cdot 143+93... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, eight teams played: each team played once with each other. In the next round, teams that scored fifteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next roun... | Answer: 5.
Solution. In total, the teams played $\frac{8 \cdot 7}{2}=28$ games, in each of which 2 or 3 points were at stake. Therefore, the maximum total number of points that all teams can have is $28 \cdot 3=84$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 15 \leq... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2018 \cdot 2019 \cdot 2020 \cdot 2021+k$ a square of a natural number | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2019$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a football tournament, six teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round? | Answer: 3
Solution. In total, the teams played $\frac{6 \cdot 5}{2}=15$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $15 \cdot 3=45$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 12 \leqslant 45$,... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. For what least natural $k$ is the expression $2016 \cdot 2017 \cdot 2018 \cdot 2019 + k$ a square of a natural number? | Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2017$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Dasha wrote the numbers $9,10,11, \ldots, 22$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the rema... | Answer: 203.
Solution. The sum of numbers from 9 to 22 is 217. If at least one number is erased, the sum of the remaining numbers does not exceed 208. Let's sequentially consider the options:
- if the sum is 208, then Dasha could have erased only the number 9; then the remaining numbers can be divided into two groups... | 203 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In the surgical department, there are 4 operating rooms: I, II, III, and IV. In the morning, they were all empty. At some point, an operation began in operating room I, then after some time in operating room II, then after some more time in operating room III, and finally in operating room IV.
All four oper... | Answer: Only the duration of the operation in Operating Room IV can be determined.
Solution. First, let's prove that the durations of the operations in Operating Rooms I, II, and III cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 70, 39, 33, 10 or 56, 5... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2019 \pi}{43} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alp... | -2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to -1540? | Answer: -5.
Solution. Applying the polynomial formula, we get
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_{... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. Given an isosceles triangle $K L M(K L=L M)$ with the angle at the vertex equal to $114^{\circ}$. Point $O$ is located inside triangle $K L M$ such that $\angle O M K=30^{\circ}$, and $\angle O K M=27^{\circ}$. Find the measure of angle $\angle L O M$. | Answer: $150^{\circ}$.
Solution. Let $L H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $M O$ and segment $L H$. Note that $K S=S M$. For example, since in triangle $K S M$ the median SH coincides with the height.
 \pi}{47}+\ldots+\operatorname{tg} \frac{2021 \pi}{47} \cdot \operat... | Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alp... | -2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-3 x+a x^{2}\right)^{8}$ be equal to $70$? | Answer: -4.
Solution. Applying the polynomial formula, we get
$$
\left(1-3 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-3 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-3)^{n_{... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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