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1. Compute the sum:
$$
\sqrt{2018-\sqrt{2019 \cdot 2017}}+\sqrt{2016-\sqrt{2017 \cdot 2015}}+\cdots+\sqrt{2-\sqrt{3 \cdot 1}}
$$ | Answer: $\frac{\sqrt{2019}-1}{\sqrt{2}}$.
Solution: Using the known formula $\sqrt{a-\sqrt{(a+1)(a-1)}}=\frac{\sqrt{a+1}-\sqrt{a-1}}{\sqrt{2}}$, valid for $a \geq 1$, we get that each of the roots can be represented as:
$$
\begin{gathered}
\sqrt{2018-\sqrt{2019 \cdot 2017}}=\frac{\sqrt{2019}-\sqrt{2017}}{\sqrt{2}} \\... | \frac{\sqrt{2019}-1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,530 |
3. Solve the system of equations:
$$
\left\{\begin{array}{l}
x y+5 y z-6 x z=-2 z \\
2 x y+9 y z-9 x z=-12 z \\
y z-2 x z=6 z
\end{array}\right.
$$ | Answer: $(-2,2,1 / 6),(x, 0,0),(0, y, 0)$, where $\boldsymbol{x}, \boldsymbol{y}$ are any real numbers.
Solution: From the third equation of the system, we find $y z-2 x z-6 z=0$, from which $z(y-2 x-6)=0$. Therefore, $z=0$ or $y=2 x+6$.
1) Let $z=0$, then the first and second equations reduce to the equality $x y=0$... | (-2,2,1/6),(x,0,0),(0,y,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,532 |
4. Let $\mathrm{A}$ be the set of all sixteen-digit natural numbers, each of which satisfies two conditions: it is a square of an integer and in its decimal representation, the digit in the tens place is 1. Prove that all numbers in set A are even and that set A contains more than $10^{6}$ numbers. | Solution: Let $m$ be an arbitrary number from set A. By condition, $m=n^{2}$, where $n \in N$. Suppose $a$ is the last digit of the number $n$, then $n=10 k+a$, where $k \in N$. Consequently, $m=n^{2}=100 k^{2}+20 a k+a^{2}$. The sum $100 k^{2}+20 a k$ ends with the digit 0 and has an even number of tens. For the tens ... | 24\cdot10^{5} | Number Theory | proof | Yes | Yes | olympiads | false | 8,533 |
5. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ and two planes $\alpha$ and $\beta$. Plane $\alpha$ is perpendicular to the line $A_{1} C_{1}$, and plane $\beta$ is parallel to the line $C D_{1}$. Determine the smallest possible angle between the planes $\alpha$ and $\beta$.
( 7 points) | Answer: $\frac{\pi}{6}$.
Solution: First, let's prove one auxiliary statement. Let planes $\alpha$ and $\beta$ intersect along a line $l$ and form a dihedral angle $\varphi$. OPEN REGIONAL INTERUNIVERSITY OLYMPIAD 2018-2019 MATHEMATICS (11th GRADE) INCLUSIVE STAGE VARIANT 2 (ANSWERS) | \frac{\pi}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,534 |
1. Compute the sum:
$$
\sqrt{2020-\sqrt{2021 \cdot 2019}}+\sqrt{2018-\sqrt{2019 \cdot 2017}}+\cdots+\sqrt{2-\sqrt{3 \cdot 1}}
$$ | Answer: $\frac{\sqrt{2021}-1}{\sqrt{2}}$.
Solution: a similar solution to this problem is present in Variant 1 under the same number. | \frac{\sqrt{2021}-1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,535 |
3. Solve the system of equations:
$$
\left\{\begin{array}{l}
3 x y-5 y z-x z=3 y \\
x y+y z=-y \\
-5 x y+4 y z+x z=-4 y
\end{array}\right.
$$ | Answer: $(\mathbf{2},-\mathbf{1} / \mathbf{3},-3),(x, 0,0),(0,0, z)$, where $x, z-$ are any real numbers.
Solution: a similar solution to this problem is present in variant 1 under the same number. | (2,-\frac{1}{3},-3),(x,0,0),(0,0,z) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,537 |
4. Let $\mathrm{B}$ be the set of all fourteen-digit natural numbers, each of which satisfies two conditions: it is a square of an integer and in its decimal representation, the digit in the tens place is 5. Prove that all numbers in the set $\mathrm{B}$ are even and that the set $\mathrm{B}$ contains more than $10^{5}... | Solution: A similar solution to a similar problem is present in version 1 under the same number, however, we will point out an important difference: in each consecutive hundred, there are four numbers whose squares have the digit 5 in the tens place. These are numbers of the form $100 l+16,100 l+34,100 l+66,100 l+84, \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,538 |
5. All edges of a regular quadrilateral pyramid $S A B C D$ with base $A B C D$ have equal length. Plane $\alpha$ is perpendicular to line $S A$, and plane $\beta$ is parallel to line $C D$. Determine the smallest possible angle between planes $\alpha$ and $\beta$. (7 points) Answer: $\frac{\pi}{6}$. | Solution: A similar solution to a similar problem is present in variant 1 under the same number; however, for this problem, let's note the following: Triangle $A S C$ is a right triangle. Let plane $\alpha$ pass through point $S$, then line $S C \in \alpha$. If we draw plane $\beta$ through line $C D$ and assume that a... | \frac{\pi}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,539 |
1. Find all values of $n$ for which the sum
$$
1!+2!+3!+\cdots+n!, \text { where } n!=1 \cdot 2 \cdot 3 \cdots n
$$
is a perfect square. | Answer: $n=1 ; 3$.
Solution.
$1!=1=1^{2}, \quad 1!+2!=3, \quad 1!+2!+3!=9=3^{2}, \quad 1!+2!+3!+4!=33$,
$1!+2!+3!+4!+5!=33+120=153$.
In the further steps, the last digit will always be 3, as 33 will always be added to numbers ending in zero. Therefore, there will be no more perfect squares. Thus, the condition of t... | 1;3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,540 |
2. For the quadratic trinomial $p(x)=(a+1) x^{2}-(a+1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$. | # Answer: 16175.
Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the min... | 16175 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,541 |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions:
$a^{3}-2022 a+1011=0, \quad b^{3}-2022 b+1011=0, \quad c^{3}-2022 c+1011=0$. | Answer: 2.
## Solution.
The cubic equation $t^{3}-2022 t+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t+1011: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2022 \\
a b c=-1011
\end{array}\right.
$$
We find ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,542 |
5. Through the vertex $M$ of some angle, a circle is drawn, intersecting the sides of the angle at points $N$ and $K$, and the bisector of this angle at point $L$. Find the sum of the lengths of segments $M N$ and $M K$, if the area of $M N L K$ is 25, and the angle $L M N$ is $30^{\circ}$. | Answer: $10 \sqrt[4]{3}$.
