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Problem 5. There is a deck of 1024 cards, each with a different set of digits from 0 to 9, and all sets are distinct (including an empty card). We will call a set of cards complete if each digit from 0 to 9 appears exactly once on them.
Find all natural $k$ for which there exists a set of $k$ cards with the following ... | Solution. Answer: 512.
For each card, consider another card that complements it to a complete set (for example, for the card 3679, such a card would be 012458). It is clear that all 1024 cards can be divided into 512 non-overlapping pairs of cards that complement each other to a complete set. Next, we will prove that ... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In a certain company, there are 100 shareholders, and any 66 of them collectively own at least $50\%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? (The percentage of shares in the company owned by a shareholder can be non-integer.) | Answer: $25 \%$.
Solution. Consider any shareholder $A$. Divide the other 99 shareholders into three groups $B, C, D$ of 33 shareholders each. By the condition, $B$ and $C$ together have at least $50 \%$ of the company's shares, similarly for $C$ and $D$, and for $B$ and $D$. Adding all this up and dividing by two, we... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. In triangle $A B C$, the angle at vertex $B$ is $120^{\circ}$, point $M$ is the midpoint of side $A C$. Points $E$ and $F$ are chosen on sides $A B$ and $B C$ respectively such that $A E=E F=F C$. Find $\angle E M F$. | Answer: $90^{\circ}$.
Solution. Note that the sum of angles $A$ and $C$ is $60^{\circ}$. Let $A E=E F=F C=u$.
On the line $E M$, place a point $G$ such that point $M$ is the midpoint of segment $E G$. Triangles $A M E$ and $C M G$ are equal by two sides and the angle between them, so $\angle M C G=\angle M A E=\angle... | 90 | Geometry | 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 |
5. A cube with a side of 5 is made up of 125 smaller cubes with a side of 1. How many small cubes does a plane perpendicular to one of the cube's diagonals and passing through its midpoint intersect? | Answer: 55.
Let's introduce a coordinate system such that the cube is located in the first octant (the set of points with non-negative coordinates) and the mentioned diagonal extends from the origin $O$. The midpoint of the cube's diagonal has coordinates $(5 / 2, 5 / 2, 5 / 2)$, so the specified plane is given by the... | 55 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all pos... | Answer: 30.
Solution. Consider any arbitrary quartet of consecutive people. If there were at least 3 knights in it, the leftmost of them would definitely lie, which is impossible. If there were at least 3 liars, the leftmost of them would definitely tell the truth, which is also impossible. Therefore, in every quartet... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live knights who always tell the truth and liars who always lie. The population of the island is 1000 people, distributed across 10 villages (with no fewer than two people in each village). One day, every islander claimed that all their fellow villagers are liars. How many liars live on the islan... | Answer: 990.
In one village, at least two knights cannot live, because otherwise the knights would lie. Also, in the village, they cannot all be liars, since then these liars would tell the truth. Therefore, in each village there is exactly one knight, and there are 10 knights in total, and 990 liars. | 990 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Petya climbed up a moving upward escalator, counting 75 steps, and then descended the same escalator (i.e., moving against the direction of the escalator), counting 150 steps. During the descent, Petya walked three times faster than during the ascent. How many steps are there on the stopped escalator? | Answer: 120.
For convenience, let's introduce a unit of time during which Petya took one step while ascending the escalator. We will measure all speeds in steps per unit of time. Petya's speed while ascending is 1 step per unit of time, and his speed while descending is 3 steps per unit of time. Let the speed of the e... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For a convex quadrilateral $A B C D$, it is known that $A B=B C=C A=$ $C D, \angle A C D=10^{\circ}$. A circle $\omega$ is circumscribed around triangle $B C D$ with center $O$. Line $D A$ intersects circle $\omega$ at points $D$ and $E$. Find the measure of angle $E O A$, express your answer in degrees. | Answer: 65.
Fig. 2: to the solution of problem 3.
Fig. 2. Since triangle $A D C$ is isosceles, and we know the angle at its vertex, the angles at its base are $\angle D A C = \angle C D A =... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Anya writes a natural number, and Boris replaces one of its digits with a digit differing by 1. What is the smallest number Anya should write to ensure that the resulting number is guaranteed to be divisible by 11? | Answer: 909090909.
