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1. (17 points) When walking uphill, the tourist walks 3 km/h slower, and downhill 3 km/h faster, than when walking on flat ground. Climbing the mountain takes the tourist 8 hours, while descending the mountain takes 4 hours. What is the tourist's speed on flat ground? | # Answer: 9
Solution. Let $x$ km/h be the tourist's speed on flat terrain. According to the problem, we get the equation $8(x-3)=4(x+3)$. From this, we find $x=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) Find the area of the triangle cut off by the line $y=3 x+1$ from the figure defined by the inequality $|x-1|+|y-2| \leq 2$. | Answer: 2.
Solution. The figure defined by the given inequality is a square.
The side of the square is $2 \sqrt{2}$ (this can be found using the Pythagorean theorem). The given line passes t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) A section of a regular triangular pyramid passes through the midline of the base and is perpendicular to the base. Find the area of the section if the side of the base is 6 and the height of the pyramid is 8. | Answer: 9.
Solution. The section MNP passes through the midline of the base of the pyramid $MN$ and is perpendicular to the base. Therefore, the height $PH$ of the triangle $MNP$ is parallel to the height of the pyramid $DO$.
 Find the area of the triangle cut off by the line $y=2x+2$ from the figure defined by the inequality $|x-2|+|y-3| \leq 3$. | Answer: 3.
Solution. The figure defined by the given inequality is a square.
The side of the square is $3 \sqrt{2}$ (this value can be found using the Pythagorean theorem). The given line pa... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) There are two circles: one with center at point $A$ and radius 6, and another with center at point $B$ and radius 3. Their common internal tangent touches the circles at points $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at point $E$. Find $C D$, if $A E=10$. | Answer: 12
Solution. Triangles $A C E$ and $B D E$ are similar (they have vertical angles and a right angle each) with a similarity coefficient of 2. From triangle $A C E$, using the Pythagorean theorem, we find $C E=8$. Therefore, $D E=4$, and $C D=12$. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Find the largest root of the equation
$$
\left|\cos (\pi x)+x^{3}-3 x^{2}+3 x\right|=3-x^{2}-2 x^{3}
$$ | # Answer: 1
Solution. It is obvious that 1 is a root of the equation (when $x=1$, both sides of the equation are equal to zero). If $x>1$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=5 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its im... | Answer: -10 Dptr $u \approx 7.1$ cm
Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain that the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{5}{10}=5 \mathrm{~cm} . \quad$ The power of the lens: $D=\frac{1}{d}-\frac{1}{f}=\... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Find the smallest root of the equation
$$
\sin (\pi x)+\tan x=x+x^{3}
$$ | Answer: 0
Solution. Obviously, 0 is a root of the equation (when $x=0$, both sides of the equation are equal to zero). If $x<0$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=20 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its i... | Answer: 5 Dpt $i \approx 14.1$ cm
Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{20}{10}=20 \mathrm{~cm} . \quad$ The optical power of the lens: $D=\frac{1}{d}-\frac{1}{f... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let in triangle $A B C$
$$
\cos (2 \angle A-\angle B)+\sin (\angle A+\angle B)=2 \text {. }
$$
Find the side $B C$, if $A B=4$. | Answer: 2.
Solution. Each term in the left part of the equation does not exceed 1. Therefore, the equality will hold only if each of them equals 1. We solve the corresponding equations, denoting $\alpha=\angle A, \beta=\angle B:$
$$
2 \alpha-\beta=2 \pi n, n \in \mathbb{Z} ; \quad \alpha+\beta=\frac{\pi}{2}+2 \pi k, ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The infantry column stretched out over 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked 2 km $400 \mathrm{m}$ during this time. How far did the sergeant travel during this time? | Answer: 3 km $600 \mathrm{~m}$
Solution. Let the speed of the column be $x$ km/h, and the sergeant travels $k$ times faster, i.e., at a speed of $k x$ km/h. To reach the end of the column, Kim traveled $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction, $t_{2}=\frac{1}{k x+x}$ hours (meeting he... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) A beam of light with a diameter of $d_{1}=5 \mathrm{~cm}$ falls on a thin diverging lens with an optical power of $D_{p}=-6$ Diopters. On a screen positioned parallel to the lens, a bright spot with a diameter of $d_{2}=20 \mathrm{~cm}$ is observed. After replacing the thin diverging lens with a thin con... | # Answer: 10 Dptr
Solution. The optical scheme corresponding to the condition:
(3 points)
The path of the rays after the diverging lens is shown in black, and after the converging lens in ... | 10 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 6 (10 points)
A cylinder with a mass of $M=1$ kg was placed on rails inclined at an angle $\alpha=30^{\circ}$ to the horizontal (the side view is shown in the figure). What is the minimum mass $m$ of the load that needs to be attached to the thread wound around the cylinder so that it starts rolling upward... | # Solution and evaluation criteria:
The moment of forces relative to the point of contact of the cylinder with the plane: $m g($ ( $-R \sin \alpha)=M g R \sin \alpha$ (5 points) $m\left(1-\f... | 1 | Other | math-word-problem | Yes | Yes | olympiads | false |
6. An ideal gas was expanded in such a way that during the process, the pressure of the gas turned out to be directly proportional to its volume. As a result, the gas heated up by $\Delta T=100^{\circ} \mathrm{C}$, and the work done by the gas was $A=831$ J. Determine the amount of substance that participated in this p... | Answer: 2 moles
Solution. From the condition $p=\alpha V$ (3 points). The work of the gas is equal to the area under the graph of the given process, constructed in coordinates $p-V$:
$A=\frac{p_{1}+p_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha V_{1}+\alpha V_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha}{2... | 2 | Other | math-word-problem | Yes | Yes | olympiads | false |
7. Two small balls with charges $Q=-20 \cdot 10^{-6}$ C and $q=50 \cdot 10^{-6}$ C are located at the vertices $A$ and $B$ of a mountain slope (see figure). It is known that $AB=2$ m, $AC=3$ m. The masses of the balls are the same and equal to $m=200$ g each. At the initial moment of time, the ball with charge $q$ is r... | Answer: $5 \mathrm{M} / \mathrm{c}$
Solution. The law of conservation of energy for this situation:
$k \frac{Q q}{A B}+m g \cdot A B=\frac{m v^{2}}{2}+k \frac{Q q}{B C}(5$ points $)$.