## Solution.
Let $L P$ and $L Q$ be the perpendiculars to $M N$ and $M K$ respectively.
Point $\mathrm{L}$ lies on the angle bisector and is therefore equidistant from the sides of the angle, which means $L P = L Q$.
Right triangles $N P L$ and $L Q K$ are congruent (by leg and hypotenuse)... | 10\sqrt[4]{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,543 |
1. Solve the equation in integers
$$
2025^{x}-100 x y+3-y^{2}=0
$$ | Answer: $(0 ; 2),(0 ;-2)$.
## Solution.
Consider $x, y \in Z$.
$$
2025^{x}-100 x y+3=y^{2} .
$$
1) $x=0 \Rightarrow 2025^{0}-0+3=y^{2} \Rightarrow 4=y^{2} \Rightarrow y= \pm 2$.
2) $x \neq 0 \Rightarrow 2025^{x}-100 x y+3=y^{2} \Rightarrow$ no solutions, because the number on the left side ends with the digit 8, wh... | (0;2),(0;-2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,544 |
2. For the quadratic trinomial $p(x)=(a-1) x^{2}-(a-1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$. | Answer: 16177.
Solution. Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the minim... | 16177 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,545 |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three distinct real numbers satisfying the conditions:
$a^{3}-2020 a+1010=0, \quad b^{3}-2020 b+1010=0, c^{3}-2020 c+1010=0$. | Answer: 2.
Solution.
The cubic equation $t^{3}-2020 t+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t+1010: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2020 \\
a b c=-1010
\end{array}\right.
$$
We find the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,546 |
5. Through the vertex $M$ of some angle, a circle is drawn, intersecting the sides of the angle at points $N$ and $K$, and the bisector of this angle at point $L$. Find the sum of the lengths of segments $M N$ and $M K$, if the area of $M N L K$ is 49, and the angle $L M N$ is $30^{\circ}$. | Answer: $14 \sqrt[4]{3}$.
## Solution.
Let $L P$ and $L Q$ be the perpendiculars to $M N$ and $M K$ respectively.
Point $\mathrm{L}$ lies on the angle bisector and is therefore equidistant from the sides of the angle, which means $L P = L Q$.
Right triangles $N P L$ and $L Q K$ are congruent (by leg and hypotenuse)... | 14\sqrt[4]{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,547 |
1. Find $x$ and $y$ that satisfy the following equation:
$$
(x-y)^{2}+(y-2 \sqrt{x}+2)^{2}=\frac{1}{2}
$$ | Answer: $x=1, y=\frac{1}{2}$.
## Solution:
Let $u=(\sqrt{x}-1)^{2}+1=x-2 \sqrt{x}+2 \geq 1, v=y-2 \sqrt{x}+2$.
Then
$$
y-x=v-u \quad \text { and the original equation becomes }
$$
$$
\begin{gathered}
(v-u)^{2}+v^{2}=\frac{1}{2} \\
v^{2}-2 u v+u^{2}-\frac{1}{2}=0
\end{gathered}
$$
Considering the last equation as ... | 1,\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,548 |
2. To walk 2 km, ride 3 km on a bicycle, and drive 20 km by car, Uncle Vanya needs 1 hour 6 minutes. If he needs to walk 5 km, ride 8 km on a bicycle, and drive 30 km by car, it will take him 2 hours 24 minutes. How much time will Uncle Vanya need to walk 4 km, ride 5 km on a bicycle, and drive 80 km by car? | # Answer: 2 hours 54 minutes. (2.9 hours.)
## Solution:
Let $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ be the walking speed, cycling speed, and driving speed, respectively. Then, according to the problem, we can set up the system:
$$
\left\{\begin{array} { c }
{ 2 x + 3 y + 2 0 z = 6 6 } \\
{ 5 x + 8 y + 3 0 z = 1 4 4... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,549 |
3. Find all values of $m$ for which any solution of the equation
$$
2019 \cdot \sqrt[3]{3.5 x-2.5}+2018 \cdot \log _{2}(3 x-1)+m=2020
$$
belongs to the interval $[1 ; 3]$. | Answer: $m \in[-8072 ;-2017]$.
## Solution:
Consider $f(x)=2019 \cdot \sqrt[3]{3.5 x-2.5}+2018 \cdot \log _{2}(3 x-1)+m-2020$.
On the domain of definition $D_{f}=\left(\frac{1}{3} ;+\infty\right)$, the function is monotonically increasing, as the sum of monotonically increasing functions. The equation $f(x)=0$ has a... | \in[-8072;-2017] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,550 |
5. In a regular triangular pyramid, a section is made which is a square. Find the volume of the pyramid if the side of the base is $a$, and the side of the square in the section is $b$. | Answer: $V=\frac{a^{3} \cdot \sqrt{2 b^{2}+2 a b-a^{2}}}{12(a-b)}$.
## Solution:
Let $H$ be the height of the pyramid, and $l$ be the length of the lateral edge. Then the volume of the pyramid is
$$
V=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot H
$$
Express the height $H$ in terms of $a$ and $b$.
Since $H^{2}... | \frac{^{3}\cdot\sqrt{2b^{2}+2-^{2}}}{12(-b)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,551 |
1. Solve the equation
$$
(x-2020)^{2}+(x-2020)^{10}=2(x-2020)^{12}
$$ | Answer: $\{2019 ; 2020 ; 2021\}$.
## Solution:
Let $t=x-2020$, then the original equation can be rewritten as
$$
t^{2}+t^{10}=2 t^{12}
$$
Therefore, $t=0$ or $t= \pm 1$. We will show that there are no other roots:
1) if we assume that $t^{2}>1$, then $t^{10}>t^{2}$ and $t^{12}>t^{2}$.
2) if we assume that $t^{2}<1... | {2019;2020;2021} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,552 |
2. To walk 4 km, ride 6 km on a bicycle, and drive 40 km by car, Uncle Vanya needs 2 hours and 12 minutes. If he needs to walk 5 km, ride 8 km on a bicycle, and drive 30 km by car, it will take him 2 hours and 24 minutes. How much time will Uncle Vanya need to walk 8 km, ride 10 km on a bicycle, and drive 160 km by car... | # Answer: 5 hours 48 minutes. (5.8 hours)
## Solution:
Let $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ be the walking speed, cycling speed, and driving speed, respectively. Then, according to the problem, we can set up the following system:
$$
\left\{\begin{array} { l }
{ 4 x + 6 y + 4 0 z = 1 3 2 , } \\
{ 5 x + 8 y + ... | 5.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,553 |
3. Find all values of $m$ for which any solution of the equation
$$
2018 \cdot \sqrt[5]{6.2 x-5.2}+2019 \cdot \log _{5}(4 x+1)+m=2020
$$
belongs to the interval $[1 ; 6]$. | # Answer: $m \in[-6054 ;-2017]$.
## Solution:
Consider $f(x)=2018 \cdot \sqrt[5]{6.2 x-5.2}+2019 \cdot \log _{5}(4 x+1)+m-2020$.
