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10.5. What is the maximum number of 4-element subsets that can be selected from a set of 8 elements such that the intersection of any three of the selected subsets contains no more than one element?
|
Answer. Eight.
Solution. We will provide two different examples of choosing eight 4-element subsets in a set $\mathrm{X}$ of eight elements, satisfying the condition of the problem. Both examples are constructed as geometric objects.
Example 1. We will consider the elements of $\mathrm{X}$ as the vertices of a unit cube, six of the chosen subsets will be the faces of this cube, and the remaining two will be the vertices of two inscribed regular tetrahedra with edge length $\sqrt{2}$. No two vertices of these tetrahedra lie on the same edge of the cube.
If among any three of the chosen subsets, both tetrahedra are included, or two parallel faces, the intersection is empty. If it is two adjacent faces and a tetrahedron, the intersection of the faces gives an edge, which intersects the tetrahedron at exactly one vertex. If it is three pairwise adjacent faces, their intersection contains the unique vertex of the trihedral angle formed by these faces.
Example 2. We will consider the elements of $\mathrm{X}$ as the vertices of a regular octagon. Number the vertices clockwise from 1 to 8, and mark the quadrilateral $\mathrm{M}$ with vertices numbered $1,2,3,5$. The chosen subsets will be $\mathrm{M}$ and seven other quadrilaterals obtained from $\mathrm{M}$ by rotations of angles $\frac{2 \pi k}{8}, k=1,2, \ldots, 6,7$. If the intersection of any three of them contained at least two vertices, i.e., some side or diagonal of the octagon other than the main ones, then by rotating all three quadrilaterals back to coincide with $\mathrm{M}$, we would obtain three different segments of the same length connecting the vertices of $\mathrm{M}$. The latter is impossible because the lengths of the sides and diagonals of $\mathrm{M}$, measured in sides, are $1,1,2,2,3,4$, and there are no more than two equal lengths among them. If the main diagonal of length 4 is mentioned, it is clear that it is contained in only two of the eight considered quadrilaterals.
We will prove that if in an 8-element set $\mathrm{X}$, nine 4-element subsets are arbitrarily chosen, then the intersection of some three of them contains more than one element. The sum of the cardinalities of the chosen subsets is 36, so one of the elements of $\mathrm{X}$, denoted by $\boldsymbol{x}$, is contained in at least $k \geq 5$ of them. Remove $\boldsymbol{x}$ from these $k$ subsets and consider the $k \geq 5$ resulting 3-element subsets in the 7-element set Y, which is equal to X without $\boldsymbol{x}$. The sum of the cardinalities of the resulting sets is at least 15, so one of the elements of Y, denoted by $\boldsymbol{y}$, is contained in at least three of the $k$ resulting 3-element subsets. Therefore, the pair of elements $\boldsymbol{x}$ and $\boldsymbol{y}$ is contained in at least three of the nine chosen 4-element subsets from $\mathrm{X}$.
Grading Criteria. ($\cdot$) Example for eight subsets: 3 points. ($\cdot$) Proof of the estimate: 4 points. ($\cdot$) Lack of justification for the example: minus 1 point.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. At the Olympiad, students from gymnasiums, lyceums, and regular schools met. Some of them stood in a circle. Gymnasium students always lie to regular school students, lyceum students lie to gymnasium students, and regular school students lie to lyceum students. In all other cases, the students tell the truth. Each one said to their right neighbor: “I am a gymnasium student.” How many students from regular schools were in this circle?
|
Answer: There were no ordinary school students in the circle.
Solution: Let's assume there was an ordinary school student in the circle. Consider his left neighbor. This neighbor could not be another ordinary school student or a lyceum student, because they would tell the truth. But he also cannot be a gymnasium student, because then he would have lied, and he said the truth. Therefore, the neighbor of the ordinary school student cannot be anyone. Consequently, ordinary school students did not join the circle. Criteria: only the answer - 0 points, answer with verification - 1 point. For the claim that to the right of a gymnasium student there must be a gymnasium student or a lyceum student (in reality, it can be either a lyceum student or a gymnasium student) - deduct 3 points.
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. In the bus, there are single and double seats. In the morning, 13 people were sitting in the bus, and there were 9 completely free seats. In the evening, 10 people were sitting in the bus, and 6 seats were completely free. How many seats are there in the bus?
|
Answer: 16.
Solution: If in the morning passengers sat on 6 double seats (i.e., as densely as possible), then they occupied 7 seats, with 9 seats still free. In total: 16 seats. If they did not sit as densely, then they would have occupied more seats. That is, there are no fewer than 16 seats in the bus, on the one hand.
On the other hand, if in the evening each passenger took a seat, then 10 seats were occupied, with 6 still free, a total of 16 seats. If some sat together, there would have been fewer seats. Therefore, there are no more than 16 seats.
Thus, there are no more, but also no fewer than 16 seats. Therefore, there are exactly 16 seats.
Criteria: only the answer, answer with verification - 1 point, estimation from only one side - 3 points, no points deducted for the lack of an example.
|
16
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. In the country, there are 15 cities, some of which are connected by roads. Each city is assigned a number equal to the number of roads leading out of it. It turned out that there are no roads between cities with the same number. What is the maximum number of roads that can be in the country?
|
Answer: 85.
Solution: Let's order the city numbers in non-increasing order:
$$
a_{1} \geq a_{2} \geq \cdots \geq a_{15}
$$
Notice that the number of cities with number $15-i$ is no more than $i$. Indeed, if there are at least $i+1$ such cities, then they cannot be connected to each other, and thus can be connected to no more than $15-(i+1)=14-i$ cities, but they should be connected to $15-i$ cities, which is a contradiction.
Therefore, the maximum number is no more than 14, the next two are no more than 13, the next three are no more than 12, the next four are no more than 11, and the remaining five are no more than 10. The total sum of the numbers is no more than $14 + 2 \times 13 + 3 \times 12 + 4 \times 11 + 5 \times 10 = 170$. Since each road contributes to the numbers of exactly two cities, the total number of roads is no more than $170 \div 2 = 85$.
We will show that this can be achieved. Divide the country into 5 regions. In the first region, there is one city, in the second region, there are two, in the third region, there are three, and so on. Let the cities be connected if and only if they lie in different regions. Then the first city is connected to all others (has number 14), the cities in the second region are connected to all except each other (and have number 13), the cities in the third region are connected to $15-3=12$ cities, the cities in the fourth region are connected to $15-4=11$ cities, and the cities in the fifth region are connected to $15-5=10$ cities.
Criteria: evaluation - 4 points, example - 2 points
|
85
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Given a convex quadrilateral $A B C D$ with side $A D$ equal to 3. Diagonals $A C$ and $B D$ intersect at point $E$, and it is known that the areas of triangles $A B E$ and $D C E$ are both 1. Find the side $B C$, given that the area of $A B C D$ does not exceed 4.
|
# Answer: 3.
Solution: Triangles $A B D$ and $A C D$ have the same area, as they are composed of the common $A E D$ and equal-area $A B E$ and $D C E$. Since $A B D$ and $A C D$ have the same base $A D$, their heights to this base are equal. Therefore, $B C$ is parallel to $A D$, meaning our quadrilateral is a trapezoid.
Let $B C$ be $x$. Triangles $B E C$ and $A E D$ are similar, so $B E / E D = B C / A D = x / 3$. Additionally, the area of $B E C$ is $B E / E D$ of the area of $B C D$, since these triangles have the same base, and their heights are in the ratio of $B E$ to $E D$. Therefore, the area of $B E C$ is $x / 3$. Similarly, the area of $A E D$ is $3 / x$.
Since the area of the entire quadrilateral does not exceed 4, $x / 3 + 3 / x \leq 2$.
However, the sum of the reciprocals of positive quantities is always at least 2, and equality is achieved only when the terms are equal, i.e., in our case, $x / 3 = 3 / x$, from which $x$ equals 3.
Criteria: Only the answer - 0 points. Proved that $A B C D$ is a trapezoid - plus 2 points. Obtained the equation - plus 1 point. Not explained why the inequality is possible only when the terms are equal - deduct 1 point.
## Criteria for Determining Winners and Prize Winners of the All-Siberian Open Mathematics Olympiad for Schoolchildren (2015-2016 academic year)
According to the Regulations, the winners and prize winners of the Olympiad were determined based on the results of the Final Stage of the Olympiad. The total number of winners and prize winners was 380 out of 1578 participants, which is 24.08%. The number of winners was 85, which is 5.38%.
Based on the overall ranking of participants and considering the noticeable gaps in the scores of the top-ranked groups, the Olympiad jury developed the following criteria for determining winners and prize winners: The maximum possible score is 35 points.
## 11th Grade:
Winners:
Participants who scored more than 77% of the maximum possible points, i.e., from 27 to 35 points; Prize winners:
2nd degree - more than 62% of the maximum possible points, i.e., from 22 to 26 points 3rd degree - more than 51% of the maximum possible points, i.e., from 18 to 21 points
10th Grade:
Winners:
Participants who scored more than 85% of the maximum possible points, i.e., from 30 to 35 points;
Prize winners:
2nd degree - more than 62% of the maximum possible points, i.e., from 22 to 29 points
3rd degree - more than 51% of the maximum possible points, i.e., from 18 to 21 points
9th Grade:
Winners:
Participants who scored more than 85% of the maximum possible points, i.e., from 30 to 35 points;
Prize winners:
2nd degree - more than 65% of the maximum possible points, i.e., from 24 to 29 points
3rd degree - more than 51% of the maximum possible points, i.e., from 18 to 23 points
8th Grade:
Winners:
Participants who scored more than 82% of the maximum possible points, i.e., from 29 to 35 points;
Prize winners:
2nd degree - more than 62% of the maximum possible points, i.e., from 22 to 28 points 3rd degree - more than 48% of the maximum possible points, i.e., from 17 to 21 points
7th Grade:
Winners:
Participants who scored more than 85% of the maximum possible points, i.e., from 30 to 35 points;
Prize winners:
2nd degree - more than 71% of the maximum possible points, i.e., from 24 to 29 points
3rd degree - more than 42% of the maximum possible points, i.e., from 15 to 21 points
Co-Chair of the Mathematics Jury
A.Yu. Avdyushenko
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Find all triples of distinct natural numbers, the least common multiple of which is equal to their sum. The least common multiple of several numbers is the smallest natural number that is divisible by each of these numbers.
|
Answer. All triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n$.
Solution. Let the required numbers be $a<b<c$. By the condition, their least common multiple, which is equal to $a+b+c$, is divisible by $c$, so $c$ divides the sum $a+b<2 c$. Therefore, $a+b=c$. Next, $a+b+c=2 a+2 b$ is divisible by $b$, so $b$ divides $2 a<2 b$. Therefore, $2 a=b$ and $c=a+b=3 a$, and the natural number $a$ can be any.
On the other hand, for any triple of natural numbers of the form $a=n, b=2 n, c=3 n$, their least common multiple is $6 n$. Indeed, it must divide $3 n$, but it is not equal to $3 n$, because $3 n$ does not divide $2 n$, so it is not less than $3 n \cdot 2=6 n$. It remains to note that $6 n$ divides each of the numbers $a=n, b=2 n, c=3 n$.
Evaluation criteria. Only the example of triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n: 2$ points. Example for specific $n: 1$ point. Proved that $a+b=c: 2$ points.
Proved that $2 a=b: 2$ points.
9.4.. In triangle $\mathrm{ABC}$, point $M$ is the midpoint of side $\mathrm{BC}$, and $\mathrm{H}$ is the foot of the altitude dropped from vertex $\mathrm{B}$. It is known that angle $\mathrm{MCA}$ is twice the angle $\mathrm{MAC}$, and the length of $\mathrm{BC}$ is 10 cm. Find the length of segment $\mathrm{AH}$.
Answer. $\mathrm{AH}=5$ cm.
Solution. In the right triangle $\mathrm{BHC}$, segment $\mathrm{HM}$ is the median to the hypotenuse, so the length of $\mathrm{HM}$ is half the length of $\mathrm{BC}$, which is 5 cm.
Triangle $\mathrm{CHM}$ is isosceles with $\mathrm{HM}=\mathrm{MC}$, so angle $\mathrm{MHC}$ equals angle $\mathrm{MCH}=\mathrm{MCA}$ and is twice the angle $\mathrm{MAC}$.
Angle $\mathrm{MHC}$ is the exterior angle for triangle $\mathrm{AMH}$, its measure is equal to the sum of angles $\mathrm{MAH}$ and $\mathrm{AMH}$ and it is twice $\mathrm{MAH}=\mathrm{MAC}$. Therefore, angles $\mathrm{MAH}$ and $\mathrm{AMH}$ are equal, triangle $\mathrm{AMH}$ is isosceles, so $\mathrm{AH}=\mathrm{HM}=5$ cm.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. For non-negative numbers $a, b, c, d$, the following equalities are satisfied:
$\sqrt{a+b}+\sqrt{c+d}=\sqrt{a+c}+\sqrt{b+d}=\sqrt{a+d}+\sqrt{b+c}$. What is the maximum number of distinct values that can be among the numbers $a, b, c, d$?
|
Answer. Two.
Solution. Square the first equality, combine like terms, cancel by 2, square again, combine like terms once more, ultimately obtaining the equality $a c+b d=a b+c d$, equivalent to $(a-d)(c-b)=0$, from which either $a=d$ or $b=c$. Performing similar manipulations with the second equality, we get $a=b$ or $d=c$. Any equality from the first pair shares a common letter with any equality from the second pair, so three out of the four numbers must be equal to each other, and among the numbers $a, b, c, d$ there cannot be more than two distinct numbers. On the other hand, any set of four non-negative numbers, where three are the same and the fourth is arbitrary, clearly satisfies the condition of the problem, so among the numbers $a, b, c, d$ there can be exactly two distinct numbers.
Grading criteria. Proof that among the numbers $a, b, c, d$ there cannot be more than two distinct numbers: 5 points. Example where there are exactly two distinct numbers: 2 points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. What is the maximum number of colors needed to color all cells of a 4 by 4 square so that for every pair of different colors, there are two cells of these colors that are either in the same row or in the same column of the square?
|
Answer: In 8 colors.
Solution. If the cells were painted in 9 or more colors, there would be a color in which only one cell is painted. There are only 6 cells located in the same row or column with it, so there are no more than 6 other colors forming a pair with it, as required by the condition - a contradiction. Therefore, the cells can be painted in no more than 8 colors.
Let's provide two essentially different examples of the required coloring in 8 colors, with each color being used for exactly two cells of the square.
| 5 | 6 | 3 | 8 |
| :--- | :--- | :--- | :--- |
| 3 | 4 | 2 | 7 |
| 2 | 1 | 6 | 5 |
| 1 | 7 | 8 | 4 |
| 6 | 7 | 8 | 3 |
| :--- | :--- | :--- | :--- |
| 4 | 5 | 2 | 6 |
| 2 | 3 | 4 | 5 |
| 1 | 1 | 7 | 8 |
Grading criteria. It is proven that the cells cannot be painted in more than 8 colors (i.e., an upper bound on the number of colors is established): 3 points.
A correct example of the required coloring in 8 colors is explicitly provided: 3 points. Justification of it and description of the construction idea are not required.
Both the example and the estimate are present: 7 points.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. A square with a side of $100 \mathrm{~cm}$ was drawn on the board. Alexei crossed it with two lines parallel to one pair of sides of the square. Then Danil crossed the square with two lines parallel to the other pair of sides of the square. As a result, the square was divided into 9 rectangles, and it turned out that the lengths of the sides of the central section were 40 cm and 60 cm. Find the sum of the areas of the corner rectangles.
|
Answer: 2400.
Solution 1: Without loss of generality, we will assume that the central rectangle has a width of 60 cm and a height of 40 cm. Let $x$ and $y$ be the width and height, respectively, of the lower left rectangle. Then the upper left rectangle has sides $x$ and $(60-y)$, the upper right rectangle has sides $(40-x)$ and $(60-y)$, and the lower right rectangle has sides $(40-x)$ and $y$. The total area of the corner rectangles is
$$
x y + 60 x - x y + 40 y - x y + 2400 - 60 x - 40 y + x y = 2400.
$$
Solution 2: Suppose Alexei not only drew the lines but also cut out the entire central part bounded by these lines, and then Daniil did the same. Then, from the original square, only a rectangle of 60 by 40 remains on one side. On the other hand, everything that remains is the rectangles in question. Therefore, the area of the corner rectangles is equal to the area of the remaining rectangle and is 2400.
Criteria: Only the answer - 0 points. Answer with verification - 1 point.
In the solution, it is explicitly assumed that the cuts were made along the grid lines: if the solution relies on this (for example, by enumeration) - no more than 3 points; if the solution does not use this - do not deduct points.
|
2400
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser would give up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play?
|
Answer: 14.
Solution: On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. Thus, exactly 27 games were played, and Nikita participated in all of them. In 13 of these, his opponent was Egor, so in the remaining 14, it was Innokentiy, and this is the answer.
Criteria: Only the answer, answer with verification - 0 points.
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. On the table, 28 coins of the same size but possibly different masses are arranged in a triangular shape (see figure). It is known that the total mass of any triplet of coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle.
|
Answer: 60 g.
Solution 1: Take a rhombus made of 4 coins. As can be seen from the diagram, the masses of two non-touching coins in it are equal. Considering such rhombi, we get that if we color the coins in 3 colors, as shown in the diagram, then the coins of the same color will have the same mass. Now it is easy to find the sum of the masses of the coins on the boundary: there are 6 coins of each color there, and the sum of the masses of three differently colored coins is 10 g; therefore, the total mass of the coins on the boundary is $6 \cdot 10=60$ g.
