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4. Alexei drew 7 lines on a plane, which divided it into several parts. Then he chose two adjacent parts (adjacent parts are those that share a common side), counted how many sides each of them contains, and added these two numbers. What is the largest number he could have obtained? Explain your answer.
|
4. Answer: 10.
Solution. Consider any two adjacent parts. Temporarily remove the line containing their boundary. Then these two parts will become one large part. It has a maximum of 6 sides (each line can contain no more than one side of this part). Return the line we removed. It will be a side for both small parts (+2),
and it intersects a maximum of two sides of the large part, turning one large side into two small sides (+2).
Example. Lines containing the sides of a regular hexagon and a line passing through the midpoints of two of its adjacent sides. The two parts forming the hexagon have a total of 10 sides.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. The numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ form a geometric progression. Among them, there are both rational and irrational numbers. What is the maximum number of terms in this progression that can be rational numbers?
|
Answer: 3.
Example: let $a_{1}=1, q=\sqrt{2}$, we get the geometric progression $1, \sqrt{2}, 2, 2\sqrt{2}, 4$. Evaluation. If there are 4 rational numbers among them, then there will be two consecutive rational members of the geometric progression. This means that the common ratio of the progression (the ratio of the subsequent term to the previous one) is a rational number. But then, with a rational $a_{1}$, all terms of the progression are rational numbers, and with an irrational $a_{1}$, all terms are irrational.
Remarks. An example of a geometric progression with three rational and two irrational numbers is given - 3 points. It is proven that there cannot be more than three rational numbers, but no example is provided - 3 points. If both are present, 7 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.1. In a box, there are oranges, pears, and apples, a total of 60 fruits. It is known that there are 3 times more apples than non-apples, and there are 5 times fewer pears than non-pears. How many oranges are in the box?
|
Answer: 5.
Solution. Since there are 3 times more apples than non-apples, apples make up $\frac{3}{4}$ of the total number of fruits, i.e., there are $\frac{3}{4} \cdot 60=45$ apples. Since there are 5 times fewer pears than non-pears, pears make up $\frac{1}{6}$ of the total number of fruits, i.e., there are $\frac{1}{6} \cdot 60=10$ pears. Therefore, the total number of oranges is $60-45-10=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.4. In the cells of a $2 \times 35$ table (2 rows, 35 columns), non-zero real numbers are placed, and all numbers in the top row are distinct. For any two numbers in the same column, the following condition is satisfied: one number is the square of the other.
(a) (1 point) What is the maximum number of negative numbers that can be in this table?
(b) (3 points) What is the minimum number of distinct numbers that can be in the bottom row?
|
Answer: (a) 35. (b) 12.
Solution. (a) In any column, there can be no more than one negative number, so the total number of negative numbers is no more than 35. There can be exactly 35 if, for example, the top numbers are $-1, -2, -3, \ldots, -35$, and below them are the numbers $1^{2}, 2^{2}, 3^{2}, \ldots, 35^{2}$ respectively.
(b) We will prove that each number in the bottom row appears no more than 3 times.
- Suppose the number $a0$ appears in the bottom row. If the top number in the column with number $a$ is the square of the bottom number, then the number $a^{2}$ is written above it. If the bottom number is the square of the top number, then above it is written $-\sqrt{a}$ or $\sqrt{a}$. Since all the numbers in the top row are distinct, the number $a$ in the bottom row appears no more than 3 times.
Since each number in the bottom row appears no more than 3 times, and there are 35 numbers in total, there are at least $\frac{35}{3} > 11$, i.e., at least 12 different numbers.
It remains to provide an example of how exactly 12 different numbers can appear in the bottom row. Consider the 12 smallest prime numbers: $2 = p_{1} < p_{2} < \ldots < p_{12}$. Let the number $p_{1}^{2}$ be written in the bottom of the first three columns, and the numbers $-p_{1}, p_{1}$, and $p_{1}^{4}$ be written above them. Similarly, in the next three columns, the number $p_{2}^{2}$ is written in the bottom, and the numbers $-p_{2}, p_{2}$, and $p_{2}^{4}$ are written above them, and so on. In the last two columns, the number $p_{12}^{2}$ is written in the bottom, and the numbers $-p_{12}$ and $p_{12}$ are written above them.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. What is the greatest number of consecutive natural numbers, each of which has exactly four natural divisors (including 1 and the number itself)?
|
Answer: three numbers.
Solution. Suppose there are four consecutive numbers that satisfy the condition. Note that among four consecutive numbers, one is divisible by 4. Then, in the prime factorization of this number, there are at least two twos. If there is another prime divisor $p$ different from two, then the number of divisors of the number is at least six: $1, 2, 4, p, 2p, 4p$. If the factorization contains only twos, then for the number of divisors to be exactly four ($1, 2, 4, 8$), there must be exactly three twos. Thus, there is only one number divisible by 4 that has exactly four divisors - the number 8. Its neighbors (7 and 9) do not satisfy the condition, so the number of such numbers is no more than three.
Example of three consecutive numbers, each of which has exactly four natural divisors: $33, 34, 35$. There are other examples as well.
## Grading Criteria
+ A complete and well-reasoned solution is provided
± A generally correct but insufficiently justified solution is provided (without containing incorrect statements). For example, the correct answer and example are provided, it is noted that among four consecutive numbers one is divisible by 4, but the case $n=8$ is not mentioned
Ғ The correct answer and example are provided, but there is an error in the reasoning. For example, it is stated that any number divisible by four has more than four divisors
干 Only the correct answer and example are provided
- Only the answer "three numbers" is provided
- The problem is not solved or is solved incorrectly
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A sequence of 2016 numbers is written. Each one, except the first and the last, is equal to the sum of its neighbors. Find the sum of all 2016 numbers.
|
Answer: the sum of all numbers is zero.
Solution. Let the first number be $a$, and the second $b$. Denote the third as $x$. Then $a+x=b$ and, therefore, $x=b-a$. Now, express the fourth in terms of $a$ and $b$: $b+y=b-a$ and, therefore, $y=-a$. Continuing, we get: $a, b, b-a, -a, -b, a-b, a, b, \ldots$ This means the numbers repeat with a period of 6. Since 2016:6=336, the sum of all numbers is the sum of the first six, multiplied by 336. The sum of the first six is zero, so the sum of all is zero.
## Grading Criteria.
Only the answer: 0 points.
Specific example analysis: 1 point.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On a $5 \times 5$ board, a cross consisting of five cells (a cell and all its neighbors) is located. What is the minimum number of detectors needed to place on the board cells to accurately determine the position of the cross? (A detector indicates whether a cell belongs to the cross or not, and the detectors trigger simultaneously.)
|
Answer: 4.
Solution. There are 9 possible positions for the cross on the board, which is the same as the number of positions for the central cell of the cross. A detector has two states, so the total number of possible states for three detectors is $2^{3}=8$, and thus, they cannot distinguish between 9 positions of the cross. Examples of possible arrangements of four detectors are shown in the figures
## Grading Criteria.
Only the answer: 0 points.
Example with 4 detectors: 3 points.
Proof of the insufficiency of three detectors: 4 points.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3.7. Katya passes the time while her parents are at work. On a piece of paper, she absentmindedly drew Cheburashkas in two rows (at least one Cheburashka was drawn in each row).
Then, after some thought, between every two adjacent Cheburashkas in a row, she drew a Gena the Crocodile. And then to the left of each Cheburashka, she drew a Old Lady Shapoklyak. And finally, between every two characters in a row, she drew a Krakozyabra.
Looking carefully at the drawing, she realized that only the Krakozyabras looked good, and she angrily erased all the others. In the end, her parents saw two rows of Krakozyabras: a total of 29. How many Cheburashkas were erased?
|
Answer: 11.
Solution. The Krokodilofes are drawn exactly in the gaps between the other characters. In each of the two rows of gaps, there is 1 less gap than there are characters, so
There are 2 fewer Krokodilofes than all the other characters. Therefore, Cheburashkas, Crocodile Gens, and Old Ladies Shapoklyak total $29+2=31$. We will now consider only them, forgetting about the Krokodilofes.
Each Cheburashka has one neighboring Old Lady Shapoklyak to the left, so there are an equal number of them. Crocodile Gens are drawn exactly in the gaps between Cheburashkas, so by the previously described principle, there are 2 fewer of them than Cheburashkas.
Let's mentally add 2 Crocodile Gens. Then the total number of characters will be $31+2=33$. In this case, all three characters are now equal, so there were 11 Cheburashkas.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.3. Vitya and his mother left home at the same time and walked in opposite directions at the same speed: Vitya - to school, and his mother - to work. After 10 minutes, Vitya realized he didn't have the keys to the house, and he would return from school earlier than his mother, so he started to catch up with her, increasing his speed fivefold. How many minutes after he realized he needed to get the keys will Vitya catch up with his mother?
|
Answer: 5 minutes.
Solution. Let Vitya and his mother initially walk at a speed of $s$ meters per minute. After 10 minutes, when Vitya realized he had forgotten his keys, the distance between him and his mother became $10 s + 10 s = 20 s$. When Vitya started to catch up with his mother, he walked at a speed of $5 s$ meters per minute, so the distance between him and his mother decreased by $4 s$ meters per minute. Therefore, it would take $20 s : 4 s = 5$ minutes for the distance to reduce to zero.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
|
Answer: 6.
Solution. Let $a, b, c$ be the natural numbers in the three lower circles, from left to right. According to the condition, $14 \cdot 4 \cdot a = 14 \cdot 6 \cdot c$, i.e., $2a = 3c$. From this, it follows that $3c$ is even, and therefore $c$ is even. Thus, $c = 2k$ for some natural number $k$, and from the equation $2a = 3c$ it follows that $a = 3k$.
It must also hold that $14 \cdot 4 \cdot 3k = 3k \cdot b \cdot 2k$, which means $b \cdot k = 28$. Note that by choosing the number $k$, which is a natural divisor of 28, the natural numbers $a, b, c$ are uniquely determined. The number 28 has exactly 6 natural divisors: $1, 2, 4, 7, 14, 28$. Therefore, there are also 6 ways to place the numbers in the circles.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.8. Given an isosceles triangle $ABC (AB = BC)$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
|
Answer: 4.
Solution. Mark point $K$ on ray $B C$ such that $B E=B K$. Then $A E=C K$ as well.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common side) and the angle between them ($\angle C A E=\angle A C K$ - adjacent to the equal base angles of the isosceles triangle). Therefore, $A K=C E=13$ and $\angle A K C=\angle A E C=60^{\circ}$.
In triangle $A D K$, the angles at vertices $D$ and $K$ are $60^{\circ}$, so it is equilateral, and $D K=A K=A D=13$. Therefore, $A E=C K=D K-D C=13-9=4$.
## 8th grade
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.2. How many real numbers $x$ exist such that the value of the expression $\sqrt{123-\sqrt{x}}$ is an integer?
|
Answer: 12.
Solution. From the condition, it follows that the value $s=123-\sqrt{x} \leqslant 123$ is a square of an integer. Since $11^{2}<123<12^{2}$, this value can take one of 12 values $0^{2}$, $1^{2}, 2^{2}, \ldots, 11^{2}$. And for each of these 12 values of $s$, there is a unique value of $x=(123-s)^{2}$ (obviously, all these values of $x$ are distinct).
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.3. At exactly noon, a truck left the village and headed for the city, at the same time, a car left the city and headed for the village. If the truck had left 45 minutes earlier, they would have met 18 kilometers closer to the city. And if the car had left 20 minutes earlier, they would have met $k$ kilometers closer to the village. Find $k$.
|
Answer: 8.
Solution. We will express all distances in kilometers, time in hours, and speed in kilometers per hour. Let the distance between the village and the city be $S$, the speed of the truck be $x$, and the speed of the car be $y$. The time until the first meeting is $\frac{S}{x+y}$, so the distance from the village to the actual meeting point is $\frac{S x}{x+y}$.
In the first hypothetical scenario, the truck will travel $0.75 x$ kilometers in 45 minutes, meaning it will take another $\frac{S-0.75 x}{x+y}$ hours to reach the expected first meeting point, and the truck will travel another $\frac{(S-0.75 x) x}{x+y}$ kilometers. According to the problem, we know that
$$
18=0.75 x+\frac{(S-0.75 x) x}{x+y}-\frac{S x}{x+y}=0.75 \frac{x y}{x+y}
$$
from which we find $\frac{x y}{x+y}=18 \cdot \frac{4}{3}=24$.