On the domain of definition $D_{f}=\left(-\frac{1}{4} ;+\infty\right)$, the function is monotonically increasing, as the sum of monotonically increasing functions. The equation $f(x)=0$ ha... | \in[-6054;-2017] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,554 |
5. The base of the pyramid $S A B C D$ is a rectangle $A B C D$, with the height being the edge $S A=25$. Point $P$ lies on the median $D M$ of the face $S C D$, point $Q$ lies on the diagonal $B D$, and the lines $A P$ and $S Q$ intersect. Find the length of $P Q$, if $B Q: Q D=3: 2$. | # Answer: 10.
## Solution:
Since lines $A P$ and $S Q$ intersect, points $A, P, S, Q$ lie in the same plane. Let $R$ be the intersection point of $S P$ and $A Q$. Then
$$
\frac{R Q}{A Q}=\frac{D Q}{B Q}=\frac{2}{3} \Rightarrow \frac{R Q}{R A}=\frac{2}{5}
$$
We will prove that $\frac{R M}{R S}=\frac{2}{5}$. Note tha... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,555 |
1. Find the sum:
$$
\frac{1}{2 \sqrt{1}+1 \sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\cdots+\frac{1}{2020 \sqrt{2019}+2019 \sqrt{2020}}
$$
(7 points) | Answer: $1-\frac{1}{\sqrt{2020}}=\frac{\sqrt{2020}-1}{\sqrt{2020}}$.
Solution: Since
$$
\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{(n+1) \sqrt{n}+n \sqrt{n+1}}{(n+1)^{2} n-n^{2}(n+1)}=\frac{(n+1) \sqrt{n}+n \sqrt{n+1}}{(n+1) n(n+1-n)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
$$
then, considering the condition of the... | \frac{\sqrt{2020}-1}{\sqrt{2020}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,556 |
2. Prove that for $a<1, b<1, a+b \geq \frac{1}{2}$ the inequality
$$
(1-a)(1-b) \leq \frac{9}{16}
$$
holds.
(7 points). | Solution: Since $a0,1-b>0$. Using the well-known inequality of means, we get $\sqrt{(1-a)(1-b)} \leq \frac{(1-a)+(1-b)}{2}=1-\frac{a+b}{2} \leq \frac{3}{4}$ under the condition that $a+b \geq \frac{1}{2}$, that is, $\sqrt{(1-a)(1-b)} \leq \frac{3}{4}$. Squaring the last inequality, we obtain the required inequality $(1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,557 |
3. Solve the system of equations:
$$
\left\{\begin{aligned}
x+y-2018 & =(x-2019) \cdot y \\
x+z-2014 & =(x-2019) \cdot z \\
y+z+2 & =y \cdot z
\end{aligned}\right.
$$ | Answer: (2022, 2, 4), (2018, 0, -2).
Solution: Introduce the substitution $\tilde{x}=x-2019$. Then the system becomes $\left\{\begin{array}{l}\tilde{x}+y+1=\tilde{x} \cdot y, \\ \tilde{x}+z+5=\tilde{x} \cdot z, \\ y+z+2=y \cdot z .\end{array} \quad\right.$ Transform the system to the form: $\left\{\begin{array}{l}(\ti... | (2022,2,4),(2018,0,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,558 |
4. In the store "Everything for School," chalk is sold in packages of three grades: regular, unusual, and superior. Initially, the ratio of the quantities by grade was 3:4:6. As a result of sales and deliveries from the warehouse, this ratio changed to 2:5:8. It is known that the number of packages of superior chalk in... | # Answer: 260 packages.
Solution: Let $x$ be the initial number of packages of regular chalk, then the number of packages of unusual chalk is $\frac{4 x}{3}$. Since the latter number is an integer, then $x=3 n$, where $n \in N$. Therefore, the initial quantities of all three types of packages are $3 n, 4 n, 6 n$ respe... | 260 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,559 |
5. The distance between the centers $O_{1}$ and $O_{2}$ of circles $\omega_{1}$ and $\omega_{2}$ is $5 r$, and their radii are $r$ and $7 r$ respectively. A chord of circle $\omega_{2}$ is tangent to circle $\omega_{1}$ and is divided by the point of tangency in the ratio $1: 6$. Find the length of this chord.
( 7 poi... | Answer: $7 r \sqrt{3}$ or $\frac{7 r}{6} \sqrt{143}$.
Solution: Let $O_{1}$ be the center of the first circle $\omega_{1}$, and point $M$ be the point of tangency of this circle with the chord $A B$. From the center $O_{2}$ of the second circle $\omega_{2}$, draw $O_{2} K \perp A B$, where point $K$ is the midpoint of... | 7r\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,560 |
1. Compute:
$$
\frac{1}{2 \sqrt{1}+1 \sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\cdots+\frac{1}{2019 \sqrt{2018}+2018 \sqrt{2019}}
$$
(7 points) | Answer: $1-\frac{1}{\sqrt{2019}}=\frac{\sqrt{2019}-1}{\sqrt{2019}}$.
Solution: a similar solution to this problem is present in Variant 1 under the same number. | \frac{\sqrt{2019}-1}{\sqrt{2019}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,561 |
2. Prove that for $a<1, b<1, a+b \geq \frac{1}{3}$, the inequality
$$
(1-a)(1-b) \leq \frac{25}{36}
$$
holds. | Solution: Since $a0,1-b>0$. Using the well-known inequality of means, we get $\sqrt{(1-a)(1-b)} \leq \frac{(1-a)+(1-b)}{2}=1-\frac{a+b}{2} \leq \frac{5}{6}$ under the condition that $a+b \geq \frac{1}{3}$, that is, $\sqrt{(1-a)(1-b)} \leq \frac{5}{6}$. Squaring the last inequality, we obtain the required inequality $(1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,562 |
3. Solve the system of equations:
$$
\left\{\begin{aligned}
x+y-2018 & =(y-2019) \cdot x \\
y+z-2017 & =(y-2019) \cdot z \\
x+z+5 & =x \cdot z
\end{aligned}\right.
$$ | Answer: (3, 2021, 4), ( $-1,2019,-2)$.
Solution: a similar solution to this problem is present in variant 1 under the same number. Note: 1 point for each correct solution found by trial and error. | (3,2021,4),(-1,2019,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,563 |
4. In the store "Everything for School," chalk is sold in packages of three grades: regular, unusual, and superior. Initially, the ratio of the quantities by grade was 2:3:6. After a certain number of packages of regular and unusual chalk, totaling no more than 100 packages, were delivered to the store, and 40% of the ... | # Answer: 24 packs.