Solution 2: All coins except the central one can be divided into 9 triplets, and all internal coins except the central one can be divided into 3 triplets. Therefore, the coins on the boundary weigh as much as 9-3=6 triplets, i.e., 60 g.
Criteria: Only the answer - 0 points.
Rhombi are considered and it is shown that two coins in them have the same mass - 2 points. In addition to this, it is proven that the weights in the entire triangle are as shown in the diagram (coloring is done, but there is no further progress) - another 2 points.
|
60
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed into several bags (each bag contains more than one barrel). We will call a bag good if the number of one of the barrels in it equals the product of the numbers of the other barrels in the same bag (for example, the bag "2, 3, 6" is good, while "4, 5, 10" is not). What is the maximum possible number of good bags?
|
# Answer: 8.
Solution: In each good bag, there are no fewer than three barrels. The smallest number in each good bag must be unique, otherwise the largest number in this bag would be no less than $10 \times 11=110$, which is impossible. For the same reason, if a good bag contains a barrel with the number 1, it must also contain another barrel with a single-digit number. Therefore, the number of good bags does not exceed 8.
On the other hand, an example can be provided where there are exactly 8 good bags. We can gather 8 good bags $(2,17,34),(3,16,48),(4,15,60),(5,14,70),(6,13,78),(7,12,84),(8,11,88)$, and $(9,10,90)$. All other numbers can be placed in a separate bag that is not good.
Criteria: Example only - 3 points.
Estimate only - 3 points.
Noting that each bag must contain a single-digit number - 1 point, can be combined with the example.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. Find the smallest natural number divisible by 99, all digits of which are even.
|
Answer: 228888.
Solution. Let the sum of the digits of the desired number $X$, located in even-numbered positions (tens, thousands, etc.), be denoted by $A$, and the sum of the digits in odd-numbered positions (units, hundreds, etc.) be denoted by $B$. According to the divisibility rules for 9 and 11, $A+B$ is divisible by 9, and $A-B$ is divisible by 11. Given the evenness of the digits of the number, $A$ and $B$ are even, so $A+B$ is also divisible by 18, and $A-B$ is divisible by 22. If $A-B=0$, then $A$ and $B$ are equal and divisible by 9, and considering their evenness, $A$ and $B$ are divisible by 18. From this, it is easy to see that in this case, the desired number is not less than 228888, and the digits of the number 228888 are even and it is divisible by 9. Therefore, if there is such a number less than 228888, then for it $A \leq 18, B \leq 24$. Thus, if $A-B$ is not equal to 0, then $A-B=-22$, and either $A=0, B=22$, or $A=2, B=24$. In both cases, $A+B$ is not divisible by 9. Therefore, $X=228888$.
|
228888
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. In the math test, each of the 25 students could receive one of four grades: $2, 3, 4$ or 5. It turned out that the number of students who received fours was 4 more than those who received threes. How many people received twos, if it is known that the sum of all grades for this test is 121?
|
# 7.3. Answer: 0.
Let's assume that the 4 students who received fours skipped school, then 21 students should have scored 105 points on the test, which is only possible if all of them received fives. Therefore, in the real situation, no one received a two.
## Comments on Evaluation.
Answer only: 1 point. Answer with verification: 2 points.
#
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Find the smallest natural number ending in the digit 4 that quadruples when its last digit is moved to the beginning of the number.
|
# 7.4. Answer: 102564
If a number ending in 4 is multiplied by 4, the result is a number ending in 6, so the last digits of the desired number are 64. If such a number is multiplied by 4, the result is a number ending in 56, so the desired number ends in 564. Continuing to restore the number in this way, it turns out that the desired number should end in 102564. The smallest number ending with these digits is 102564. Comments on evaluation.
Answer only: 3 points. Answer and the observation that the number can be restored from the end: 4 points.
#
|
102564
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. Find the maximum natural number $A$ such that for any arrangement of all natural numbers from 1 to 100 inclusive in a row in some order, there will always be ten consecutively placed numbers whose sum is not less than $A$.
|
Answer: 505.
Solution: The sum of all numbers from 1 to 100 is 5050. Let's divide the 100 numbers in a row into 10 segments, each containing 10 numbers. Clearly, the sum of the numbers in one of these segments is not less than 505, so A is not less than 505.
We will show that among the numbers arranged in the following order: $100,1,99,2,98,3, \ldots, 51,50$, there are no ten consecutive numbers whose sum is greater than 505. This will prove that A is not greater than 505, and considering the above, that $A=505$.
We will use the fact that in this arrangement, all numbers in odd positions decrease monotonically, while those in even positions increase monotonically, and the sum of a number in an odd position and the one following it is always 101. Therefore, the sum of any 10 consecutive numbers, the first of which is in an odd position, is always $505=5 * 101$. The sum of a segment of 10 consecutive numbers, the first of which is in an even position, is 505 minus the rightmost number among those to the left of this segment plus the last number of this segment. The two mentioned numbers are in odd positions, differing by 10, so the first is 5 greater than the second, and thus the entire sum is 500, which is less than 505.
Grading Criteria: Proof that A is not less than 505: 3 points. No justification for the lower bound $\mathrm{A}=505: -2$ points or -1 point.
Example proving that A is not greater than 505: 4 points.
Insufficient justification of the example: -2 points.
|
505
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase?
|
Answer: 500 rubles.
Solution: The additional 100 rubles spent the second time brought the merchant an additional 20 rubles in profit. Therefore, the first time, to earn $5 \cdot 20=100$ rubles in profit, the merchant must have paid $5 \cdot 100=500$ rubles.
Second solution: Let the amount of the first purchase be $x$, then for every ruble invested in Tver, he would receive $\frac{x+100}{x}$ rubles in Moscow. Consequently, after the second purchase-sale, he would receive $\frac{x+100}{x}(x+100)=x+100+120$ rubles. Solving this, we get $x=500$.
Grading criteria: Only the answer with verification: 2 points.
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Ten numbers are written around a circle, the sum of which is 100. It is known that the sum of any three consecutive numbers is not less than 29. Indicate the smallest number $A$ such that in any set of numbers satisfying the condition, each number does not exceed $A$.
|
Answer. $A=13$
Solution. Let $X$ be the largest of the listed numbers. The remaining numbers can be divided into 3 "neighbor" triplets. The sum of the numbers in each such triplet is no less than 29, therefore, $X \leq 100 - 3 \cdot 29 = 13$. An example of a set with the maximum number 13: $13,9,10,10,9,10,10,9,10,10$.
Grading criteria. Proven that the desired number is no less than 13: 4 points. Example for 13: 3 points
|
13
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. Find the smallest natural number in which each digit occurs exactly once and which is divisible by 990.
|
Answer: 1234758690.
Solution. The number 990 is the product of coprime numbers 2, 5, 9, and 11. Any ten-digit number composed of different digits, each used once, is divisible by 9, since their sum, which is 45, is divisible by 9. According to the divisibility rule for 10, the desired number must end in 0. It remains to address the divisibility by 11.
The divisibility rule for 11 states: a number is divisible by 11 if and only if the difference between the sum of all its digits in odd positions from left to right and the sum of its digits in even positions is divisible by 11. Let's estimate the value of $\mathrm{S}$, the sum of the digits of the desired number in odd positions: it is no less than $1+2+3+4+5=15$ and no more than $5+6+7+8+9=35$. Therefore, the difference between the sum of all the digits in odd positions and the sum of the digits in even positions, equal to $2S-45$, is an odd number in the interval from -15 to 25, divisible by 11.
There are only two such numbers: -11 and 11, for which $\mathrm{S}$ is 17 and 28, respectively. It is easy to verify that for $S=17$ there are only two options: $S=1+2+3+4+7$ and $S=1+2+3+5+6$. Considering the minimality, this gives the number 1526384970.
For $\mathrm{S}=28$, we will list the smallest possible digits from left to right, as long as it is possible while maintaining the condition that the sum of the digits in odd positions can ultimately be 28, and the sum of the digits in even positions can be 17. This results in 1234, and the sum of the remaining 3 digits in the fifth, seventh, and ninth positions must equal 24, which is only possible if they are 7, 8, and 9, leading to the number in the answer. This number is smaller than the previously found 1526384970 for $S=17$. If a smaller number could be found, it would have $S=28$ and the fifth digit from the left would be less than 7, which, as we have seen, is impossible.
Grading criteria. If the answer is guessed and divisibility is verified: 1 point. Possible values of S are found: 2 points. The answer is written, and then the minimality is proven by full enumeration: 7 points.
|
1234758690
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. Let $M$ - be a finite set of numbers (distinct). It is known that among any three of its elements, there will be two whose sum belongs to $M$. What is the maximum number of elements that can be in $M$?
|
Answer: 7.
Solution: An example of a set with 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that a set $M=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ with $n>7$ numbers does not have the required property.
We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a_{4}>0$ (changing the signs of all elements does not change our property). Then $a_{1}+a_{2}>a_{1}+a_{3}>a_{1}+a_{4}>a_{1}$, i.e., none of the sums $a_{1}+a_{2}, a_{1}+a_{3}$, and $a_{1}+a_{4}$ belong to the set $M$. Moreover, the sums $a_{2}+a_{3}$ and $a_{2}+a_{4}$ cannot both belong to $M$, since $a_{2}+a_{3}>a_{2}+a_{4}>a_{2}$. This means that for at least one of the triples ( $a_{1}, a_{2}, a_{3}$ ) and ( $a_{1}, a_{2}, a_{4}$ ), the sum of any two of its elements does not belong to the set $M$.
Grading criteria: Proved that there are no more than 7 elements in M: 5 points. Example for 7 elements: 2 points.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. In the wagon, several kilograms of apple jam were loaded, of which $20 \%$ was good and $80 \%$ was bad. Every day, half of the existing bad jam rotted, and it was thrown away. After several days, it turned out that $20 \%$ of the jam in the wagon was bad and $80 \%$ was good. How many days have passed since the loading?
|
Answer: 4 days.
Solution 1: Let the initial total amount be $x$, and the final amount be $y$ kilograms of jam. Then, since the amount of good jam did not change, $0.2 x = 0.8 y$, which means $x = 4 y$. Therefore, initially, the amount of bad jam was $0.8 x = 3.2 y$, and it became $0.2 y$, meaning the mass of bad jam decreased by a factor of 16. Given that the amount of bad jam halved each day, it is clear that 4 days have passed.
Solution 2: Initially, there were 100 arbitrary units of jam: 20 good and 80 bad. In the end, the amount of bad jam became $20 \%$, and the amount of good jam became $80 \%$ of the new quantity, meaning the amount of good jam increased by a factor of 4. Therefore, the new amount of bad jam is 5 arbitrary units. Since it decreased by a factor of 2 each time and ultimately decreased by a factor of 16, 4 days have passed.
Criteria: Only the answer, answer with verification - 1 point. In a solution similar to the second one, if abstract units of measurement are not used but specific ones are used without any explanation that the units of measurement are not important - 1 point.
|
4
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. Once Alexei and Daniil were playing such a game. If a number \( x \) is written on the board, it can be erased and replaced with \( 2x \) or \( x - 1000 \). The player who gets a number not greater than 1000 or not less than 4000 loses. Both players aim to win. At some point, the boys stopped playing. Who lost if the first number was \( 2017 \)?
|
Answer: no one lost.
Solution: note that if a number is less than 2000 but greater than 1000, then by multiplying by 2, you can get a number that is less than 4000. If a number is less than 4000 but greater than 2000, then by subtracting 1000 (possibly twice), you can get a number between 1000 and 2000. Thus, the only number from which no move can be made is 2000.
We will prove that no one could have obtained 2000. Note that the initial number is not divisible by 5. If we multiply it by 2 or subtract 1000 from it, the new number will again not be divisible by 5.
Thus, Alexei and Danila could have continued their game indefinitely, and no one would have lost.
Criteria:
Noted only that from numbers less than 2000 and greater than 2000, a move can always be made - 3 points. Proved only that 2000 cannot be obtained - 3 points.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. In the list $1,2, \ldots, 2016$, two numbers $a<b$ were marked, dividing the sequence into 3 parts (some of these parts might not contain any numbers at all). After that, the list was shuffled in such a way that $a$ and $b$ remained in their places, and no other of the 2014 numbers remained in the same part where they were initially. In how many ways could $a$ and $b$ have been chosen?
|
Answer: $1+2+\ldots+1008=1009 * 504=508536$ ways.
## Solution:
Hypothesis. We will prove that the question is equivalent to counting the number of ways to split 2014 into three ordered non-negative addends $2014=x+y+z$, for which the non-strict triangle inequality holds, i.e., $x+y \geq z, x+z \geq y, y+z \geq x$.
Necessity. If we choose $a$ and $b$, then we have formed three numerical segments, in which there are 2014 numbers in total. It is clear that if the number of numbers in one of them is greater than the sum of the numbers in the other two, we will not be able to shuffle the numbers in the specified way.
Sufficiency. Choose $a=x+1, b=x+y+2$, that is, divide the sequence into parts with $x, y$ and $z$ numbers. We will prove that we can shuffle the numbers in the specified way. Assume $x \geq y \geq z$. It is known that $y+z \geq x$, and the other inequalities are automatically satisfied. Now we will swap numbers from $y$ and $z$ in pairs until the number of untouched numbers in $y$ and $z$ is exactly $x$. After that, we swap all numbers from $x$ with all untouched numbers from $y$ and $z$ and obtain the desired permutation.
Why is this possible? First, the number of numbers in $y$ and $z$ is at least $x$ by the triangle inequality, and each action reduces the number of untouched numbers by 2. But $x+y+z=$
2014, so $x$ and $(y+z)$ have the same parity, which means we will eventually get the required result.
Counting the ways: We will iterate over $x$. It changes from 0 to 1007, then $y+z \geq x$ is satisfied.
$x=0, y \geq z, z \geq y$, therefore, $z=y=1007$ - one way
$x=1,1+y \geq z, z+1 \geq y, y=2013-z$, therefore, $1007 \geq z ; z \geq 1006$ - two ways
$x=k . k+y \geq z, z+k \geq y, y=2014-k-z$. Therefore, $k+2014-k-z \geq z, z+k \geq 2014-k-z$, that is, $1007 \geq z ; z \geq 1007-k-(k+1)$ ways.
In total, $1+2+\ldots+1008=1009 * 504=508536$ ways.
Criteria:
Only the answer - 0 points
Only the hypothesis - 1 point.
Only the necessity - 1 point.
Only the sufficiency - 2 points.
Only the counting - 3 points.
These points are summed.
|
508536
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Around a round table, 15 boys and 20 girls sat down. It turned out that the number of pairs of boys sitting next to each other is one and a half times less than the number of pairs of girls sitting next to each other. Find the number of boy - girl pairs sitting next to each other.
|
Answer: 10.
Solution. Let's call a group several children of the same gender sitting in a row, with children of the opposite gender sitting to the left and right of the outermost ones. Let $X$ be the number of groups of boys, which is equal to the number of groups of girls sitting in a row. It is easy to see that the number of pairs of adjacent children in each group is one less than the number of children in that group, so the number of pairs of boys sitting next to each other is $15-X$, and the number of pairs of girls sitting next to each other is $20-X$. According to the condition, $3(15-X)=2(20-X)$, from which $X=5$. Therefore, the number of pairs of adjacent boys is 10, the number of pairs of adjacent girls is 15, and the number of mixed adjacent pairs is $15+20-(10+15)=10$.
Grading criteria. Noted that the number of pairs of adjacent children in each group is one less than the number of children in that group: 1 point. Noted that the number of pairs of boys sitting next to each other is $15-X$, and the number of pairs of girls sitting next to each other is $20-X$: 2 points. Noted that the number of groups of boys is equal to the number of groups of girls: 1 point.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. What is the maximum number of triangles with vertices at the vertices of a regular 18-gon that can be marked so that no two different sides of these triangles are parallel? The triangles can intersect and have common vertices, coinciding segments are considered parallel.
|
Answer: 5.
Solution. Estimating the number of triangles. Let's number the vertices of the 18-gon from 1 to 18 clockwise. The sides of the triangles are the sides and diagonals of the regular 18-gon. We will call a diagonal even if an even number of sides lies between its ends, and odd otherwise. The parity of a diagonal coincides with the parity of the difference in the numbers of its ends. Due to the evenness of the total number of sides of the polygon, it does not matter from which side of the diagonal to count the number of sides. We also consider the sides as odd diagonals. Two diagonals $\mathrm{AB}$ and $\mathrm{CD}$, where $\mathrm{AC}$ and $\mathrm{BD}$ do not intersect, are parallel if and only if the number of sides between A and C, and B and D, is the same, that is, the positive difference in the numbers of A and C is equal to the positive difference in the numbers of B and D. It is not difficult to notice that any odd diagonal is parallel to one of the nine sides of the 18-gon, and any even diagonal is parallel to one of the nine diagonals that cut off a triangle from the 18-gon (two sides of which are sides of the polygon). In total, there are 18 families of diagonals, any two diagonals of the same family are parallel, and any two diagonals of different families are not parallel. Nine of these families contain even diagonals and nine contain odd diagonals. As representatives of the odd families, we can take the sides with ends $1223, \ldots, 89,910$ and the diagonals $13,24, \ldots, 810,911$. Therefore, triangles with pairwise non-parallel sides, constructed on the vertices of a regular 18-gon, cannot use more than one from each of these families of diagonals, and the total number of such triangles does not exceed $18: 3=6$. Moreover, any triangle constructed on three vertices of the 18-gon can contain either three even diagonals or one even and two odd, since the sum of the parities of its three sides equals the parity of the number 18. Consequently, the total number of odd sides in any set of triangles with pairwise non-parallel sides must be even, and we will not be able to use all 18 families of diagonals for their sides. Thus, the number of triangles with pairwise non-parallel sides, constructed on the vertices of a regular 18-gon, does not exceed five.