According to the divisibility rule for 11, the remainder of a number when divided by 11 is the same as the remainder of the alternating sum of its digits when divided by 11. Therefore, changing a digit by 1 also changes the remainder by 1 when divided by 11. This means the original number gives a re... | 909090909 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles. | Answer: 28.
Fig. 3: Solution to problem 5.
Example: The arrangement of 28 tiles can be seen in Fig. 3.
Estimation. Cells not covered by tiles will be called empty. First, let's prove that no... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On the parade ground, 2018 soldiers are lined up in one row. The commander can order either all soldiers standing in even positions or all soldiers standing in odd positions to leave the formation. After this order, the remaining soldiers close up into one row. In how many ways can the commander issue a series of 8 ... | Answer: 30.
Let's add 30 imaginary people to the end of the line. We will number all the people in the line from 0 to 2047. We will write all these numbers in binary using 11 digits. This will result in sequences from 00000000000 to 11111111111. The imaginary soldiers correspond to numbers from 11111100010 to 11111111... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. When a five-digit number is multiplied by 9, the result is a number composed of the same digits but in reverse order. Find the original number. | Answer: 10989.
Let $\overline{a b c d e}$ be the original number. The condition is written as the equation $9 \cdot \overline{a b c d e}=\overline{e d c b a}$. Note that $a=1$, because if $a \geqslant 2$, then $9 \cdot \overline{a b c d e} \geqslant 9 \cdot 20000>100000>\overline{e d c b a}$.
We have $9 \cdot \overli... | 10989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. We will call a pair of numbers magical if the numbers in the pair add up to a multiple of 7. What is the maximum number of magical pairs of adjacent numbers that can be obtained by writing down all the numbers from 1 to 30 in a row in some order? | Answer: 26.
Example: $1,6,8,13,15,20,22,27,29,2,5,9,12,16,19,23,26,30,3,4$, $10,11,17,18,24,25,7,14,21,28$. It is not hard to see that in this sequence, only the pairs $(29,2),(30,3),(25,7)$ are not magical.
Evaluation: Suppose it is possible to make no less than 27 pairs magical. Then there would be no more than two... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Dima has 25 identical bricks of size $5 \times 14 \times 17$. Dima wants to build a tower from all his bricks, each time adding one more brick on top (each new brick adds 5, 14, or 17 to the current height of the tower). We will call a number $n$ constructible if Dima can build a tower of height exactly $n$. How man... | Answer: 98.
In essence, we need to find the number of different towers by height that can be built from the given set of bricks.
Mentally reduce the length, width, and height of each brick by 5: from $5 \times 14 \times 17$ to $0 \times 9 \times 12$. Then the total height of a potential tower will decrease by $25 \cd... | 98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles. | Answer: 28.
Fig. 1: to the solution of problem 6.
Example: the arrangement of 28 tiles can be seen in Fig. 1.
Estimate. Cells not covered by tiles will be called empty. To begin with, let's... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. Through the point with coordinates $(10,9)$, lines (including those parallel to the coordinate axes) have been drawn, dividing the plane into angles of $10^{\circ}$. Find the sum of the x-coordinates of the points of intersection of these lines with the line $y=101-x$. | Answer: 867. Solution: Shift the entire picture to the left by 1. We get a set of lines passing through the point $(9,9)$ and intersecting the line $y=100-x$. The picture will become symmetric with respect to the line $y=x$, so the sum of the abscissas on it is equal to the sum of the ordinates. Through the point $(9,9... | 867 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 4. A pair of natural numbers is called good if one of the numbers is divisible by the other. Numbers from 1 to 30 were divided into 15 pairs. What is the maximum number of good pairs that could result? | Solution. Answer: 13.
First, let's prove that more than 13 good pairs could not have been formed. We will call a number bad if it is prime and not less than 15. There are only four bad numbers: \(17, 19, 23, 29\). None of the other numbers from 1 to 30 can be divisible by any bad number, and the bad numbers themselves... | 13 | Number Theory | 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 |
Problem 2. In a $3 \times 3$ table, natural numbers (not necessarily distinct) are placed such that the sums in all rows and columns are different. What is the minimum value that the sum of the numbers in the table can take? | Answer: 17.