Solution: a similar solution to this problem is present in Variant 1 under the same number. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,564 |
5. The distance between the centers $O_{1}$ and $O_{2}$ of circles $\omega_{1}$ and $\omega_{2}$ is $10 r$, and their radii are $5 r$ and $6 r$ respectively. A line intersects circle $\omega_{1}$ at points $M$ and $N$ and is tangent to circle $\omega_{2}$ at point $K$, such that $M N=2 N K$. Find the length of the chor... | Answer: $2 r \sqrt{21}$ or $6 r$.
Solution: a similar problem is present in variant 1 under the same number, the difference is that there are 2 cases of the chord and circle placement:
1) the point of tangency with circle $\omega_{2}$ lies outside circle $\omega_{1} ; 2$ ) the point of tangency with circle $\omega_{2... | 2r\sqrt{21}or6r | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,565 |
3. Determine how many times the number $((2014)^{2^{2014}}-1)$ is larger than the number written in the following form: $\left.\left.\left((2014)^{2^{0}}+1\right) \cdot\left((2014)^{2^{1}}+1\right) \cdot\left((2014)^{2^{2}}+1\right) \cdot \ldots \quad \cdot\right)^{2^{2013}}+1\right)$. | Justify the solution.
Answer: 2013. | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,567 |
1. Real
numbers
$x, y, z$
satisfy
$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$
relations:
Find the maximum of the sum $a+b+c$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$. | Solution. Note that
$$
a-b+c=0
$$
Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get
$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& B-C=b \cdot(5 y+4 x-3 z+1)=0 \\
& A-C=c \cdot(1-2 x-10 y-3 z)=0
\end{aligned}
$$
Assume that all ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,569 |
4. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate). | Solution. Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$.
Let $q=10$ be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} \cdot \frac{q^{... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,571 |
5. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25(712)$ is 3.) | Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.).
As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$.
## Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Ma... | 48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,572 |
6. It is known that the lengths of the sides of a convex quadrilateral are respectively $a=4, b=5, c=6, d=7$. Find the radius $R$ of the circle circumscribed around this quadrilateral. Provide the integer part of $R^{2}$ as the answer. | Solution. By the cosine theorem, we express the length of the diagonal:
$$
l^{2}=a^{2}+b^{2}-2 a b \cos \gamma, l^{2}=c^{2}+d^{2}-2 c d \cos (\pi-\gamma)
$$
From this, we get $\cos \gamma=\frac{a^{2}+b^{2}-c^{2}-d^{2}}{2(a b+c d)}$. Since $R=\frac{l}{2 \sin \gamma^{\prime}}$, we obtain
$$
R^{2}=\frac{l^{2}}{4\left(1... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,573 |
3. $A B C D E F$ - a regular hexagon with a mirrored inner surface. A light ray exits from point $A$ and after two reflections from the sides of the hexagon (at points $M$ and $N$), it reaches point $D$ (Fig. 1). Find the tangent of angle $E A M$.
 and transport the cargo to point B, where they are instantly unloaded and return to A. The speeds of the trucks are the same, and the speed of a loaded truck is 6/7 of the speed of an empty truck.... | Answer: loading time - 13, number of loaded trucks - 5.
Criteria for determining winners and prize winners of the final stage of the 2013-14 academic year
Maximum for each task - 3 points
1st place: no less than 23 points;
2nd place: no less than 6 fully solved tasks or no less than 19 points;
3rd place: no less t... | loading\time\-\13,\\of\loaded\trucks\-\5 | Other | math-word-problem | Yes | Yes | olympiads | false | 8,578 |
1. It is known that the equation $x^{4}-8 x^{3}+a x^{2}+b x+16=0$ has (taking into account multiplicity) four positive roots. Find $a-b$. | Solution: Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of our equation (some of them may be the same). Therefore, the polynomial on the left side of the equation can be factored as:
$$
x^{4}-8 x^{3}+a x^{2}+b x+16=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$
Expanding the brack... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,579 |
2. 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 m... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,580 |
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 $1^{2}+2^{2}+4^{2}=21$ ). | 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 | 8,581 |
4. 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 $a=2, b=3$.
=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 $\quad$ of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has $\quad$ roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coe... | 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-x_{... | -1216 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,583 |
1. Oleg has 1000 rubles, and he wants to give his mother tulips for March 8, with the condition that the number of tulips must be odd, and no color shade should be repeated. In the store where Oleg went, one tulip costs 49 rubles, and there are twenty shades of flowers available. How many ways can Oleg give his mother ... | Solution. From the condition, it is obvious that the maximum number of flowers in a bouquet is 20.
1st method
Using the property of binomial coefficients
$$
\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5}+\cdots=2^{n-1}
$$
and also considering their combinatorial meaning, we get that the number of ways to ... | 2^{19} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,585 |
3. 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 relative to line $A F$, and then relative to 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^{\cir... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,587 |
4. Let $x_{1}$ and $x_{2}$ be the largest roots of the polynomials $f(x)=1-x-4 x^{2}+x^{4}$ and $g(x)=16-8 x-16 x^{2}+x^{4}$ respectively. Find $\frac{x_{1}}{x_{2}}$. | # Solution.
1st method
Notice that $g(2 x)=16 f(x)$. Then $x_{1}-$ is a root of $f(x)$ if and only if $2 x_{1}-$ is a root of $g(x)$. Therefore, $\frac{x_{1}}{x_{2}}=\frac{1}{2}$.
2nd method
Comparing the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text { and } g(x)=16-8 x-16 x^{2}+x^{4}
$$
shows t... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,588 |
5. Solve the system of equations $\left\{\begin{array}{l}x^{4}+\frac{7}{2} x^{2} y+2 y^{3}=0 \\ 4 x^{2}+7 x y+2 y^{3}=0\end{array}\right.$. | Solution. Consider the function $f(t)=t^{2}+\frac{7}{2} t y+2 y^{3}$. Under the condition that the equalities of the original system are satisfied, its roots will be $t_{1}=x^{2}$ and $t_{2}=2 x$. If $t_{1}=t_{2}$, then $x_{1}=0, x_{2}=2$. From this, we find $y_{1}=0, y_{2}=-1$. If $t_{1} \neq t_{2}$, then using Vieta'... | (0;0),(2;-1),(-\frac{11}{2};-\frac{11}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,589 |
6. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$ | Solution: Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of ... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,590 |
7. It is known that the number $\cos 6^{0}$ is a root of the equation $32 t^{5}-40 t^{3}+10 t-\sqrt{3}=0$. Find the other four roots of this equation. (Answers in the problem should be compact expressions, not containing summation signs, ellipses, etc.) | Solution: Let $t=\cos \varphi$. The equation will take the form: $32 \cos ^{5} \varphi-40 \cos ^{3} \varphi+10 \cos \varphi=\sqrt{3}$. Transform the left side:
\[
\begin{aligned}
& 2 \cos \varphi\left(16 \cos ^{4} \varphi-20 \cos ^{2} \varphi+5\right)=[\text{reduction formulas}]= \\
& =2 \cos \varphi\left(4(1+\cos 2 \... | \cos78,\cos150,\cos222,\cos294 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,591 |
8. Let $A$ and $B$ be some numerical sets, and the set $C=\{a+b \mid a \in A, b \in B\}$ represents their sum. (That is, the set $C$ consists of all possible sums of elements of sets $A$ and $B$. For example, if $A=\{0,1,2\}, B=\{1,2\}$, then $C=\{1,2,3,4\}$.)