Example. Five triangles with vertices $\{1,2,3\},\{3,4,5\},\{5,6,7\},\{7,8,9\},\{1,5,9\}$.
Grading criteria. It is shown that there are 18 pairwise non-parallel families of diagonals: 2 points. It is shown that all 9 non-parallel odd diagonals cannot be used as sides of triangles: 2 points. Example: 2 points. Justification of the example: 1 point.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. Find the maximum length of a horizontal segment with endpoints on the graph of the function $y=x^{3}-x$
|
Answer: 2.
Solution 1. A horizontal segment of length $a>0$ with endpoints on the graph of the function $y=x^{3}-x$ exists if and only if the equation $(x+a)^{3}-(x+a)=x^{3}-x$ has at least one solution for the given value of the parameter $a$. Expanding the brackets, combining like terms, and dividing by $a>0$, we obtain the quadratic equation $3 x^{2}+3 a x+a^{2}-1=0$, which is solvable when $D=12-3 a^{2} \geq 0$, from which $0<a \leq 2$. Therefore, the length of the desired segment does not exceed 2. When $a=2$, the solution to the equation is $x=-1$, from which it follows that the length 2 is achieved for the segment with endpoints $(-1,0)$ and $(1,0)$ on the graph of the function $y=x^{3}-x$.
## Solution 2.
As in Solution 1, we obtain the equation $3 x^{2}+3 a x+a^{2}-1=0$, which we consider as a quadratic equation in $a$ with parameter $x: a^{2}+3 x a+3 x^{2}-1=0$. We find its roots $a_{1,2}=\frac{-3 x \pm \sqrt{4-3 x^{2}}}{2}$, and since $a$ is positive, we consider only the one with the plus sign: $a=\frac{\sqrt{4-3 x^{2}}-3 x}{2}$. This function of $x$ is defined for $|x| \leq \frac{2}{\sqrt{3}}$ and is positive for $-\frac{2}{\sqrt{3}} \leq x \leq \frac{1}{\sqrt{3}}$. Its derivative, equal to
$a^{\prime}(x)=-\frac{3 x+3 \sqrt{4-3 x^{2}}}{2 \sqrt{4-3 x^{2}}}$, is zero at $x=-1$, is greater than zero to the left, and less than zero to the right. Therefore, its value is maximal at $x=-1$ and equals $a_{\max }=2$. Indeed, in this case, the segment of length 2 connects on the x-axis the two roots $x_{1}=-1$ and $x_{2}=1$ of the equation $x^{3}-x=0$.
Grading criteria. The answer 2 and an example of a segment of such length are provided: 1 point. Lack of an explicit example in the solution: minus 2 points.
|
2
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. Let $A$ be a set of ten distinct positive numbers (not necessarily integers). Determine the maximum possible number of arithmetic progressions consisting of three distinct numbers from the set $A$.
|
Answer: 20.
Solution: Let the elements of set $A$ be denoted as $a_{1}<a_{2}<\ldots<a_{10}$. Three numbers $a_{k}<a_{l}<a_{m}$ form a three-term arithmetic progression if and only if $a_{l}-a_{k}=a_{m}-a_{l}$. Let's see how many times each element of $A$ can be the middle term $a_{l}$ of such a progression. It is easy to see that for an element $a_{l}, l=2,3,4,5$, the first element $a_{k}$ can be chosen in no more than $l-1$ different ways - it can only be $a_{1}, a_{2}, \ldots, a_{l-1}$. Therefore, the number of such progressions with the middle term $a_{l}, l=2,3,4,5$ cannot exceed $1+2+3+4=10$. Similarly, for an element $a_{l}, l=6,7,8,9$, the third element $a_{m}$ can be chosen in no more than $10-l$ different ways - it can only be $a_{l+1}, a_{l+2}, \ldots, a_{10}$. Therefore, the number of such progressions with the middle term $a_{l}, l=6,7,8,9$ cannot exceed $4+3+2+1=10$. Any sought three-term progression must be one of these two types, so the total number of such ways cannot exceed $1+2+3+4+4+3+2+1=20$.
As an example of a set $A$ with 10 elements where the value 20 is achieved, one can take the set of all natural numbers from 1 to 10 inclusive.
Grading criteria. ($\cdot$) Proven that the number of progressions is no more than 20: 5 points. ($\cdot$) Example with 20 progressions: 2 points.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. How many cells need to be marked on an 8 by 8 grid so that each cell on the board, including the marked ones, is adjacent by side to some marked cell? Find all possible answers. Note that a cell is not considered adjacent to itself.
|
Answer: 20.
Solution: First, let's gather our strength and mark twenty cells on an 8 by 8 board as required by the problem. For example, as shown in the figure. In this case, the board naturally divides into 10 parts, as indicated by the bold lines in the figure. Each part consists of cells adjacent to the given pair of marked cells.
Now, using the constructed example, we will prove that the only answer to the problem is indeed 20 cells. Consider the division of the board into 10 parts as shown in the example. From now on, we will call these parts figures, as in the example. We will call the central cells of each figure those that are marked in the example in the figure. Recall that the chessboard has a natural coloring of cells in a checkerboard pattern, and consider the black and white parts of each figure, consisting of black and white cells of this figure, respectively. Note that the white central cell of a figure is adjacent only to black cells of this figure, and only to all of them, and the black central cell of a figure is adjacent only to white cells of this figure, and only to all of them.
Now consider any marking of cells on the board that satisfies the problem's condition, and prove that each figure contains exactly two marked cells, from which the answer to the problem will follow. Indeed, if some figure contains at least three marked cells, then it contains at least two white marked cells or at least two black marked cells, then the central cell of the opposite color of this figure will be adjacent to at least two marked cells, which contradicts the condition. And if some figure contains no more than one marked cell, then it either has no white marked cells or no black marked cells, then the central cell of the opposite color of this figure will not be adjacent to any marked cell, which also contradicts the condition. Therefore, each figure contains exactly two marked cells, so any example contains exactly 20 marked cells.
Note that the example constructed in this solution is not unique, and we did not prove that the marked cells of any example coincide with those marked by us. The board can be rotated by 90 degrees, and a new example different from the one considered will result, but still, each figure contains exactly two marked cells from the new example.
Remark. The estimate for 20 can also be done differently, by first proving that the number of extreme marked cells is at least 9, etc. This is a long and risky path with case-by-case analysis. When evaluating such claimed solutions, each step must be very carefully assessed, and if such a solution contains unconsidered cases or errors in calculations, the estimation part of the solution is not rated higher than 1 point.
Evaluation Criteria: (•) An example with 20 cells is constructed: 3 points. (•) It is proven that the number of marked cells is always 20: 4 points. (•) The idea of considering figures from the board solution for estimation (without significant progress): 1 point.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Pasha and Sasha made three identical toy cars. Sasha did one-fifth of the total work. After selling the cars, they divided the proceeds in proportion to the work done. Pasha noticed that if he gave Sasha 400 rubles, and Sasha made another such car and sold it, they would have the same amount of money. How much does one car cost?
|
Answer: 1000 rubles.
Solution: Sasha did one-fifth of the entire work, which means he made 0.6 of one car, while Pasha did the remaining 2.4. That is, the difference is 1.8 cars. If Sasha makes another car, the difference will be 0.8 of one car. Pasha gave 400 of his rubles, thereby reducing the amount of money he had by 400 rubles and increasing the amount of money Pasha had by the same amount. That is, he changed the difference in the amount of money between one and the other by 800 rubles. Thus, 0.8 of a car corresponds to 800 rubles. Therefore, a whole car costs 1000 rubles.
Criteria: Only the answer - 0 points.
Answer with verification - 1 point.
Solution, in which 0.8 of a car corresponds to 400 rubles - no more than 2 points.
|
1000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. From identical isosceles triangles, where the angle opposite the base is $45^{\circ}$ and the lateral side is 1, a figure was formed as shown in the diagram. Find the distance between points $A$ and $B$.
|
Answer: 2.
Solution: Let's denote the points $K, L, M$, as shown in the figure. We will construct an isosceles triangle $A K C$ equal to the original one. Connect vertex $C$ to other points as shown in the figure.
In the original triangles, the angle at the vertex is $45^{\circ}$. Therefore, the other two angles are each $62.5^{\circ}$. Then the angle $\angle C K L = 360^{\circ} - 62.5^{\circ} - 62.5^{\circ} - 45^{\circ} - 45^{\circ} - 62.5^{\circ} = 62.5^{\circ}$. Therefore, triangles $A K C$ and $K L C$ are equal by two sides and the angle between them. Similarly, it can be proven that triangles $L M C$ and $M B C$ are equal to the original ones. Then the angle $\angle A C B$ is $4 \times 45^{\circ} = 180^{\circ}$, which means points $A, C, B$ lie on the same line. Therefore,
$A B = A C + C B = 2$.
## Criteria:
A suitable construction is made, but it is not justified that points $A, C, B$ lie on the same line (or point $C$ is the midpoint of segment $A B$, but it is not justified that the resulting triangles are equal to the original ones) - 1 point.
The solution refers to the fact that a regular octagon has a center and other obvious properties of the octagon - no more than 1 point should be deducted.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. On a plane, there are points $A, B, C, D, X$. Some segment lengths are known: $A C=2$, $A X=5$, $A D=11$, $C D=9$, $C B=10$, $D B=1$, $X B=7$. Find the length of the segment $C X$.
|
Answer: 3.
Solution: Note that $A D=11=2+9=A C+C D$. Therefore, points $A, C, D$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Note that $C B=10=9+1=C D+D B$. Therefore, points $C, D, B$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Therefore, all four points lie on the same line $C D$, with $C$ between $A D$ and $D$ between $C$ and $B$, i.e., in the order $A, C, D, B$. Note that $A B=A D+D B=11+1=5+7=A X+X B$. Therefore, points $A, X, B$ lie on the same line $A B$ (since the triangle inequality becomes an equality).
Thus, all points lie on the segment $A B$. Therefore, $C X=A X-A C=5-2=3$.
## Criteria:
Answer only - 0 points.
Answer with verification - 1 point.
Solution based on the assumption that all points lie on one line - no more than 3 points.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. The company receives apple and grape juices in standard barrels and produces a cocktail (mixture) of these juices in standard cans. Last year, one barrel of apple juice was enough for 6 cans of cocktail, and one barrel of grape juice was enough for 10. This year, the proportion of juices in the cocktail (mixture) has changed, and now a standard barrel of apple juice is enough for 5 cans of cocktail. How many cans of cocktail will now be enough for a standard barrel of grape juice?
|
Answer: 15 cans.
Solution. Last year, one barrel of apple juice was enough for 6 cans of cocktail, which means each can contained $1 / 6$ of a barrel of apple juice. Similarly, one barrel of grape juice was enough for 10 cans, which means each can contained $1 / 10$ of a barrel of grape juice. Therefore, the capacity of a can is $1 / 6 + 1 / 10 = 4 / 15$ of a standard barrel. After the proportion change in the new year, each can now contains $1 / 5$ of a barrel of apple juice, which means the amount of grape juice in each can has become $4 / 15 - 1 / 5 = 1 / 15$ of a barrel. Therefore, a barrel of grape juice now suffices for 15 cans of cocktail.
Grading criteria. Clear explanation that the capacity of a can is 4/15 of a barrel: 3 points.
For not excluding an incorrect answer when the solution and answer are correct, minus 2 points.
For lack of explanation, from -2 to 0 points.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. For what minimum $\boldsymbol{n}$ in any set of $\boldsymbol{n}$ distinct natural numbers, not exceeding 100, will there be two numbers whose sum is a prime number?
|
Answer. $\boldsymbol{n}=51$.
Solution. The sum of two even natural numbers is always even and greater than two, hence it cannot be a prime number. Therefore, the example of a set of all fifty even numbers not exceeding 100 shows that the minimum $\boldsymbol{n}$ is not less than 51.
On the other hand, let's divide all natural numbers from 1 to 100 into 50 pairs, the sum of the numbers in each of which is the prime number 101: 1 and 100, 2 and 99, $\ldots, 50$ and 51. If the selected set contains no fewer than 51 numbers, then by the pigeonhole principle, at least two of them will fall into one pair, and their sum will be a prime number.
Grading criteria. Example showing that $\boldsymbol{n}$ is greater than 50: 2 points. Proof that in any set of 51 distinct natural numbers not exceeding 100, there will be two whose sum is a prime number: 5 points.
Idea of the example, but not explicitly provided +1 point.
|
51
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. In each cell of a 5 by 5 table, a letter is written such that in any row and in any column there are no more than three different letters. What is the maximum number of different letters that can be in such a table
|
Answer: 11.
Solution: If each row contains no more than two different letters, then the total number of letters does not exceed $10=5 * 2$. Further, we can assume that the first row contains exactly three different letters. If each of the remaining rows has at least one letter in common with the first, then the total number of letters does not exceed $3+4 * 2=11$. Suppose there is a row, which we can consider the second, with three different letters, distinct from the letters in the first row. Then, in each column, apart from the letters in the first and second rows, there can be no more than one new letter, totaling no more than $3+3+5 * 1=11$. An example of arranging 11 different letters: along the main diagonal of the table from the bottom left corner to the top right, the first five different letters are written, along the next lower diagonal - the next four, in the top left corner - the tenth, and in the remaining cells - the eleventh letter.
Grading criteria. The maximum of 11 is proven: 5 points. Example for 11: 2 points. Any incorrect answer and an attempt to prove it: 0 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Let $a^{2}+b^{2}=c^{2}+d^{2}=1$ and $a c+b d=0$ for some real numbers $a, b, c, d$. Find all possible values of the expression $a b+c d$.
|
Answer: 0.
Solution: Let's first assume $b \neq 0$. From the second equation, express $d=\frac{-a c}{b}$ and substitute it into the equation $c^{2}+d^{2}=1$. Eliminating the denominator, we get $c^{2}\left(a^{2}+b^{2}\right)=b^{2}$, from which, given $a^{2}+b^{2}=1$, we obtain $b= \pm c$. Substituting this into the equation $a^{2}+b^{2}=c^{2}+d^{2}$, we get $a= \pm d$. Given the equation ${ }^{a c}+b d=0$, we have either $b=c, a=-d$, or $b=-c, a=d$. In both cases, $a b+c d=0$.
If $b=0$, then from $a c+b d=0$ it follows that ${ }^{a=0}$ or ${ }^{c=0}$. The first is impossible due to $a^{2}+b^{2}=1$, so $c=0$, from which we again get $a b+c d=0$.
Grading criteria. Only the answer even with the corresponding $a, b, c, d$: 0 points. Failure to consider the case ${ }^{b=0}$ (if necessary in the solution): minus 1 point.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. In a set $X$ of 17 elements, a family of $N$ distinct non-empty subsets is selected such that each element of the set $X$ is contained in exactly two subsets from this family. What is the maximum value of $N$? Find the number of all possible different types of such families for the maximum $N$. Two families of subsets have different types if one cannot be obtained from the other by permuting the elements of $X$.
|
Answer. The maximum $N$ is 25, and there exist two different types of families of 25 subsets that satisfy the condition of the problem.
Solution. Consider an arbitrary family of $N$ distinct non-empty subsets such that each element of the set $X$ is contained in exactly two subsets of this family. If there are $x$ one-element subsets, then the total sum $\mathrm{S}$ of the cardinalities of the subsets in this family is at least $x + 2(N - x) = 2N - x$. On the other hand, each element of the set $X$ is contained in exactly two subsets of this family, so $\mathrm{S} = 34$, which means $\mathrm{N}$ does not exceed $(34 + x) / 2$. Finally, the number of one-element subsets $x$ does not exceed the number of elements in $X$, which is 17, so $N$ does not exceed 25. Note that for $N = 25$, the number of one-element subsets in the family can be 16 or 17.
1) If $x = 17$, then the remaining 8 "large" subsets must be two- or more-element subsets. If there are $y$ two-element subsets among them, then the total number of elements in the 8 large subsets is 17 and is at least $2y + 3(8 - y) = 24 - y$, so $y$ is at least 7. If it is greater than 7, then the three- and more-element subsets in total would contain one element, which is impossible. Therefore, the only possibility is 7 pairwise disjoint two-element subsets and one three-element subset. Together with the 17 one-element subsets, they form the first example of the family of subsets sought in the problem. An example of a family of this type: 17 one-element subsets $\{1\}, \{2\}, \ldots, \{17\}$, 7 two-element subsets $\{1, 2\}, \{3, 4\}, \ldots, \{13, 14\}$, and one three-element subset $\{15, 16, 17\}$.
2) If $x = 16$, then the remaining 9 subsets must be two- or more-element subsets. If there are $y$ two-element subsets among them, then the total number of elements in the 9 large subsets is 18 and is at least $2y + 3(9 - y) = 27 - y$, so $y$ is at least 9. Therefore, all large subsets are two-element subsets. In total, they contain 18 elements, so two of them intersect at one element (which does not lie in any one-element subset), while the rest are pairwise disjoint. Together with the 16 one-element subsets, they form the second, different from the first, example of the family of subsets sought in the problem. An example of a family of this type: 16 one-element subsets $\{1\}, \{2\}, \ldots, \{16\}$, 9 two-element subsets $\{1, 2\}$, $\{3, 4\}, \ldots, \{13, 14\}, \{15, 17\}, \{16, 17\}$.
In the course of the reasoning, we proved that there are no other families different from the ones found.