Solution. Note that in each row and column, the sum of the numbers is no less than 3. Then, the doubled sum of all numbers in the table, which is equal to the sum of the sums of the numbers in the rows and columns, is no less than \(3+4+\ldots+8=33\), so the simple sum of the numbers in the table is no les... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, cont... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Polycarp has 2 boxes, the first of which contains $n$ coins, and the second is empty. In one move, he can either transfer one coin from the first box to the second, or remove exactly $k$ coins from the first box, where $k$ is the number of coins in the second box. For which $n$ can Polycarp make the first bo... | Solution. Let's call a move where one coin is moved from the first box to the second box a move of the first type, and a move where coins are removed from the first box a move of the second type. Suppose a total of $x$ moves of the first type and $y$ moves of the second type were made. Then, the second box contains no ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Lёnya has cards with digits from 1 to 7. How many ways are there to glue them into two three-digit numbers (one card will not be used) so that their product is divisible by 81, and their sum is divisible by 9? | Answer: 36.
Solution. If one of the numbers is not divisible by 9, then the other is not either, since their sum is divisible by 9. But then the product cannot be divisible by 81, a contradiction. Therefore, both numbers are divisible by 9.
Then the sum of the digits in each number is divisible by 9, and thus the tot... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\le... | 3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\le... | 4080401 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal? | 5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal?
Answer: 432.
Estimate. We color the cells of the table ... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all pos... | Answer: 30.
Solution. Consider any four consecutive people. If there were at least 3 knights among them, the leftmost knight would definitely lie, which is impossible. If there were at least 3 liars, the leftmost liar would definitely tell the truth, which is also impossible. Therefore, in every group of four consecut... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. Masha and the Bear ate a basket of raspberries and 60 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 raspberries 6 times faster than Masha, and pies only 3 times faster. How many pies did the Bear... | Answer: 54 pies. Solution: Divide the raspberries into 3 equal parts. The bear ate each part 6 times faster than Masha, but there are two parts, so he spent only 3 times less time on the raspberries than Masha. This means Masha ate pies in one-third of the time the bear did. Since she eats three times slower, she ate 9... | 54 | Algebra | 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 |
5. On the table, there are 2018 playing cards (2018 stacks, each with one card). Petya wants to combine them into one deck of 2018 cards in 2017 operations. Each operation consists of merging two stacks. When Petya merges stacks of $a$ and $b$ cards, Vasily Ivanovich pays Petya $a \cdot b$ rubles. What is the maximum a... | Answer: $\frac{2017 \cdot 2018}{2}=2035153$
Let's mentally connect the cards in one stack with invisible threads, each with each. Then the operation of combining two stacks with $a$ and $b$ cards adds exactly $a \cdot b$ threads. In the end, each card will be connected by a thread to each other, and the number of thre... | 2035153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. In each cell of a square table of size $200 \times 200$, a real number not exceeding 1 in absolute value was written. It turned out that the sum of all the numbers is zero. For what smallest $S$ can we assert that in some row or some column, the sum of the numbers will definitely not exceed $S$ in absolut... | Answer: 100.
Solution. First, we show that $S<40000$.
$$
This means that one of the numbers $A$ or $D$ in absolute value exceeds 10000. However, each of the corresponding squares contains only 10000 cells, and the numbers in them do not exceed 1 in absolute value. Contradiction.
## Criteria
Any correct solution to ... | 100 | Combinatorics | 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 |
8.5. Through the point with coordinates $(2,2)$, lines (including two parallel to the coordinate axes) are drawn, dividing the plane into angles of $18^{\circ}$. Find the sum of the abscissas of the points of intersection of these lines with the line $y=2016-x$. | Answer: 10080. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(2,2)$, 10 lines are drawn, and the line $y=2016-x$ intersects all of them. For each point on the line $y=2016-x$, the sum of the coordinates is 2016, so... | 10080 | 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 |
Problem 4. Positive numbers $a, b, c, d$ are greater than 1. Find the smallest possible value of the expression
$$
\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)
$$ | Answer: 67.