It is known that $C=\left\{0,1,2, \ldots, 2^{2828}\right\}... | Solution. 1) Correct: if set $A$ or set $B$ is infinite, then set $C$ will also be infinite. Therefore, we can denote by $a, b, c$ the maximum elements of these sets, respectively, and note for solving part 3 that $a+b=c$. Separately, we note that such sets exist: for example, $A=\{0, \ldots, a\}, B=\{0, \ldots, c-a\}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,592 |
2. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{c}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
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 by $b_{2}$ and subtract from it equation (b) multiplied by $b_{1}$. The result is
$$
a_{2} \cdot \Delta=b_{2}
$$
Here $\Delta=b_{2} b_{3}-b_{1} b_{4}$. Similarly, from (c) and (d) we find that
$$
a_{3}... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,594 |
3. Solve the equation $2^{x}+2^{y}=2^{x y-1}$ in integers. | Solution. Let $x=y$ first. Then the equation becomes $2^{x}+2^{x}=2^{x^{2}-1}$.
From this, $2^{x+1}=2^{x^{2}-1} \Leftrightarrow x+1=x^{2}-1 \Leftrightarrow x=-1$ or $x=2$.
Now let the numbers $x$ and $y$ be different. We can assume that $x<y$. Let $y=x+n, n \in \mathbb{N}$. Then $2^{x}+2^{x+n}=2^{x y-1} \Leftrightarr... | (-1,-1),(2,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,595 |
5. Let $O$ be the point of intersection of the medians of triangle $ABC$. Find the length of the median drawn from vertex $A$, if $\angle BAC=35^{\circ}, \angle BOC=145^{\circ}, BC=a$. | Solution. Let the length of the desired median $AD$ be $m$. On the line $AD$ outside the triangle, mark a point $F$ such that $OD = DF = m / 3$. The quadrilateral $OBFC$ is a parallelogram because its diagonals are bisected by their point of intersection (by the condition $BD = DC$, and $OD = DF$ by construction). In a... | \frac{\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,597 |
6. Find the area of triangle $ABC$, the vertices of which have coordinates $A(0,0), B(1424233,2848467), C(1424234,2848469)$. Round your answer to the hundredths. | Solution. Note that points $B$ and $C$ lie on the line $y=2x+1$. Their abscissas differ by 1, hence $BC=\sqrt{5}$. The length of the height of triangle $ABC$, drawn from vertex $A$, is equal to the distance $h$ from point $A$ to the line $y=2x+1$, which, in turn, is equal to $1 / \sqrt{5}$. The desired area
$$
S_{ABC}... | 0.50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,598 |
7. Let's consider all 100-digit natural numbers, in the decimal representation of which only the digits 1, 2 appear. How many of them are divisible by 3? | Solution. Each 100-digit natural number can be obtained by appending two digits to the right of a 98-digit number. Let $x$ be some 98-digit number. We will examine which two digits (each equal to 1 or 2) need to be appended to the number $x$ so that the resulting 100-digit number is divisible by 3. We will use the fact... | \frac{4^{50}+2}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,599 |
8. On the Cartesian plane, consider a circle of radius $R$ centered at the origin. Specify at least one value of $R$ for which exactly 32 integer points lie on such a circle (a point is called integer if its abscissa and ordinate are integers).
Hint. A natural number $x$ can be represented as the sum of the squares of... | Solution. If each of the numbers $a$ and $b$ can be represented as the sum of two squares, then, as noted in the hint, their product can also be represented in such a form. Moreover, the product is usually representable as the sum of two squares in more ways than each of the factors. For example, the number 5 can be re... | \sqrt{1105} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,600 |
1. The function $y=f(x)$ is defined on the set $(0,+\infty)$ and takes positive values on it. It is known that for any points $A$ and $B$ on the graph of the function, the areas of the triangle $A O B$ and the trapezoid $A B H_{B} H_{A}$ are equal to each other $\left(H_{A}, H_{B}\right.$ - the bases of the perpendicul... | # Solution:
Let $M$ be the intersection point of segments $O B$ and $A H_{A}$. Since the areas of triangle $A O B$ and trapezoid $A B H_{B} H_{A}$ are equal, then
the areas of triangles $A M... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,601 |
2. Let $x_{1}$ and $x_{2}$ be the largest roots of the polynomials
$$
\begin{gathered}
f(x)=1-x-4 x^{2}+x^{4} \\
\text { and } \\
g(x)=16-8 x-16 x^{2}+x^{4}
\end{gathered}
$$
respectively. Find $\frac{x_{2}}{x_{1}}$. | Solution:
Notice that $f(-2)>0, f(-1)<0, f(0)>0, f(1)<0$. Therefore, the polynomial $f(x)$ has 4 real roots. Similarly, from the inequalities $g(-4)>0, g(-2)<0, g(0)>0, g(2)<0$ it follows that the polynomial $g(x)$ has 4 real roots.
Comparison of the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text { ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,602 |
3. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
# | # Solution:
Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness ... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,603 |
4. It is known that the number $\cos 6^{0}$ is a root of the equation $32 t^{5}-40 t^{3}+10 t-\sqrt{3}=0$. Find the other four roots of this equation. (Answers in the problem should be compact expressions, not containing summation signs, ellipses, and radicals.)
# | # Solution:
$\cos 5 x=16 \cos ^{5} x-20 \cos ^{3} x+5 \cos x$. Then the other roots of the equation will be the numbers $\cos 66^{\circ}, \cos 78^{\circ}, \cos 138^{\circ}, \cos 150^{\circ}$.
Answer: $\cos 66^{\circ}, \cos 78^{\circ}, \cos 138^{\circ}, \cos 150^{\circ}$ | \cos66,\cos78,\cos138,\cos150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,604 |
5. Find the area of a geometric figure, the coordinates of the vertices of which are the solutions to the system of equations
$$
\left\{\begin{array}{l}
x^{4}+\frac{7}{2} x^{2} y+2 y^{3}=0 \\
4 x^{2}+7 x y+2 y^{3}=0
\end{array}\right.
$$ | # Solution:
The system of equations has only three real solutions: $(0 ; 0),(2 ;-1),\left(-\frac{11}{2} ;-\frac{11}{2}\right)$. The area of the triangle with vertices at these coordinates is $16 \frac{1}{2}$.