Grading criteria. For each found example of one of the two types of families: 1 point each. Proof of the maximality of $N = 25$: 3 points. Proof that there are exactly 2 types of maximal families: 2 points.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Large sandglasses measure an hour, and small ones measure 11 minutes. How can you use these sandglasses to measure a minute?
|
Solution: We will run the large hourglass twice in a row and the small one eleven times in a row. A minute will be measured between the second time the large hourglass finishes (120 minutes) and the 11th time the small one finishes (121 minutes).
Criteria: Any correct example - 7 points.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. Given a triangle $\mathrm{ABC}$ with angle $\mathrm{BAC}$ equal to $30^{\circ}$. In this triangle, the median $\mathrm{BD}$ was drawn, and it turned out that angle $\mathrm{BDC}$ is $45^{\circ}$. Find angle $\mathrm{ABC}$.
|
Answer: $45^{\circ}$
Solution: Draw the height $C H$. Then $H D=A D=C D$ as the median to the hypotenuse. Moreover, $\angle H C D=\angle C H A-\angle H A C=60^{\circ}$, so triangle $C H D$ is equilateral, which means $\angle H D C=60^{\circ}$ (from which it follows, in particular, that $H$ lies between $A$ and $B$). Then $\angle H D B=\angle H D C-\angle B D C=15^{\circ}$. Furthermore, $\angle A B D=\angle B D C-\angle B A D=15^{\circ}$. Therefore, triangle $H B D$ is isosceles, and $H B=H D=H C$, which implies that $B H C$ is an isosceles right triangle, and thus $\angle A B C=45^{\circ}$.
Criteria: The height $C H$ is drawn - 1 point.
It is additionally proven that $\angle H D B=15^{\circ}$ - another 2 points.
No points are deducted for the lack of proof that $H$ lies between $A$ and $B$.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. In the city of Omsk, a metro has been built, represented by a straight line. On this same line is the house where Nikita and Egor live. Every morning they leave the house for classes at the same time, after which Egor runs to the nearest metro station at a speed of 12 km/h, while Nikita walks along the metro line to another station at a speed of 6 km/h. Despite this, every day Nikita manages to arrive on time for the first lesson, while Egor does not, even though he is not delayed anywhere. Find the greatest possible speed of the metro trains, given that it is constant and equal to an integer. (Assume that the school is located directly on a certain metro station, different from the given ones).
|
Answer: 23 km/h
Solution: Obviously, this is only possible if the subway train first arrives at the nearest station A, where Egor runs to, and then goes to station B, where Nikita is heading.
Let $v$ be the speed of the subway, $S$ be the distance between two adjacent stations, and $R$ be the distance between this subway train and the nearest station at the moment when Nikita and Egor leave home simultaneously.
Egor does not catch the subway, which means $R / v < (S / 2) / 6$
From this, we get: $S v / 12 - S1 / 10$ hours and does not make it.
The subway takes 4.75 km / 23 km/h = 4.75 / 23 hours to reach station B. Nikita walks to station B in $1.23 / 6$ hours < 4.75 / 23 and makes it!
Criteria: Only the answer - 3 points.
Only a complete example with full justification - 3 points
Deduct 2 points for a correct example without justification.
|
23
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Vikentiy walked from the city to the village, and at the same time Afanasiy walked from the village to the city. Find the distance between the village and the city, given that the distance between the pedestrians was 2 km twice: first, when Vikentiy had walked half the way to the village, and then, when Afanasiy had walked a third of the way to the city.
|
Answer: 6 km.
Solution. Let the distance between the village and the city be denoted as $S$ km, the speeds of Vikentiy and Afanasy as $x$ and $y$, and calculate the time spent by the travelers in the first and second cases. In the first case, we get: $\frac{S / 2}{x}=\frac{S / 2-2}{y}$, in the second case $\frac{2 S / 3+2}{x}=\frac{S / 3}{y}$. From here, eliminating $x$ and $y$, we have $S^{2}-2 S-24=0$, from which $S=6$ km.
Instructions. Answer with verification: 1 point. Formulation of equations: 3 points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. From points A and B towards each other with constant speeds, a motorcyclist and a cyclist started simultaneously from A and B, respectively. After 20 minutes from the start, the motorcyclist was 2 km closer to B than the midpoint of AB, and after 30 minutes, the cyclist was 3 km closer to B than the midpoint of AB. How many minutes after the start did the motorcyclist and the cyclist meet
|
Answer: In 24 minutes.
Solution: In 10 minutes, the motorcyclist travels $1 / 4$ of the distance from A to B plus 1 km, while the cyclist travels $1 / 6$ of the distance from A to B minus 1 km. Therefore, in 10 minutes, both of them, moving towards each other, cover $1 / 4 + 1 / 6 = 5 / 12$ of the distance from A to B. This means they will cover the distance from A to B, that is, meet, in $10 * 12 / 5 = 24$ minutes.
Grading Criteria. Correct answer with verification: 3 points. Correctly formulated system of equations: 3 points.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. From two cities, the distance between which is 105 km, two pedestrians set out simultaneously towards each other at constant speeds and met after 7.5 hours. Determine the speed of each of them, knowing that if the first walked 1.5 times faster, and the second 2 times slower, they would have met after $8 \frac{1}{13}$ hours.
|
Answer: 6 and 8 km per hour.
Solution: Let their speeds be $x$ and $y$ km per hour, respectively. From the condition, we get: $\frac{15}{2}(x+y)=105, \frac{105}{13}\left(\frac{3}{2} x+\frac{1}{2} y\right)=105$, from which $x=6, y=8$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Several consecutive natural numbers are written on the board. It is known that $48 \%$ of them are even, and $36 \%$ of them are less than 30. Find the smallest of the written numbers.
|
Answer: 21.
Solution. $\frac{48}{100}=\frac{12}{25}, \frac{36}{100}=\frac{9}{25}$ - these are irreducible fractions, so the total number of numbers is divisible by 25. If there were 50 or more, then, by the condition, there would be at least 2 fewer even numbers than odd numbers, which is impossible for consecutive natural numbers. Therefore, there are 25, and exactly 9 of them are less than 30. Thus, the first one is 21.
|
21
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. In a family of 4 people. If Masha's scholarship is doubled, the total income of the entire family will increase by $5 \%$, if instead the mother's salary is doubled - by $15 \%$, if the father's salary is doubled - by $25 \%$. By what percentage will the family's total income increase if the grandfather's pension is doubled?
|
Answer: by $55 \%$.
Solution: When Masha's scholarship is doubled, the family's total income increases by the amount of this scholarship, so it constitutes $5 \%$ of the income. Similarly, the salaries of Masha's mother and father constitute $15 \%$ and $25 \%$. Therefore, the grandfather's pension constitutes $100-5-15-25=55 \%$, and if it is doubled, the family's income will increase by $55 \%$.
Criterion: Only the answer, answer with verification - 0 points.
|
55
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. There are 100 boxes numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all of them. What is the minimum number of notes that need to be sent to definitely find out where the prize is?
|
Answer: 99.
Solution: To be able to definitively determine which of the 100 boxes contains the prize, it is necessary to have the possibility of receiving at least 100 different answers to one set of questions. Since the host's answers for different prize positions can only differ by the number of "yes" responses, it is necessary to have the possibility of receiving at least 100 different counts of "yes." Therefore, at least 99 questions are required (from 0 to 99 "yes").
Example with 99 questions: Let the $k$-th question be: "Is the number of the box containing the prize less than or equal to $k$?" Then, if the number of "yes" answers is zero, the prize is in the hundredth box; if one, it is in the 99th box, and so on.
Criteria: Only the evaluation -3 points, only the example -3 points. Only the answer - 0 points.
|
99
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. On a certain island, only knights, who always tell the truth, and liars, who always lie, live. One day, 1001 inhabitants of this island stood in a circle, and each of them said: "All ten people following me in a clockwise direction are liars." How many knights could there be among those standing in the circle?
|
Answer: 91 knights.
Solution. Note that all people cannot be liars, because then it would mean that each of them is telling the truth. Therefore, among these people, there is at least one knight. Let's number all the people so that the knight is the 1001st in the sequence. Then the 10 people with numbers from 1 to 10 are liars (this follows from the knight's statement). Moreover, the 10 people with numbers 991-1000 are also liars, as they made an incorrect statement (there is a knight among the ten people standing after them). Therefore, since these 10 people stand in a row along our circle after person number 990, he told the truth and, consequently, is also a knight. Repeating the reasoning for him, we get that people with numbers 980-989 are liars, and so on. In the end, we get that the knights are people with numbers $1001, 990, 979, \ldots, 11$, that is, there are a total of $1001 / 11 = 91$ of them.
For completeness of the reasoning, note that we have proven that if the arrangement is possible, it looks exactly like this. It is not difficult to check that there are no contradictions, and the people could indeed stand this way.
Criteria. Only the answer - 0 points.
Only the answer with an example of the arrangement - 2 points.
The case where there are no knights is not considered - do not deduct points.
Not verified that the obtained arrangement fits - do not deduct points.
|
91
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. We will call a number remarkable if it can be decomposed into the sum of 2023 addends (not necessarily distinct), each of which is a natural composite number. Find the largest integer that is not remarkable.
|
Answer. $4 \times 2023+3=8095$.
Solution. Replace 2023 with $n$ and solve the problem in the general case for a sum of $n \geqslant 2$ composite addends. We will prove that the answer is $4 n+3$, from which we will obtain the answer to the original problem.
Claim 1. The number $4 n+3$ is not remarkable.
Proof of Claim 1. Note that $4 n+3$ is odd, so in its decomposition into a sum, there must be at least one odd addend. The smallest odd composite number is 9, and the smallest even composite number is 4. Therefore, the sum of $n$ addends is at least
$$
9+4(n-1)=4 n+5>4 n+3,
$$
from which it follows that $4 n+3$ cannot be decomposed into a sum of $n$ composite addends, that is, it is not remarkable.
Claim 2. Any natural number greater than $4 n+3$ is remarkable. Proof of Claim 2. Consider the number $a>4 n+3$. If it is even, then it can be written as
$$
a=(4 n-4)+b=b+\underbrace{4+4+\ldots+4}_{n-1},
$$
where $b \geqslant 8-$ is some even number that is composite. If $a$ is odd, then
$$
a=4 n+1+c=9+(4 n-8)+c=9+c+\underbrace{4+4+\ldots+4}_{n-2}
$$
where $c \geqslant 4-$ is some even number that is composite.
In total, the number $4 n+3$ is not remarkable, and all larger numbers are remarkable. From this, we obtain the answer to the question of the problem.
Criteria. Proof of Claim 1 or an equivalent statement - 3 points. Proof of Claim 2 or an equivalent statement - 2 points for the even and odd cases.
|
8095
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. Let $n$ be a natural number not ending in 0, and $R(n)$ be the four-digit number obtained from $n$ by reversing the order of its digits, for example $R(3257)=7523$: Find all natural four-digit numbers $n$ such that $R(n)=4n+3$.
|
Answer: 1997.
Solution. Consider the decimal representation of the original four-digit number $n=\overline{a b c d}$, then $R(n)=\overline{d c b a}=4 n+3$ is also a four-digit number. Therefore, $4 a \leq 9$, so $a=1$ or $a=2$. Moreover, the number $R(n)=\overline{d c b a}=4 n+3$ is odd and ends in $a$, hence $a=1$. If the number $4 n+3$ ends in 1, then $n$ must end in 2 or 7, that is, $d=2$ or $d=7$. The first is impossible, since $d \geq 4 a \geq 4$, so $d=7$. Substituting the found values into the equation from the condition: $R(n)=4 \cdot \overline{a b c d}+3=4031+400 b+40 c=7001+100 c+10 b$, from which $13 b-2 c=99$. From this, $b \geq \frac{99}{13}=7 \frac{8}{13}$, that is, $b=8$ or $b=9$. When $b=8$, the left side of the equation $13 b-2 c=99$ is even, which is impossible, so $b=9$ and $c=9$.
Grading criteria. Answer: 2 points. Proof that there are no other answers: 5 points.
|
1997
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. Two circles intersect at points A and B. A tangent to the first circle is drawn through point A, intersecting the second circle at point C. A tangent to the second circle is drawn through point B, intersecting the first circle at point D. Find the angle between the lines $\mathrm{AD}$ and $\mathrm{BC}$.
|
Answer. The lines are parallel, the angle is $0^{\circ}$.
Solution. Mark point $\mathrm{P}$ on the extension of $\mathrm{CA}$ beyond point A and point M on the extension of DB beyond point B. The inscribed angle ABD in the first circle, which subtends the chord AD, is equal to the angle PAD between the chord AD and the tangent AC to the first circle. The inscribed angle CAB in the second circle, which subtends the chord BC, is equal to the angle CBM between the chord BC and the tangent BD to the second circle. Then, the angle BAD between the secant line AB and the line AD is equal to the difference of $180^{\circ}$ and the sum of angles PAD and CAB, which is equal to the difference of $180^{\circ}$ and the sum of angles ABD and CBM, that is, the angle ABC between the secant line AB and the line BC. Therefore, the lines AD and BC are parallel.
Grading criteria. Only answer: 0 points.
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. There are two ingots of different copper and tin alloys weighing 6 and 12 kg respectively. From each of them, a piece of the same weight was cut off and the first piece was alloyed with the remainder of the second ingot, and the second piece - with the remainder of the first ingot, after which the ratio of copper to tin in the two resulting new ingots turned out to be the same. Find the weight of each of the cut-off pieces.
|
Answer: 4 kilograms.
Solution: Let the weight of each of the cut pieces be $x$ kg, and the proportions of tin in the first and second ingots be $a \neq b$ respectively. Then, the proportion of tin after remelting in the first ingot will be $\frac{b x + a(6 - x)}{6}$, and in the second ingot $\frac{a x + b(12 - x)}{12}$, and they are equal according to the condition. Solving the obtained equation, we find $x = 4$ for all $a \neq b$.
Grading criteria. Correct answer with verification for arbitrary $a \neq b: 1$ point. Analysis of a specific case: 0 points.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. In the tournament, each of the six teams played against each other exactly once. In the end, the teams scored 12, 10, 9, 8, 7, and 6 points respectively. a) How many points were awarded for a win in a match, if 1 point was awarded for a draw, and 0 points for a loss? The answer, of course, should be a natural number. b) Find the number of wins, draws, and losses for each team and prove the uniqueness of these numbers. c) Provide an example of a corresponding tournament.
|
Answer. a) 4 points.
b) The first team had three wins, the second team had two wins and two draws, the third team had two wins and one draw, the fourth team had two wins, the fifth team had one win and three draws, the sixth team had one win and two draws. The rest of the matches were lost by the teams.
c) One example that meets the condition is as follows: the first team won against the fourth, fifth, and sixth teams, the second team won against the first and third teams, the third team won against the first and sixth teams, the fourth team won against the second and third teams, the fifth team won against the fourth team, and the sixth team won against the fourth team. In addition, the fifth team drew with the sixth, third, and second teams, and the sixth and second teams also drew with each other.
Solution. a) Let $n$ be the number of points awarded for a win in a match. There were a total of 15 matches played in the tournament, of which $x$ ended in a win for one of the teams, and the remaining $15-x$ ended in draws. In drawn matches, the participants collectively score 2 points, while in the others, they score $n$ points, so the total number of points scored by all teams in the tournament is $n x + 2(15-x) = (n-2)x + 30 = 12 + 10 + 9 + 8 + 7 + 6 = 52$ points, from which $(n-2)x = 22$. There were a total of 15 games, so $x$ can be 1, 2, or 11. In the first two cases, $n$ equals 24 or 13, which exceeds the maximum number of points scored by the teams, which is impossible because a win in a match would then be impossible in principle, meaning all matches would end in draws, which contradicts the assumption about $x$. In the remaining case, $x$ equals 11, then $n$ equals 4, which is the number of points awarded for a win.
b) We will find the number of wins, draws, and losses for each team. There were a total of four draws and 11 matches that ended in a win for one of the teams. Each team played five games, so it was impossible to score 6 or more points solely from draws, meaning the sixth and fifth teams each won once and drew two and three times, respectively.
If the fourth team had won no more than once, it would have had at least four draws, together with the fifth and sixth teams, there would already be at least $(2+3+4)/2 = 4.5$ draws, which exceeds the total of 4. Therefore, the fourth team won twice and lost three times.
For the same reasons, the third team had two wins and one draw, and the second team had two wins and two draws. For all teams except the first, there are at least $(2+3+1+2)/2 = 4$ draws, which equals their total number. Therefore, the first team had three wins and two losses.
c) Let's provide an example of a corresponding tournament. The fifth team drew three times, while the first and fourth teams had no draws. Therefore, the fifth team drew with the sixth, third, and second teams. In addition to these draws, the sixth and second teams each had one draw, so they drew with each other. All draws are thus uniquely determined. However, the decisive matches cannot be uniquely determined. One example that meets the condition is as follows: the first team won against the fourth, fifth, and sixth teams, the second team won against the first and third teams, the third team won against the first and sixth teams, the fourth team won against the second and third teams, the fifth team won against the fourth team, and the sixth team won against the fourth team.
Grading criteria. Correct and justified answer in part a): 3 points. Found the number of wins and draws for each team with justification: 2 points. Provided any valid example of a tournament in part c): 2 points.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya?
|
Solution: If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adjacent boats is 3 km, and the next boat will have to travel exactly this distance when the previous one has already reached Sovunya. The speed of the boat is 10 km/h, which means 1 km in 6 minutes, so 3 km will be covered by the boat in 18 minutes, which is the answer.
Criteria: Correctly found the distance of 3 km - 3 points.