Solution. From the properties of logarithms, it follows that $\log _{a} b \cdot \log _{b} c \cdot \log _{c} d \cdot \log _{d} a=1$. Also, all these four factors are positive, since all numbers $a, b, c, d$ are greater than 1.
Transform and estimate the given expression
$$
\begin{gathered}
S=\log _{a}\lef... | 67 | Algebra | 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 |
6. Tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to the plane $ABC$. The line $DE$ intersects the plane $ABC$ at point $F$, whi... | Answer: $40^{\circ}$.
Note that point $E$ is equidistant from points $A, B, C$, so its projection onto the plane $A B C$ coincides with the projection of point $O$ onto this plane and is the center of the circumscribed circle of triangle $A B C$.
Consider triangles $A D E$ and $B D E$. They have a pair of equal sides... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Buses from Moscow to Voronezh depart every hour, at 00 minutes. Buses from Voronezh to Moscow depart every hour, at 30 minutes. The trip between the cities takes 8 hours. How many buses from Voronezh will the bus that left from Moscow meet on its way? | Answer: 16.
It is clear that all buses from Moscow will meet the same number of buses from Voronezh, and we can assume that the bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Oryol at 4:30, 5:30, ..., 18:30, 19:30 and only them. There are 16 such buses.
$\pm$ Correct rea... | 16 | Logic and Puzzles | 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 |
Problem 4. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex? | Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, cont... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let's call a natural number an almost palindrome if it can be transformed into a palindrome by changing one of its digits. How many nine-digit almost palindromes exist? (20 points) | Solution: After replacing one digit in the almost palindrome, we get a number of the form $\overline{a b c d e d c b a}$. Let's divide the digits into pairs by place value: 1 and 9, 2 and 8, 3 and 7, 4 and 6. In any almost palindrome, the digits in three of the specified pairs of place values must be the same, and in e... | 3240000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. If in an acute scalene triangle three medians, three angle bisectors, and three altitudes are drawn, they will divide it into 34 parts. | Write the answer in digits in ascending order, without spaces (for example, 12345).
## Mathematics $8-9$ grade
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 2. (20 points)
Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.
# | # Solution.
Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1... | 100 | Number Theory | 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 |
5. In the language of the "Tekimar" tribe, there are only 7 letters: A, E, I, K, M, R, T, but the order of these letters in the alphabet is unknown. A word is defined as any sequence of seven different letters from the alphabet, and no other words exist in the language. The chief of the tribe listed all existing words ... | Solution: The total number of words in the tribe's language is $7!=5040$. Note that the number of words starting with any particular letter is the same for any first letter, which is $7!\div 7=6!=720$. If a word starts with the first letter in the alphabet, the numbering of any such word starts from number 1 and ends a... | 3745 | 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. Find the number of triples of natural numbers $m, n, k$, which are solutions to the equation $m+$ $\sqrt{n+\sqrt{k}}=2023$. (20 points) | Solution: For the left side to be an integer, the numbers $k$ and $n+\sqrt{k}$ must be perfect squares, and since $n+\sqrt{k} \geq 2$, it follows that $\sqrt{n+\sqrt{k}} \geq 2$ and thus $m \leq 2021$. Since $1 \leq m \leq$ 2021, $\sqrt{n+\sqrt{k}}$ can take any value from 2 to 2022 - this value uniquely determines the... | 27575680773 | 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 |
1. A printing house determines the cost of printing a book as follows: it adds the cost of the cover to the cost of each page, and then rounds the result up to the nearest whole number of rubles (for example, if the result is 202 rubles and 1 kopeck, it is rounded up to 203 rubles). It is known that the cost of a book ... | Solution: Let's convert all costs into kopecks. Let one page cost $x$ kopecks, and the cover cost $100 y$ kopecks. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
13300<100 y+104 x \leq 13400 \\
18000<100 y+192 x \leq 18100
\end{array}\right.