In the second scenario, the car will travel $y / 3$ kilometers in 20 minutes, meaning it will take another $\frac{S-y / 3}{x+y}$ hours to reach the expected second meeting point, and the truck will travel $\frac{(S-y / 3) x}{x+y}$ kilometers during this time. The value we need to find is
$$
\frac{S x}{x+y}-\frac{(S-y / 3) x}{x+y}=\frac{x y / 3}{x+y}=24 / 3=8
$$
Another solution. Instead of the scenario "the truck left 45 minutes earlier," we will consider the scenario "the car left 45 minutes later" - it is clear that the meeting point will be the same in this case.
We will solve the problem graphically. Along the $O x$ axis, we will plot the coordinates of the vehicles such that the city corresponds to the coordinate 0 and the village to the coordinate $p>0$. Along the $O t$ axis, we will plot time (the common start of movement in the original scenario will correspond to $t=0$). We will represent the car's movement with a red line and the truck's movement with a blue line; in addition, we will mark two alternative graphs of the car's movement (with a start 45 minutes later and 20 minutes earlier). The meeting point in the original scenario (location and time) will be denoted by $A$, and in the alternative scenarios by $B$ and $C$:
By Thales' theorem,
$$
\frac{45}{20}=\frac{A B}{A C}=\frac{\Delta x_{1}}{\Delta x_{2}}
$$
Since $\Delta x_{1}=18$ kilometers according to the problem, then $\Delta x_{2}=8$ kilometers.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.6. In a grid table with 5 rows and 6 columns, each cell contains either a cross, a zero, or a star. It is known that:
- in each column, the number of zeros is not less than the number of crosses;
- in each column, the number of zeros is not less than the number of stars;
- in each row, the number of crosses is not less than the number of zeros;
- in each row, the number of crosses is not less than the number of stars.
How many stars can be in such a table? List all possible options.
|
Answer: 6.
Solution. Since in each column there are no fewer O's (noughts) than X's (crosses), then in the entire table there are no fewer O's than X's. Similarly, considering the rows, it follows that in the entire table there are no fewer X's than O's. Therefore, there are an equal number of X's and O's in the entire table, and also there are an equal number of them in each row and each column.
From the condition, it also follows that in each column there cannot be 0 or 1 O's, so there are no fewer than 2. Since the number of X's and O's in the column is equal, there cannot be 3 or more, i.e., there are exactly 2. Therefore, in each column there is exactly one star, i.e., there are a total of 6.
It is also worth noting that the described construction indeed exists. For example, in the following table all conditions are met:
| x | $\times$ | $\times$ | | | O |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | O | 0 | $\times$ | $\times$ | $x$ |
| $*$ | * | $\times$ | $\times$ | 0 | O |
| ○ | 0 | * | $*$ | $\times$ | $x$ |
| $\times$ | $x$ | 0 | 0 | $*$ | ＊ |
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3.1. Condition:
Danil took a white cube and numbered its faces with numbers from 1 to 6, writing each one exactly once. It turned out that the sum of the numbers on one pair of opposite faces is 11. What can the sum of the numbers on none of the remaining pairs of opposite faces NOT be?
## Answer options:
$\square 5$
$\square 6$
$\square 7$
$\square 9$
|
# Solution.
If the sum of the numbers on opposite faces is 11, then the numbers on these faces are 5 and 6. The remaining numbers can be paired in three ways: $(1,2)$ and $(3,4); (1,3)$ and $(2,4); (1,4)$ and $(2,3)$, i.e., the sums that CAN be obtained are: $3,7,4,6$ and 5. Therefore, the sums that CANNOT be obtained are 8, 9, and 10.
|
9
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
# 5.1. Condition:
Polina came to the cafeteria and saw that 2 puff pastries cost 3 times more than 2 pies. Polina didn't have enough money for 2 puff pastries, but she did have enough for 1 pie and 1 puff pastry. After the purchase, she wondered how many times more money she spent buying 1 puff pastry and 1 pie instead of 2 pies. Help Polina answer this question.
|
Answer: 2
## Solution.
2 puff pastries are 3 times more expensive than 2 pies, so one puff pastry is 3 times more expensive than one pie, which means one puff pastry costs as much as 3 pies. Therefore, 1 puff pastry and 1 pie cost as much as 4 pies. Then they are 2 times more expensive than 2 pies.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. If the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic polynomial $g(x)=(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$.
|
Answer: 6.
Solution: We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=$ $4 f(-2)$.
Comment: Correct answer without justification - 0 points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Six people - liars and knights - sat around a table. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, 3 people said: "I have one coin," while the other 3 said: "I have no coins." What is the maximum number of knights that could have been sitting at the table?
|
Answer: 4.
Solution: After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 6. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs from the actual total by $6-3=3$, at least 2 people must have lied. Therefore, there are no more than 4 knights at the table. Suppose the knights sitting at the table pass the coins as follows (arrows indicate the direction of the coin transfer; the number of coins after the transfer is in parentheses): $\leftrightarrow K(1)-L(0) \rightarrow L(1) \rightarrow L(1) \rightarrow L(1) \rightarrow K(2) \leftrightarrow$. In this case, all knights tell the truth, while the liars lie, saying they have 0 coins.
Comment: It is proven that there are no more than 4 knights at the table - 5 points. It is proven that there can be 4 knights at the table - 2 points.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. On the board, there are $N$ prime numbers (not necessarily distinct). It turns out that the sum of any three numbers on the board is also a prime number. For what largest $N$ is this possible
|
Answer: $N=4$.
Solution. Consider the remainders when the $N$ written numbers are divided by 3. All three remainders cannot occur, because in this case, the sum of three numbers with different remainders will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Therefore, there can be no more than 2 possible remainders. Also, note that there cannot be 3 numbers with the same remainder, because the sum of these three numbers will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Thus, $N$ cannot be greater than $2 \cdot 2=4$.
$N$ can equal 4. For example, if the numbers written on the board are $3,3,5,5$ (sums $11,13)$.
Remark. There are other examples. For instance, $3,3,5,11$ (sums 19, 17, 11). Comment. It is proven that $N$ is not greater than $4-5$ points. An example with 4 numbers is provided - 2 points.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. Each cell of a $7 \mathrm{x} 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the number of red cells and not less than the number of green cells. How many green cells can there be in such a table?
|
Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones. In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones. Thus, in the table, there is an equal number of red and yellow cells.
2) Suppose that in some column, there are more yellow cells than red ones. Since in each of the other columns, there are no fewer yellow cells than red ones, then in the entire table, there would be more yellow cells than red ones, but this is not the case (see 1). Therefore, in each of the eight columns, there are an equal number of red and yellow cells.
3) Since in each column, there are no fewer yellow cells than green ones, the cases where in each column: a) 1 yellow, 1 red, 5 green cells and b) 2 yellow, 2 red, 3 green cells are excluded. The only remaining case is when in each column there are 3 red, 3 yellow, and 1 green cell. Then, in total, there are 8 green cells in the table. This case is possible. For example, see the table.
| | $\bar{\sim}$ | Y | $\mathbf{R}$ | $\bar{R}$ | $\cdots$ | $j \pi$ | $\mathbf{R}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\bar{\kappa}$ | $\bar{\kappa}$ | | | $\mathbf{R}$ | ऊ | $\mathbf{R}$ | |
| | | | $\vec{x}$ | | | | |
| | | $=$ | | | | | |
| | $\overline{\mathbf{R}}$ | $\mathbf{R}$ | 3 | $\mathbf{R}$ | $\boldsymbol{A}$ | $\kappa$ | |
| | $\mathcal{\Psi}$ | $\mathbb{w}$ | Y | $\mathbf{R}$ | y | | |
| | | | | | | | |
## Grading Criteria
7 points - A complete and well-reasoned solution is provided.
6 points - A generally correct solution is provided, containing minor gaps or inaccuracies.
4 points - It is proven that there can only be 8 green cells, but an example is not provided.
2 points - Only the correct answer and an example are provided.
0 points - Only the answer is provided.
0 points - The problem is not solved or is solved incorrectly.
Internet resources: $\underline{\text { http://www.problems.ru, https://olimpiada.ru. }}$
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: $9^{\circ}$.
Solution. In triangle $D F E$, the angles are $7 \alpha, 8 \alpha$ and $45^{\circ}$.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha + 8 \alpha + 45^{\circ} = 180^{\circ}$, from which $\alpha = \frac{1}{15} \cdot 135^{\circ} = 9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
|
Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then $\angle A C O=\angle O A C=90^{\circ}-67^{\circ}=23^{\circ}$ (here we used the fact that triangle $A C D$ is a right triangle: angle $A C D$, which subtends the diameter, is a right angle).
Thus, $x=32^{\circ}-23^{\circ}=9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Three motorcyclists start simultaneously from one point on a circular highway in the same direction. The first motorcyclist caught up with the second for the first time after making 4.5 laps from the start, and 30 minutes before that, he caught up with the third motorcyclist for the first time. The second motorcyclist caught up with the third for the first time three hours after the start. How many laps per hour does the first motorcyclist make?
#
|
# Solution
Let $x, y, z$ be the speeds of the first, second, and third motorcyclists in circles per hour, respectively.
Express the time it takes for the first motorcyclist to catch up with the second, the first to catch up with the third, and the second to catch up with the third in different ways.
We obtain the system of equations
$$
\left\{\begin{array}{c}
\frac{1}{x-y}=\frac{4.5}{x} \\
\frac{1}{x-z}=\frac{4.5}{x}-\frac{1}{2} \\
\frac{1}{y-z}=3
\end{array}\right.
$$
Solving the system of equations, we get the answer $x=3$.
## Grading Criteria
Only the correct answer - 0 points.
Individual expressions for the time of individual motorcyclists are obtained - 1 point.
One, two, or three equations of the system are correctly formulated, or if another method was chosen, depending on the progress in the solution - 2-4 points.
The problem is solved, but contains one or two arithmetic errors, resulting in an incorrect answer - 5-6 points.
The correct answer is obtained with justification - 7 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. It is known that $a^{2}+2=b^{4}, b^{2}+2=c^{4}, c^{2}+2=a^{4}$. What is the value of the product $\left(a^{2}-1\right)\left(b^{2}-1\right)\left(c^{2}-1\right)$? Find all possible values and prove that there are no others.
|
Solution: Subtract 1 from both sides of each equation, we get:
$$
\begin{gathered}
a^{2}+1=b^{4}-1=\left(b^{2}-1\right)\left(b^{2}+1\right) \\
b^{2}+1=\left(c^{2}-1\right)\left(c^{2}+1\right), \quad c^{2}+1=\left(a^{2}-1\right)\left(a^{2}+1\right)
\end{gathered}
$$
Multiply the left and right parts of all three equations, and after canceling the positive number $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)$, we get $\left(a^{2}-1\right)\left(b^{2}-1\right)\left(c^{2}-1\right)=1$. Therefore, the desired product can only be 1. Since the transformations involved an inequivalent step (multiplying the left and right parts of several equations is only a transition to a consequence), it is necessary to show that the number 1 can indeed be obtained. This case will occur, for example, when $a=b=c=\sqrt{2}$.
Answer: 1.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and justified answer with a confirming example of numbers $a, b, c$ | 7 points |
| Correct and justified answer without a confirming example of numbers $a, b, c$ | 6 points |
| Example of numbers $a, b, c$ satisfying the condition of the problem, for which the value of the desired expression is 1 | 1 point |
| Algebraic calculations that did not lead to a solution, as well as an answer without justification or an incorrect answer | 0 points |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. What is the smallest value that the GCD $(x, y)$ can take, where $x$ and $y$ are natural numbers, if the LCM $(x, y)=(x-y)^{2}$?
|
Solution:
When $x=4$ and $y=2$, LCM $(x, y)=(x-y)^{2}=4$, and GCD $(x, y)=2$. Thus, 2 is achievable.