Only the answer or the answer with verification - 1 point.
|
18
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. Find the largest four-digit number, all digits of which are different, and which is divisible by each of its digits. Of course, zero cannot be used.
|
Solution: Since the number of digits is fixed, the number will be larger the larger the digits in its higher places are. We will look for the number in the form $\overline{98 a}$. It must be divisible by 9. Therefore, the sum $a+b$ must give a remainder of 1 when divided by 9. At the same time, this sum does not exceed 13, as it consists of different single-digit addends, which are less than 8. Therefore, this sum is either 1 or 10. The largest variant that fits these conditions is 9873, but it does not work since it is not divisible by 8. The next highest in seniority, 9864, clearly works.
Solution: Only proven that the number 9864 works - 2 points. The idea to look for the answer in the form $\overline{98 a b}-1$ - 1 point.
|
9864
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. Point $A$ is located midway between points $B$ and $C$. The square $A B D E$ and the equilateral triangle $C F A$ are in the same half-plane relative to the line $B C$. Find the angle between the lines $C E$ and $B F$.
|
Solution: Let $M$ be the point of intersection of segments $B F$ and $C E$. Note that $\angle C F B = 90^{\circ}$, since $F A$ is a median equal to half the side to which it is drawn. From this, it follows that $\angle F B C = 180^{\circ} - 90^{\circ} - \angle F C B = 30^{\circ}$. From the isosceles right triangle $C A E$, it follows that $\angle E C A = 45^{\circ}$.
Then, in triangle $C M B$, the angle $\angle C M B = 180^{\circ} - \angle M B C - \angle M C B = 180^{\circ} - 30^{\circ} - 45^{\circ} = 105^{\circ}$, which is the answer.
Criteria: The angle $\angle C B F$ is found - 2 points.
The angle $\angle E C A$ is found - 1 point.
It doesn't matter whether the angle is found to be $105^{\circ}$ or $75^{\circ}$.
|
105
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 21 games, and the second - 10. How many games did the third player play?
|
Answer: 11.
Solution: According to the problem, the first player played 21 games, so there were at least 21 games in total. Out of any two consecutive games, the second player must participate in at least one, which means there were no more than \(2 \cdot 10 + 1 = 21\) games. Therefore, a total of 21 games were played, and the first player participated in all of them. In 10 games, he played against the second player, and in the remaining 11 games, he played against the third player. An example of such a tournament: the first player plays against the second player in games with even numbers, and against the third player in games with odd numbers.
Grading Criteria: If the answer is guessed and an example of the tournament is provided: 1 point. If it is noted that out of any two consecutive games, the second player must participate in at least one: 2 points. If an example of the tournament is not provided: minus 1 point.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. On the board, 10 natural numbers are written, some of which may be equal, and the square of each of them divides the sum of all the others. What is the maximum number of different numbers that can be among the written ones?
|
Answer. Four
Solution. An example for four different numbers: 1,1,1,2,2,3,5,5,5,5.
Let among the listed numbers there are exactly $n \geq 2$ different ones, the maximum of which we denote by $x$, and the sum of all numbers by $\mathrm{S}$. Then the sum of all numbers except the maximum does not exceed $(9-n) x + x - 1 + x - 2 + \ldots + x - (n-1) = 9x - \frac{n(n-1)}{2}$, which is not less than $x^2$, since it is divisible by it. Therefore, the inequality $x^2 - 9x + \frac{n(n-1)}{2} \leq 0$ has solutions in natural numbers, and $\frac{9 - \sqrt{81 - 2n(n-1)}}{2} \leq x \leq \frac{9 + \sqrt{81 - 2n(n-1)}}{2}$. The positivity of the discriminant gives us $n \leq 6$, and $x \leq 8, S - x \leq 72 - 1 = 71$. Consider each case of the maximum number separately.
a) $x=8$, then $S - x \leq 71$ and is divisible by 64, so $S = 72$. Then $72 - k$ is divisible by $k^2, k \leq 7$ only for $k=1$, so in this case $n \leq 2$.
b) $x=7$, then $S - x \leq 62$ and is divisible by 49, so $S = 56$. Then $56 - k$ is divisible by $k^2, k \leq 6$ only for $k=1$, so in this case $n \leq 2$.
c) $x=6$, then $S - x \leq 53$ and is divisible by 36, so $S = 42$. Then $42 - k$ is divisible by $k^2, k \leq 5$ only for $k=1,2$, so in this case $n \leq 3$.
d) $x=5$, then $S - x \leq 44$ and is divisible by 25, so $S = 30$. Then $30 - k$ is divisible by $k^2, k \leq 4$ for $k=1,2,3$, so in this case $n \leq 4$. The desired set must contain 10 natural numbers with a sum of 30, among which there must be $1,2,3,5$. Now it is not difficult to construct an example for $n=4: 1,1,1,2,2,3,5,5,5,5$. This example is not unique.
Grading criteria. It is proven that $n \leq 4: 5$ points. Example for $n=4: 2$ points. Estimate $n \leq 6: 2$ points.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. The bathtub fills up from the hot water tap in 17 minutes, and from the cold water tap in 11 minutes. After how many minutes from opening the hot water tap should the cold water tap be opened so that by the time the bathtub is full, there is one third more hot water than cold water?
|
Answer. In 5 minutes.
Solution. Let the volume of the bathtub be $V$, then by the time it is filled, there should be $\frac{3}{7} V$ of cold water and $\frac{4}{7} V$ of hot water in it. The filling rates of cold and hot water are $\frac{V}{11}$ and $\frac{V}{17}$, respectively. Therefore, the desired time is the difference between the time to fill $\frac{4}{7} V$ of the bathtub with hot water and the time to fill $\frac{3}{7} V$ of the bathtub with cold water, that is, $\frac{4}{7} V / \frac{V}{17} - \frac{3}{7} V / \frac{V}{11} = \frac{68 - 33}{7} = 5$ minutes.
Instructions. Answer with verification: 2 points. Formulation of equations: 3 points.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Find the smallest natural number $n$ such that in any set of $n$ distinct natural numbers, not exceeding 1000, it is always possible to select two numbers, the larger of which does not divide the smaller one.
|
Answer. $n=11$.
Solution. Among the first 10 powers of two $1=2^{0}, 2=2^{1}, 4=2^{2}, \ldots, 512=2^{9}$, in each pair of numbers, the larger number is divisible by the smaller one, hence $n \geq 11$.
On the other hand, let there be some set of $n \geq 11$ numbers where the larger number of each pair is divisible by the smaller one. Arrange all the numbers in ascending order; from the assumption, it follows that each subsequent number is at least twice the previous one. Therefore, the largest number is at least $2^{10}=1024$ times larger than the first, meaning it is greater than 1000 - a contradiction.
Instructions. Proving $n \geq 11$ with an example: 3 points. Proving only $n \leq 11$ (the second part): 4 points.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. In each cell of a 10 by 10 table, a minus sign is written. In one operation, it is allowed to simultaneously change the signs to their opposites in all cells of some column and some row (plus to minus and vice versa). What is the minimum number of operations required to make all the signs in the table pluses?
|
Answer. In 100 operations.
Solution. There are 19 cells in total in the row and column passing through a given cell, so if we perform operations on all pairs of rows and columns of the table (a total of $10 \times 10=100$ operations), each sign in the table will change 19 times, turning from minus to plus, so 100 operations are sufficient.
We will call the operation of changing the signs in all cells of a certain column and a certain row an operation relative to the cell at the intersection of these row and column. Cells relative to which we performed operations will be called red, and the rest - blue. Rows and columns containing an even number of red cells will be called even, and those containing an odd number of red cells - odd. Suppose it is possible to change all the signs in the table in fewer than 100 operations, then consider some blue cell A in row X and column Y. For the sign in A to change, row X and column Y together must contain an odd number of red cells, we can assume row X is even, and column Y is odd. Note that at the intersection of rows and columns of the same parity, there should be a red cell, and at the intersection of rows and columns of different parity - a blue cell, otherwise the sign in this cell will not change after all operations. Therefore, the number of red cells in each even row is equal to the number of even columns, and the number of blue cells is equal to the number of odd columns in the table. There is at least one even row X, so there is an even number of odd columns in the table. But the number of red cells in each odd row (odd!) is equal to the number of odd columns, which is an even number - a contradiction with the fact that there is at least one odd column. Therefore, it is impossible to manage with fewer than 100 operations.
Instructions. Correct example for 100 operations with explanations: 2 points. Proof of the minimality of $100: 5$ points.
|
100
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. Anya wrote down 100 numbers in her notebook. Then Sonya wrote down in her notebook all the pairwise products of the numbers written by Anya. Artem noticed that there were exactly 2000 negative numbers in Sonya's notebook. How many zeros did Anya initially write down in her notebook?
|
Answer: 10 zeros.
Solution: Let Anya write down $n$ positive numbers, $m$ negative numbers, and $100-n-m$ zeros in her notebook. Then, by the condition, $n m=2000$, since a negative number can only be obtained by multiplying numbers of different signs.
Let's list all the divisors of the number $2000=2^{4} * 5^{3}$:
| | $2^{0}$ | $2^{1}$ | $2^{2}$ | $2^{3}$ | $2^{4}$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $5^{0}$ | 1 | 2 | 4 | 8 | 16 |
| $5^{1}$ | 5 | 10 | 20 | 40 | 80 |
| $5^{2}$ | 25 | 50 | 100 | 200 | 400 |
| $5^{3}$ | 125 | 250 | 500 | 1000 | 2000 |
They can be paired such that the product of each pair is 2000. Let's write them in pairs and list their sums:
| 1 | 2 | 4 | 5 | 8 | 10 | 16 | 20 | 25 | 40 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2000 | 1000 | 500 | 400 | 250 | 200 | 125 | 100 | 80 | 50 |
| 2001 | 1002 | 504 | 405 | 258 | 210 | 141 | 120 | 105 | 90 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
Only in one of these cases is the sum less than 100, so this is the case that is realized. Thus, there are 90 non-zero numbers and 10 zeros.
Criteria: Only the answer - 0 points.
Only the answer with verification - 1 point.
Not all cases of divisors are considered - no more than 3 points.
No explanation of why there are no other divisors - deduct 1 point.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone has first arrived and then left?
|
Answer: 120
Solution: Consider any pair of friends. Their "friendship" was counted exactly once, as it was included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Similarly, each "enmity" will be counted exactly once by the person who left earlier. Therefore, after everyone has left, the sum of the numbers on the door will be increased by the total number of "enmities." In total, the overall sum of the numbers on the door will be equal to the sum of the total number of friendships and enmities, which is precisely the number of pairs of people who arrived, i.e., $16 * 15 / 2 = 120$.
Criteria: only answer - 0 points.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. In a school, there are 1000 students and 35 classes. On the forehead of each student, the number of students in their class is written. What can the sum of the reciprocals of these numbers be? List all the options and prove that there are no others. Recall that the reciprocal of a number $a$ is the number $1 / a$.
|
Answer: 35.
Solution: Let there be a people in the class, then the sum of the fractions corresponding to the numbers from this class is 1 (a fractions, each equal to 1/a). There are 35 classes in total. Therefore, the total sum is 35.
Criteria:
Answer - 0 points.
Answer with examples - 0 points.
The idea of partitioning into classes and calculating sums separately for each class without further progress - 5 points.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Arseny sat down at the computer between 4 and 5 PM, when the hour and minute hands were pointing in opposite directions, and got up from it on the same day between 10 and 11 PM, when the hands coincided. How long did Arseny sit at the computer?
|
Solution: Let's see where the hands will be 6 hours after Arseny sat down at the computer. The minute hand will go around the clock 6 times and return to its place. The hour hand will move exactly half a circle. Therefore, the angle between the hands will change by 180 degrees, i.e., the hands will coincide. It is obvious that there is exactly one moment between 22 and 23 hours when the hands coincide, so this will be the time when Arseny gets up from the computer. Thus, Arseny sat at the computer for 6 hours.
## Criteria:
Only the answer - 1 point.
If in a similar solution it is not noted that such a moment between 22 and 23 is exactly one, deduct 1 point.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. Several seventh-graders were solving problems. The teacher does not remember how many children there were and how many problems each of them solved. However, he remembers that, on the one hand, each solved more than a fifth of what the others solved, and on the other hand, each solved less than a third of what the others solved. How many seventh-graders could there be? Find all the options and prove that there are no others.
|
Answer: 5 seventh-graders.
Solution: Let one seventh-grader solve $a$ problems, and the rest solve $S - a$. Then
$$
\begin{gathered}
a < (S-a) / 3 \\
3a < S - a \\
4a < S \\
a < S / 4
\end{gathered}
$$
Similarly,
$$
\begin{gathered}
(S-a) / 5 < a \\
S - a < 5a \\
S < 6a \\
S / 6 < a
\end{gathered}
$$
Thus, if there are 4 or fewer students, and each solved less than a quarter of all problems, then together they solved fewer than all the problems, which is impossible. Similarly, if there are 6 or more students, and each solved more than a fifth of all problems, then together they solved more than all the problems, which is also impossible. Therefore, there must be 5 students in total.
## Criteria:
Answer - 0 points.
Solved under the assumption that each solved more than a fifth of all problems and less than a quarter of all problems - 3 points.
|
5
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. In a competition, there are 2018 Dota teams, all of different strengths. In a match between two teams, the stronger one always wins. All teams paired up and played one game. Then they paired up differently and played another game. It turned out that exactly one team won both games. How could this be
|
Solution: Let's number the teams in ascending order of strength from 1 to 2018. In the first round, we will have the matches 1 - 2, $3-4, \ldots, 2017$ - 2018, and in the second round - $2018-1, 2-3, 4$ - 5, ..., 2016 - 2017. It is obvious that only the team with the number 2018 will win in both rounds.
## Criteria:
Any correct example without explanation - 7 points.
|
2018
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Eleven of the best football teams each played one match against each other. It turned out that each team scored 1 goal in their first match, 2 goals in their second match, ..., 10 goals in their tenth match. What is the maximum number of matches that could have ended in a draw?
|
Solution: According to the condition, each team scored 1 goal in the first game. In the case of a draw, it also conceded 1 goal. Then for the other team, this match was also the first. Since the number of teams is odd, they cannot be paired. Therefore, at least one of the teams played a non-draw in their first match. The same is true for the second, third, ..., tenth matches. Since each match is counted twice, there were at least 5 non-draw matches, and then the number of draws was no more than $11 \cdot 10 / 2 - 5 = 50$.
Let's provide an example of a tournament with 50 draws. Number the teams from 1 to 11. Form 5 pairs of adjacent teams (leaving the 11th team without a pair). First, the teams in the pairs play against each other, and then the 11th team plays against the 1st (with a score of $1:2$). Again, we pair the teams, this time excluding the 1st, and let the teams in the pairs play again (resulting in 5 draws). Now arrange the teams in a circle with a step of 2, specifically in the order 1357911246810 (each team has two new neighbors). Applying the same match scheme, we get another 11 games, one of which is non-draw. Then apply this scheme for steps of 3, 4, and 5.
We can even write out a specific table:
| All play the first game, all draws | $12,34,56,78,910$ |
| :--- | :--- |
| 11 - first game, 1 - second game | 111 |
| All play the second game, all draws | $23,45,67,89,1011$ |
| All play the third game, all draws | $13,57,911,24,68$ |
| 10 - third game, 1 - fourth game | 101 |
| All play the fourth game, all draws | $35,79,112,46,810$ |
| All play the fifth game, all draws | $14,710,25,811,36$ |
| 9 - fifth game, 1 - sixth game | 91 |
| All play the sixth game, all draws | $47,102,58,113,69$ |
| All play the seventh game, all draws | $15,92,610,37,114$ |
| 8 - seventh game, 1 - eighth game | 81 |
| All play the eighth game, all draws | $59,26,103,711,48$ |
| All play the ninth game, all draws | $16,115,104,93,82$ |
| 7 - ninth game, 1 - tenth game | 71 |
| All play the tenth game, all draws | $611,510,49,38,27$ |
## Criteria:
Only the answer - 0 points.
Only the estimate that there are no more than $50-2$ points.
Only an example with 50 draws - 3 points.
There is an estimate and a correct idea for the example, but there are inaccuracies - 5 or 6 points depending on the size of the inaccuracies.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. In triangle $\mathrm{ABC}$, the bisectors of angles $\mathrm{BAC}$ and $\mathrm{BCA}$ intersect sides ВС and АВ at points К and Р, respectively. It is known that the length of side АС is equal to the sum of the lengths of segments АР and СК. Find the measure of angle $\mathrm{ABC}$.
|
Answer: $60^{\circ}$.
Solution 1. Let the angles of triangle ABC be denoted by the corresponding letters $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$, and the intersection point of the angle bisectors by I. Reflect points P and K with respect to the angle bisectors AK and CP, respectively; their images will be points $\mathrm{P}^{\prime}$ and $\mathrm{K}^{\prime}$ on side $\mathrm{AC}$ such that $\mathrm{AP}^{\prime}=\mathrm{AP}, \mathrm{CK}^{\prime}=C K$. Since $\mathrm{AP}^{\prime}+\mathrm{CK}^{\prime}=\mathrm{AP}+\mathrm{CK}=\mathrm{AC}$, points $\mathrm{P}^{\prime}$ and $\mathrm{K}^{\prime}$ coincide, and we denote this point as M. The measures of angles AMI=API=APC $=180^{\circ}-\mathrm{A}-$ C/2, CMI=CKI=CKA $=180^{\circ}-\mathrm{C}-\mathrm{A} / 2$. Their sum is $360^{\circ}-3(\mathrm{~A}+\mathrm{C}) / 2=360^{\circ}-$ $270^{\circ}+3 \mathrm{~B} / 2=90^{\circ}+3 \mathrm{~B} / 2=180^{\circ}$, from which $\mathrm{B}=60^{\circ}$. Therefore, the measure of angle $\mathrm{ABC}$ is 60.