$$
Subtracting the first inequality from the second, we ... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Maxim came up with a new way to divide numbers by a two-digit number $N$. To divide any number $A$ by the number $N$, the following steps need to be performed:
1) Divide $A$ by the sum of the digits of the number $N$;
2) Divide $A$ by the product of the digits of the number $N$;
3) Subtract the second result from th... | Solution: Let $N=\overline{x y}=10 x+y$, where $x, y$ are digits. We need to find all pairs $(x, y)$ such that $\frac{A}{10 x+y}=\frac{A}{x+y}-\frac{A}{x y}$, which means $\frac{1}{10 x+y}=\frac{1}{x+y}-\frac{1}{x y}$. Note that $y \neq 0$. Transform the equation to the form
$$
\begin{gathered}
x y(x+y)=x y(10 x+y)-(x... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (20 points)
It is known that the only solution to the equation
$$
\pi / 4=\operatorname{arcctg} 2+\operatorname{arcctg} 5+\operatorname{arcctg} 13+\operatorname{arcctg} 34+\operatorname{arcctg} 89+\operatorname{arcctg}(x / 14)
$$
is a natural number. Find it. | # Solution.
Consider the equation:
$$
\operatorname{arcctg} a-\operatorname{arcctg} b=\operatorname{arcctg} y \text {. }
$$
Let $\alpha=\operatorname{arcctg} a, \beta=\operatorname{arcctg} b$. Now $y=\operatorname{ctg}(\alpha-\beta)=\frac{1+\operatorname{ctg} \alpha \operatorname{ctg} \beta}{\operatorname{ctg} \beta... | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Before you is a segment-digit. For displaying time on electronic clocks, each digit uses seven segments, each of which can be lit or not; the lit segments form the digit as shown in the figure
## 8723456789
That is, to display zero, six segments are used, to display one - two segments, and so on. On electronic clo... | Solution: Let's write down the number of segments required to display each digit:
\section*{8723456789
$$
\begin{array}{llllllllll}
6 & 2 & 5 & 5 & 4 & 5 & 6 & 3 & 7 & 6
\end{array}
$$
Consider several types of displayed time:
1) The last digit of the minutes is not 9. Then the next displayed time differs by only t... | 630 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Masha chose a natural number $n$ and wrote down all natural numbers from 1 to 6 n on the board. Then, Masha halved half of these numbers, reduced a third of the numbers by a factor of three, and increased all the remaining numbers by a factor of six. Could the sum of all the resulting numbers match the sum of the or... | Solution: Yes, this could have happened. Let $n=2$, then the sum of all numbers from 1 to 12 is
$$
1+2+3+\ldots+12=78
$$
Suppose Masha halved the numbers 2, 3, 5, 9, 10, 11, reduced the numbers 4, 6, 8, 12 by a factor of three, and increased the numbers 1, 7 by a factor of six. Then the sum of the resulting numbers w... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In how many ways can the numbers from 1 to 9 be arranged in a $3 \times 3$ table (each number appearing exactly once) such that in each column from top to bottom and in each row from left to right, the numbers are in increasing order? (20 points) | Solution: Let's number the cells of the table as shown in the figure. It is clear that the number 1 is in the upper left cell, and the number 9 is in the lower right cell.
| 1 | $a_{2}$ | $a_{3}$ |
| :---: | :---: | :---: |
| $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{7}$ | $a_{8}$ | 9 |
By the condition, $a_{5}>a_{2}, a_{5... | 42 | Combinatorics | 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 |
1. How many units are in the number $S=9+99+999+\cdots+\overbrace{9 \ldots 90}^{1000}$? | $$
\begin{aligned}
& S=10-1+100-1+1000-1+\cdots+10^{1000}-1=10\left(1+10+100+\cdots+10^{999}\right)- \\
& -100=10 \frac{10^{1000}-1}{9}-1000=\underbrace{111 \ldots 10}_{999 \text { ones }}-1000 \Rightarrow
\end{aligned}
$$
the number has 998 ones. Answer: $\{998\}$. | 998 | Number Theory | 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 |
1. Solve the equation $n+S(n)=1964$, where $S(n)$ is the sum of the digits of the number $n$. | 1. We have $S(n) \leq 9 * 4=36 \Rightarrow n \geq 1964-36=1928 \Rightarrow n=1900+10 k+l$, where $2 \leq k \leq 9$ $0 \leq l \leq 9 ; 1900+10 k+l+10+k+l=1964,11 k+2 l=54,2 \leq 2 l \leq 18,36 \leq 11 k \leq 54$ $3 \frac{3}{11} \leq k \leq 4 \frac{10}{11} \Rightarrow k=4, l=5, n=1945$. Answer: $\{1945\}$. | 1945 | Number Theory | 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 |