As a result, we get: $v=\sqrt{\frac{2 k Q q}{m \cdot A B}+2 \cdot g \cdot A B-\frac{2 k Q q}{m \cdot B C}}=\sqrt{\frac{2 \cdot 9 \cdo... | 5 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Petya was given several physics problems and several math problems for homework. All the problems solved by Petya constitute $5 \%$ of the total number of physics problems and $20 \%$ of the total number of math problems. What percentage of the total number of problems did Petya solve? | Solution. Answer: $4 \%$.
Let Pete solve $N$ problems. This constitutes $5 \%$ ( $\frac{1}{20}$ of) the physics problems, so the total number of physics problems was $20 \mathrm{~N}$. Similarly, the number of math problems was $5 N$, so the total number of problems assigned was $20 N + 5 N = 25 N$. The problems solved... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. We will call a natural number odd-powered if all its prime divisors enter its factorization with an odd exponent. What is the maximum number of odd-powered numbers that can occur consecutively?
## Answer: 7. | Solution. Note that among any eight consecutive natural numbers, there will definitely be a number that is divisible by 4 but not by 8. Then the number 2 will appear in its factorization to the second power.
Example of seven odd-power numbers: $37,38,39,40,41,42,43$.
Problem 4/1. On side $A C$ of triangle $A B C$, a ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 2. On an island, there live knights who always tell the truth, and liars who always lie. One day, 30 inhabitants of this island sat around a round table. Each of them said one of two phrases: "My left neighbor is a liar" or "My right neighbor is a liar." What is the smallest number of knights that can be at the ta... | # Answer: 10.
Solution. Next to each liar, there must be at least one knight (otherwise, if liars sit on both sides of a liar, the statement made by the liar would definitely be true). Therefore, among any three consecutive residents, there is at least one knight. If we divide all the people sitting at the table into ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.2. Masha and the Bear ate a basket of raspberries and 40 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate both raspberries and pies 3 times faster than Masha. How many pies did Masha eat, if they ate t... | Answer: 4 pies. Solution: The bear ate his half of the raspberries three times faster than Masha. This means Masha ate pies for three times less time than the bear. Since she eats three times slower, she ate 9 times fewer pies than the bear. Dividing the pies in the ratio of $9: 1$, we see that Masha got the 10th part,... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. For positive numbers $x$ and $y$, it is known that
$$
\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}=1
$$
What values can the product $x y$ take? List all possible options and prove that there are no others. | Answer: 1.
Solution. Note that for each positive $y$ the function
$$
f_{y}(x)=\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}
$$
strictly decreases on the ray $(0 ;+\infty)$, since the denominators of all three fractions increase. Therefore, the function $f_{y}$ takes each value no more than once. Moreover, ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Answer: 4.
Solution. Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. The miserly knight has 5 chests of gold: the first chest contains 1001 gold coins, the second - 2002, the third - 3003, the fourth - 4004, the fifth - 5005. Every day, the miserly knight chooses 4 chests, takes 1 coin from each, and places them in the remaining chest. After some time, there were no coins l... | Answer: in the fifth.
Solution. Let's look at the remainders of the number of coins in the chests when divided by 5. Each day, the number of coins in each chest either decreases by 1 or increases by 4, which means the remainder when divided by 5 always decreases by 1 (if it was 0, it will become 4). Since after some t... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Petya has a $3 \times$ 3 table. He places chips in its cells according to the following rules:
- no more than one chip can be placed in each cell;
- a chip can be placed in an empty cell if the corresponding row and column already contain an even number of chips (0 is considered an even number).
What is th... | Solution. Answer: 9.
Let $a, b, c$ denote the left, middle, and right columns of the table, and $1, 2, 3$ denote the bottom, middle, and top rows of the table. Petya can fill all 9 cells of the table with chips, for example, in the following order: $a 1, b 2, c 3, a 2, b 3, c 1, a 3, b 1, c 2$. Obviously, it is not po... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1011 of them such that the sum of any two does not equal 2021 or 2022. How many ways are there to do this? | Solution. Answer: 1.
Among the numbers from 1 to 2021, we can select 1010 non-overlapping pairs of numbers that sum to 2022. Specifically: $(1,2021),(2,2020), \ldots,(1009,1013),(1010,1012)$; only the number 1011 remains unpaired.
According to the condition, from each pair, we can choose no more than one number. Ther... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. In the Kurchatov School, exactly 2 people sit at each desk. It is known that exactly $70 \%$ of the boys have a boy as a desk partner, and exactly $40 \%$ of the girls have a girl as a desk partner. How many times more boys are there than girls? | Answer: 2 times.
Solution. Let the number of boys be $x$, and the number of girls be $y$. Note that $30\%$ of the boys sit at desks with girls and $60\%$ of the girls sit at desks with boys. Since exactly 2 people sit at each desk, then $0.3 x = 0.6 y$, from which $x = 2 y$. Thus, there are 2 times more boys than girl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Dodson, Williams, and their horse Bolivar want to get from city A to city B as quickly as possible. Along the road, there are 27 telegraph poles that divide the entire journey into 28 equal segments. Dodson takes 9 minutes to walk one segment, Williams takes 11 minutes, and either of them can ride Bolivar to... | Solution. Let the distance from A to B be taken as a unit, and time will be measured in minutes. Then Dodson's speed is $1 / 9$, Williams' speed is $1 / 11$, and Bolivar's speed is $-1 / 3$.
Let the desired post have the number $k$ (i.e., the distance from city A is $k / 28$). Then Dodson will arrive in time
$$
\frac... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. How many lines exist that pass through the point $(0,2019)$ and intersect the parabola $y=x^{2}$ at two points with integer coordinates on the $y$-axis?