Solution 2. Let the lengths of sides $\mathrm{AB}, \mathrm{BC}, \mathrm{AC}$ of the triangle be $c, a, b$ respectively, and find the lengths of all segments used in the problem using the property of the angle bisector. Then $A P=\frac{b c}{a+b}, C K=\frac{a b}{b+c}$, from which $A P+C K=\frac{b c}{a+b}+\frac{a b}{b+c}=\frac{b^{2} c+b c^{2}+a^{2} b+a b^{2}}{a b+a c+b c+b^{2}}=A C=b$. Simplify both sides by $b$, multiply by the denominator, combine like terms, and we get: $b^{2}=a^{2}+c^{2}-a c$. By the cosine rule, $b^{2}=a^{2}+c^{2}-2 a c \cdot \operatorname{Cos} B$, from which $\operatorname{Cos} B=\frac{1}{2}$ and $B=\angle A B C=60^{\circ}$.
Grading criteria. In the first solution ($\bullet$) The idea of reflecting points P and K with respect to the angle bisectors AK and CP is present: 1 point, ($\cdot$) It is shown that the images under such reflection will give a common point on side AC: another 1 point. In the second solution ($\cdot$) AP and CK are found: 3 points. ($\cdot$) The equality $b^{2}=a^{2}+c^{2}-a c$ is obtained: another 2 points. ($\cdot$) From this, $\operatorname{Cos} B=\frac{1}{2}$ is derived: 2 points.
($\cdot$) A special case with a certain arrangement of radii is considered: 2 points.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. On some cells of a rectangular board of size 101 by 99, there is one turtle each. Every minute, each of them simultaneously crawls to one of the cells adjacent to the one they are in, by side. In doing so, each subsequent move is made in a direction perpendicular to the previous one: if the previous move was horizontal - to the left or right, then the next one will be vertical - up or down, and vice versa. What is the maximum number of turtles that can move around the board indefinitely such that at each moment, no more than one turtle is in any cell?
|
Answer: 9800.
Solution. Examples of unlimited movement on the board of 9800 turtles.
Example 1. Place the reptiles in the cells of a rectangle consisting of cells at the intersection of the 98 lower horizontals and 100 leftmost verticals. They will all move in the same way: first all to the right, then all up, then all to the left, and then all down.
After 4 moves, the situation will coincide with the initial one, so the movement can continue in a permitted manner indefinitely.
Example 2. The initial placement of turtles is the same as in Example 1, but the movement is organized such that they are divided into 2x2 square cells, in each of which they simultaneously move clockwise.
We will prove that it is impossible to properly place more than 9800 turtles on a 101x99 board. Suppose the contrary, that the turtles are placed on the board in some way such that they have the possibility to move around the board indefinitely in the manner specified in the condition and not end up on the same cell in a quantity of two or more simultaneously. Color the cells of the board in a checkerboard pattern, with the lower left cell, as usual, being black. This will result in 5000 black and 4999 white cells. Among the black cells, we will call the cells with both vertical and horizontal coordinates being odd "odd" cells, and those with both coordinates being even "even" cells. Note that for white cells, one of these coordinates is even, and the other is odd. There will be $51 \cdot 50=2550$ odd black cells and $50 \cdot 49=2450$ even black cells. Clearly, at any moment, the number of turtles on even black cells should not exceed 2450.
Note that when two consecutive moves are performed, the parity of both coordinates of each turtle changes, so after two moves, all turtles from odd black cells will move to even black cells and vice versa, while turtles from white cells will move back to white cells. Consequently, at any time, the number of turtles on all black cells does not exceed $2 \cdot 2450=4900$.
Finally, when one move is performed, turtles from black cells move to white cells and vice versa, so the total number of turtles on the entire board cannot exceed $2 \cdot 4900=9800$.
Remark. In fact, we have proven that if there are more than 9800 turtles on the board, they will be able to make no more than two moves without ending up on the same cell in a quantity of two or more simultaneously.
Grading criteria. Proof of the maximality of the number 9800: 6 points. Construction of an example for 9800 turtles: 1 point. Attempt to prove any other estimate except 9800: 0 points.
|
9800
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. In each cell of a $5 \times 5$ table, one of the numbers $1,2,3,4,5$ is written in such a way that each row, each column, and each of the two diagonals of the table contains each of the numbers from 1 to 5. What is the maximum value that the sum of the five numbers written in the cells marked with dots on the diagram can take?
|
Answer: 22.
Solution. If all 4 numbers marked with a dot and not in the top right corner are different, then the sum of all numbers marked with a dot does not exceed $5+5+4+3+2=19$. Let's further assume that among these numbers there are two equal ones, and consider three possible arrangements of their positions. Denote the columns of the table from left to right by the letters $a, b, c, d, e$, and the rows by the numbers from 1 to 5. The diagonal from corner $a 1$ to corner $e 5$ will be called the main diagonal, and the diagonal from corner $a 5$ to corner $e 1$ will be called the secondary diagonal. The pair of equal numbers will be denoted by $a$.
1) The numbers in cells $a 4$ and $d 1$ are equal. The remaining numbers equal to $a$ cannot be placed in the 1st and 4th rows and in columns $a$ and $d$, so on the secondary diagonal, $a$ can only be in cell $c 3$. The two remaining numbers $a$ can only be in cells $b 5$ and $e 2$.
2) The numbers in cells $b 3$ and $c 2$ are equal. The remaining numbers equal to $a$ cannot be placed in the 2nd and 3rd rows and in columns $b$ and $c$, so on the diagonals, the number $a$ can only be in the corner cells and cell $d 4$.
The corner cells cannot contain two equal numbers, as any pair of corner cells lies either in the same row, column, or diagonal, so one $a$ is in cell $d 4$. It excludes cells on the main diagonal, row 4, and column $d 4$, so the two remaining numbers $a$ can only be in cells $a 5$ and $e 1$ on the secondary diagonal, which is impossible.
3) The numbers in cells $a 4$ and $c 2$ are equal. The remaining numbers equal to $a$ cannot be placed in the 2nd and 4th rows and in columns $a$ and $c$, so on the diagonals, the numbers $a$ can only be in the corner cells $e 1$ and $e 5$. These cells lie in the same column, which is impossible.
4) The numbers in cells $a 4$ and $b 3$ are equal. The remaining numbers equal to $a$ cannot be placed in the 3rd and 4th rows and in columns $a$ and $b$, and on the main diagonal, the number $a$ must be in cell $e 5$. Then on the secondary diagonal, it is in cell $d 2$, and on the first row, it is in cell $c 1$.
The conducted analysis shows that three identical numbers or two pairs of equal numbers among the 4 numbers marked with a dot and not in the top right corner cannot exist because the permissible configurations of cases 1) and 4) cannot occur simultaneously. Therefore, the sum of these four numbers does not exceed $5+5+4+3=17$, and the sum of the five numbers marked with dots does not exceed $17+5=22$. The upper bound has been obtained.
Example of number placement satisfying the condition with a sum of 22:
| 3 | 4 | 1 | 2 | 5 |
| :--- | :--- | :--- | :--- | :--- |
| 5 | 1 | 3 | 4 | 2 |
| 4 | 5 | 2 | 1 | 3 |
| 2 | 3 | 4 | 5 | 1 |
| 1 | 2 | 5 | 3 | 4 |
Grading criteria. Any attempt to prove an incorrect answer: 0 points. Correct answer with an example: 3 points. Proof of the maximum sum of 22: 4 points. If the enumeration in the proof is incomplete: points are reduced proportionally.
|
22
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. A bus with programmers left Novosibirsk for Pavlodar. When it had traveled 70 km, Pavel Viktorovich set off from Novosibirsk in a car along the same route, and caught up with the programmers in Karasuk. After that, Pavel drove another 40 km, while the bus traveled only 20 km in the same time. Find the distance from Novosibirsk to Karasuk, given that both the car and the bus traveled at constant speeds.
|
Solution. Since by the time the car has traveled 40 km, the bus has traveled half that distance, its speed is exactly half the speed of the car. However, when the bus has traveled 70 km after the car's departure, the car will have traveled 140 km and will just catch up with the bus. According to the problem, this happened in Karasuk, so 140 km is the answer.
Criteria. Only the answer -1 point.
Answer with verification (for example, for specific speeds) - 2 points. Proven that the car's speed is twice the speed of the bus - 3 points.
|
140
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. In triangle $A B C$, a point $D$ is marked on side $A C$ such that $B C = C D$. Find $A D$, given that $B D = 13$, and angle $C A B$ is three times smaller than angle $C B A$.
|
Solution. Let $\angle C A B=x$. Then $\angle C B A=3 x$ and $\angle A C B=180^{\circ}-4 x$. According to the problem, triangle $B C D$ is isosceles, so $\angle C D B=\angle C B D=\left(180^{\circ}-\angle B C D\right) / 2=2 x$. Therefore, $\angle D B A=\angle A B C-\angle D B C=3 x-2 x=x=\angle D A B$. Hence, triangle $A B D$ is isosceles, and $A D=B D=13$.
|
13
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There are 5 pieces of transparent glass of the same square shape and the same size. Each piece of glass is conditionally divided into 4 equal parts (right triangles) by its diagonals, and one of these triangles is painted with an opaque paint of its individual color, different from the colors of the painted parts of the other glasses. Then all these glasses are stacked on top of each other (with precise alignment of edges and vertices) with the painted parts facing up. How many different ways are there to stack the glasses so that the entire stack ends up completely opaque in the vertical direction.
(12 points)
|
Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then the layouts (with a fixed vertical order of glasses) will differ from each other by the rotation of each of the subsequent 4 upper glasses by $0^{\circ}, 90^{\circ}, 180^{\circ}$, and $270^{\circ}$ relative to the fixed bottom glass. Therefore, we will get a total of $4^{4}=256$ layout options with a fixed vertical order of glasses. Adding now all possible vertical permutations of the five glasses (5! = 120 options), we get the total number of possible glass layouts in a stack: $5!4^{4}=120 \cdot 256=30720$ pieces. But not all these layouts meet the condition of the problem. The condition of the problem is satisfied only by those layouts in which the entire stack turns out to be vertically opaque. Consider columns of triangles located above each of the four fixed triangles of the bottom glass. The condition of the problem will be met when each of these four columns is opaque. One of these columns is already opaque by default (the one standing on the bottom shaded triangle). Consider the ordered set (vector) of possible angles of rotation (for definiteness, clockwise) of the shaded triangles on each of the 4 upper glasses relative to the bottom shaded triangle: $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)$, where $\alpha_{k} \in\left\{0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}\right\}, k=1,2,3,4$.
All "columns" of glass triangles will be opaque if in this set of rotation angles there is at least one angle $90^{\circ}$, at least one angle $180^{\circ}$, and at least one angle $270^{\circ}$ (while the angle $0^{\circ}$ does not have to appear, although it will not hurt if it does appear). To count the total number of such sets (and thus the number of layouts with a fixed vertical order of glasses), we will divide them into four groups:
- sets in which all rotation angles are different, i.e., all $\alpha_{k}$ are pairwise different; there are 4! = 24 such sets (due to permutations);
- sets in which the rotation angles $180^{\circ}$ and $270^{\circ}$ appear once each, and the rotation angle $90^{\circ}$ appears twice; there are $6 \cdot 2=12$ such sets (due to permutations) (here 6 is the number of positions in the vector for two $90^{\circ}$ angles, and 2 is the two permutations of the angles $180^{\circ}$ and $270^{\circ}$ in the remaining two positions);
- sets in which the rotation angles $90^{\circ}$ and $270^{\circ}$ appear once each, and the rotation angle $180^{\circ}$ appears twice; similarly, there are $6 \cdot 2$ = 12 such sets (due to permutations);
- sets in which the rotation angles $90^{\circ}$ and $180^{\circ}$ appear once each, and the rotation angle $270^{\circ}$ appears twice; similarly, there are $6 \cdot 2=12$ such sets (due to permutations).
In the end, we get a total of $24+12+12+12=60$ ordered sets of rotation angles that meet the required condition. This is the total number of layouts with a fixed vertical order of glasses, in which the final stack turns out to be opaque. Finally, by permuting the glasses in $5!=120$ ways, we get the total number of $120 \cdot 60=7200$ layouts in which the final stack turns out to be opaque.
Answer: 7200 ways.
|
7200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find a natural number that has six natural divisors (including one and the number itself), two of which are prime, and the sum of all its natural divisors is 78.
|
Solution: The desired natural number $n$ can be represented as $n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}, 12,1+p_{2}+p_{2}^{2}>7$, then no factorization of the number 78 into a product of two natural factors fits $78=1 \cdot 78,78=2 \cdot 39,78=3 \cdot 26,78=6 \cdot 13$ up to the order).
2) $\alpha_{1}=2, \alpha_{2}=1, n=p_{1}^{2} \cdot p_{2}$. Since the sum of all divisors is 78, then $\left(1+p_{1}+p_{1}^{2}\right)\left(1+p_{2}\right)=78$. Since $13,1+p_{1}+p_{1}^{2}>3$, we have
a) $1+p_{1}+p_{1}^{2}=6,1+p_{2}=13$, there are no natural solutions.
b) $1+p_{1}+p_{1}^{2}=13,1+p_{2}=6$, or $p_{1}=3, p_{2}=5, n=3^{2} \cdot 5=45$.
Answer: 45.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The automatic line for processing body parts initially consisted of several identical machines. The line processed 38880 parts daily. After the production was modernized, all the machines on the line were replaced with more productive but also identical ones, and their number increased by 3. The automatic line began to process 44800 parts per day. How many parts did each machine process daily initially?
|
Solution: Let $x$ be the number of machines before modernization, $y$ be the productivity of each machine, i.e., the number of parts processed per day, and $z$ be the productivity of the new machines. Then we have $x y = 38880 = 2^{5} \cdot 3^{5} \cdot 5, (x+3) z = 44800 = 2^{8} \cdot 5^{2} \cdot 7, x > 1, y \frac{38880 \cdot 3}{5920} = \frac{2^{5} \cdot 3^{6} \cdot 5}{2^{5} \cdot 5 \cdot 37} = \frac{3^{6}}{37} = 19 \frac{26}{37}, \quad x \geq 20$. Since $(x+3) z = 2^{8} \cdot 5^{2} \cdot 7$, $x$ is not divisible by 3, and $x = 2^{\alpha} \cdot 5^{\beta}$, where $\alpha \in \{0,1,2,3,4,5\}, \beta \in \{0,1\}$.
1) $\beta=0, x=2^{\alpha}, \alpha \in \{5\}$. When $\alpha=5$ we have $x=32, x+3=35, y=3^{5} \cdot 5=1215, z=2^{8} \cdot 5=1280$
2) $\beta=1, x=2^{\alpha} \cdot 5, \alpha \in \{2,3,4,5\}$.
When $\alpha=2$ we have $x=20, x+3=23$, which is impossible.
When $\alpha=3$ we have $x=40, x+3=43$, which is impossible.
When $\alpha=4$ we have $x=80, x+3=83$, which is impossible.
When $\alpha=5$ we have $x=160, x+3=163$, which is impossible.
Answer: 1215.
|
1215
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The numbers $u, v, w$ are roots of the equation $x^{3}-3 x-1=0$. Find $u^{9}+v^{9}+w^{9}$. (12 points)
|
Solution. $\quad$ According to Vieta's theorem, $u+v+w=0, uv+vw+uw=-3, uvw=1$. Consider the sequence $S_{n}=u^{n}+v^{n}+w^{n}$. We have $S_{0}=3, S_{1}=0$. Let's find $S_{2}$: $S_{2}=u^{2}+v^{2}+w^{2}=(u+v+w)^{2}-2(uv+vw+uw)=6. \quad$ Since $\quad u^{3}=3u+1, v^{3}=$ $3v+1, w^{3}=3w+1, \quad$ then $S_{3}=u^{3}+v^{3}+w^{3}=3S_{1}+S_{0}$. Further, $u^{n}=3u^{n-2}+u^{n-3}, v^{n}=$ $3v^{n-2}+v^{n-3}, w^{n}=3w^{n-2}+w^{n-3}$, then $S_{n}=3S_{n-2}+S_{n-3}$. We find $S_{3}=3, S_{4}=18, S_{5}=$ $15, S_{6}=57, S_{7}=63, S_{9}=246$
## Answer: 246
|
246
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In the laboratory, there are flasks of two sizes (volume $V$ and volume $V / 3$) in a total of 100 pieces, with at least 2 flasks of each size. The lab assistant randomly selects two flasks in sequence, and fills the first one with a 70% salt solution, and the second one with a 40% salt solution. Then, he pours the contents of these two flasks into one dish and determines the percentage of salt in it. For what minimum number of large flasks $N$ will the event "the percentage of salt in the dish is between $50\%$ and $60\%$ inclusive" occur less frequently than the event "when two fair coins are tossed, one head and one tail appear (in any order)"? Justify your answer. (16 points)
|
Solution. Let $N$ be the number of large flasks in the laboratory, $N=2,3, \ldots, 98$, $n=100-N$ be the number of small flasks in the laboratory, $n=2,3, \ldots, 98$, and $\mathrm{P}(A)$ be the probability of the event $A=\{$ the salt content in the dish is between $50 \%$ and $60 \%$ inclusive\}. It is necessary to find the smallest $N$ such that $\mathrm{P}(A)45<N<55$. Therefore, $N_{\text {min }}=46$.