Assume that GCD $(x, y)=1$ (the numbers are coprime). Then LCM $(x, y)=x y$. From the condition, we get that $x y=x^{2}-2 x y+y^{2}$. We obtain $x^{2}-3 x y+y^{2}=0$. Solving the obtained quadratic equation with respect to $x$, we get the discriminant $5 y^{2}$. The square root of the discriminant $\sqrt{5} \cdot y$ is an irrational number, and for $x$ to be a natural number, it must be rational. This is a contradiction. Therefore, GCD $(x, y) \neq 1$.
Answer: 2.
| Criteria | Points |
| :--- | :---: |
| Complete solution of the problem. | 7 |
| Proved that GCD $(x, y) \neq 1$. Correct answer, but no example of numbers for GCD $(x, y)=2$ provided. | 5 |
| Correct answer with an example. | 2 |
| Only the correct answer, without evaluation and example. | 0 |
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A natural number is written on the board. Nikolai noticed that he can append a digit to the right of it in two ways so that the resulting number is divisible by 9. In how many ways can he append a digit to the right of the given number so that the resulting number is divisible by 3?
|
Answer: 4 ways.
Solution. Note that the difference between the two numbers "noticed" by Nikolai is less than 10, but is divisible by 9. Therefore, this difference is 9. This is only possible if the appended digits are 0 and 9. Then it is easy to see that for divisibility by 3, in addition to these two digits, the digits 3 and 6 can also be appended. In total, there are 4 ways.
## Criteria
4 6. A complete and justified solution is provided.
In the absence of a correct solution, the highest applicable criterion from those listed below is used
2 6. It is proven that Nikolai could only append 0 or 9 at the beginning.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. An archipelago consists of several small islands and one large island. It was decided to build bridges between the islands so that the large island would be connected to each small island by two bridges, and any two small islands would be connected by one bridge.
By November 1, all the bridges between the small islands and several (at least one) bridges leading to the large island - a total of 28 bridges - had been built. How many islands are there in the archipelago?
Answer: 8 islands
|
Solution. Let's number the small islands of the archipelago. If a bridge connects islands with numbers $a$ and $b$, we write the smaller of these two numbers on this bridge.
Suppose the number of small islands in the archipelago is no more than six. Then there are no more than 5 bridges with the number 1, no more than 4 bridges with the number 2, and so on. No more than 12 bridges lead to the large island. Therefore, the total number of bridges built is no more than
$$
5+4+3+2+1+12=2728
$$
which also contradicts the condition.
Thus, there are 7 small islands in the archipelago, and a total of 8 islands.
## Criteria
4 points. A complete and justified solution is provided.
06 . The problem is not solved or solved incorrectly.
In other cases, sum the following criteria (but do not exceed 4 points):
2 points. It is proven that the number of islands is no more than eight (if there is an arithmetic error in the proof, then 16 .)
2 6. It is proven that the number of islands is no less than eight (if there is an arithmetic error in the proof, then 1 point).
16 . The correct answer is provided.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. For what values of $a$ do the equations
$$
x^{2}+a x+1=0 \quad \text { and } \quad x^{2}+x+a=0
$$
have at least one common root?
|
Solution. Let $x_{0}$ be the common root of the equations. Substitute into the equations, equate the expressions for $x_{0}^{2}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}^{2} + a x_{0} + 1 = 0, \\
x_{0}^{2} + x_{0} + a = 0
\end{array}\right. \\
& (a-1) x_{0} = a-1 \quad a x_{0} + 1 = x_{0} + a \\
&
\end{aligned}
$$
We get two cases. When $a=1$, the equations coincide, $x^{2} + x + 1 = 0$ has no roots. When $x_{0}=1$, by substitution we get $a=-2$.
Answer. When $a=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let $F(x)$ and $G(x)$ be polynomials of degree 2021. It is known that for all real $x$, $F(F(x)) = G(G(x))$ and there exists a real number $k, k \neq 0$, such that for all real $x$, $F(k F(F(x))) = G(k G(G(x)))$. Find the degree of the polynomial $F(x) - G(x)$.
|
Solution. Since $F(F(x))=G(G(x))$, then $F(k G(G(x)))=G(k G(G(x)))$. This equality holds for all $x$, i.e., for more than 2022 values of the variable (Since the polynomial $G(x)$ takes an infinite number of values). Therefore, the polynomials coincide, their difference has degree 0.
Answer. The degree of $F(x)-G(x)$ is 0, the polynomials coincide.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 6. CONDITION
On an 8x8 chessboard, 64 checkers numbered from 1 to 64 are placed. 64 students take turns approaching the board and flipping only those checkers whose numbers are divisible by the ordinal number of the current student. A "Queen" is a checker that has been flipped an odd number of times. How many "Queens" will be on the board after the last student has stepped away from it?
|
Solution. Obviously, each checker is flipped as many times as its number has divisors. Therefore, the number of "queens" will be the number of numbers from 1 to 64 that have an odd number of divisors, and this property is only possessed by perfect squares. Thus, the numbers of the "queens" remaining on the board will be $1, 4, 9, 16, 25, 36, 49$, and 64, which totals 8.
Answer: 8.
## Grading Criteria for Problem 6.
The answer is correct, but there are remarks on the clarity of its presentation and justification. 5 points.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. What is the largest number of different natural numbers that can be chosen so that the sum of any three of them is a prime number?
|
Solution. One example of four numbers that satisfy the condition of the problem is $1,3,7,9$. Indeed, the numbers $1+3+7=11, 1+3+9=13, 1+7+9=17, 3+7+9=19$ are prime.
Suppose it was possible to choose five numbers. Consider the remainders of these numbers when divided by 3. If there are three identical remainders among them, then the sum of the corresponding numbers is divisible by 3. If there are no three identical remainders, then each of the remainders 0, 1, or 2 must be present. Then the sum of three numbers with different remainders when divided by 3 is divisible by 3. This sum is not equal to 3, since all numbers are different and natural. Therefore, this sum is a composite number.
There are other examples as well, such as $3,7,9,31$, etc.
Answer: 4 numbers
Criteria. 7 points - correct and justified solution; 5 points - correct estimate without example; 2 points - correct answer and example provided; 0 points - only the correct answer provided.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
#
|
# Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get that $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this equation are the numbers 2 and -2, so the only root of the original equation is the number 2.
## Grading Criteria
- Only the correct answer is provided - 1 point.
- The correct solution process, but the extraneous root is not discarded - 3 points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Each of the equations $a x^{2}-b x+c=0$ and $c x^{2}-a x+b=0$ has two distinct real roots. The sum of the roots of the first equation is non-negative, and the product of the roots of the first equation is 9 times the sum of the roots of the second equation. Find the ratio of the sum of the roots of the first equation to the product of the roots of the second equation.
|
# Solution.
From the condition, it follows that the coefficients $a, c \neq 0$.
By Vieta's theorem, from the condition it follows that $\frac{c}{a}=9 \frac{a}{c}$. Hence, $c^{2}=9 a^{2}$, which means $\left[\begin{array}{l}c=3 a, \\ c=-3 a .\end{array}\right.$
1 case.
$c=3 a$.
We get the equations $a x^{2}-b x+3 a=0$ and $3 a x^{2}-a x+b=0$.
By the condition $\frac{b}{a} \geq 0, b^{2}-12 a^{2}>0, a^{2}-12 a b>0$.
That is, $\frac{b}{a} \geq 0, \frac{b^{2}}{a^{2}}>12,1-12 \frac{b}{a}>0$. Making the substitution $\frac{b}{a}=t$, we get the system of inequalities:
$\left\{\begin{array}{l}t \geq 0, \\ t^{2}>12, \\ t0, a^{2}+12 a b>0$.
These inequalities are satisfied for any $a$ and $b$ such that $\frac{b}{a} \geq 0$.
The sum of the roots of the first equation is $\frac{b}{a}$. The product of the roots of the second equation is $-\frac{b}{3 a}$. The ratio is -3.
Answer: -3.
## Recommendations for checking.
Only the value $3-2$ points found.
Values $\pm 3-3$ points found.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A circle is inscribed in triangle $A B C$. Two points $E$ and $F$ are marked on the largest side of the triangle $A C$ such that $A E=A B$, and $C F=C B$. Segment $B E$ intersects the inscribed circle at points $P$ and $Q$, with $B P=1, P Q=8$. What is the length of segment $E F$?
|
Solution.
Let the sides of the triangle be $AB=c, AC=b, BC=a$. Then $EF=a+c-b$.
Let $G$ be the point of tangency of the inscribed circle with side $BC$. Then $BG^2 = BQ \cdot BP = 9$. Therefore, $BG = 3$.
Since the segments of tangents drawn from a point to a circle are equal, we have $2BG + 2b = a + b + c$.
Therefore, $a + c - b = 2BG = 6$.
Answer: 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.7. Vanya goes to the swimming pool every Wednesday and Friday. After one of his visits, he noticed that he had already gone 10 times this month. What will be the date of the third visit in the next month if he continues to go on Wednesdays and Fridays?
|
Answer: 12.
Solution. Vanya visited the swimming pool 10 times in a month, which means he visited it 5 times on Wednesdays and 5 times on Fridays.
If the first visit to the swimming pool in this month fell on a Friday, then from the first Friday to the last Friday he visited (including the first and last Friday), 29 days have passed. This means that in this month, there are no more than two days left, so the next Wednesday will fall into the next month.
If the first visit to the swimming pool in this month fell on a Wednesday, then from the first Wednesday to the last Wednesday he visited (including the first and last Wednesday), 29 days have passed. In the month, there are no more than two days left, so the last Friday falls exactly on the 31st of the month.
Thus, the Wednesdays in the next month will fall on the 5th, 12th, 19th, and 26th, and the Fridays on the 7th, 14th, 21st, and 28th. The third visit will be on the 12th.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.4. In parallelogram $A B C D$, the angle at vertex $A$ is $60^{\circ}$, $A B=73$, and $B C=88$. The bisector of angle $A B C$ intersects segment $A D$ at point $E$, and ray $C D$ at point $F$. Find the length of segment $E F$.
Answer, option 1. 9.
Answer, option 2. 13.
Answer, option 3. 12.
Answer, option 4. 15.
|
Solution 1. Lines $A D$ and $B C$ are parallel, therefore
$$
\angle A B C=180^{\circ}-\angle B A D=120^{\circ}
$$
(angles $A B C$ and $B A D$ are consecutive interior angles). Then $\angle A B E=60^{\circ}$ and $\angle C B E=60^{\circ}$, since $B E$ is the bisector of angle $A B C$. In triangle $A B E$, the angles at vertices $A$ and $B$ are both $60^{\circ}$. Therefore, the angle at vertex $E$ is also $60^{\circ}$, making this triangle equilateral. Thus, $B E=A B=78$. Also, in triangle $B C F$, the angles at vertices $B$ and $C$ are both $60^{\circ}$. Therefore, this triangle is also equilateral, so $B F=B C=87$. Thus, $E F=B F-B E=87-78=9$.
|
9
|
Geometry
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
3. Given a triangle $A B C$, where $\angle B A C=60^{\circ}$. Point $S$ is the midpoint of the angle bisector $A D$. It is known that $\angle S B A=30^{\circ}$. Find DC/BS.
|
3. Answer: 2. Solution: We have $\mathrm{BS}=\mathrm{AS}=\mathrm{SD}$, therefore, triangle $\mathrm{ABD}$ is a right triangle (in it, the median is equal to half the side), then $\angle A C D=30^{\circ}$, hence $A D=D C$, from which we get that the required ratio is 2.