Answer: 9. | Solution. A vertical line obviously does not fit. All lines, different from the vertical one and passing through the point $(0,2019)$, are given by the equation $y=k x+2019$ for some $k$. Let such a line intersect the parabola at points $\left(a, a^{2}\right)$ and $\left(b, b^{2}\right)$, where $a^{2}$ and $b^{2}$ are ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Answer: 4.
Solution. Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.3. A country has the shape of a square and is divided into 25 identical square counties. In each county, either a knight-count, who always tells the truth, or a liar-count, who always lies, rules. One day, each count said: "Among my neighbors, there are an equal number of knights and liars." What is the maxim... | Answer: 8.
Solution. First, note that counties with exactly three neighboring counties must be governed by lying counts (Fig. ??). Therefore, the corner counties are also governed by liars, since both of their neighbors are liars.
| | $L$ | $L$ | $L$ | |
| :--- | :--- | :--- | :--- | :--- |
| $L$ | | | | $L$ |
|... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Buses from Moscow to Oryol depart at the beginning of each hour (at 00 minutes). Buses from Oryol to Moscow depart in the middle of each hour (at 30 minutes). The journey between the cities takes 5 hours. How many buses from Oryol will the bus that left from Moscow meet on its way? | Answer: 10.
It is clear that all buses from Moscow will meet the same number of buses from Orel, and we can assume that a bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Orel at $7:30, 8:30, \ldots, 15:30, 16:30$ and only them. There are 10 such buses.
$\pm$ Correct reaso... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Nезнayka, Doctor Pilulkin, Knopochka, Vintik, and Znayka participated in a math competition. Each problem in the competition was solved by exactly four of them. Znayka solved strictly more than each of the others - 10 problems, while Nезнayka solved strictly fewer than each of the others - 6 problems. How many probl... | Answer: 10.
Each of Dr. Pill, Knopochka, and Vintik, according to the condition, solved from 7 to 9 problems. Therefore, the total number of solved problems ranges from $10+6+3 \cdot 7=37$ to $10+6+3 \cdot 9=43$. Note that this number should be equal to four times the number of problems. Among the numbers from 37 to 4... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. If the width of a rectangle is increased by $30 \%$, and the height is decreased by $20 \%$, its perimeter will not change. Will the perimeter decrease or increase, and by what percentage, if instead the width of the original rectangle is decreased by $20 \%$, and the height is increased by $30 \%$? | Answer. It will increase by $10 \%$. Solution. Let the width of the original rectangle be $s$, and the height be $h$. In the first case, the modified width and height will be 1.3 and 0.8 $s$ respectively. According to the condition, $2(s+h)=2(1.3 s+0.8 h)$, from which $h=1.5 s$. This means that the original perimeter i... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a school tic-tac-toe tournament, 16 students participated, each playing one game against every other student. A win was worth 5 points, a draw -2 points, and a loss -0 points. After the tournament, it was found that the participants collectively scored 550 points. What is the maximum number of participant... | Solution. A total of $\frac{16 \cdot 15}{2}=120$ games were played in the tournament. In each game, either 5 points were awarded (in case of a win-loss) or 4 points (in case of a draw). If all games had ended in a draw, the total number of points for all participants would have been $120 \cdot 4=480$, which is 70 point... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Several numbers were written on the board, their arithmetic mean was equal to $M$. They added the number 15, after which the arithmetic mean increased to $M+2$. After that, they added the number 1, and the arithmetic mean decreased to $M+1$. How many numbers were on the board initially? (Find all options and prove t... | Answer: 4.
Let there be $k$ numbers in the original list with a sum of $S$. Then, by the condition,
$$
\frac{S+15}{k+1}-\frac{S}{k}=2, \quad \frac{S+15}{k+1}-\frac{S+16}{k+2}=1.
$$
By bringing to a common denominator and transforming in an obvious way, we get that these equations are equivalent to the following two:... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many solutions in natural numbers does the equation
$$
(2 x+y)(2 y+x)=2017^{2017} ?
$$ | Answer: 0.
Note that the sum of the numbers $A=2x+y$ and $B=2y+x$ is divisible by 3. Since the number on the right side is not divisible by 3, neither $A$ nor $B$ are divisible by 3. Therefore, one of these two numbers gives a remainder of 2 when divided by 3, and the other gives a remainder of 1. Thus, their product ... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Dodson, Williams, and their horse Bolivar want to get from city A to city B as quickly as possible. Along the road, there are 27 telegraph poles that divide the entire journey into 28 equal segments. Dodson takes 9 minutes to walk one segment, Williams takes 11 minutes, and either of them can ride Bolivar to... | Solution. Let the distance from A to B be taken as a unit, and time will be measured in minutes. Then Dodson's speed is $1 / 9$, Williams' speed is $1 / 11$, and Bolivar's speed is $-1 / 3$.
Let the desired post have the number $k$ (i.e., the distance from city A is $k / 28$). Then Dodson will arrive in time
$$
\frac... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In a toy store, 125 plush bears are sold in $k$ different colors and six different sizes. For what largest $k$ can we assert that there will be at least three identical bears? (i.e., matching both in color and size) (20 points) | Solution: If $k=10$, then there are 60 varieties of teddy bears in the store. If there are no more than two of each type, then there are no more than 120 teddy bears in the store - a contradiction. Therefore, there will be at least three identical teddy bears among them.