Answer: when $N=46$.
|
46
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The fraction $\frac{1}{5}$ is written as an infinite binary fraction. How many ones are there among the first 2022 digits after the decimal point in such a representation? (12 points)
|
Solution. The smallest number of the form $2^{n}-1$ divisible by 5 is 15. Then
$$
\frac{1}{5}=\frac{3}{15}=\frac{3}{16-1}=\frac{3}{2^{4}-1}=\frac{3}{16} \cdot \frac{1}{1-2^{-4}}=\left(\frac{1}{16}+\frac{1}{8}\right)\left(1+2^{-4}+2^{-8}+2^{-12}+\cdots\right)=
$$
$\left(2^{-3}+2^{-4}\right)\left(1+2^{-4}+2^{-8}+2^{-12}+\cdots\right)=0,001100110011 \ldots=0,(0011) \quad$ in $\quad$ binary. Since $2022=4 \cdot 505+2$, among the first 2022 digits after the decimal point, there will be 1010 ones.
## Answer: 1010.
|
1010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. There is a cube fixed on legs, and six different paints. In how many ways can all the faces of the cube be painted (each in one color, not all paints have to be used) so that adjacent faces (having a common edge) are of different colors? (16 points)
|
Solution. Let's consider 4 cases of coloring a cube.
1) The top and bottom faces are the same color, and the left and right faces are the same color. We choose the color for the top and bottom faces in 6 ways, then the color for the left and right faces in 5 ways, then the color for the front face in 4 ways, and the color for the back face in 4 ways. In total, $6 \cdot 5 \cdot 4 \cdot 4=480$ ways.
2) The top and bottom faces are the same color, and the left and right faces are different colors. We choose the color for the top and bottom faces in 6 ways, then the color for the left face in 5 ways, then the color for the right face in 4 ways, then the color for the front face in 3 ways, and the color for the back face in 3 ways.
In total, $6 \cdot 5 \cdot 4 \cdot 3 \cdot 3=1080$ ways.
3) The top and bottom faces are different colors, and the left and right faces are the same color. We choose the color for the left and right faces in 6 ways, then the color for the top face in 5 ways, then the color for the bottom face in 4 ways, then the color for the front face in 3 ways, and the color for the back face in 3 ways.
In total, $6 \cdot 5 \cdot 4 \cdot 3 \cdot 3=1080$ ways.
4) The top and bottom faces are different colors, and the left and right faces are different colors. We choose the color for the top face in 6 ways, then the color for the bottom face in 5 ways, then the color for the left face in 4 ways, then the color for the right face in 3 ways, then the color for the front face in 2 ways, and the color for the back face in 2 ways.
In total, $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 2=1440$ ways.
Finally, we have $480+1080+1080+1440=4080$.
## Answer: 4080.
|
4080
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Point $M$ lies on the leg $A C$ of the right triangle $A B C$ with a right angle at $C$, such that $A M=2, M C=16$. Segment $M H$ is the altitude of triangle $A M B$. Point $D$ is located on the line $M H$ such that the angle $A D B$ is $90^{\circ}$, and points $C$ and $D$ lie on the same side of the line $A B$. Find the length of segment $B L$, if $L$ is the intersection of $B D$ and $A C$, and the tangent of angle $A C H$ is 1/18. (16 points)
|
Solution. 1. A circle can be circumscribed around quadrilateral $A B C D$ with diameter $A B$ (angles $A D B$ and $A C B$ are right angles). Then $\angle A B D=\angle A C D$,
$$
\angle H A D=90^{\circ}-\angle A B D, \angle A D H=\angle A B D=\angle A C D
$$
Triangles $A C D$ and $A D M$ are similar, and $\frac{A D}{A C}=\frac{A M}{A D}=$ $\frac{M D}{D C}, A D^{2}=A C \cdot A M=2 \cdot 18=36, A D=6$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then the layouts (with a fixed vertical order of glasses) will differ from each other by the rotation of each of the subsequent four upper glasses by $0^{\circ}, 90^{\circ}, 180^{\circ}$, and $270^{\circ}$ relative to the fixed bottom glass. Therefore, we will get a total of $4^{4}=256$ layout options with a fixed vertical order of glasses. Adding now all possible vertical permutations of the five glasses ($5!=120$ options), we get the total number of possible glass layouts in a stack: $5!4^{4}=120 \cdot 256=30720$ pieces. But not all these layouts meet the condition of the problem. The layouts that meet the condition of the problem are only those in which the entire stack turns out to be vertically opaque. Consider the columns of triangles located above each of the four fixed triangles of the bottom glass. The condition of the problem will be met when each of these four columns is opaque. One of these columns is already opaque by default (the one standing on the bottom shaded triangle). Consider the ordered set (vector) of possible angles of rotation (for definiteness, clockwise) of the shaded triangles on each of the 4 upper glasses relative to the bottom shaded triangle: $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)$, where $\alpha_{k} \in\left\{0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}\right\}, k=1,2,3,4$.
All "columns" of glass triangles will be opaque if in this set of rotation angles there is at least one angle $90^{\circ}$, at least one angle $180^{\circ}$, and at least one angle $270^{\circ}$ (while the angle $0^{\circ}$ does not have to appear, although it will not hurt if it does appear). To count the total number of such sets (and thus the number of layouts with a fixed vertical order of glasses), we will divide them into four groups:
- sets in which all rotation angles are different, i.e., all $\alpha_{k}$ are pairwise different; there are a total of $4!=24$ such sets (due to permutations);
- sets in which the rotation angles $180^{\circ}$ and $270^{\circ}$ appear once each, and the rotation angle $90^{\circ}$ appears twice; there are a total of $6 \cdot 2=12$ such sets (here 6 is the number of positions in the vector for two angles of $90^{\circ}$, and 2 is the two permutations of the angles $180^{\circ}$ and $270^{\circ}$ in the remaining two positions);
- sets in which the rotation angles $90^{\circ}$ and $270^{\circ}$ appear once each, and the rotation angle $180^{\circ}$ appears twice; similarly, there are a total of $6 \cdot 2=12$ such sets (due to permutations);
- sets in which the rotation angles $90^{\circ}$ and $180^{\circ}$ appear once each, and the rotation angle $270^{\circ}$ appears twice; similarly, there are a total of $6 \cdot 2=12$ such sets (due to permutations).
In the end, we get a total of $24+12+12+12=60$ ordered sets of rotation angles that meet the required condition. This is the total number of layouts with a fixed vertical order of glasses, in which the final stack turns out to be opaque. Finally, by permuting the glasses in $5!=120$ ways, we get the total number of $120 \cdot 60=7200$ layouts in which the final stack turns out to be opaque.
|
Answer: 7200 ways.
#
|
7200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then different layouts (with a fixed vertical order of glasses) will differ from each other by the rotation of each of the subsequent 4 upper glasses by $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$ relative to the fixed bottom glass. Therefore, we will get a total of $3^{4}=81$ layout options with a fixed vertical order of glasses. Taking into account all possible vertical permutations of the five glasses (5! = 120 options), we get the total number of possible glass layouts in a stack: 5! $3^{4}=120 \cdot 81=9720$ pieces. But not all these layouts meet the condition of the problem. The condition of the problem is satisfied by all layouts except those in which the entire stack turns out to be completely vertically opaque. Let's first find the number of such layouts.
For this, consider columns of isosceles triangles (parts located above each of the three fixed isosceles triangles of the bottom glass). Let's determine the number of such layouts in which each of these four columns is opaque. One of these columns is already opaque by default (the one standing on the bottom shaded isosceles triangle). Consider the ordered set (vector) of possible rotation angles (for definiteness, clockwise) of the shaded isosceles triangles on each of the 4 upper glasses relative to the bottom shaded equilateral triangle: $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)$, where $\alpha_{k} \in\left\{0^{\circ}, 120^{\circ}, 240^{\circ}\right\}$, $k=1,2,3,4$.
All "columns" of glass isosceles triangles will be opaque if this set of rotation angles necessarily includes at least one $120^{\circ}$ and at least one $240^{\circ}$ (while the angle $0^{\circ}$ is not required to appear, although it will not hurt if it does appear). The other two places in the set can be occupied by any of the three possible values $\left\{0^{\circ}, 120^{\circ}, 240^{\circ}\right\}$, i.e., $0^{\circ}$ and $120^{\circ}$, or $0^{\circ}$ and $240^{\circ}$, or $120^{\circ}$ and $240^{\circ}$, as well as $0^{\circ}$ and $0^{\circ}$, or $120^{\circ}$ and $120^{\circ}$, or $240^{\circ}$ and $240^{\circ}$. Therefore, to count the total number of such sets (and thus the number of layouts with a fixed vertical order of glasses), we will divide them into the following six groups:
- If the two additional values are $0^{\circ}$ and $120^{\circ}$, then we get sets in which the rotation angles $0^{\circ}$ and $240^{\circ}$ appear once, and the angle $120^{\circ}$ appears twice. The total number of such sets (due to permutations) will be $6 \cdot 2=12$ (here 6 is the number of ways to choose places in the vector for two angles of $120^{\circ}$, and 2 is the number of ways to permute the angles $0^{\circ}$ and $240^{\circ}$ in the two remaining places);
- If the two additional values are $0^{\circ}$ and $240^{\circ}$, then we get sets in which the rotation angles $0^{\circ}$ and $120^{\circ}$ appear once, and the angle $240^{\circ}$ appears twice. Analogously to the previous case, the total number of such sets (due to permutations) will be $6 \cdot 2=12$;
- If the two additional values are $120^{\circ}$ and $240^{\circ}$, then we get sets in which the rotation angles $120^{\circ}$ and $240^{\circ}$ appear twice each. The total number of such sets (due to permutations) will be $6 \cdot 1=6$ (here 6 is the number of ways to choose two places out of four for placing two angles of $120^{\circ}$, and 1 is the only way to place two angles of $240^{\circ}$ in the two remaining places);
- If the two additional values are $0^{\circ}$ and $0^{\circ}$, then we get sets in which the rotation angles $120^{\circ}$ and $240^{\circ}$ appear once, and the angle $0^{\circ}$ appears twice. The total number of such sets (due to permutations) will be $6 \cdot 2=12$ (here 6 is the number of ways to choose two places out of four for placing two angles of $0^{\circ}$, and 2 is the number of ways to permute the angles $120^{\circ}$ and $240^{\circ}$ in the two remaining places);
- If the two additional values are $120^{\circ}$ and $120^{\circ}$, then we get sets in which the rotation angle $240^{\circ}$ appears once, and the angle $120^{\circ}$ appears three times. The total number of such sets will be $4 \cdot 1=4$ (here 4 is the number of ways to choose one place out of four for placing the angle $240^{\circ}$, and 1 is the only way to place the identical angles $120^{\circ}$ in the three remaining places);
- If the two additional values are $240^{\circ}$ and $240^{\circ}$, then we get sets in which the rotation angle $120^{\circ}$ appears once, and the angle $240^{\circ}$ appears three times. Analogously to the previous case, the total number of such sets will be $4 \cdot 1=4$.
In the end, we get a total of $12+12+6+12+4+4=50$ ordered sets of rotation angles that satisfy the required condition. This is the total number of layouts with a fixed vertical order of glasses in which the final stack turns out to be opaque. By permuting the glasses (in the vertical direction) in $5!=120$ ways, we get the total number of $120 \cdot 50=6000$ layouts in which the final stack turns out to be opaque. Finally, by subtracting the number of 6000 unwanted layouts from the total number of different layouts 9720, we finally get $9720-6000=3720$ ways to lay out a partially transparent stack of five glasses.
|
Answer: 3720 ways.
#
|
3720
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$
|
# Solution
Notice that 345 and $5 y^{2}$ are divisible by 5, so $3 x^{2}$ must also be divisible by 5. Therefore, $\quad x=5 t, t \in Z$. Similarly, $y=3 n, n \in Z$. After simplification, the equation becomes $5 t^{2}+3 n^{2}=23$. Therefore, $t^{2} \leq \frac{23}{5}$, $n^{2} \leq \frac{23}{3}$ or $|t| \leq 2,|n| \leq 2$. By trying the corresponding values of $t, n$, we get that $|t|=2,|n|=1$ or
$$
\left\{\begin{array}{c}
x_{1}=-10 \\
y_{1}=-3
\end{array},\left\{\begin{array}{c}
x_{2}=-10 \\
y_{2}=3
\end{array},\left\{\begin{array}{l}
x_{3}=10 \\
y_{3}=-3
\end{array},\left\{\begin{array}{c}
x_{4}=10 \\
y_{4}=3
\end{array}\right.\right.\right.\right.
$$
The maximum value of the expression $x+y$ is $10+3=13$.
Answer. 13.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The car traveled half of the distance at a speed of 60 km/h, then one third of the remaining distance at a speed of 120 km/h, and the remaining distance at a speed of 80 km/h.
Find the average speed of the car during this trip. Give your answer in km/h.
|
Solution: Let x hours be the time the car traveled at a speed of 60 km/h, then $60 x=\frac{s}{2}$. Let y hours be the time the car traveled at a speed of 120 km/h, then $120 y=\frac{s}{6}$. Let z hours be the time the car traveled at a speed of 80 km/h, then $80 z=\frac{s}{3}$. By definition $v_{cp}=\frac{s}{t_{\text{total}}}$. Then $v_{cp}=\frac{s}{x+y+z}=\frac{s}{\frac{s}{120}+\frac{s}{720}+\frac{s}{240}}=\frac{1}{\frac{1}{120}+\frac{1}{6 \cdot 120}+\frac{1}{2 \cdot 120}}=\frac{6 \cdot 120}{10}=72$.
Answer: 72.
|
72
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.
On a coordinate line, 16 points are marked and numbered from left to right. The coordinate of any point, except for the extreme points, is equal to half the sum of the coordinates of the two adjacent points. Find the coordinate of the fifth point if the first point has a coordinate of 2 and the sixteenth point has a coordinate of 47.
#
|
# Solution
Solution. Let $a, b$ and $c$ be the coordinates of three consecutive points (from left to right). Then $b=\frac{a+c}{2}$, which means the second point is the midpoint of the segment with endpoints at the neighboring points. This condition holds for any triple of consecutive points, meaning the distances between any neighboring points are the same. The distance between the extreme points is 47-2=45, and there are 15 equal intervals between them, so the distance between neighboring points is $45: 15=3$. There are 4 equal intervals of length 3 between the fifth and the first point, so the coordinate of the fifth point is $2+3 \cdot$ $4=14$.
Answer: 14
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.
In a 6-liter vessel, 4 liters of a 70% (by volume) sulfuric acid solution are poured, and in a second vessel of the same capacity, 3 liters of a 90% solution of sulfuric acid are poured. A certain amount of the solution is transferred from the second vessel to the first so that it becomes an $r-\%$ solution of sulfuric acid. Find the greatest integer value of $\mathrm{r}$ for which the problem has a solution.
|
# Solution.
Let $x$ liters of the solution be transferred from the second vessel to the first. Since it follows from the condition that $0 \leq x \leq 2$, to find the amount of pure acid in the new solution, we obtain the equation $2.8 + 0.9x = (4 + x) \frac{r}{100}$, from which $x = \frac{4r - 280}{90 - r}$. Now, considering that $x \in [0; 2]$, we conclude that the problem has a solution only when $r \in \left[70; \frac{230}{3}\right]$. Therefore, the largest integer $r$ for which the problem has a solution is 76.
Answer: 76
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.
Find the roots of the equation $f(x)=8$, if $4 f(3-x)-f(x)=3 x^{2}-4 x-3$ for any real value of $x$. In your answer, specify the product of the found roots. #
|
# Solution:
Notice that when $x$ is replaced by $3-x$, the expression $3-x$ changes to $x$. That is, the pair $f(x)$ and $f(3-x)$ is invariant under this substitution. Replace $x$ with $3-x$ in the equation given in the problem. We get:
$4 f(x) - f(3-x) = 3(3-x)^2 - 4(3-x) - 3 = 3x^2 - 14x + 12$. Express $f(x)$ from the system: $\left\{\begin{array}{l}4 f(3-x) - f(x) = 3x^2 - 4x - 3 \\ 4 f(x) - f(3-x) = 3x^2 - 14x + 12\end{array}\right.$. Multiply the second equation of the system by 4 and add it to the first, we get: $15 f(x) = 15x^2 - 60x + 45$; $f(x) = x^2 - 4x + 3$. Form the equation: $x^2 - 4x + 3 = 8 ; x^2 - 4x - 5 = 0$; its roots are $x_1 = -1 ; x_2 = 5$. The product of the roots is ( -5 ).
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.
In triangle $A B C$, the bisector $A L$ ( $L \in B C$ ) is drawn, and $M$ and $N$ are points on the other two bisectors (or their extensions) such that $M A=M L$ and $N A=N L, \angle B A C=50^{\circ}$.
Find the measure of $\angle M A N$ in degrees.
|
# Solution
We will use the auxiliary statements.
If the bisector $B K$ in triangle $A B C$ intersects the circumscribed circle at point $W$, then:
1) $A W=C W$ (since $\angle C A W=\angle C B W=\angle A B W=\angle A C W$, that is, triangle $A W C$ is isosceles and $A W=C W$).
2) Point $W$ lies on the perpendicular bisector of segment $A C$.
3) The bisector of the angle of a non-isosceles triangle and the perpendicular bisector of the opposite side intersect on the circumscribed circle.
Next, consider triangle $A B L (A B \neq B L)$, $M N$ is the perpendicular bisector of $A L$ and $B N$ is the bisector of angle $B$. Therefore, points $A, B, L, N$ lie on the same circle.
Similarly, points $A, C, L, M$ lie on the same circle.
Then:
$$
\begin{gathered}
\angle M A N=\angle L A M+\angle L A N=\angle L C M+\angle N B L= \\
=\frac{\angle C+\angle B}{2}=90^{\circ}-\frac{1}{2} \angle A=90^{\circ}-\frac{1}{2} \cdot 50^{\circ}=65^{\circ}
\end{gathered}
$$
Answer. 65.