Criteria: correct solution - 7 points, proved that the triangle is right, but no further progress - 3 points, otherwise -0 points.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. For what least natural value of \( b \) does the equation
$$
x^{2}+b x+25=0
$$
have at least one root?
|
Answer: 10
Solution. The equation has at least one root if and only if the discriminant $D=b^{2}-4 \cdot 25=$ $=b^{2}-100$ is greater than or equal to 0. For positive $b$, this condition is equivalent to the condition $b \geqslant 10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. It is known that the area of the shaded region of the figure is $\frac{32}{\pi}$, and the radius of the smaller circle is 3 times smaller than the radius of the larger circle. What is the length of the smaller circle?
|
# Answer: 4
Solution. Let the radius of the smaller circle be $R$, then the radius of the larger circle is $3R$. The area of the smaller circle is $S_{1}=\pi R^{2}$, and the area of the larger circle is $S_{2}=\pi(3R)^{2}=9\pi R^{2}$. Therefore, the area of the shaded part is $S_{2}-S_{1}=8\pi R^{2}$. We have $8\pi R^{2}=\frac{32}{\pi}$, from which $(\pi R)^{2}=4$ and $\pi R=2$. Thus, the length of the smaller circle is $2\pi R=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many four-digit numbers exist that are divisible by 17 and end in 17?
|
1. Let's denote the four-digit number as $\overline{x y 17}$. Then the number $\overline{x y 17}-17$ is also divisible by 17. But $\overline{x y 17}-17=100 \cdot \overline{x y}+17-17=100 \cdot \overline{x y}$. Since the numbers 100 and 17 are coprime, the two-digit number $\overline{x y}$ is divisible by 17. By enumeration, we find all two-digit numbers divisible by 17: 17, 34, 51, 68, 85. There are five of them. Therefore, there will also be five such numbers: 1717, 3417, 5117, 6817, 8517.
Answer: 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
|
Answer: 6.
Solution. Let $a, b, c$ be the natural numbers in the three lower circles, from left to right. According to the condition, $14 \cdot 4 \cdot a = 14 \cdot 6 \cdot c$, i.e., $2a = 3c$. From this, it follows that $3c$ is even, and therefore $c$ is even. Thus, $c = 2k$ for some natural number $k$, and from the equation $2a = 3c$ it follows that $a = 3k$.
It must also hold that $14 \cdot 4 \cdot 3k = 3k \cdot b \cdot 2k$, which means $b \cdot k = 28$. Note that by choosing the number $k$, which is a natural divisor of 28, the natural numbers $a, b, c$ are uniquely determined. The number 28 has exactly 6 natural divisors: $1, 2, 4, 7, 14, 28$. Therefore, there are also 6 ways to place the numbers in the circles.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.8. Given an isosceles triangle $ABC (AB = BC)$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
|
Answer: 4.
Solution. Mark point $K$ on ray $B C$ such that $B E=B K$. Then $A E=C K$ as well.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common side) and the angle between them ($\angle C A E=\angle A C K$ - adjacent to the equal base angles of the isosceles triangle). Therefore, $A K=C E=13$ and $\angle A K C=\angle A E C=60^{\circ}$.
In triangle $A D K$, the angles at vertices $D$ and $K$ are $60^{\circ}$, so it is equilateral, and $D K=A K=A D=13$. Therefore, $A E=C K=D K-D C=13-9=4$.
## 8th grade
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. The teacher drew the graph of the function $y=\frac{k}{x}$ and three lines with a slope of $k$ ( $k$ is not equal to zero). Petya wrote down the abscissas of all six points of intersection and multiplied them. Prove that the result does not depend on the choice of the number $k$.
|
10.1. A line parallel to the line $y=k x$ has the equation $y=k x+b$ The abscissas of its intersection points with the hyperbola are both roots of the equation $\frac{k}{x}=k x+b, \quad$ equivalent to the equation $k x^{2}+b x-k=0$. The product of the roots of this equation is -1. Multiplying three such products, we get the answer: -1.
|
-1
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
2. The plane departed from Perm on September 28 at noon and arrived in Kirov at 11:00 AM (all departure and arrival times mentioned in the problem are local). At 7:00 PM the same day, the plane departed from Kirov to Yakutsk and arrived there at 7:00 AM. Three hours later, it departed from Yakutsk to Perm and returned there at 11:00 AM on September 29. How much time did the plane spend in the air? Answer: 12 hours.
|
# Solution.
The plane was absent in Perm for 23 hours. Out of these, it was stationed in Kirov for 8 hours (from 11 to 19) and for 3 hours in Yakutsk. In total, out of these 23 hours, it was stationed $8+3=11$ (hours), i.e., the plane was in the air for $23-11=12$ (hours).
## Grading Criteria.
- Correct solution -7 points.
- Correct answer without justification -2 points.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On a glade, 25 gnomes gathered. It is known that 1) every gnome who put on a hat also put on shoes; 2) 12 gnomes came without a hat; 3) 5 gnomes came barefoot. Which gnomes are more and by how many: those who came in shoes but without a hat, or those who put on a hat?
|
Answer. There are 6 more gnomes who put on a cap.
## Solution.
From condition 2, it follows that $25-12=13$ gnomes came in a cap.
From condition 1, we get that exactly 13 gnomes came both in a cap and in shoes.
From condition 3, it follows that a total of $25-5=20$ gnomes came in shoes.
Thus, $20-13=7$ gnomes came in shoes but without a cap.
Therefore, those who put on a cap (13 gnomes) are 6 more than those who came in shoes but without a cap (7 gnomes).
## Grading Criteria.
- Correct solution - 7 points.
- Correct answer with incomplete justification - 3-4 points.
- Only the correct answer without justification - 2 points.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The difference of the squares of two numbers is 6, and if each of these numbers is decreased by 2, then the difference of their squares becomes equal to 18. What is the sum of these numbers?
|
Answer: -2.
## Solution.
Given:
\[
\begin{aligned}
& a^{2}-b^{2}=6 \\
& (a-2)^{2}-(b-2)^{2}=18
\end{aligned}
\]
There are different ways to proceed.
## Method 1.
\((a-2)^{2}-(b-2)^{2}=a^{2}-4a+4-b^{2}+4b-4=a^{2}-b^{2}-4(a-b)\). Since from the first condition \(a^{2}-b^{2}=6\), we get \(6-4(a-b)=18\). Hence, \(a-b=-3\), and from the first equation, we get \(a+b=-2\).
## Method 2.
\[
\begin{aligned}
& a^{2}-b^{2}=(a-b)(a+b)=6 \\
& (a-2)^{2}-(b-2)^{2}=(a-b)(a+b-4)=18
\end{aligned}
\]
Clearly, all factors in the given equations are not zero. Dividing the second equation by the first and denoting the desired sum \(a+b=x\), we get \(\frac{x-4}{x}=3\), from which \(x=-2\).
## Grading Criteria.
- Correct solution - 7 points.
- Obtained the equation \(\frac{a+b-4}{a+b}=3\) but made an arithmetic error in further solution - 4 points.
- Correctly found the difference \(a-b\), but no further progress - 3 points.
- Made a sign error in finding \(a+b\) or \(a-b\) - no more than 3 points.
- Only the correct answer without justification - 2 points.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. On the board, the numbers $\sqrt{2}$ and $\sqrt{5}$ are written. It is allowed to add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that it is possible to write the number 1 on the board.
|
Solution: The simplest way is to provide a sequence of numbers that will lead to the number 1. For example, the following sequence works:
$$
\begin{gathered}
\sqrt{2}+\sqrt{5}, \quad 2 \sqrt{2}+\sqrt{5}, \quad 3 \sqrt{2}+\sqrt{5}, \quad \sqrt{2}-\sqrt{5}, \quad 2 \sqrt{2}-\sqrt{5}, \quad 3 \sqrt{2}-\sqrt{5} \\
(\sqrt{2}+\sqrt{5})(\sqrt{2}-\sqrt{5})=-3, \quad(3 \sqrt{2}+\sqrt{5})(3 \sqrt{2}-\sqrt{5})=13 \\
13-3=10, \quad 10-3=7, \quad 7-3=4, \quad 4-3=1
\end{gathered}
$$
Note: The sequence of operations is not unique.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct sequence of listed numbers | 7 points |
| Correct sequence is not present, but a method to obtain an integer is found | 3 points |
| Examples of sequences that do not lead to the answer or examples of impossible sequences | 0 points |
|
1
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
9.6. Let \(a\) and \(b\) be positive numbers. Find the minimum value of the fraction \(\frac{(a+b)(a+2)(b+2)}{16ab}\). Justify your answer.
|
Solution: By the inequality between the arithmetic mean and the geometric mean of two numbers, the following three inequalities hold:
$$
a+b \geqslant 2 \sqrt{a b}, \quad a+2 \geqslant 2 \sqrt{2 a}, \quad b+2 \geqslant 2 \sqrt{b}
$$
Multiplying the left and right sides of these three inequalities, we get
$$
(a+b)(a+2)(b+2) \geqslant 8 \sqrt{a b \cdot 2 a \cdot 2 b}
$$
which is equivalent to the inequality $\frac{(a+b)(a+2)(b+2)}{16 a b} \geqslant 1$. The value 1 is achieved at the set $a=b=2$.
Note: The fraction given in the problem is a special case of the fraction $\frac{(a+b)(a+c)(b+c)}{8 a b c}$. Its minimum (for positive variables) is also 1 and is achieved when $a=b=c$. The proof of this fact does not differ from the one provided in the solution.
## Answer: 1.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| The equation $\frac{(a+b)(a+2)(b+2)}{16 a b}=t$ (or an equivalent one) is investigated as a quadratic (in one of the variables) with parameters, but the investigation is incomplete or incorrect | 4 points |
| The case $a=b$ is correctly investigated, but it is not justified that the minimum of the fraction is achieved precisely under this assumption | 3 points |
| The set at which the minimum value of the fraction is achieved $(a=b=2)$ is indicated, but its optimality is not proven | 1 point |
| Correct answer without justification and/or any calculations from which the solution process is not clear | 0 points |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Matvey decided to start eating properly and every day he ate one bun less and one pear more than the previous day. In total, during the time of proper nutrition, he ate 264 buns and 187 pears. How many days was Matvey on a proper diet?
|
Answer: 11 days.
Solution: If we "reverse" the sequence of the number of buns, while leaving the pears unchanged, the total number of buns and pears will not change, and the difference between the number of buns and pears eaten each day will become constant. Since $264-187=77=7 \cdot 11$, the correct diet lasted either 7, 11, or 77 days. Since the number of days is odd, the total number of buns and pears eaten is equal to the product of the number of days and the number of buns and pears eaten on the middle day. But 264 is not divisible by 7, so the number of days cannot be 7 or 77.
Criteria: Full solution - 7 points. If the reason why there cannot be a different number of days is not justified, but the rest is correct - 5 points. Only the answer - 1 point. The problem assumes that the event has already occurred, so the student is not required to provide an example of how Matvey could have eaten this way.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. Option 1.
Along a road, 10 lampposts were placed at equal distances, and the distance between the outermost posts was $k$ meters. Along another road, 100 lampposts were placed at the same distances, and the distance between the outermost posts was $m$ meters. Find the ratio $m: k$.
|
Answer: 11.
Solution: Let the distance between adjacent posts be $x$ meters. Then, in the first case, the distance between the outermost posts is $(10-1) x=9 x=k$ meters. And in the second case, $(100-1) x=99 x=m$ meters. Therefore, the desired ratio is $m: k=(99 x):(9 x)=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.3. Sasha had 47 sticks. Using them all, he formed several letters "B" and "V" as shown in the figure. What is the maximum number of letters "B" that Sasha could have formed?
|
Answer: 8.
Solution. To form the letter "Б", 4 sticks are needed, and to form the letter "В", 5 sticks are needed.
- Sasha could not have formed at least 12 letters "Б" because it would require no less than 48 sticks.
- If Sasha had formed 11 letters "Б", he would have $47-11 \cdot 4=3$ sticks left. This would not be enough even for one letter "В".
- If Sasha had formed 10 letters "Б", he would have $47-10 \cdot 4=7$ sticks left. One letter "В" could be formed, but there would be extra sticks left.
- If Sasha had formed 9 letters "Б", he would have $47-9 \cdot 4=11$ sticks left. Two letters "В" could be formed, but there would be one extra stick left.
- If Sasha had formed 8 letters "Б", he would have $47-8 \cdot 4=15$ sticks left. This is exactly enough for 3 letters "В".
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.6. Zhenya took a $3 \times 3$ board and placed a column of blue and red cubes on each cell. Then he drew a diagram of the resulting arrangement: he labeled the number of cubes of both colors in each column (the order of the cubes is unknown).
What is the maximum number of blue cubes Zhenya can see if he looks at the construction from the front? (For example, if a column of 8 cubes is in front of a column of 5, then all 5 cubes of the closer column and only the top 3 cubes of the farther column will be visible.)
|
Answer: 12.