Answer: $16 \frac{1}{2}$ | 16\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,605 |
6. Points $F$ and $G$ are chosen on the sides $AB$ and $BC$ of rectangle $ABCD$, respectively. A perpendicular $FK$ is dropped from point $F$ to side $CD$. A perpendicular $GH$ is dropped from point $G$ to side $AD$. The intersection point of $FK$ and $GH$ is denoted as $E$. Find the area of triangle $DFG$, given that ... | # Solution:
Let $A D=a, D C=b, H D=x$, and $D K=y$.
$$
\begin{aligned}
S_{D F G}=S_{A B C D} & -S_{A F D}-S_{F G B}-S_{D G C}=a b-\frac{1}{2} a y-\frac{1}{2}(b-y)(a-x)-\frac{1}{2} b x \\
&... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,606 |
1. Solve the equation $2^{x}+2^{y}=2^{x y-1}$ in integers. | Solution. Let $x=y$ first. Then the equation becomes $2^{x}+2^{x}=2^{x^{2}-1}$.
From this, $2^{x+1}=2^{x^{2}-1} \Leftrightarrow x+1=x^{2}-1 \Leftrightarrow x=-1$ or $x=2$. Now let the numbers $x$ and $y$ be different. We can assume that $x<y$. Let $y=x+n, n \in \mathbb{N}$. Then $2^{x}+2^{x+n}=2^{x y-1} \Leftrightarro... | (-1,-1),(2,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,607 |
3. Solve the equation $\sin ^{3} x+6 \cos ^{3} x+\frac{1}{\sqrt{2}} \sin 2 x \sin \left(x+\frac{\pi}{4}\right)=0$. | Solution. Transform the equation
$$
\begin{aligned}
& \sin ^{3} x+\sin x \cos x(\sin x+\cos x)+6 \cos ^{3} x=0 \\
& \sin ^{3} x+\sin ^{2} x \cos x+\sin x \cos ^{2} x+6 \cos ^{3} x=0
\end{aligned}
$$
If $\cos x=0$, then substituting into the equation, we get $\sin x=0$, which is impossible. Divide by $\cos ^{3} x$ :
... | -\operatorname{arctg}2+\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,609 |
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 ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics
$$
\begin{cases... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,610 |
5. On the Cartesian plane, consider a circle of radius $R$ centered at the origin. Specify at least one value of $R$ for which exactly 32 integer points lie on such a circle (a point is called integer if its abscissa and ordinate are integers).
Hint. A natural number $x$ can be represented as the sum of the squares of... | Solution. If each of the numbers $a$ and $b$ can be represented as the sum of two squares, then, as noted in the hint, their product can also be represented in such a form. Moreover, the product is typically representable as the sum of two squares in more ways than each of the factors. For example, the number 5 can be ... | \sqrt{1105} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,611 |
6. At the vertices of a square with a side length of 4, there are four cities. These cities need to be connected by roads such that it is possible to travel from any city to any other. Propose at least one variant of such roads with a total length of less than 11.
Hint. The following statement may be useful in solving... | Solution. Suppose our road system has no intersections, that is, there is one road connecting the vertices of the square $E, F, G$, and $H$ sequentially. Then its length will be no less than three sides of the square (the letter "П" in Fig. (a)), that is, no less than 12 (a road connecting adjacent vertices of the squa... | 4(\sqrt{3}+1)<10.92 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,612 |
7. Find the area of triangle $ABC$, the vertices of which have coordinates $A(0,0), B(1424233,2848467), C(1424234,2848469)$. Round your answer to the hundredths. | Solution. Note that points $B$ and $C$ lie on the line $y=2x+1$. Their abscissas differ by 1, hence $BC=\sqrt{5}$. The length of the height of triangle $ABC$, drawn from vertex $A$, is equal to the distance $h$ from point $A$ to the line $y=2x+1$, which, in turn, is equal to $1 / \sqrt{5}$. The desired area
$$
S_{ABC}... | 0.50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,613 |
8. In an acute-angled triangle $A B C$, a point $Q$ is chosen on side $A C$ such that $A Q: Q C=1: 2$. From point $Q$, perpendiculars $Q M$ and $Q K$ are dropped to sides $A B$ and $B C$ respectively. It is given that $B M: M A=4: 1, B K=K C$. Find $M K: A C$. | Solution. Draw the heights $A K_{1}$ and $C M_{1}$. The idea of the solution is as follows: we will show that triangles $M_{1} B K_{1}$, $M B K$, and $A B C$ are similar to each other; from this, it will be easy to find the required ratio.
Let's denote the lengths:
$A Q=x, Q C=2 x, C K=z, K B=z, B M=4 y, M A=y$.
Fro... | \frac{2}{\sqrt{10}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,614 |
1. (2 points) A square of 90 by 90 is divided by two horizontal and two vertical lines into 9 rectangles. The sides of the central rectangle are 34 and 42. Find the total area of the four corner rectangles. | 1. 2688 .
+(2 points) - the solution is correct
-(0 points) - there are errors in the solution, including arithmetic errors | 2688 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,615 |
2. (3 points) Solve the equation $4 x^{2}+\frac{16 x^{2}}{|x+2|^{2}}=9$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. $x=\frac{1 \pm \sqrt{17}}{4}$
+ (3 points) - solution is correct
+- (2 points) - the main idea of transformations in the solution is correct (up to reducing the equation to a quadratic), but there are arithmetic errors further on
-+ (1 point) - only partial transformations are carried out in the solution. | \frac{1\\sqrt{17}}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,616 |
4. (4 points) Oleg usually arrives on a business trip by the 11 o'clock train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine t... | 4. 10 hours 55 minutes.
$+(4$ points) - solution is correct (by any method)
+- (3 points) - solution is correct, but there are arithmetic errors
-+ (2 points) - there are reasonable ideas in solving the problem, but the problem is not solved in general | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,618 |
5. (4 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$,
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \l... | 5. 2 .
$+(4$ points) - solution is correct
+- (3 points) - the idea of the recurrence relation is proven, but there are arithmetic errors in the solution
-+ (2 points) - the idea of the recurrence relation is formulated, but the problem is not completed | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,619 |
6. (5 points) Find all solutions to the system of equations
$$
\left\{\begin{array}{l}
\frac{y^{3}}{x^{2}}+\frac{x}{y}=1 \\
\frac{2 x}{y^{2}}+\frac{4 y}{x}=1
\end{array}\right.
$$ | 6. $x_{1}=-8, \quad y_{1}=-4, \quad x_{2}=-54, \quad y_{2}=-18$
+ (5 points) - solution is correct.
+- (3 points) - solution is generally correct, but there are unfortunate arithmetic errors
-+ (2 points) - only one of the two solutions is found in the solution. | x_{1}=-8,\quady_{1}=-4,\quadx_{2}=-54,\quady_{2}=-18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,620 |
1. There are 101 cells. Two players take turns writing one of the digits from 0 to 9 in these cells from left to right. If the sum of all the digits written after filling all the cells is divisible by 11, the player who moved first wins; if it is not divisible by 11, the second player wins. Which player will win with t... | # Solution:
The winning strategy will be for the first player. Let's describe it:
To ensure the first player wins, we need the sum of all digits to be divisible by 11 on the 101st step, meaning the remainder when divided by 11 should be "0". We can achieve this if and only if the sum of all digits after the 100th ste... | thefirstplayer | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,621 |
3. Simplify the fraction $\frac{2 x^{6}+5 x^{4}-3 x^{3}+2 x^{2}-12 x-14}{4 x^{6}-4 x^{4}-6 x^{3}-3 x^{2}+25 x-28}$. After simplification, the degrees of the polynomials in the numerator and the denominator should be reduced.