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Find all values of the parameter $a$, for each of which the solution set of the inequality $\frac{x^{2}+(a+1) x+a}{x^{2}+5 x+4} \geq 0$ is the union of three non-overlapping intervals. In your answer, specify the sum of the three smallest integer values of $a$ from the obtained interval.
|
# Solution
Let's factorize the numerator and the denominator of the left part of the inequality, it will take the form: $\frac{(x+1)(x+a)}{(x+1)(x+4)} \geq 0$. There are five possible cases for the placement of the number ($-a$) relative to the numbers (-4) and (-1). In each case, the inequality is solved using the interval method on the number line (see Fig.2). Let's list the results of the investigation: 1 )
$-a > -1 ; a < 4 \quad x \in (-\infty ; -4) \cup (-4 ; -a] \cup (-1 ; +\infty)$ - does not fit. 2 )
$-a = -1 ; a = 1 \quad x \in (-\infty ; -4) \cup (-1 ; +\infty)$ - does not fit. 3 )
$-4 < -a < -1 ; 1 < a < 4 \quad x \in (-\infty ; -4) \cup (-4 ; -a] \cup (-1 ; +\infty)$ - does not fit. 4 )
$-a = -4 ; a = 4 \quad x \in (-\infty ; -4] \cup (-1 ; +\infty)$ - does not fit. 5 )
$-a < -4 ; a > 4 \quad x \in (-\infty ; -a] \cup (-4 ; -1) \cup (-1 ; +\infty)$ - fits. Thus, the answer to the problem satisfies $a \in (1 ; +\infty)$. The smallest integers that fall within this interval are 2, 3, and 4. $2 + 3 + 4 = 9$.
Answer: 9.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.
Calculate the number $8^{2021}$, find the sum of the digits in this number, and write down the result. Then, in the newly written number, find the sum of the digits and write down the result again. These actions were repeated until a single-digit number was obtained. Find this number.
|
# Solution.
Consider the natural powers of 8. Notice that even powers of the number 8 give a remainder of 1 when divided by 9, while odd powers (including the number \(8^{2021}\)) give a remainder of 8. Indeed, let's analyze the powers of 8:
\[
\begin{gathered}
8^{2}=(9-1)^{2}=9 n+1, n \in N \\
8^{3}=(9 n+1) \cdot 8=9 k+8, k \in N \\
8^{4}=(9 k+8) \cdot 8=9 k \cdot 8+8^{2}=9 k \cdot 8+9 n+1=9 \cdot s+1, s \in N
\end{gathered}
\]
Assuming that \(8^{2 m-1}=9 t+8, t \in N, m \in N, m \geq 2\), we get:
\[
\begin{aligned}
& 8^{2 m}=8^{2 m-1} \cdot 8=(9 t+8) \cdot 8=9 t \cdot 8+8^{2}=9 t \cdot 8+9 n+1=9 q+1, q \\
& \quad \in N
\end{aligned}
\]
\[
8^{2 m+1}=8^{2 m} \cdot 8=(9 q+1) \cdot 8=9 q \cdot 8+8
\]
Since an integer when divided by 9 gives the same remainder as the sum of its digits, the sum of the digits of the number \(8^{2021}\) and the sum of the digits of subsequent summation results, when divided by 9, have the same remainder of 8.
Answer. 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. We will call a ticket with a number from 0001 to 2014 excellent if the difference between some two adjacent digits of its number is 5. Find the number of excellent tickets.
|
Solution. The number of excellent tickets from 0001 to 2014 is equal to the number of excellent tickets from 0000 to 2014. First, let's calculate the number of non-excellent tickets from 0000 to 2014.
The number of non-excellent tickets from 0000 to 1999 can be found as follows. Let $a_{1} a_{2} a_{3} a_{4}$ be the number of a non-excellent ticket. Then, $a_{1}$ can be chosen as two digits: 0 and 1.
$a_{2}$ can be chosen as any digit except those forming pairs 0-5, 1-6, i.e., any of 9 digits.
$a_{3}, a_{4}$ can be chosen as any digit except those forming pairs 0-5, 1-6, 2-7, 3-8, 4-9, i.e., any of 9 digits.
In total, there will be $2 \cdot 9 \cdot 9 \cdot 9=1458$ tickets.
In tickets from 2000 to 2014, there are fourteen non-excellent tickets (all except 2005).
Thus, we get 1458+14=1472 - non-excellent, therefore, 2015-1472=543 excellent tickets from 0000 to 2014, which means there are also 543 excellent tickets from 0001 to 2014.
Answer: 543.
|
543
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (15 points) Pavel caught 32 crayfish and decided to sell them at the market. When part of his catch was bought, it turned out that the buyer paid 4.5 rubles less for each one than the number of crayfish that remained on the counter. At the same time, the boy earned the maximum amount of money possible. How much money did Pavel earn? How many crayfish did he sell?
|
Solution: Let $x$ be the number of crayfish left on the counter, then (32-x) crayfish were bought, ($x-4.5$) - the cost of one crayfish.
$(32-x)(x-4.5)$ - the cost of all crayfish.
$y=(32-x)(x-4.5)$
$x_{B}=18.25$.
$x_{1}=19, y_{1}=(32-19)(19-4.5)=13 * 14.5=188.5$.
$x_{2}=18, y_{2}=(32-18)(18-4.5)=14 * 13.5=189$. The maximum number is 14.
Answer: a) the maximum amount of money is 189 rubles;
b) sold 14 crayfish.
| Points | |
| :--- | :--- |
| 15 | The correct answer is obtained with justification. |
| 10 | An arithmetic error is made in an otherwise correct solution. |
| 5 | The problem is solved by trial and error, finding values of the variable that satisfy the condition. |
| 2 | There are some guesses. |
| 0 | The solution does not meet any of the criteria listed above. |
|
189
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (20 points) In an acute-angled triangle $\mathrm{ABC}$, a point $\mathrm{D}$ is chosen on side $\mathrm{BC}$ such that $\mathrm{CD}: \mathrm{DB}=2: 1$, and a point $\mathrm{K}$ is chosen on segment $\mathrm{AD}$ such that $\mathrm{AK}=\mathrm{CD}+\mathrm{DK}$. A line is drawn through point $\mathrm{K}$ and vertex $\mathrm{B}$, intersecting side $\mathrm{AC}$ at point $\mathrm{E}$. Triangle $\mathrm{AEK}$ is isosceles ($\mathrm{AE}=\mathrm{EK}$). Find the measure of angle $\mathrm{ADC}$ in degrees.
|
Solution: Extend SV beyond point V so that BN = BD. Draw NM || BE. NM intersects AD at point L. Draw segment $\mathrm{CH} \perp \mathrm{AD}$. Extend it to $\mathrm{P}$ such that $\mathrm{HP}=\mathrm{HC}, \mathrm{PN}|| \mathrm{AD}$.
In triangle DLN, segment $\mathrm{BK}$ is the midline, therefore, $\mathrm{DK}=\mathrm{KL}$ and thus $\mathrm{AL}=\mathrm{AK}-\mathrm{LK}=\mathrm{AK}-\mathrm{DK}=\mathrm{CD}$.
Triangle САР is isosceles (AH is the median and altitude), therefore $\angle \mathrm{PAD}=\angle \mathrm{DAC}$, but $\angle \mathrm{DAC}=\angle \mathrm{AKE}$ (by condition), $\angle \mathrm{AKE}=\angle \mathrm{ALM}$ (corresponding), $\angle \mathrm{ALM}=\angle \mathrm{NLD}$ (vertical), $\angle \mathrm{NLD}=\angle \mathrm{NLP}$ (alternate interior), hence, LN || AP, so ALNP is a parallelogram.
In the right triangle $\mathrm{CPN}$, the length $\mathrm{CN}=2 \mathrm{CD}=2 \mathrm{AL}=2 \mathrm{PN}$, i.e., $\angle \mathrm{PCN}=30^{\circ}$. Therefore, $\angle \mathrm{ADC}=60^{\circ}$.
Answer: $60^{0}$.
| Points | Grading Criteria |
| :---: | :---: |
| 20 | A well-reasoned and correctly executed solution to the problem. |
| 15 | The solution is correct and well-reasoned but contains an arithmetic error or is insufficiently justified. |
| 7 | The solution is correctly started, some intermediate results are obtained, but the further solution is incorrect or missing. |
| 0 | The solution does not meet the above requirements. |
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A farmer initially placed his produce in boxes with a capacity of 8 kg each, but one box was not fully loaded. Then the farmer repackaged all the produce into boxes with a capacity of 6 kg each, which required 8 more boxes, but in this case, one box was also not fully loaded. When all the produce was placed in boxes of 5 kg each, all the boxes were fully loaded, but it required an additional 5 boxes. How many kilograms did the farmer's produce weigh? Give your answer as a number without specifying the unit.
(5 points)
|
Solution. Let $x$ kg be the weight of the farmer's produce. Then $\quad 8(n-1)<x<8 n, \quad 6(n+7)<x<6(n+8)$, $5(n+13)=x, \Rightarrow 8(n-1)<5(n+13)<8 n, \quad 6(n+7)<5(n+13)<6(n+8)$,
$\Rightarrow 21 \frac{2}{3}<n<23, \quad n=22, \quad x=35 \cdot 5=175$.
Answer: 175.
|
175
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Solve the equation $8 \sin ^{4}(\pi x)-\sin ^{2} x=\cos ^{2} x-\cos (4 \pi x)$. In your answer, specify the sum of the roots that belong to the interval $[-1 ; 2]$.
(5 points)
|
Solution. Considering the basic trigonometric identity, we get
$8 \sin ^{4}(\pi x)-1+\cos (4 \pi x)=0 \quad \Rightarrow \quad 8 \sin ^{4}(\pi x)-2 \sin ^{2}(2 \pi x)=0 \quad \Rightarrow$
$\left(2 \sin ^{2}(\pi x)-2 \sin (\pi x) \cos (\pi x)\right)\left(2 \sin ^{2}(\pi x)+2 \sin (\pi x) \cos (\pi x)\right)=0 \Rightarrow$
$\sin ^{2}(\pi x)\left(\sin ^{2}(\pi x)-\cos ^{2}(\pi x)\right)=0 \Rightarrow \sin ^{2}(\pi x)(\cos (2 \pi x))=0$.
Therefore, $\left[\begin{array}{l}\sin (\pi x)=0, \\ \cos (2 \pi x)=0,\end{array} \Rightarrow\left[\begin{array}{l}\pi x=\pi k, k \in Z, \\ 2 \pi x=\frac{\pi}{2}+\pi n, n \in Z,\end{array} \Rightarrow\left[\begin{array}{l}x=k, k \in Z, \\ x=\frac{1}{4}+\frac{n}{2}, n \in Z,\end{array}\right.\right.\right.$ then the roots of the equation belonging to the interval $[-1 ; 2]$, will be $-1,1,2,-0.75,-0.25,0.25,0.75,1.25,1.75$. Their sum is 5.
Answer: 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the country of Landia, which breeds an elite breed of horses, an annual festival is held to test their speed, in which only one-year-old, two-year-old, three-year-old, and four-year-old horses can participate. For each horse that meets the speed standard, the festival organizers pay a fixed amount of money to the stud farm where the horse was raised: 1 landric for a one-year-old, 2 landrics for a two-year-old, 3 landrics for a three-year-old, and 4 landrics for a four-year-old. Each stud farm participating in the festival annually presents four new horses (of any age combination as they wish) for testing, which have not previously participated in the tests, as well as personally all horses (not older than four years) that previously participated in the tests at a younger age and met the standard. What is the maximum amount of money a stud farm can earn in the first six years of its participation in the festival?
(12 points)
|
Solution. A four-year-old horse can earn a maximum of 4 landricks over its entire participation in festivals. If the horse starts participating in festivals at 1 year old, it can participate for another 3 years after that. In the case of winning every year, it will earn 1+2+3+4=10 landricks over 4 years. If the horse starts participating in festivals at 2 years old, it can earn a maximum of $2+3+4=9$ landricks over the 3 possible years of participation. If the horse starts participating in festivals at 3 years old, it can earn a maximum of $3+4=7$ landricks. Therefore, the most optimal strategy is as follows. In the first year, the stable enters 4 one-year-old horses. The maximum winnings amount to 4 landricks. In the second year, the stable enters 4 new one-year-old horses and 4 two-year-olds that participated and won in the first year. The maximum winnings amount to $4+4 \cdot 2=12$ landricks. In the third year, the stable enters 4 new one-year-old horses, 4 two-year-olds that participated in the second year, and 4 three-year-olds that participated in the previous 2 years. The maximum bonus is $4+4 \cdot 2+4 \cdot 3=24$ landricks. In the fourth year, there is no point in entering one-year-old horses, as they will only be able to participate for 2 more years. Therefore, it makes sense to enter 4 new two-year-old horses. The winnings will amount to $4 \cdot 2+4 \cdot 2+4 \cdot 3+4 \cdot 4=44$ landricks. Horses that start participating in the 5th year will only participate once after that, so it makes sense to enter new three-year-olds. The winnings will amount to $4 \cdot 3+4 \cdot 3+4 \cdot 3+4 \cdot 4=52$ landricks. In the sixth year, it makes sense to enter only four-year-olds. The winnings will amount to $4 \cdot 4+4 \cdot 4+4 \cdot 4+4 \cdot 4=64$ landricks. In total, over 6 years of participation in festivals, the stable can earn a maximum of $4+12+24+44+52+64=200$ landricks.
Answer: 200.
|
200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$.
|
Solution. The number $N$ can be represented as
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot 2 \cdot 2 \cdot \ldots \cdot 2 \cdot(1 \cdot 2 \cdot \ldots \cdot 2017 \cdot 2018)}{2018!}=(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot 2^{2018}
\end{aligned}
$$
We obtained the product of odd numbers and a power of two.
Answer: 2018.
|
2018
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency
|
Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$
$$
\begin{aligned}
& S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-\text { abscissa of the point of tangency } A, \\
& A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad \text { - intersection point }
\end{aligned}
$$
with the y-axis. Let $C(0, c)$. The equation of the tangent line to the graph of the function $f(x)=\sqrt{x-3}$ is $y=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+f\left(x_{0}\right)$. Point $C$ lies on the tangent line, so we substitute its coordinates into the equation of the tangent line: $c=-f^{\prime}\left(x_{0}\right) x_{0}+f\left(x_{0}\right)$.
Then $A B=x_{0}, \quad B C=f\left(x_{0}\right)-c=f^{\prime}\left(x_{0}\right) x_{0}, \quad S_{A B C}=\frac{1}{2} A B \cdot B C=\frac{1}{2} f^{\prime}\left(x_{0}\right)\left(x_{0}\right)^{2}=\frac{x_{0}^{2}}{2 \sqrt{x_{0}-3}}$. To find the extrema of the function $\quad S_{A B C}=S\left(x_{0}\right)=\frac{x_{0}^{2}}{4 \sqrt{x_{0}-3}}$ we find the zeros of the derivative of this function $S^{\prime}\left(x_{0}\right)=\frac{3\left(x_{0}^{2}-4 x_{0}\right)}{8 \sqrt{\left(x_{0}-3\right)^{3}}}$. Since $x_{0} \geq 3$, the only extremum point, and specifically, the point of minimum for this function is the point $\quad x_{0}=4, \quad S_{\min }=S(4)=\frac{4^{2}}{4 \sqrt{4-3}}=4$.
Answer: 4.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$.
|
Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Let $O E=x, A E=y$. Then we arrive at the system $\left\{\begin{array}{c}x^{2}+y^{2}=2, \\ (1+\sqrt{3}-y)^{2}+x^{2}=4\end{array}\right.$. Solving the system, we get $\quad y=1, x=1$. Then $\quad \angle D A C=\angle B C A=45^{\circ}$, $B C=\sqrt{6}, \angle F C A=\operatorname{arctg} \frac{1}{\sqrt{3}}=30^{\circ}$,
$\angle A B E=30^{\circ}, \quad A B=2$.
Answer: 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A vessel with a capacity of 10 liters is filled with air containing $24\%$ oxygen. A certain volume of air was pumped out of the vessel and the same volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again the same volume of argon was added. In the new mixture, $11.76\%$ oxygen was found. How many liters of the mixture were released each time from the vessel? Give your answer as a number without specifying the unit.
(5 points)
|
Solution. Let $x$ liters of the mixture be released each time from the vessel. Then, the first time, the amount of oxygen left in the vessel is $2.4 - 0.24x$. The percentage of oxygen in the mixture after adding argon is $(2.4 - 0.24x) \times 10$. The second time, the amount of oxygen left in the vessel is $2.4 - 0.24x - (2.4 - 0.24x) \times 0.1x$. The percentage of oxygen in the mixture after adding argon in this case is $10(2.4 - 0.24x - (2.4 - 0.24x) \times 0.1x)$. According to the condition, $10(2.4 - 0.24x - (2.4 - 0.24x) \times 0.1x) = 11.76$. Solving the equation, we arrive at the quadratic equation $x^2 - 20x + 51 = 0$, with roots $x_1 = 3$, $x_2 = 17$. The second root does not fit the condition.
Answer: 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In how many ways can a rectangular board of size $2 \times 18$ be covered with identical rectangular tiles of size $1 \times 2$? The tiles must be placed so that they fit entirely on the board and do not overlap.
(12 points)
|
Solution. Let there be a board of size $2 \times$ n. Denote the number of ways to tile it with tiles of size $1 \times 2$ by $P_{n}$. Then the following recurrence relation holds: $P_{n}=P_{n-1}+P_{n-2}$. Since $P_{1}=1, P_{2}=2$, performing sequential calculations using the recurrence relation, we arrive at the answer $P_{18}=4181$.
Answer: 4181.
|
4181
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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