Solution. Let's understand the maximum number of blue cubes Zhenya can see in each of the three rows: left, middle, and right.
Left row. The first column consists of 5 cubes (2 red and 3 blue), so it completely blocks the second column, as well as 5 out of 7 cubes in the last column.
Thus, Zhenya sees all the cubes in the first column (3 of which are blue), as well as 2 cubes from the last column (both can be blue). That is, in this row, he will see a maximum of $3+2=5$ blue cubes.
Middle row. The first column consists of 4 cubes (2 red and 2 blue), so it completely blocks the last column, as well as 4 out of 5 cubes in the second column.
Thus, Zhenya sees all the cubes in the first column (2 of which are blue), as well as 1 cube from the second column (it can be blue). That is, in this row, he will see a maximum of $2+1=3$ blue cubes.
Right row. The first column consists of 3 cubes (2 red and 1 blue), so it blocks 3 out of 6 cubes in the second column. Meanwhile, the second column completely blocks the last column.
Thus, Zhenya sees all the cubes in the first column (1 of which is blue), as well as 3 cubes from the second column (all three can be blue). That is, in this row, he will see a maximum of $1+3=4$ blue cubes.
In total, Zhenya will see a maximum of $5+3+4=12$ blue cubes.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.7. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, it is allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal?
|
Answer: 11.
Solution. Suppose $N$ moves were made, after which the number of coins in all stacks became equal.
Let's slightly change the rules. Suppose initially there were not 9, 7, 5, and 10 coins in the stacks, but $N+9, N+7, N+5$, and $N+10$ respectively; and we will perform the moves as follows: instead of adding one coin to three stacks, we will take one coin from a stack (the one into which we did not add a coin during the original move). Note that the final result will not change! (In fact, instead of adding coins to three stacks, we add them to all four and then take one away.)
Under the new rules, answering the question is much easier. In one move, we take one coin from any stack, and our goal is to make the number of coins in all stacks equal as quickly as possible. It is easy to understand that for this, we need to leave $N+5$ coins in each stack. For this, we need to take 4 coins from the first stack, 2 from the second, 0 from the third, and 5 from the fourth. In total, we need to make $4+2+0+5=11$ moves.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's number, if it is known that after he ran away, 3 people remained in the line? (After each command, one or several players ran away, after which the line closed, and there were no empty spaces between the remaining players.)
|
Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
|
Answer: 3.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals are bisected by their intersection point $L$, it is a parallelogram (in particular, $AC = DX$). Therefore, $DX \parallel AC$. Since $AC \parallel ED$ by the condition, the points $X, D, E$ lie on the same line.
Since $AC \parallel EX$, then $\angle EAX = \angle CAX = \angle AXE$, i.e., triangle $AEX$ is isosceles, $EA = EX$. Then
$$
ED = EX - XD = EA - AC = 15 - 12 = 3
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Dad is preparing gifts. He distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest gifts, there are 50 candies. How many bags are the candies distributed into? How many candies are in the smallest gift?
|
Answer: 10 packages, 5 candies.
Solution. Let's number the gifts from the smallest to the largest, from 1 to $n$. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, the third from the last gift has no more than 15 candies $(16+17+18=51>50)$.
Remove the three largest and the three smallest gifts. In the remaining gifts, there will be $115-20-$ $50=45$ candies, and each has between 9 and 14 candies. Three packages are clearly not enough $(14+13+12=39)$, and five would be too many $(9+10+11+12+13=55)$. Therefore, 45 candies are distributed in 4 packages. This is possible: $47=9+11+12+13$. Note that the fourth package cannot have more than 9 candies: $10+11+12+13=46>45$.
If the fourth package has 9 candies, then the third has no more than 8, the second no more than 7, so the first package has no less than $20-8-7=5$ candies. But no more than that, since $6+7+8=21$.
Criteria. Only the answer - 0 points. Only correct estimates for the third from the beginning and the third from the end package - 3 points. Only a justified answer for the number of packages - 5 points. Full solution - 7 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4 Let $n$ - be a natural number greater than 10. What digit can stand immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$? Provide all possible answers and prove that there are no others.
|
Solution. Method 1. $n^{2}+n=(n+0.5)^{2}-0.2510$ ), the digit immediately after the decimal point is no less than 4. That is, it is 4.
Method 2. The required digit is the last digit in the number $\left[10 \sqrt{n^{2}+n}\right]=$ $\left[\sqrt{100 n^{2}+100 n}\right]$. Note that $100 n^{2}+100 n100 n^{2}+80 n+16=(10 n+4)^{2}$. Then $\left[\sqrt{100 n^{2}+100 n}\right]=$ $10 n+4$, and the last digit of this number is 4.
Method 3. We have $n+0.4$ solved incorrectly, which may have led to an incorrect answer | 5 points |
| One of the two statements is proven: a) the required digit is less than 5; b) the required digit is greater than 3; plus an example is provided showing that the digit 4 is possible | 4 points |
| One of the two statements is proven: a) the required digit is less than 5; b) the required digit is greater than 3 | 3 points |
| The problem is correctly reduced to a system of inequalities or to an equation with an integer part | 2 points |
| The correct answer is illustrated with an example showing that the digit 4 is possible | 1 point |
| The correct answer without justification or an incorrect answer | 0 points |
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find the sum $\sin x + \sin y + \sin z$, given that $\sin x = \tan y$, $\sin y = \tan z$, $\sin z = \tan x$.
|
Answer: 0.
First solution. From $\sin x = \tan y$, we get $\sin x \cos y = \sin y$. Therefore, $|\sin x| \cdot |\cos y| = |\sin y|$. This means $|\sin x| \geq |\sin y|$, and the inequality becomes an equality only if either $\sin y = \sin x = 0$, or $|\cos y| = 1$ (which again implies $\sin y = \sin x = 0$). Similarly, from the remaining equations, we obtain the inequalities $|\sin y| \geq |\sin z|$ and $|\sin z| \geq |\sin x|$. Thus, $|\sin x| \geq |\sin y| \geq |\sin z| \geq |\sin x|$. Therefore, $|\sin x| = |\sin y| = |\sin z|$. Since all inequalities have become equalities, we have $\sin x = \sin y = \sin z = 0$, and $\sin x + \sin y + \sin z = 0$.
Second solution. If one of the sines is zero, then the tangent equal to it is also zero, which means the sine in the numerator of the tangent is zero. Consequently, the other sines and tangents are also zero. In this case, $\sin x + \sin y + \sin z = 0$.
Suppose none of the sines are zero. Multiplying all three equations, we get $\sin x \sin y \sin z = \tan y \tan z \tan x = \frac{\sin x \sin y \sin z}{\cos x \cos y \cos z}$. Since $\sin x \sin y \sin z \neq 0$, we have $\cos x \cos y \cos z = 1$. This is only possible if $|\cos x| = |\cos y| = |\cos z| = 1$, which means the sines are zero, and the considered case is impossible.
Comment. A correct answer without justification - 0 points.
An answer obtained by considering an example - 1 point.
At least one case is incorrectly considered (or omitted) - no more than 3 points.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (7 points) 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." That child said to their right neighbor: "The majority of us are girls," and that one to their right neighbor: "The majority of us are boys," and so on, until the last child said to the first: "The majority of us are boys." How many boys are at the table?
|
Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. A group of boys sitting next to each other is followed by a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so their number is even. Since the statement "most of us are boys" was heard seven times, six of the statements "most of us are girls" were false, and there were six groups.
The alternation of true and false statements means that there were two children in each group. Only the first and last child sitting next to each other said the same thing, so there were three people in their group. These are boys, as they are in the majority. In total, there were $2+2+2=6$ girls and $2+2+3=7$ boys sitting at the table.
The diagram shows exactly how the children were seated at the table. The first speaker is outlined in a frame.
Answer. 7.
Maximum points - 35.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, you are allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal?
#
|
# Answer: in 11 moves.
Solution 1. Consider the differences between the number of coins in each of the other stacks and the number of coins in stack No. 3. In one move, either each of these differences increases by 1 (if all stacks except 3 are chosen for adding coins), or exactly 1 of these differences decreases by 1 (in other cases). Initially, these differences are 4, 2, and 5. Thus, to reduce all these differences to 0, it will take no less than $4+2+5=11$ moves.
11 moves are possible: 4 times we choose all stacks except No. 1, 2 times all stacks except No. 2, and 5 times all stacks except No. 4. In the end, there will be 16 coins in each stack.
Solution 2. Let us say we have made $s$ moves, of which $s_{\mathrm{k}}$ moves involve adding a coin to all stacks except $\mathrm{k}(\mathrm{k} \in\{1,2,3,4\})$. Then, from the condition that the number of coins in all stacks is equal after all moves, we get: $9+\mathrm{s}-\mathrm{s}_{1}=7+\mathrm{s}-\mathrm{s}_{2}=5+\mathrm{s}-\mathrm{s}_{3}=10+\mathrm{s}-\mathrm{s}_{4}$. Therefore, $s_{1}=4+s_{3} \geq 4, s_{2}=2+s_{3} \geq 2$ and $s_{4}=5+s_{3} \geq 5$. Thus, $s \geq s_{1}+s_{2}+s_{4} \geq 4+2+5=11$. An example of 11 moves is the same as in Solution 1.
Criteria. Only the answer - 1 point. Answer with an example - 3 points. Only proof that the number of moves is no less than 11 - 4 points.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. 4.1. In a right triangle $ABC$ (right angle at $C$), the bisector $BK$ is drawn. Point $L$ on side $BC$ is such that $\angle CKL = \angle ABC / 2$. Find $KB$, if $AB = 18, BL = 8$.
|
Answer: 12.
## Solution.
Note that $\angle L K B=\angle C K B-\angle C K L=\angle C A B+\angle A B K-\angle C K L$ (the last equality holds because $\angle C K B$ is an exterior angle of triangle $A B K$). Since $\angle C K L=\angle A B C / 2=\angle A B K$, we have that $\angle L K B=\angle C A B$. From the fact that $\angle A B K=\angle K B L$, we conclude that triangles $K L B$ and $A K B$ are similar by two angles. We obtain the equality of ratios $\frac{K B}{B L}=\frac{A B}{K B}$. Therefore, $K B=\sqrt{B L \cdot A B}=\sqrt{18 \cdot 8}=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In the rebus
$$
\mathbf{K}\mathbf{O}>\mathbf{H}>\mathbf{A}>\mathbf{B}>\mathbf{U}>\mathbf{P}>\mathbf{y}>\mathbf{C}
$$
different letters represent different digits. How many solutions does the rebus have?
|
Answer: 0.
Solution. From the rebus, it follows that $\mathbf{P}>\mathbf{O}>\mathbf{P}$. This cannot be! Evaluation. 7 points for the correct solution.
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Seryozha decided to start eating properly and every day he ate one fewer chocolate cookie and one more sugar-free cookie than the previous day. In total, during the time of proper nutrition, he ate 264 chocolate cookies and 187 sugar-free cookies. How many days was Seryozha on a proper diet?
|
Answer: 11 days.
Solution 1. The total number of cookies eaten each day is the same. In total, Seryozha ate $264+187=451=11 \cdot 41$ cookies. Since the proper diet lasted more than one day and he ate more than one cookie each day, either he ate for 11 days, 41 cookies each day, or for 41 days, 11 cookies each day. But since the number of sugar-free cookies increased by 1 each day, in the second case, it would have exceeded 11 on one of the days. Therefore, this case is impossible.
Solution 2. If we "reverse" the sequence of the number of chocolate cookies, while leaving the sugar-free cookies unchanged, the total number of cookies will not change, but the difference between the number of chocolate cookies and sugar-free cookies eaten each day will become constant. Since $264-187=77=7 \cdot 11$, the proper diet lasted either 7, 11, or 77 days. Since the number of days is odd, the total number of cookies eaten is the product of the number of days and the number of cookies eaten on the middle day. But 264 is not divisible by 7, so the number of days cannot be 7 or 77.