# | # Solution:
Let's find the greatest common divisor of the polynomials in the numerator and the denominator using the Euclidean algorithm.
For this, we will divide the denominator by the numerator with a remainder:
$$
\begin{gathered}
4 x^{6}-4 x^{4}-6 x^{3}-3 x^{2}+25 x-28= \\
=2 \cdot\left(2 x^{6}+5 x^{4}-3 x^{3}+2... | \frac{x^{3}+2x+2}{2x^{3}-3x+4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,622 |
4. Solve the equation $8 \cos ^{5} x-5 \cos x-2 \cos 3 x=1$.
# | # Solution:
We need to figure out how to express $\cos ^{5} x$ in terms of $\cos 5 x$. Then the original equation will be reduced to $\cos 5 x + \cos 3 x = 2$. The last equality is only possible if the condition $\left\{\begin{array}{l}\cos 5 x=1 \\ \cos 3 x=1\end{array} \Leftrightarrow x=2 \pi n, n \in Z\right.$ is s... | 2\pin,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,623 |
5. It is known that positive numbers $x, y, z$ satisfy the system:
$$
\left\{\begin{array}{c}
x^{2}+y^{2}+x y=529 \\
x^{2}+z^{2}+\sqrt{3} x z=441 \\
z^{2}+y^{2}=144
\end{array}\right.
$$
Find the value of the expression $\sqrt{3} x y+2 y z+x z$. | # Solution:
Consider triangle $ABC$ with a point $O$ chosen inside it such that $AO=y, OB=z, OC=x$, $\angle AOB=90^{\circ}, \angle AOC=120^{\circ}, \angle BOC=150^{\circ}$ (see figure). The conditions of the system represent the cosine theorem (including the Pythagorean theorem) for triangles $AOB, AOC, BOC$.
From th... | 224\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,624 |
6. Let's fix 10 natural numbers $n_{1}, n_{2}, \ldots, n_{10}$ and denote their sum by $n=n_{1}+\cdots+n_{10}$. Suppose now that on a board, $n$ numbers $a_{1}, \ldots, a_{n}$ are written in a row, each of which is either 0 or 1. These numbers (in the order they are written) divide into 10 groups:
$$
\underbrace{a_{1}... | # Solution:
The desired number of sets is found by summing the number of sets with a given number of $\mathrm{k}$ non-zero groups:
For $\mathrm{k}=0$, such a set is unique.
For $\mathrm{k}=2 \sum_{1 \leq i<j \leq 10}\left(2^{n_{i}}-2\right) \cdot\left(2^{n_{j}}-2\right)$
For $\mathrm{k}=4 \sum_{1 \leq i<j<l<s \leq ... | 2^{n-1}+\frac{1}{2}\cdot(2^{n_{1}}-2)\cdot(2^{n_{2}}-2)\cdot\ldots\cdot(2^{n_{10}}-2) | Combinatorics | proof | Yes | Yes | olympiads | false | 8,625 |
7. Let $a_{n, m}$ denote the number obtained by writing down all natural numbers from $n$ to $m$ in sequence, where $n$ and $m$ are natural numbers, and $n > m \geq 1$. For example, the number $a_{4,2}=432$, and the number $a_{11,7}=1110987$. Prove that among such numbers there is a number divisible by 2022. | # Solution:
Consider numbers of the form $a_{n, 1}$, where $n$ is odd. Since there are infinitely many such numbers, there will be two numbers $a_{n, 1}$ and $a_{k, 1}$, with $n > k$, that have the same remainder when divided by 2022. Then the difference $a_{n, 1} - a_{k, 1}$ is divisible by 2022. In this case, $a_{n,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,626 |
8. Solve the equation $x^{2}+y^{2}+1=6 x y$, where $x$ and $y$ are natural numbers
# | # Solution:
We will prove an auxiliary statement.
Statement. Let the pair of natural numbers $(x_{0}, y_{0})$ satisfy the original equation
$$
x^{2} + y^{2} + 1 = 6xy
$$
Then
1) $x_{0} \neq y_{0}$
2) The equation (1) has another solution in natural numbers $\left(6y_{0} - x_{0}, y_{0}\right)$.
## Proof:
1) By se... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,627 |
1. Oleg has 550 rubles, and he wants to give his mother tulips for March 8, and there must be an odd number of them, and no color shade should be repeated. In the store where Oleg came, one tulip costs 49 rubles, and there are eleven shades of flowers available. How many ways are there for Oleg to give his mother flowe... | Solution. From the condition, it is obvious that the maximum number of flowers in a bouquet is 11.
1st method
Using the property of binomial coefficients
$$
\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5}+\cdots=2^{n-1}
$$
and also considering their combinatorial meaning, we get that the number of ways to ... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,628 |
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 $p$. Justify your solution. | 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 | 8,629 |
3. Come up with a system of two equations with two unknowns $x$ and $y$, so that its solutions are only the following three pairs of numbers: $x=y=1, x=y=2$ and $x=3, y=4$. In the equations of the system, in addition to numbers and the unknowns $x$ and $y$ themselves, you are allowed to use parentheses, the equality si... | Solution. For example, $\left\{\begin{array}{c}(x-1)(x-2)(x-3)=0 \\ (|x-1|+|y-1|)(|x-2|+|y-2|)(|x-3|+|y-4|)=0\end{array}\right.$ | {\begin{pmatrix}(x-1)(x-2)(x-3)=0\\(|x-1|+|y-1|)(|x-2|+|y-2|)(|x-3|+|y-4|)=0\end{pmatrix}.} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,630 |
4. The function $y=f(x)$ is defined on the set $(0,+\infty)$ and takes positive values on it. It is known that for any points $A$ and $B$ on the graph of the function, the areas of the triangle $A O B$ and the trapezoid $A B H_{B} H_{A}$ are equal to each other $\left(H_{A}, H_{B}-\right.$ the bases of the perpendicula... | Solution. Let $M$ be the intersection point of segments $O B$ and $A H_{A}$. Since the areas of triangle $A O B$ and trapezoid $A B H_{B} H_{A}$ are equal, the areas of triangles $A M O$ and trapezoid $M B H_{B} H_{A}$ are also equal. From this, it follows that the areas of triangles $A O H_{A}$ and trapezoid $B O H_{B... | f(x)=\frac{}{x},>0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,631 |
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 $C E F$. Justify your solution. | 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 | 8,632 |
6. Let $x_{1}$ and $x_{2}$ be the largest roots of the polynomials $f(x)=1-x-4 x^{2}+x^{4}$ and $g(x)=16-8 x-16 x^{2}+x^{4}$ respectively. Find $\frac{x_{1}}{x_{2}}$. Justify your solution.