If $k=11$, then there may not be three identica... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, such that each, when added to the product of the others, equals two. | Solution. Let $x_{1} x_{2} x_{3} x_{4}=p$. Then our system of equations will take the form:
$$
x_{i}+\frac{p}{x_{i}}=2, \quad i=1,2,3,4
$$
The case when one of the sought numbers is zero leads to a contradiction. Indeed, substituting zero for $x_{1}$, we get
$$
x_{2} x_{3} x_{4}=2, x+2=x+3=x+4=2 \text {. }
$$
Thus,... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On a piece of paper, 25 points are marked - the centers of the cells of a $5 \times 5$ square. The points are colored with several colors. It is known that no three points of the same color lie on any straight line (vertical, horizontal, or at any angle). What is the minimum number of colors that could have been use... | Solution: If no more than two colors are used, then, in particular, in the first row there will be at least three points of the same color - they lie on the same straight line. An example of using three colors (identical numbers denote points painted in the same color):
## Answer: 3 colors
| 1 | 2 | 2 | 3 | 3 |
| :--... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Daria Dmitrievna is preparing a test on number theory. She promised to give each student as many problems as the number of addends they create in the numerical example
$$
a_{1}+a_{2}+\ldots+a_{n}=2021
$$
where all numbers $a_{i}$ are natural numbers, greater than 10, and are palindromes (do not change if their dig... | Solution: A student cannot receive one problem since 2021 is not a palindrome. Suppose he can receive two problems, then at least one of the numbers $a_{1}, a_{2}$ is a four-digit number. If it starts with 2, then the second digit is 0 and the number itself is 2002. In this case, the second number is 19, which is not a... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. for $a<0.5$
## Correct answer: for $a<1$
## Question 3
Correct
Points: 1.00 out of a maximum of 1.00
Question 4
Correct
Points: 1.00 out of a maximum of 1.00
## Question 5
Correct
Points: 1.00 out of a maximum of 1.00
Worker $A$ and worker $B$ can complete the work in 7.2 days, worker $A$ and worker $C$ in... | Answer: 6
Correct answer: 6
A piece of copper-tin alloy has a total mass of 12 kg and contains 45% copper. How much pure tin should be added to this piece of alloy to obtain a new alloy containing 40% copper?
Answer: 1.5
Correct answer: 1.5
Find a two-digit number that is three times the product of its digits. If ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x$ be a natural number. Solve the equation
$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x} \cdots+\frac{1}{x}=3
$$ | Solution. Multiply both sides of the equation by $x$. We get
$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$
The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1$, and $a_{n}=1$. The progression has $(x-1)$ terms.
Since $S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n$, the equation c... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation $\sqrt{6-x}+\sqrt{x-4}=x^{2}-10 x+27$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. Given $4 \leq x \leq 6$. Find the maximum of the function $f(x)=\sqrt{6-x}+\sqrt{x-4}$
$f^{\prime}(x)=-\frac{1}{\sqrt{6-x}}+\frac{1}{\sqrt{x-4}} \Rightarrow f^{\prime}(x)=0 ; \sqrt{4-x}=\sqrt{6-x} \Rightarrow x=5$. When $x=5$, $f(x)$ has a maximum value of 2. On the other hand, $x^{2}-10 x+27=(x-5)^{2}+2 \Rightarro... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Solve the inequality $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} \geq 0$. In the answer, indicate the sum of all natural numbers that are solutions to the inequality. | 4. $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} * \frac{\left|x^{2}-3 x\right|+\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|+\left|x^{2}-2 x\right|} \geq 0 \Leftrightarrow \frac{\left(x^{2}-3 x\right)^{2}-\left(x^{2}-2\right)^{2}}{\left(x^{2}-x-2\right)^{2}-\left(x^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Find all solutions to the equation $2017^{x}-2016^{x}=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. It is obvious that $x=1$ will be a solution to the given equation. Let's show that there are no others. We have: $\left(\frac{2017}{2016}\right)^{x}=1+\frac{1}{2016^{x}} ;\left(\frac{2017}{2016}\right)^{x}-1=\frac{1}{2016^{x}}$ From this, it is clear that the function on the left side is increasing, while the functi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Three circles with radii $1,2,3$ touch each other externally. Find the radius of the circle passing through the three points of tangency of these circles. | 6. Let the centers of the given circles be denoted as $O_{1}, O_{2}, O_{3}$. Let $K, M, N$ be the points of tangency of the given circles. Then $O_{1} K=O_{1} N=1, O_{2} K=O_{2} M=2, O_{3} M=O_{3} N=3$.
 set off in pursuit of the raft. The motorboat caught up with the raft and immediately turned back to point $A$. After 3.6 hours, the motorboat arrived at point $A$, while the raft reached po... | 7. Let the speed of the river current be $x$ km/h, $t$-the time elapsed from the moment the boat left until the moment of the meeting. Then, by the condition: 1) the distance traveled by the raft until the meeting, 2) the distance traveled by the motorboat until the meeting, 3) the distance traveled by the motorboat af... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Calculate
the sum: $\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\cdots+\cos 160^{\circ}+\cos 180^{\circ}$ | 2.