Criteria. Full solution - 7 points. Not justified why there cannot be a different number of days, but the rest is correct - 5 points. Only the answer - 1 point. The problem assumes that the event has already occurred, so the student is not required to provide an example of how Seryozha could have eaten this way.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Three bear cubs were dividing three pieces of cheese weighing 10 g, 12 g, and 15 g. A fox came to help them. She can simultaneously bite and eat 1 g of cheese from any two pieces. Can the fox leave the bear cubs equal pieces of cheese?
|
Answer: She can.
Solution. Let's provide one of the possible examples of how the fox could do it. For convenience, let's record the results of the fox's "work" in a table.
| 10 | 12 | 15 |
| :---: | :---: | :---: |
| 9 | 12 | 14 |
| 8 | 12 | 13 |
| 7 | 12 | 12 |
| 7 | 11 | 11 |
| 7 | 10 | 10 |
| 7 | 9 | 9 |
| 7 | 8 | 8 |
| 7 | 7 | 7 |
How one might guess the solution. First, try to equalize only two pieces of cheese, and then all three.
## Grading Criteria.
- Correct algorithm (regardless of length, written in words or in a table) - 7 points.
- Generally correct algorithm with missing steps - **5** points.
- Shows how to equalize two pieces, but no further progress - 2 points.
- Only the answer "yes" or "she can" - **0** points.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the underwater kingdom, there live octopuses with seven and eight legs. Those with 7 legs always lie, while those with 8 legs always tell the truth. One day, a conversation took place between three octopuses.
Green octopus: "We have 21 legs together." Blue octopus (to the green one): "You're lying!" Red octopus: "Both of you are lying!"
1) Could the green octopus have told the truth? Why?
2) How many legs did each octopus have? (Justify your answer.)
|
Answer. 1) Could not.
2) The green octopus has 7 legs, the blue one has 8 legs, and the red one has 7 legs.
Solution.
1) If the green octopus had told the truth, then each octopus would have 7 legs. This means that the green octopus, according to the condition, should have lied. We get a contradiction, so the green octopus lied.
2) Since the green octopus lied, it has 7 legs. The blue octopus told the truth about the green one, so it has 8 legs. The red octopus lied because not both but only one in front of him lied, so the red one has 7 legs.
## Grading Criteria.
- Correct solution for both parts - 7 points.
- Correct solution for the first part + partial solution for the second - 4-5 points.
- Correct solution for the first part - 3 points.
- Partially correct but incomplete reasoning - 1-2 points.
- Only answers for all parts - $\mathbf{0}$ points.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 11.5. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars always lie, and truth-tellers always tell the truth. What is the minimum number of liars in the presidium for the described situation to be possible? (Two members of the presidium are considered neighbors if one of them sits to the left, right, in front, or behind the other).
|
Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The figure shows how eight liars can be seated in the presidium so that the condition of the problem is satisfied.
Comments. Without an example of seating the liars - 5 points.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. CONDITION
In what minimum number of colors should natural numbers be painted so that any two numbers, the difference between which is 3, 4, or 6, are of different colors?
|
Solution. Let the number $n=1$ be color $A$, then the numbers 4, 5, and 7 must be painted in another color. Let $n=4$ be color $B$, then from $7-4=3$, it follows that the number $n=7$ is of the third color $C$. Therefore, at least 3 colors are required. The coloring $A A A B B B C C C A A A B B B \ldots$ is the desired one.
Answer. 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by one. What is the maximum number of viewers who could have voted after moment $T$?
(O. Dmitriev, R. Zhenodarov)
|
Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. Thus, the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y = x - n - 1$. The maximum possible value of $x$ is 10, and the minimum possible value of $n$ is 1; therefore, the maximum value of $y$ (on the first such step) is 8.
With each subsequent step, the value of $x$ decreases by 1, and the value of $n$ increases by 1. Consequently, on the second step, the value of $y$ does not exceed 6, on the third step - 4, and so on. Since any rating is not less than 0, the number of steps does not exceed 5.
It remains to show that five steps are possible. Suppose that at moment $T$ the rating is 10 (with 1 voter), then the second viewer gives 8 points, the third - 6, the fourth - 4, the fifth - 2, and the sixth - 0. Then the rating sequentially takes the values $9, 8, 7, 6,$ and 5.
Comment. Only the answer - 0 points.
Only an example is provided where 5 viewers voted - 2 points.
It is proven with full justification that more than 5 viewers could not have voted (but the example is missing or incorrect) - 5 points.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by one. What is the maximum number of viewers who could have voted after moment $T?$
(O. Dmitriev, R. Zhenodarov)
|
# Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. This means the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y = x - n - 1$. The maximum possible value of $x$ is 10, and the minimum possible value of $n$ is 1; therefore, the maximum value of $y$ (on the first such step) is 8.
With each subsequent step, the value of $x$ decreases by 1, and the value of $n$ increases by 1. Consequently, on the second step, the value of $y$ does not exceed 6, on the third step - 4, and so on. Since any rating is not less than 0, the number of steps does not exceed 5.
It remains to show that five steps are possible. Suppose the rating at moment $T$ is 10 (with one voter), then the second viewer gives 8 points, the third - 6, the fourth - 4, the fifth - 2, and the sixth - 0. Then the rating sequentially takes the values $9, 8, 7, 6,$ and 5.
Comment. Only the answer - 0 points.
Only an example is provided where 5 viewers voted - 2 points.
It is proven with full justification that more than 5 viewers could not have voted (but the example is missing or incorrect) - 5 points.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were worked with planned productivity, and then it decreased by $30 \%$. By what percentage did the overall labor productivity for the shift increase as a result of extending the working day?
|
1. Answer: by 8 percent. Solution. Let's take 1 for the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times 100 \% = 108 \%$ of the initial, meaning it increased by $8 \%$. Grading criteria. Correct solution - 7 points. Initial and final productivity per shift correctly expressed in conditional units, but the percentage ratio is expressed incorrectly - 4 points. In all other cases - 0 points.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Triangle $A B C$ is similar to the triangle formed by its altitudes. Two sides of triangle $A B C$ are 4 cm and 9 cm. Find the third side.
|
4. Answer. 6 cm. Solution. Let $A B=9, B C=4, A C=x, S$ - the area of triangle $A B C$. Then the heights of the triangle are $2 S / 4, 2 S / 9, 2 S / x$. By the triangle inequality, side $A C$ can be either the largest or the middle in length. Applying the triangle inequality to the triangle formed by the heights, we see that among the quantities $2 S / 4, 2 S / 9, 2 S / x$, the quantity $2 S / x$ can also be either the largest or the middle. However, the larger the side, the smaller the height drawn to it, and the height to the largest side can only be the smallest. Therefore, the side $x$ (and the height drawn to it) can only be the middle. This allows us to write the proportion $9: x = 2 S / 4: 2 S / x$ and $x=6$. Grading criteria. Full solution - 7 points. The proportion is written correctly, but not justified (in any way) - 4 points. In all other cases (including an answer without a solution) - 0 points.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. Six different natural numbers from 6 to 11 are placed on the faces of a cube. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 36, and the second time it was 33. What number is written on the face opposite the one with the number 10? Justify your answer.
|
Solution: The sum of the numbers on all 6 faces of the die is 51. Therefore, the sum of the numbers on the top and bottom faces during the first roll is 15, and during the second roll, it is 18. The number 15 can be obtained (as the sum of two different integers from the interval [6; 11]) in two ways: 15 = 9 + 6 = 8 + 7. The number 18 also in two ways: 18 = 7 + 11 = 8 + 10. If the numbers 7 and 8 lie on opposite faces, then both of the latter representations are impossible. Therefore, of the two specified representations of the number 15, the first one holds, and in the die, the number 9 lies opposite the number 6. Now, if during the second roll, the numbers on the top and bottom faces were 8 and 10, it is clear that the number 10 lies opposite the number 8. And if during the second roll, the numbers on the top and bottom faces were 7 and 11, then on one pair of opposite side faces are the numbers 6 and 9, and on the other - the remaining 8 and 10, and again, the number 10 lies opposite the number 8.
Answer: 8.
Note: The problem can be easily reduced to a regular die as follows: remove 5 dots from each face. Then the die will become regular, and the condition will read: "The first time, the sum of the numbers on the four side faces was 16, the second time - 13. What number is written on the face opposite the one where the number 5 is written?". Further reasoning does not change fundamentally.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| Correct answer, supported by a correct example of a die (any of the two possible) | 3 points |
| Impossibility of some situations is justified (for example, that 10 cannot lie opposite 11 or that the sum during the first roll cannot be realized as 6 + 9 + 10 + 11), but not all impossible situations are analyzed | 2 points |
| The problem is correctly reduced to a regular die | |
| Answer without justification (or incorrect answer) | 1 point |
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other?
|
Answer: 8.
Solution: Note that it is impossible to lay out all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor card with the number 1. Therefore, both cards 5 and 7 must be at the edges, and the card with the number 1 must be adjacent to each of them, which is impossible.
It is possible to select 8 cards and arrange them in a row according to the requirements of the problem, for example: $9,3,6,2,4,8,1,5$.
Comment: Proof that it is impossible to lay out all the cards - 4 points.
Any correct example of arranging 8 cards - 3 points.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent?
|
Answer: 5 cupcakes.
Solution: The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote P, K, and B as the costs of a cake, a cupcake, and a bagel, respectively, we get the equation: $(P + 2K + 3L) + (3P + 5) = 12K$, from which it follows that $4P + 4B = 10K$, or $2P + 2B = 5K$.
Remark: From the condition, we can also calculate all the cost ratios, specifically, $P : K : B = 7 : 4 : 3$.
Comment: Only the correct answer without explanation -1 point.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. What is the minimum number of 3-cell corners that need to be painted in a $6 \times 6$ square so that no more corners can be painted? (Painted corners must not overlap.)
|
Answer: 6.
Solution. Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted. Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get that at least $9 \cdot 2 = 18$ cells are painted. Therefore, at least 6 corners are painted.
Figure 3 shows how to paint 6 corners so that no more corners can be painted.
Comment. It is proven that the number of painted corners is no less than $6-4$ points.
An example with 6 painted corners is drawn - 3 points.
Figure 3
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. From points $A$ and $B$, two cars set off towards each other simultaneously with constant speeds. One hour before the first car arrived at $B$ and four hours before the second car arrived at $A$, they met. Find the ratio of the speeds of the cars.
|
3. Answer: the speed of the first car is twice as high.
Let the speed of one car be $k$ times the speed of the other. Since they started at the same time, one car will travel a distance $k$ times greater. After this, the segments will switch, and the slower car, whose speed is $k$ times less, will have to cover a distance $k$ times greater. It is clear that it will take $k^{2}$ times more time for this. In our case, this number is 4, from which $k=2$. That is, the speed of the first car (which has less time left to travel) is twice the speed of the second.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On his birthday, Piglet baked a large cake weighing 10 kg and invited 100 guests. Among them was Winnie-the-Pooh, who is fond of sweets. The birthday boy announced the rule for dividing the cake: the first guest cuts a piece of the cake that is $1 \%$, the second guest cuts a piece of the cake that is $2 \%$ of the remaining part, the third guest cuts a piece of the cake that is $3 \%$ of the remaining part, and so on. What place in line should Winnie-the-Pooh take to get the largest piece of cake?
|
Solution. The first guests in the queue receive increasingly larger pieces of the pie because the remaining part of the pie is large at the initial stages of division. However, since the remaining part of the pie decreases, there will come a point when guests start receiving smaller pieces of the pie. At which guest will this happen?
Let's compare the pieces of the pie received by the $(n-1)$-th and $n$-th guests. Suppose the mass of the remaining part of the pie when it was the $(n-1)$-th guest's turn is $m$ kg. Then the $(n-1)$-th guest received $\frac{n-1}{100} \cdot m$ kg, and the remaining part of the pie became $m - \frac{n-1}{100} \cdot m$ kg, which simplifies to $\frac{101-n}{100} \cdot m$ kg. Therefore, the $n$-th guest received $\frac{101-n}{100} \cdot m \cdot \frac{n}{100} = \frac{n(101-n)}{10000} \cdot m$ kg. The difference between the pieces received by the $n$-th and $(n-1)$-th guests is $\frac{n(101-n)}{10000} \cdot m - \frac{n-1}{100} \cdot m = \frac{-n^2 + n + 100}{10000} \cdot m$. This difference is positive if $n^2 - n - 100 < 0$. Solving the quadratic inequality, we find that the largest natural solution is $n=10$. Therefore, the 10th guest in line will receive the largest piece of the pie, so Winnie-the-Pooh should take the 10th place in the queue.