4. The denominator of an irreducible fraction is less than the square of the numerator by one. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $\frac{1}{3}$. If 3 is subtracted from both the numerator and the denominator, the fraction will be less than $\frac{1}{1... | 4. Let $\frac{m}{n}$ be the desired irreducible fraction, where $m, n$ are single-digit numbers. According to the problem, we have $n=m^{2}-1, \frac{m+2}{n+2}>\frac{1}{3} \Rightarrow 3 m+6>m^{2}+1, m^{2}-3 m-5<0$, then $\frac{3-\sqrt{29}}{2}<m<\frac{3+\sqrt{29}}{2} \Rightarrow$ $\Rightarrow 0<m<\frac{3+5.4}{2} \Rightar... | 38 | 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 |
7. By the property of absolute value, replacing $x$ with $-x$ does not change this relation. This means that the figure defined by the given inequality is symmetric with respect to the OY axis. Therefore, it is sufficient to find the area of half of the figure for $x \geq 0$. In this case, we obtain the inequality $\le... | Answer: $\{30\}$.
Fig. 2 | 30 | 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. Since you have to approach each apple and return to the basket, the number of meters walked will be equal to twice the sum of the first hundred numbers, or 101 taken a hundred times, i.e., $10100 \mathrm{M}$. | Answer: $\{10100$ meters $\}$ | 10100 | Number Theory | 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 |
3. Find the area of the region defined by the inequality: $|y-| x-2|+| x \mid \leq 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. Consider three intervals of change for $x$: 1) $x \leq 0$; 2) $0 \leq x \leq 2$; 3) $x \geq 2$.
1) $x \leq 0$; then $|y+x-2|-x \leq 4, |y+x-2| \leq 4+x \Rightarrow -4 \leq x \leq 0$. Squaring the inequality, we get $(y+x-2)^{2} \leq (4+x)^{2}$, from which $(y-6)(y+2x+2) \leq 0$
2) $0 \leq x \leq 2$, then $|y+x-2| \l... | 32 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. Find all roots of the equation
$1-\frac{x}{1}+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\frac{x(x-1)(x-2)(x-3)}{4!}-\frac{x(x-1)(x-2)(x-3)(x-4)}{5!}+$
$+\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!}=0 . \quad($ (here $n!=1 \cdot 2 \cdot 3 . . . n)$
In the Answer, indicate the sum of the found roots. | 4. Note that by substituting the numbers $1,2,3,4,5,6$ sequentially into the equation, we will get the equality: $0=0$. This means these numbers are roots of the given equation. Since the equation is of the sixth degree, there are no other roots besides the numbers mentioned above. Answer: $\{21\}$ | 21 | Algebra | 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 |
6. Two acute angles $\alpha$ and $\beta$ satisfy the condition $\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin}(\alpha+\beta)$. Find the sum of the angles $\alpha+\beta$ in degrees. | 6. Using the formula for $\operatorname{Sin}(\alpha+\beta)$, we get:
$\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin} \alpha \operatorname{Cos} \beta+\operatorname{Cos} \alpha \operatorname{Sin} \beta \Leftrightarrow \operatorname{Sin} \alpha(\operatorname{Sin} \alpha-\operatorname{Cos} \... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find $\max x^{2} y^{2} z$ subject to the condition that $x, y, z \geq 0$ and $2 x+3 x y^{2}+2 z=36$. | 7. From the inequality for the arithmetic mean and geometric mean of three numbers, we have:
$\sqrt[3]{2 x \cdot 3 x y^{2} \cdot 2 z} \leq \frac{2 x+3 x y^{2}+2 z}{3}=12 \Rightarrow 3 x y^{2} \cdot 2 z \cdot 2 x \leq 12^{3}, x^{2} y^{2} z \leq 144$
This inequality becomes an equality if the numbers are equal, i.e., $... | 144 | 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 |
1. Masha has an integer number of times more toys than Lena, and Lena has the same number of times more toys than Katya. Masha gave Lena 3 toys, and Katya gave Lena 2 toys. After that, the number of toys the girls had formed an arithmetic progression. How many toys did each girl originally have? Indicate the total numb... | 1. Let initially Katya has $a$ toys, then Lena has $k a$, and Masha has $\kappa^{2} a$, where $\kappa-$ is a natural number $\geq 2$. After the changes: Katya has $a-2$, Lena has $a \kappa+5$, and Masha has $a \kappa^{2}-3$. Then these numbers form an arithmetic progression in some order. Let's consider the possible ca... | 105 | Algebra | 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 |