$\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\cos 80^{\circ}+\cos 100^{\circ}+\cos 120^{\circ}+\cos 140^{\circ}+\cos 160^{\circ}+$ $\cos 180^{\circ}=\cos 20+\cos 40+\frac{1}{2}+\sin 10^{\circ}-\sin 10^{\circ}-\frac{1}{2}-\cos 40^{\circ}-1-\cos 20=-1$ Answer: $\{-1\}$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two pedestrians set out simultaneously from $A$ to $B$ and from $B$ to $A$. When the first had walked half the distance, the second had 24 km left to walk, and when the second had walked half the distance, the first had 15 km left to walk. How many kilometers will the second pedestrian have left to walk after the fi... | 3. Let $v_{1}, v_{2}$ be the speeds of the first and second pedestrians, $S$ be the distance from $A$ to $B$, and $x$ be the distance the second pedestrian still needs to walk when the first pedestrian finishes the crossing. From the problem statement, we get the system $\frac{S}{2} \frac{v_{2}}{v_{1}}+24=S, \frac{S}{2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Solve the inequality: $\frac{\left(\left|x^{2}-2\right|-7\right)(|x+3|-5)}{|x-3|-|x-1|}>0$. Write the largest integer that is a solution to the inequality in your answer. | 5. $\frac{\left(\left|x^{2}-2\right|-7\right)\left(\left|x^{2}-2\right|+7\right)(|x+3|-5)(|x+3|+5)}{(|x-3|-|x-1|)(|x-3|+|x-1|)}>0 \Leftrightarrow \frac{\left(\left(x^{2}-2\right)^{2}-49\right)\left((x+3)^{2}-25\right)}{(x-3)^{2}-(x-1)^{2}}>0$ $\Leftrightarrow \frac{\left(x^{2}-2-7\right)\left(x^{2}-2+7\right)(x+3-5)(x+... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Let the cost of a pound of rice be x coins, a pound of brown sugar be y coins, and a pound of refined sugar be z coins. We obtain the system:
$$
\left\{\begin{array}{l}
4 x+\frac{9}{2} y+12 z=6 ; \\
12 x+6 y+6 z=8
\end{array}\right. \text { Subtract the first equation from twice the second equation and }
$$
expres... | Answer: 4 pounds sterling. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality: $\log _{|x-1|}\left(\frac{x-2}{x}\right)>1$. In the answer, indicate the largest negative integer that is a solution to this inequality. | 1. Find the domain: $(x-2) x>0$. From this, $x<0$ or $x>2$. If $|x-1|<1$, then $0<x<2$. But if $|x-1|>1$, then $x<0$ or $x>2$. Therefore, the inequality is equivalent to the system:
$\left\{\begin{array}{c}|x-1|>1 \\ \frac{x-2}{x}>|x-1|\end{array} \Leftrightarrow\right.$ a) $\left\{\begin{array}{c}x<0 \\ \frac{x-2}{x}... | -1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. Solve the equation in integers: $5 x^{2}-2 x y+2 y^{2}-2 x-2 y=3$. In the answer, write the sum of all solutions $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Transforming the equation, we get: $(x+y)^{2}-2(x+y)+1+y^{2}-4 x y+4 x^{2}=4 \Leftrightarrow(x+y-1)^{2}+(y-2 x)^{2}=4$. Since $x$ and $y$ are integers, this equality is possible only in four cases:
$\left\{\begin{array}{c}x+y-1=0 \\ y-2 x=2\end{array}, \quad\left\{\begin{array}{c}x+y-1=2 \\ y-2 x=0\end{array}, \quad... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Determine which of the numbers is greater: (1000!) $)^{2}$ or $1000^{1000}$. Write 1 in the answer if the first is greater or equal, 2 if the second is greater. | 8. Let's show that the left number is greater than the right one. For this, we will write the first number as follows: $(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \ldots \cdot 1000)(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \ldots \cdot 1000)=(1 \cdot 1000)(2 \cdot 999)(3 \cdot 998)(4 \cdot 997) \ldots(k(10... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. What is the maximum area that a triangle with sides $a, b, c$ can have, given that the sides are within the ranges: $0 < a \leq 1, 1 \leq b \leq 2, 2 \leq c \leq 3$? | 9. Among the triangles with two sides $a, b$, which satisfy the conditions $0<a \leq 1, 1 \leq b \leq 2$. The right triangle with legs $a=1, b=2$ has the largest area. Its area $S=1$. On the other hand, the third side $c=\sqrt{5}$ as the hypotenuse of the right triangle satisfies the condition that $2 \leq c \leq 3$ an... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given a triangle $\mathrm{PEF}$ with sides $\mathrm{PE}=3, \mathrm{PF}=5, \mathrm{EF}=7$. On the extension of side $\mathrm{FP}$ beyond point $\mathrm{P}$, a segment $\mathrm{PA}=1.5$ is laid out. Find the distance $d$ between the centers of the circumcircles of triangles $\mathrm{EPA}$ and $\mathrm{EAF}$. In the an... | 3. By the cosine theorem for angle EPF, we find that $\operatorname{Cos} E P F=-\frac{1}{2}$. Therefore, the angle $\mathrm{EPF}=120^{\circ}$; hence the adjacent angle will be $60^{\circ}$. Draw a perpendicular from point E to PF, meeting at point D. Since angle EPF is obtuse, point D will lie outside triangle EPF. In ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Find all integers $n$ for which the equality $\frac{n^{2}+3 n+5}{n+2}=1+\sqrt{6-2 n}$ holds. | 5. We have: $\frac{n^{2}+3 n+5}{n+2}-1=\sqrt{6-2 n} ; \frac{n^{2}+2 n+3}{n+2}=\sqrt{6-2 n} ; n+\frac{3}{n+2}=\sqrt{6-2 n}$. Since for integer $n$, the right-hand side can be either an integer or irrational, while the left-hand side is either an integer or rational, the given equality is possible only if $\frac{3}{n+2}$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the system
$$
\left\{\begin{array}{l}
\operatorname{tg}^{3} x+\operatorname{tg}^{3} y+\operatorname{tg}^{3} z=36 \\
\operatorname{tg}^{2} x+\operatorname{tg}^{2} y+\operatorname{tg}^{2} z=14 \\
\left(\operatorname{tg}^{2} x+\operatorname{tg} y\right)(\operatorname{tg} x+\operatorname{tg} z)(\operatorname{tg} ... | 6. Let $\operatorname{tg} x=x_{1}, \operatorname{tg} y=x_{2}, \operatorname{tg} z=x_{3}$.
Then $\left\{\begin{array}{l}x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=36 \\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=14 \\ \left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)=60\end{array}\right.$. Further, let $\left\{\begin{arra... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the integer part of the number $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$ | 9. Let $a=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$. It is easy to see that $a>\sqrt{6}>2 \Rightarrow$ $a>2$. Let's find an upper bound for this number. At the end of the expression for $a$, replace $\sqrt{6}$ with $\sqrt{9}$. Then we get that $a<\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}}=\sqrt{6+\sqrt{6+\sqrt{6+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1.