Answer: 10.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On the website of the football club "Rostov," a poll is being conducted to determine which of the $m$ football players the website visitors consider the best at the end of the season. Each visitor votes once for one player. The website displays the rating of each player, which is the percentage of votes cast for them, rounded to the nearest whole number. After several visitors have voted, the total rating of the nominees was $95 \%$. What is the smallest $m$ for which this is possible?
|
Solution. Let $a$ be the greatest loss of a percentage point due to rounding when determining a footballer's rating. Then, according to rounding rules, $aa m \geqslant 5$, which means $0.5 m>5$ or $m>10$.
We will show that a solution exists when $m=11$. For example, let $m=11$ and 73 visitors voted, with 33 of them voting for one footballer, while the remaining 40 visitors evenly distributed their votes among 10 footballers, giving each 4 votes. In this case, the total rating is
$\frac{33}{73} \cdot 100 \%+10 \cdot \frac{4}{73} \cdot 100 \approx 45 \%+10 \cdot 5 \%=95 \%$.
Thus, the minimum $m=11$. Answer: 11.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 10.1
For all real $x$ and $y$, the equality $f\left(x^{2}+y\right)=f(x)+f\left(y^{2}\right)$ holds. Find $f(-1)$.
## Number of points 7
|
Answer 0.
Solution
Substituting $x=0, y=0$, we get $f(0)=f(0)+f(0)$, that is, $f(0)=0$.
Substituting $x=0, y=-1$, we get $f(-1)=f(0)+f(1)$, that is, $f(-1)=f(1)$.
Substituting $x=-1, y=-1$, we get $f(0)=f(-1)+f(1)$. Therefore, $2 f(-1)=0$.
#
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board?
(S. Berlov)
#
|
# Answer. Two.
Solution. Note that no two squares of natural numbers differ by 1, because $x^{2}-y^{2}=(x-y)(x+y)$, where the second bracket is greater than one. Therefore, the numbers $a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwise their product $a^{2} \cdot b(b+2)$ would also be a square, and thus the number $b(b+2)$ would be a square as well. Similarly, among the numbers $(a+2)b$ and $(a+2)(b+2)$, at most one can be a square. In total, there are no more than two squares on the board.
Two squares can be obtained, for example, with $a=2$ and $b=16$: then $a(b+2)=6^{2}$ and $(a+2)b=8^{2}$.
Remark. There are other examples as well, such as $a=6$ and $b=96$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation of the number $a+k b$ also be 15? (A.S. Golyanov)
|
# Answer. $k=6$.
Solution. By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
We will use the following known fact: the decimal representation of a rational number $r$ is purely periodic with (not necessarily minimal) period length $T$ if and only if $r$ can be written as $\frac{m}{10^{T}-1}$ for some integer $m$.
Applying this to the problem, this means that $a=\frac{m}{10^{30}-1}$ and $b=\frac{n}{10^{30}-1}$. We also know that the numbers $a-b=\frac{m-n}{10^{30}-1}$ and $a+k b=\frac{m+k n}{10^{30}-1}$ can be written as decimal fractions with a period length of 15, meaning they can be written as common fractions with the denominator $10^{15}-1$. Therefore, their difference $(k+1) b=\frac{(k+1) n}{10^{30}-1}$ can also be written with the denominator $10^{15}-1$. Thus, the number $(k+1) n$ is divisible by $10^{15}+1$, while the number $n$ is not (otherwise, $b$ would be written as a fraction with a period length of 15). Therefore, the number $k+1$ is divisible by some prime divisor of the number $10^{15}+1$. The smallest such divisor is 7. Indeed, the number $10^{15}+1$ gives a remainder of 1 when divided by 2 and 5, and a remainder of 2 when divided by 3. On the other hand, it is divisible by $10^{3}+1=7 \cdot 143$. Thus, $k+1 \geqslant 7$ and $k \geqslant 6$.
In a certain sense, the minimal example of numbers satisfying the condition for $k=6$ is obtained by setting $a-b=\frac{1}{10^{15}-1}$ and $a+6 b=\frac{2}{10^{15}-1}$. Then $a=\frac{8}{7\left(10^{15}-1\right)}$ and $b=\frac{1}{7\left(10^{15}-1\right)}$. It is clear that the lengths of the minimal periods of the numbers $a-b$ and $a+6 b$ are 15. Furthermore, the lengths of the minimal periods of the numbers $a$ and $b$ are greater than 15 and are divisible by 15 (since $10^{T}-1$ must divide $10^{15}-1$). On the other hand, since $10^{30}-1 \vdots 7\left(10^{15}-1\right)$, the numbers $a$ and $b$ are periodic with a period length of 30. Therefore, the lengths of their minimal periods are 30.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Three brigades, working together, must complete a certain job. It is known that the first and second brigades together can complete it 36 minutes faster than the third brigade. In the time it takes for the first and third brigades to complete the job together, the second brigade can complete only half of the job. In the time it takes for the second and third brigades to complete the job together, the first brigade can complete $\frac{2}{7}$ of the job. How long will it take for all three brigades to complete the job together?
|
Answer: 1 hour 20 minutes.
Solution.
Let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ be the productivity of the first, second, and third teams, respectively, that is, the part of the work that a team completes in 1 hour. Using the first condition of the problem, we form an equation. Since the productivity of the first and second teams working together is $(x+y)$, they will complete the entire work in $\frac{1}{x+y}$ hours. One third of the third team will complete the work in $\frac{1}{z}$ hours. We have: $\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z}$.
Similarly, using the other conditions of the problem, we obtain the system: $\left\{\begin{array}{l}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ \frac{1}{x+z}=\frac{1}{2 y} \\ \frac{1}{y+z}=\frac{2}{7 x}\end{array}\right.$.
$\left\{\begin{array}{c}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ x+z=2 y \\ 2 y+2 z=7 x\end{array} \Leftrightarrow\left\{\begin{array}{c}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ z=2 y-x \\ 2 y+2(2 y-x)=7 x\end{array} \Leftrightarrow\left\{\begin{array}{c}\frac{1}{x+y}+\frac{3}{5}=\frac{1}{z} \\ z=2 x \\ y=1.5 x\end{array} \Leftrightarrow\left\{\begin{array}{c}\frac{1}{x+1.5 x}+\frac{3}{5}=\frac{1}{2 x} \\ z=2 x \\ y=1.5 x\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{1}{6} \\ y=\frac{1}{4} \\ z=\frac{1}{3}\end{array}\right.\right.\right.\right.\right.$
Thus, all three teams together complete the work in $\frac{1}{x+y+z}$ hours, that is, in 1 hour 20 minutes.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. All gnomes are divided into liars and knights. Liars always lie, and knights always tell the truth. On each cell of a $4 \times 4$ board, there is a gnome. It is known that among them, there are both liars and knights. Each gnome stated: “Among my neighbors (by side) there are an equal number of liars and knights.” How many liars are there in total?
|
Answer: 12 liars.
Solution: Any dwarf standing on the side of the square but not in a corner cannot be telling the truth, because they have three neighbors, and among them, there cannot be an equal number of knights and liars. Therefore, these eight dwarfs are liars. Thus, the dwarfs standing in the corners are also liars because both of their neighbors are liars, not a knight and a liar. Therefore, all dwarfs except the four central ones are liars. Since there are knights among the dwarfs, at least one of the four central ones is a knight. Two of his neighbors are already definitely liars, so the two remaining ones are knights. For any of these two neighboring knights to have an equal number of knights and liars among their neighbors, the fourth central dwarf must also be a knight. Thus, there are 12 liars in total. In the diagram below, knights are denoted by the letter $\mathrm{P}$, and liars by the letter $\mathrm{L}$.
| $L$ | $L$ | $L$ | $L$ |
| :--- | :--- | :--- | :--- |
| $L$ | P | P | $L$ |
| $L$ | P | P | $L$ |
| $L$ | $L$ | $L$ | $L$ |
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. A car, moving at a constant speed, traveled from point A to point B in 3 hours. To reduce the travel time on the return trip, the driver left point B at a speed 25% higher, and upon reaching the midpoint of the journey between A and B, increased the speed by another 20%. How long will the return trip take?
|
Answer: 2 hours 12 minutes. Solution. Let the distance from A to B be $a$ (km), and the speed of movement from A to B be $v$ (km/h). Then $\frac{a}{v}=3$. Let $\mathrm{C}$ be the midpoint of the path between A and B. Then the time of movement on the return trip from B to C is $\frac{a / 2}{v \cdot 1.25}=\frac{2}{5} \frac{a}{v}=\frac{6}{5}$ (hour), and the time of movement from C to A is $\frac{a / 2}{v \cdot 1.25 \cdot 1.2}=\frac{1}{3} \frac{a}{v}=1$ (hour). Therefore, the time of the return trip $\frac{6}{5}+1=\frac{11}{5}$ (hour).
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There are candies in five bags. The first has 2, the second has 12, the third has 12, the fourth has 12, and the fifth has 12. Any number of candies can be moved from any bag to any other bag. What is the minimum number of moves required to ensure that all bags have an equal number of candies?
|
Answer: 4.
There are 50 candies in total and they should be 10 each. In four bags, there are 12 each, which is more than 10, and these bags participate in the redistributions to reduce to 10. Therefore, there are no fewer than 4 redistributions.
From the second to the fifth bag, two candies each go to the first.
Criteria. 2 points for an example of redistributions. 5 points for the estimate.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$ if it is known that $\mathrm{a}_{1}=1, \mathrm{a}_{2}=1$, $\mathrm{a}_{3}=-1$
|
Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_{n-3}^{2} \cdot a_{n-4} \cdot a_{n-5}=a_{n-4} \cdot a_{n-5}= \\
& =\left(a_{n-4}\right) \cdot a_{n-5}=\left(a_{n-5} \cdot a_{n-7}\right) \cdot a_{n-5}=a_{n-5}^{2} \cdot a_{n-7}=a_{n-7}
\end{aligned}
$$
That is, the sequence is periodic with a period of 7. Therefore,
$$
\mathrm{a}_{2019}=\mathrm{a}_{288 \cdot 7+3}=\mathrm{a}_{3}=-1
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The sides of the quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ have the following lengths: $A B=9, B C=2$, $C D=14, D A=5$. Find the length of the diagonal $\boldsymbol{A} \boldsymbol{C}$, if it is known that it is an integer.
|
Solution. Apply the triangle inequality to $\triangle ABC$ and to $\triangle ACD$: for the first, we will have that $AB + BC > AC$, that is, $AC < AB + BC = 11$; for the second, we will have that $AC + CD > AD$, that is, $AC > CD - DA = 9$. Therefore, $9 < AC < 11$, from which $AC = 10$.
Answer: $AC = 10$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Unlucky Emelya was given several metal balls, from which he broke 3 of the largest ones (their mass was $35 \%$ of the total mass of all the balls), then lost 3 of the smallest ones, and brought the remaining balls (their mass was $8 / 13$ of the unbroken ones) home. How many balls were given to Emelya
|
Answer: 10 balls.
## Solution:
Let the total mass of all the balls be M. Of these, Emelya broke balls with a total mass of $\frac{35}{100} M = \frac{7}{20} M$, did not break balls with a total mass of $\frac{13}{20} M$, brought home balls with a total mass of $\frac{8}{13} \cdot \frac{13}{20} M = \frac{8}{20} M$, and lost balls with a total mass of $M - \frac{8}{20} M - \frac{7}{20} M = \frac{5}{20} M$.
Notice that Emelya brought home more than 3 balls, because the three heaviest balls together weigh $\frac{7}{20} M$, while he brought home already $\frac{8}{20} M$.
However, Emelya could not have brought home 5 or more balls, because otherwise their average mass would be no more than $\frac{1}{5} \cdot \frac{8}{20} M = \frac{2}{25} M$, while the average mass of the three lightest balls is $\frac{1}{3} \cdot \frac{5}{20} M = \frac{1}{12} M$, and $\frac{1}{12} M > \frac{2}{25} M$ (i.e., the average mass of the three lightest is greater than the average mass of the next few by weight).