# | # Solution.
1st method
Notice that $g(2 x)=16 f(x)$. Then $x_{1}-$ is a root of $f(x)$ if and only if $2 x_{1}-$ is a root of $g(x)$. Therefore, $\frac{x_{1}}{x_{2}}=\frac{1}{2}$.
2nd method
Comparing the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text { and } g(x)=16-8 x-16 x^{2}+x^{4}
$$
shows t... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,633 |
7. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$ | Solution. Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of ... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,634 |
8. Let $A$ and $B$ be some numerical sets, and the set $C=\{a+b \mid a \in A, b \in B\}$ represents their sum. (In other words, the set $C$ consists of all possible sums of elements from sets $A$ and $B$. For example, if $A=\{0,1,2\}, B=\{1,2\}$, then $C=\{1,2,3,4\}$.) It is known that $C=\left\{0,1,2, \ldots, 2^{2828}... | Solution. 1) Correct: if set $A$ or set $B$ is infinite, then set $C$ will also be infinite. Therefore, we can denote by $a, b, c$ the maximum elements of these sets, respectively, and note for solving part 3 that $a+b=c$. Separately, we note that such sets exist: for example, $A=\{0, \ldots, a\}, B=\{0, \ldots, c-a\}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,635 |
1. Solve the inequality $2^{\log _{2}^{2} x}-12 \cdot x^{\log _{0.5} x}<3-\log _{3-x}\left(x^{2}-6 x+9\right)$. | Solution.
$$
\begin{gathered}
2^{\log _{2}^{2} x}-12 \cdot x^{\log _{0.5} x} < 0. \text{ Let } y = x^{\log _{2} x}. \text{ Then (1) } \Leftrightarrow y-\frac{12}{y} < 0, \text{ so } y \in (0,4).
\end{gathered}
$$
From here, $x^{\log _{2} x} < 4 \Leftrightarrow \log _{2}^{2} x < 2 \Leftrightarrow -\sqrt{2} < \log _{2}... | x\in(2^{-\sqrt{2}},2)\cup(2,2^{\sqrt{2}}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,636 |
2. $\quad$ Solve the equation $\sqrt{\frac{2 t}{1+t^{2}}}+\sqrt[3]{\frac{1-t^{2}}{1+t^{2}}}=1$. | Solution. Let's make the substitution
$$
t=\operatorname{tg} \frac{\alpha}{2}, \alpha \in(-\pi, \pi) .
$$
Then
$$
\frac{2 t}{1+t^{2}}=\sin \alpha, \frac{1-t^{2}}{1+t^{2}}=\cos \alpha
$$
and the original equation will take the form
$$
\sqrt{\sin \alpha}+\sqrt[3]{\cos \alpha}=1
$$
If $\cos \alpha<0$, then the left ... | \in{0,1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,637 |
3. Let's consider all 100-digit natural numbers, in the decimal representation of which only the digits 1 and 2 appear. How many of them are divisible by 3? | Solution: Each 100-digit natural number can be obtained by appending two digits to the right of a 98-digit number. Let $x$ be some 98-digit number. We will examine which two digits (each equal to 1 or 2) need to be appended to the number $x$ so that the resulting 100-digit number is divisible by 3. We will use the fact... | \frac{4^{50}+2}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,638 |
4. Solve the equation $2^{x}+2^{y}=6^{t}$ in integers. | Solution. Let $x=y$ at first. In this case, the original equation will take the form
$$
2^{x+1}=6^{t}
$$
If $t>0$, then the right-hand side (1) is divisible by three, while the left-hand side is not. Therefore, $t \leq 0$. If we assume that $t<0$, then the left-hand side (5) is divisible by 4. If $t$ is odd, then the... | (x,y,)=(-1,-1,0),(1,2,1),(2,5,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,639 |
5. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{aligned}
a_{1} b_{1}+a_{2} b_{3} & =1 \\
a_{1} b_{2}+a_{2} b_{4} & =0 \\
a_{3} b_{1}+a_{4} b_{3} & =0 \\
a_{3} b_{2}+a_{4} b_{4} & =1
\end{aligned}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_... | Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Multiply equation (a) of the original system
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & \text { (a) } \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1\end{cases}
$$
by $b_{2}$ ... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,640 |
6. The base of the pyramid $T A B C D$ is a trapezoid $A B C D (B C \| A D)$. The distances from points $A$ and $B$ to the plane $T C D$ are $r_{1}$ and $r_{2}$, respectively. The area of triangle $T C D$ is $S$. Find the volume of the pyramid $T A B C D$. | Solution: The volume of the pyramid $T A B C D$ is equal to the sum of the volumes of the pyramids $T B C D$ and $T A B D: V_{T A B C D}=V_{T B C D}+V_{T A B D}$. Moreover, $V_{T A B D}=V_{T A C D}$, because the pyramids $T A B D$ and $T A C D$ have a common height (from vertex $T$), and the areas of their bases are eq... | \frac{S(r_{1}+r_{2})}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,641 |
7. Given a triangle $ABC$. On side $AC$, a point $Q$ is chosen in such a way that the length of the segment $MK$, where $M$ and $K$ are the feet of the perpendiculars dropped from point $Q$ to sides $AB$ and $BC$ respectively, is minimized. In this case, $QM=1$, $QK=\sqrt{2}$, and $\angle B=45^{0}$. Find the area of tr... | Solution. Let the lengths of the perpendiculars dropped from point $Q$ to the base $AC$ be denoted as $d_{1}$ and $d_{2}$; let $\angle B=\beta$. The quadrilateral $MBKQ$ is inscribed in a circle, and $BQ$ is its diameter. Using the formula for the radius of the circumcircle of triangle $MBK$, we have
$$
BQ = \frac{MK}... | \frac{25}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,642 |
8. Find all non-negative integers $a$ and $b$ that satisfy the equation
$$
a^{2}+b^{2}=841 \cdot(a b+1)
$$ | Solution. Let the pair of numbers $(a, b)$ satisfy the equation of the problem:
$$
a^{2}+b^{2}=k^{2}(a b+1), k=29
$$
Suppose one of the numbers, for example $a$, is zero. Then, obviously, $b=k$. Therefore, we will further consider such solutions ( $a, b$ ) of the equation (1), for which
$$
a \neq 0, b \neq 0
$$
Mor... | Thesolutions(,b)(underthecondition\leqb)thoseonlythosepairsof(a_{n},b_{n}),whichforeachn\in\mathbb{N}calculatedtheformulas:a_{n}=b_{n-1},b_{n}=k^{2}b_{n-1}-a_{n-1},a_{0}=0,b_{0}=k;herek= | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,643 |
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