The pedestrian's speed is 5 km/h. There are more oncoming trams than overtaking ones, because the speed of the former relative to the pedestrian is greater than that of the latter. If we assume that the pedestrian is standing still, the speed of the oncoming trams is the sum of the tram's own speed $V+$ the ... | Answer: $11 \mathrm{km} /$ hour. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A professor from the Department of Mathematical Modeling at FEFU last academic year gave 6480 twos, thereby overachieving the commitments taken at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a result, a new record was se... | 1. The professors of the Department of Mathematical Modeling at DVFU in the previous academic year gave 6480 twos, thereby over-fulfilling the obligations they had taken on at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a r... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find natural numbers $n$ such that for all positive numbers $a, b, c$, satisfying the inequality
$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there exists a triangle with sides $a, b, c$. | 2. Find natural numbers $n$ such that for all positive numbers $a, b, c$ satisfying the inequality
$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there exists a triangle with sides $a, b, c$.
$\square$ Since $(a b+b c+c a) \leq\left(a^{2}+b^{2}+c^{2}\right)$, then $n>5$. If $n \geq 7$, then the numbers $a=1, b... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. A rectangle is drawn on the board. It is known that if its width is increased by $30 \%$, and its length is decreased by $20 \%$, then its perimeter remains unchanged. How would the perimeter of the original rectangle change if its width were decreased by $20 \%$, and its length were increased by $30 \%$? | Answer. It will increase by $10 \%$.
Solution. Let's denote the width of the original rectangle as $a$, and the height as $b$. In the first case, the modified width and length will be $1.3a$ and $0.8b$ respectively. According to the condition, $2(a+b) = 2(1.3a + 0.8b)$, from which we get $b = 1.5a$. This means the ori... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Several teams held a hockey tournament - each team played against each other once. 2 points were awarded for a win, 1 point for a draw, and no points for a loss. The team "Squirrels" won the most games and scored the fewest points. What is the minimum number of teams that could have participated in the tournament | Answer: 6 teams.
Solution. In a tournament with $n$ teams, the "Squirrels" scored less than the average number, i.e., no more than $n-2$ points. Therefore, some team must have scored more than the average, i.e., no less than $n$ points. For this, they had to win at least one match. Consequently, the "Squirrels" must h... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On the glade, there are 6 fewer bushes than trees. Birds flew in and sat both on the bushes and on the trees. They sat so that there were an equal number of birds on all the trees and an equal number of birds on all the bushes, but there were at least 10 more birds on a tree than on a bush. There were a total of 128... | Answer: 2.
Solution: There are no fewer than 7 trees, as there are 6 more trees than shrubs. There are no fewer than 11 birds on one tree, as there are at least 10 more birds on one tree than on one shrub. There cannot be 12 or more trees, as then there would be $12 \cdot 11=$ 132 or more birds on the trees. Therefore... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. There are 28 students in the class. 17 have a cat at home, and 10 have a dog. 5 students have neither a cat nor a dog. How many students have both a cat and a dog? | Answer: 4.
Solution: The number of students with cats or dogs is $28-5=23$. If we add $17+10=27$, it exceeds 23 due to those who have both a cat and a dog, who were counted twice. The number of such students is $27-23=4$.
Comment: Correct solution - 20 points. Solution started, with some progress - 5 points. Solution... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On graph paper, large and small triangles are drawn (all cells are square and of the same size). How many small triangles can be cut out from the large triangle? The triangles cannot be rotated or flipped (the large triangle has a right angle at the bottom left, the small triangle has a right angle at the top right)... | Answer: 12.
Solution. See the figure.
Comment. Correct drawing with 12 triangles - 20 points. Incorrect answer with drawing - 1 point. Any answer without drawing - 0 points. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On a certain day, there were several nuts in the bag. The next day, as many nuts were added to the bag as there already were, but eight nuts were taken away. On the third day, again as many nuts were added as there already were, but eight were taken away. The same thing happened on the fourth day, and after that, th... | Answer: 7 nuts.
Solution. At the beginning of the fourth day, there were $(0+8): 2=4$ nuts in the bag. Therefore, at the beginning of the third day, there were $(4+8): 2=6$, and at the very beginning $-(6+8): 2=7$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Natural numbers divisible by 3 were painted in two colors: red and blue, such that the sum of a blue and a red number is red, and the product of a blue and a red number is blue. In how many ways can the numbers be colored so that the number 546 is blue? | Answer: 7.
Solution: We will prove that all blue numbers are divisible by the same number. First, we will prove that the product of two blue numbers is blue. Assume the statement is false. Let $a$ and $b$ be two such blue numbers that their product is a red number. Let $d$ be some red number (we can take $d = ab$). Th... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Find the smallest natural $n>1$, for which the sum of no two natural powers is a perfect square of a natural number.
# | # Answer.4.
Solution. For $n=2: 2^{1}+2^{1}=2^{2}$. For $n=3: 3^{3}+3^{2}=6^{2}$. Let $n=4$. Suppose there exist such $m, k$ that $4^{m}+4^{k}=a^{2}$.
Any power of 4 gives a remainder of 1 when divided by 3 (since $\left.4^{m}=(3+1)^{m}\right)$. Therefore, $4^{m}+4^{k}$ will give a remainder of 2 when divided by 3. B... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A little squirrel collected 15 nuts in the forest weighing $50, 51, \ldots, 64$ grams. He knows the weight of each nut. Using a balance scale, the little squirrel tries to prove to his friends that the first nut weighs 50 g, the second 51 g, the third 52 g, and so on (initially, the friends know nothing about the we... | Answer. One weight.
Solution. Note that it is impossible to do without weights altogether, since if the weight of each nut is halved, the results of all weighings will not change. We will show that the baby squirrel can manage with the help of one 1 g weight. First, he will convince his friends that the nuts weigh $n,... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In a $5 \times 5$ square, color as many cells black as possible, so that the following condition is met: any segment connecting two black cells must necessarily pass through a white cell. | Answer. We will prove that it is impossible to color more than 9 cells. Divide the square into 9 parts:
In each part, no more than one cell can be colored.
9 cells can be colored as follows:... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find all natural numbers $a$ for which the number
$$
\frac{a+1+\sqrt{a^{5}+2 a^{2}+1}}{a^{2}+1}
$$
is also a natural number. | Answer. $a=1$.