4. Find the minimum value of the sum
$$
\left|x-1^{2}\right|+\left|x-2^{2}\right|+\left|x-3^{2}\right|+\ldots+\left|x-10^{2}\right|
$$ | 4. Grouping the terms, we get $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)+\left(\left|x-2^{2}\right|+\left|x-9^{2}\right|\right)+\left(\left|x-3^{2}\right|+\left|x-8^{2}\right|\right)+\left(\left|x-4^{2}\right|+\left|x-7^{2}\right|\right)+\left(\left|x-5^{2}\right|+\left|x-6^{2}\right|\right)$. It is known... | 275 | Algebra | 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 |
7. Solve the system
$\left\{\begin{array}{l}a+c=4 \\ a d+b c=5 \\ a c+b+d=8 \\ b d=1\end{array}\right.$
In the answer, write the sum of all solutions of the given system. | 7. We have
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=$
$=x^{4}+4 x^{3}+6 x^{2}+5 x+2=(x+1)(x+2)\left(x^{2}+x+1\right)$. Since the polynomial $x^{2}+x+1$ does not have
real roots, the identity $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2... | 14 | 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.
Since the pedestrian covers 1 km in 10 minutes, his speed is 6 km/h. There are more oncoming trams than overtaking ones because, relative to the pedestrian, the speed of the former is greater than that of the latter. If we assume the pedestrian is standing still, the speed of the oncoming trams is the sum of... | Answer: 15 km $/$ hour.
# | 15 | 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)
$$
there exists a triangle with sides $a, b, c$.
$\square$ Since $(a b+b c+c a) \leq\left(a^{2}+b^{2}+c^{2}\right)$, then $n>5$. If $n \geq 7$, then the numbers $a=1, b... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have? | 3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have?
$\square$ We will prove by induction that the polynomial $P_{n}(x)=x^{n}+\ldots$ has exactly $n$ distinct rea... | 2018 | Algebra | 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 |
4. Misha, over the course of a week, picked an apple each day and weighed it. Each apple weighed a different amount, but the weight of each apple was a whole number of grams and ranged from 221 grams to 230 grams (inclusive). Misha also calculated the average weight of all the apples he picked, and it was always a whol... | Answer: 230 grams.
Solution: Each apple weighed 220 grams plus an integer from 1 to 10. From the numbers 1 to 10, we need to choose 7 numbers such that their sum is divisible by 7. One of these numbers is 5, so we need to choose 6 from the numbers $1,2,3,4,6,7,8,9,10$. The smallest sum of six of these numbers is 23, a... | 230 | Number Theory | 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. Senya thought of two numbers, then subtracted the smaller from the larger, added both numbers and the difference, and got 68. What was the larger of the numbers Senya thought of? | Answer: 34.
Solution. The subtrahend plus the difference equals the minuend. Therefore, the doubled minuend equals 68.
Comment. Correct solution - 20 points. The result is obtained based on examples and the pattern is noticed but not explained - 15 points. The result is obtained based on one example - 10 points. The ... | 34 | 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 |
2. The little squirrel has several bags of nuts. In two bags, there are 2 nuts each, in three bags, there are 3 nuts each, in four bags, there are 4 nuts each, and in five bags, there are 5 nuts each. Help the little squirrel arrange the bags on two shelves so that there are an equal number of bags and nuts on each she... | Solution. For example, $5+5+5+4+4+2+2=27$ nuts in 7 bags - the first shelf, $5+5+4+4+3+3+3=27$ nuts in 7 bags - the second shelf.
Comment. Correct solution - 20 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Solution incorrect or absent - 0 points. | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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