Solution. Let $a+1=b, \sqrt{a^{5}+2 a^{2}+1}=c$. Write the polynomial $c^{2}-b^{2}=$ $a^{5}+2 a^{2}+1-(a+1)^{2}=a^{5}+a^{2}-2 a$, and divide it by $a^{2}+1$ using long division, we get a remainder of $-(a+1)$. For $a>1$, the absolute value of the remainder is less than $a^{2}+1$, so the remainder cannot... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (9th grade) In how many ways can the numbers $1,2,3,4,5,6$ be written in a row so that for any three consecutive numbers $a, b, c$, the quantity $a c-b^{2}$ is divisible by 7? Answer: 12. | Solution. Each subsequent member will be obtained from the previous one by multiplying by a certain fixed value (modulo 7). This value can only be 3 or 5. And the first term can be any, so there are 12 options in total. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A line parallel to the selected side of a triangle with an area of 16 cuts off a smaller triangle with an area of 9. Find the area of the quadrilateral, three vertices of which coincide with the vertices of the smaller triangle, and the fourth lies on the selected side. Choose the answer option with the number close... | Answer: 12. (C)
 | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A line parallel to the selected side of a triangle with an area of 27 cuts off a smaller triangle with an area of 3. Find the area of the quadrilateral, three vertices of which coincide with the vertices of the smaller triangle, and the fourth lies on the selected side. Choose the answer option with the number close... | Answer: 9. (B)
 | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Calculate
$$
\frac{y^{2}+x y-\sqrt[4]{x^{5} y^{3}}-\sqrt[4]{x y^{7}}}{\sqrt[4]{y^{5}}-\sqrt[4]{x^{2} y^{3}}} \cdot(\sqrt[4]{x}+\sqrt[4]{y})
$$
where $x=3, \underbrace{22 \ldots 2}_{2013} 3, y=4, \underbrace{77 \ldots .7}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 8. (D)
Options.
 | 8 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Calculate
$$
(\sqrt[4]{x}+\sqrt[4]{y}) \cdot \frac{x^{2}+x y-\sqrt[4]{x^{7} y}-\sqrt[4]{x^{3} y^{5}}}{\sqrt[4]{x^{5}}-\sqrt[4]{x^{3} y^{2}}}
$$
where $x=1, \underbrace{11 \ldots 1}_{2013} 2, y=3, \underbrace{88 \ldots 8}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 5. (B)
Options.
 | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. Calculate
$$
\frac{\sqrt[4]{x^{5} y^{3}}+\sqrt[4]{x y^{7}}-x y-y^{2}}{\sqrt[4]{x^{2} y^{3}}-\sqrt[4]{y^{5}}} \cdot(\sqrt[4]{x}+\sqrt[4]{y})
$$
where $x=2, \underbrace{44 \ldots .4}_{2013} 5, y=1, \underbrace{55 \ldots .5}_{2014}$. Choose the answer option with the number closest to the one you found. | Answer: 4. (E)
Options.
 | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Determine the number of different values of $a$ for which the equation
$$
\left(3-a^{2}\right) x^{2}-3 a x-1=0
$$
has a unique solution | Answer: 2. (C)
Options.
 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Determine how many different values of $a$ exist for which the equation
$$
\left(a^{2}-5\right) x^{2}-2 a x+1=0
$$
has a unique solution | Answer: 2. (C)
Options.
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|}
\hline \mathbf{A} & 0 & \mathbf{B} & 1 & \mathbf{C} & 2 & \mathbf{D} & 3 & \mathbf{E} \\
\hline
\end{array}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality $\frac{\sqrt{\frac{x}{\gamma}+(\alpha+2)}-\frac{x}{\gamma}-\alpha}{x^{2}+a x+b} \geqslant 0$.
In your answer, specify the number equal to the number of integer roots of this inequality. If there are no integer roots, or if there are infinitely many roots, enter the digit 0 in the answer sheet.
... | # Answer: 7.
Solution. Given the numerical values, we solve the inequality:
$$
\frac{\sqrt{x+5}-x-3}{x^{2}-15 x+54} \geqslant 0 \Leftrightarrow \frac{\sqrt{x+5}-x-3}{(x-6)(x-9)} \geqslant 0
$$
Since $\sqrt{x+5} \geqslant x+3 \Leftrightarrow\left[\begin{array}{c}-5 \leqslant x \leqslant-3, \\ x+5 \geqslant(x+3)^{2}\e... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. Find all integer values of $a$, not exceeding 15 in absolute value, for each of which the inequality
$$
\frac{4 x-a-4}{6 x+a-12} \leqslant 0
$$
is satisfied for all $x$ in the interval $[2 ; 3]$. In your answer, specify the sum of all such $a$. | Answer: -7.
Solution. Solving the inequality using the interval method, we find that its left side equals 0 at $x=1+\frac{a}{4}$ and is undefined at $x=2-\frac{a}{6}$. These two values coincide when $a=\frac{12}{5}$. In this case, the inequality has no solutions.
For $a>\frac{12}{5}$, the solution to the inequality i... | -7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5.1. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{2} 18-\log _{3} 12\right)-\log _{3} 16-4 \log _{2} 3
$$
do not exceed 8. | Solution. Let $a=\log _{2} 3$. Then the condition of the problem will turn into the inequality
$$
x^{2}+2\left(a-\frac{1}{a}\right) x-4\left(a+\frac{1}{a}+2\right) \leqslant 0
$$
Considering that $a \in\left(\frac{3}{2}, 2\right)$, we get $x \in\left[-2 a-2, \frac{2}{a}+2\right]$. Since $-6<-2 a-2<-5.3<\frac{2}{a}+2<... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.2. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{2} 20-\log _{5} 8\right)-\log _{2} 5-9 \log _{5} 2
$$
do not exceed 6. | Solution. Let $a=\log _{2} 5$. Then the condition of the problem will turn into the inequality
$$
x^{2}+\left(a-\frac{3}{a}+2\right) x-\left(a+\frac{9}{a}+6\right) \leqslant 0
$$
Considering that $a \in(2,3)$, we get $x \in\left[-a-3, \frac{3}{a}+1\right]$. Since $-6<-a-3<-5,2<\frac{3}{a}+1<3$, the integer solutions ... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.4. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function
$$
y=x^{2}+x\left(\log _{5} 2-\log _{2} 10\right)-\log _{2} 25-3 \log _{5} 2
$$
do not exceed 7. | Solution. Let $a=\log _{2} 5$. Then the condition of the problem will turn into the inequality
$$
x^{2}-\left(a-\frac{1}{a}+1\right) x-\left(2 a+\frac{3}{a}+7\right) \leqslant 0 .
$$
Considering that $a \in(2,3)$, we get $x \in\left[-\frac{1}{a}-2, a+3\right]$. Since $-3<-\frac{1}{a}-2<-2,5<a+3<6$, the integer soluti... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. Specify the integer closest to the larger root of the equation
$$
\operatorname{arctg}\left(\left(\frac{2 x}{7}+\frac{7}{8 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{2 x}{7}-\frac{7}{8 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. Let $y=\left(\frac{2 x}{7}-\frac{7}{8 x}\right)^{2}$, then $\left(\frac{2 x}{7}+\frac{7}{8 x}\right)^{2}=y+1$ and the equation will take the form
$$
\operatorname{arctg}(y+1)-\operatorname{arctg} y=\frac{\pi}{4}
$$
Since $0 \leqslant \operatorname{arctg} y<\operatorname{arctg}(y+1)<\frac{\pi}{2}$, the last ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. Specify the integer closest to the smaller root of the equation
$$
\operatorname{arctg}\left(\left(\frac{5 x}{26}+\frac{13}{10 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{5 x}{26}-\frac{13}{10 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{5 x}{26}=\frac{13}{10 x} \Longleftrightarrow|x|=\frac{13}{5}$.
Answer: -3 . | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. Specify the integer closest to the smaller root of the equation
$$
\operatorname{arctg}\left(\left(\frac{7 x}{10}-\frac{5}{14 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{7 x}{10}+\frac{5}{14 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.
Answer: -1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. Indicate the integer closest to the larger root of the equation
$$
\operatorname{arctg}\left(\left(\frac{3 x}{22}-\frac{11}{6 x}\right)^{2}\right)-\operatorname{arctg}\left(\left(\frac{3 x}{22}+\frac{11}{6 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.
Answer: 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. Indicate the integer closest to the smaller root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{2 x}{7}+\frac{7}{8 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{2 x}{7}-\frac{7}{8 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{2 x}{7}=\frac{7}{8 x} \Longleftrightarrow |x|=\frac{7}{4}$.
Answer: -2. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. Indicate the integer closest to the larger root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{5 x}{26}+\frac{13}{10 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{5 x}{26}-\frac{13}{10 x}\right)^{2}\right)=-\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{5 x}{26}=\frac{13}{10 x} \Longleftrightarrow|x|=\frac{13}{5}$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.7. Indicate the integer closest to the larger root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{7 x}{10}-\frac{5}{14 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{7 x}{10}+\frac{5}{14 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{7 x}{10}=\frac{5}{14 x} \Longleftrightarrow |x|=\frac{5}{7}$.
Answer: 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.8. Provide the integer closest to the smaller root of the equation
$$
\operatorname{arcctg}\left(\left(\frac{3 x}{22}-\frac{11}{6 x}\right)^{2}\right)-\operatorname{arcctg}\left(\left(\frac{3 x}{22}+\frac{11}{6 x}\right)^{2}\right)=\frac{\pi}{4}
$$ | Solution. The equation is equivalent to $\frac{3 x}{22}=\frac{11}{6 x} \Longleftrightarrow |x|=\frac{11}{3}$.
Answer: -4. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. How many solutions in integers does the equation
$6 y^{2}+3 x y+x+2 y-72=0$ have? | Answer: 4.
## Solution:
Factorize:
$(3 y+1)(2 y+x)=72$.
The first factor must give a remainder of 1 when divided by 3. The number 72 has only 4 divisors that give such a remainder: $-8,-2,1,4$. They provide 4 solutions.
Problem 7a. How many solutions in integers does the equation
$6 y^{2}+3 x y+x+2 y+180=0$ have?... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A trip to St. Petersburg is being organized for 30 schoolchildren along with their parents, some of whom will be driving cars. Each car can accommodate 5 people, including the driver. What is the minimum number of parents that need to be invited on the trip?
ANSWER: 10 | Solution: No more than 4 students can fit in a car, so 8 cars will be needed, i.e., 10 drivers. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A trip to Nizhny Novgorod is being organized for 50 schoolchildren along with their parents, some of whom will be driving cars. Each car can accommodate 6 people, including the driver. What is the minimum number of parents that need to be invited on the trip?
ANSWER: 10 | Solution: No more than 5 students can fit in a car, so 10 cars will be needed, i.e., 10 drivers. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Find the largest natural number that cannot be represented as the sum of two composite numbers.
ANSWER: 11 | Solution: Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. Odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we can see that 11 cannot be represented in such a way.
Lomonosov Moscow State University
## Scho... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Five runners ran a relay. If the first runner ran twice as fast, they would have spent $5 \%$ less time. If the second runner ran twice as fast, they would have spent $10 \%$ less time. If the third runner ran twice as fast, they would have spent $12 \%$ less time. If the fourth runner ran twice as fast, they would ... | Solution: If each ran twice as fast, they would run 50% faster. This means that if the 5th ran faster, the time would decrease by $50-5-10-12-15=8 \%$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation
$$
\log _{3}(x+2) \cdot \log _{3}(2 x+1) \cdot\left(3-\log _{3}\left(2 x^{2}+5 x+2\right)\right)=1
$$ | Problem 4.
Answer: $x=1$.
Solution. For admissible values $x>-\frac{1}{2}$ we have: $\log _{3}(x+2)>0$.
If $\log _{3}(2 x+1)0$ and the left side of the equation is negative.
Thus, all three factors on the left side of the equation are positive, and the roots should be sought only among those $x$ for which
$$
\left... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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