Thus, the only possible case is that Emelya brought home 4 balls, which means he was given 10 balls in total.
## Criteria:
Correct answer with incorrect reasoning - 1 point.
Fractions of broken/lost/brought home balls are recorded - plus 1 point.
Proved why he could not bring home 3 or fewer balls - plus 2 points. Proved why he could not bring home 5 or more balls - plus 3 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On the road between cities A and B, there are some poles in certain places, and two numbers are written on each pole: how many kilometers from the pole to city A, and how many kilometers from the pole to city B. A tourist walking from one city to the other saw a pole where one of the numbers was three times the other. After walking 40 kilometers from the first pole, he saw another pole where one of the numbers was three times the other. The tourist walked another 10 kilometers and saw another pole. What is the ratio of the larger number to the smaller number on this pole?
|
Answer: 7.
Solution: Without loss of generality, we can assume that the tourist saw the first post when he was closer to city A. Let the distance from the first post to A be $x$ kilometers, then the distance from it to B is $3x$. If the numbers on the second post are $y$ and $3y$, then $x + 3x = y + 3y$, since the sum of the numbers on the posts is equal to the distance between the cities. Thus, $x = y$, meaning the pairs of numbers on the first and second posts are the same. After 40 kilometers, the distances could not have remained the same, so they must have switched places. Therefore, the tourist is farther from city A by $3x - x = 2x$, that is, $2x = 40, x = 20$. So, on the third post, the numbers are $60 + 10 = 70$ and $20 - 10 = 10$, and their ratio is 7.
Comment: Only the answer - 1 point.
Describing the numerical configuration that leads to the correct answer - 2 points.
Note: A solution that contains only the answer (even if it explains why it is achievable) cannot be complete. It is necessary to explain why there cannot be other possible answers.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. At competitions, an athlete's performance is evaluated by 7 judges, each of whom gives a score (an integer from 0 to 10). To obtain the final score, the best and worst scores from the judges are discarded, and the arithmetic mean is calculated. If the average score were calculated based on all seven scores, the athletes would be ranked in the exact reverse order. What is the maximum number of athletes that could have participated in the competition?
|
Answer: 5.
Solution: Suppose there are no fewer than six dancers. Let $A, a, S_{A}$ be the best score, the worst score, and the sum of all non-discarded scores of the winner, respectively, and $B, b, S_{B}$ be the same for the last athlete. Instead of averages, the dancers can be ranked by the sum of all scores or the sum of all except the extreme ones. From the condition, it follows that such sums are different for all and are in reverse order. Since the sums are integers and there are at least six dancers, the inequalities must hold:
$$
S_{A}-S_{B} \geq 5 \text { and }\left(B+b+S_{B}\right)-\left(A+a+S_{A}\right) \geq 5
$$
Adding these inequalities, we get $B+b-A-a \geq 10$.
Thus, $b \geq A+a+(10-B) \geq A$, i.e., the worst score of the last athlete is not less than the best score of the winner. But then each score of the last athlete is not less than that of the winner, i.e., $S_{B} \geq S_{A}$. This is a contradiction.
Example of score distribution among athletes: $0-0-0-0-0-0-10,0-0-0-0-0-1-8$, $0-0-0-0-1-1-6,0-0-0-1-1-1-4,0-0-1-1-1-1-2$.
Comment: example - 3 points, evaluation - 4 points.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3 In a pond with carp, 30 pikes were released, which began to gradually eat each other. A pike is considered full if it has eaten three other pikes (hungry or full), and each full pike ate exactly one carp for dessert (hungry pikes did not eat carp). What is the maximum number of carp that could have been eaten? Justify your answer.
|
Solution. The maximum number of pikes eaten is 29, so no more than $29: 3=9.6666 \ldots$ pikes can be satiated, which means no more than 9 pikes (and, accordingly, no more than 9 roaches can be eaten). 9 roaches can be eaten as follows (the method is not unique). Choose 9 pikes and denote them as $A_{1}, A_{2}, \ldots, A_{9}$. First, pike $A_{1}$ eats three of the unselected (becomes satiated) and a roach, then pike $A_{2}$ eats pike $A_{1}$, two unselected, and a roach, and so on until pike $A_{9}$, which eats pike $A_{8}$, two unselected, and a roach.
Answer: 9 roaches.
| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| Correct example of how 9 pikes can be satiated, without proof that 9 is the maximum number | 3 points |
| Justification that 10 pikes cannot be satiated (in the absence of an example showing 9 pikes can be satiated) | 2 points |
| Rougher estimates and examples of satiation of fewer than 9 pikes | 0 points |
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.6 On a horizontal line, points $A$ and $B$ are marked, the distance between which is 4. Above the line, two semicircles with a radius of 2 are constructed, centered at points
A and B. Additionally, one circle, also with a radius of 2, is constructed, for which the point of intersection of these semicircles is the lowest point - see the figure. Find the area of the figure obtained by subtracting from the area of the circle the parts common to it and the two semicircles (the shaded area in the figure). Justify your answer.
|
Solution. Method 1. Let the midpoint of segment $A B$ be $E$, the center of the circle be $O$, and the second points of intersection of the circle and the semicircles be $C$ and $D$ (see figure). Since the radii of the semicircles and the radius of the circle are equal, segments $A D, E O$, and $B C$ will be equal to each other. Then points $D, O, C$ lie on the same line (parallel to line $A B$), and quadrilateral $A B C D$ will be a rectangle with sides 4 and 2. The area of this rectangle is 8. We will show that this rectangle has the same area as the desired figure. Indeed, the figures marked with the Roman numeral I (and also with the number II) are equal due to symmetry. $D C$ is a diameter, which divides the area of the circle in half. The figure in question is a semicircle plus figure I plus figure II. The rectangle consists of these same figures, only arranged differently.
To the condition of problem 7.6, the area of the rectangle is 8. We will show that this rectangle has the same area as the desired figure. Indeed, the figures marked with the Roman numeral I (and also with the number II) are equal due to symmetry. $D C$ is a diameter, which divides the area of the circle in half. The figure in question is a semicircle plus figure I plus figure II. The rectangle consists of these same figures, only arranged differently.
To the solution of problem 7.6.
Method 1.
Method 2.
Method 2. Consider the square $A B C D$ lying on the same side of line $A B$ as the figure of interest. Draw semicircles with centers at points $C$ and $D$ with a radius of 2, as shown in the figure. The points of tangency of the semicircles lie exactly on the drawn circle and divide it into 4 equal parts. If we now connect these points sequentially, a square inscribed in the circle is formed. Our figure will result from this square if we replace two segments shaded in a "striped" pattern with two such segments shaded in a "checkerboard" pattern, so its area is equal to the area of the square. The diagonal of this square is equal to the length of $A B$, that is, equal to 4. Then the area of the square (and the area of our figure) is 8.
Answer: 8.
| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| There are arithmetic errors in the correct solution, possibly leading to an incorrect answer | 6 points |
| There is an idea to find a figure with the same composition, the area of which is easy to find | 3 points |
| The correct approach is indicated (for example, subtract the areas of the common parts of the circle and semicircles from the area of the circle), but this plan is not implemented | 1 point |
| Reasoning and calculations from which the solution approach is not clear | 0 points |
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Variant 1.
On the coordinate plane, the graphs of three reduced quadratic trinomials are drawn, intersecting the y-axis at points $-15, -6, -27$ respectively. For each trinomial, the coefficient of $x$ is a natural number, and the larger root is a prime number. Find the sum of all roots of these trinomials.
|
Answer: -9.
## Solution.
The parabola intersects the $O y$ axis at a point whose ordinate is equal to the free term. Therefore, our trinomials have the form $x^{2}+a x-15, x^{2}+b x-6, x^{2}+c x-27$. Let's denote their larger roots by $p, q, r$ respectively. By Vieta's theorem, the second root of the first trinomial is $-a-p$, then $p(-a-p)=-15$ or $p(p+a)=15=3 \cdot 5$. Similarly, $q(q+b)=6=2 \cdot 3, r(r+c)=27=3 \cdot 9$.
Since $p+a>p, q+b>q, r+c>c$ and $p, q, r$ are prime, then $p=3, q=2, r=3$ and $p+a=5$, $q+b=3, r+c=9$. Therefore, the sum of the roots is $3+(-5)+2+(-3)+3+(-9)=-9$.
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (7 points) In rectangle $A B C D$, side $A B$ is equal to 6, and side $B C$ is equal to 11. Bisectors of the angles from vertices $B$ and $C$ intersect side $A D$ at points $X$ and $Y$ respectively. Find the length of segment $X Y$.
## Answer: 1.
|
Solution. Angles $A X B$ and $X B C$ are equal as alternate interior angles when lines $A D$ and $B C$ are parallel and line $B X$ is the transversal. Angles $X B C$ and $X B A$ are equal since $B X$ is the bisector of angle $A B C$. We obtain that $\angle A X B = \angle X B A$, which means triangle $A X B$ is isosceles, $A B = A X = 6$; $X D = A D - A X = 11 - 6 = 5$. Similarly, we get that $A Y = 5$. Then $X Y = A D - A Y - X D = 11 - 5 - 5 = 1$.
Criteria. Any correct solution: 7 points.
Proved that $A Y = 5$, but the length of segment $X Y$ is not found or found incorrectly: 4 points.
Proved that triangle $A B X$ is isosceles and no further progress: 2 points.
Only the correct answer is provided: 1 point.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
|
Answer: 6.
Solution. Let $a, b, c$ be the natural numbers in the three lower circles, from left to right. According to the condition, $14 \cdot 4 \cdot a = 14 \cdot 6 \cdot c$, i.e., $2a = 3c$. From this, it follows that $3c$ is even, and therefore $c$ is even. Thus, $c = 2k$ for some natural number $k$, and from the equation $2a = 3c$ it follows that $a = 3k$.
It must also hold that $14 \cdot 4 \cdot 3k = 3k \cdot b \cdot 2k$, which means $b \cdot k = 28$. Note that by choosing the number $k$, which is a natural divisor of 28, the natural numbers $a, b, c$ are uniquely determined. The number 28 has exactly 6 natural divisors: $1, 2, 4, 7, 14, 28$. Therefore, there are also 6 ways to place the numbers in the circles.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.8. Given an isosceles triangle $ABC (AB = BC)$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
|
Answer: 4.
Solution. Mark point $K$ on ray $B C$ such that $B E=B K$. Then $A E=C K$ as well.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common side) and the angle between them ($\angle C A E=\angle A C K$ - adjacent to the equal base angles of the isosceles triangle). Therefore, $A K=C E=13$ and $\angle A K C=\angle A E C=60^{\circ}$.
In triangle $A D K$, the angles at vertices $D$ and $K$ are $60^{\circ}$, so it is equilateral, and $D K=A K=A D=13$. Therefore, $A E=C K=D K-D C=13-9=4$.
## 8th grade
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Alla thought of a three-digit number, in which there is no digit 0, and all digits are different. Bella wrote down the number in which the same digits are in reverse order. Galia subtracted the smaller number from the larger one. What digit stands in the tens place of the resulting difference?
#
|
# Answer: 9
Solution. Since we are subtracting a smaller number from a larger one, the digit in the hundreds place of the first number is greater than that of the second. Then, in the units place, conversely, the digit of the first number is smaller than that of the second. Therefore, when subtracting, we will have to borrow a unit from the tens place. Initially, the same digits were in the tens place, but now the digit of the minuend is one less than that of the subtrahend. Therefore, we will have to borrow a unit from the hundreds place. In this case, in the tens place, the minuend ends up being 9 more than the subtrahend. Thus, the resulting digit is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Peter has 5 rabbit cages (the cages are in a row). It is known that there is at least one rabbit in each cage. We will call two rabbits neighbors if they sit either in the same cage or in adjacent ones. It turned out that each rabbit has either 3 or 7 neighbors. How many rabbits are sitting in the central cage?
|
Answer: 4 rabbits.
Solution. Let's number the cells from 1 to 5 from left to right.
Notice that the neighbors of the rabbit in the first cell are all the rabbits living in the first two cells. The neighbors of the rabbit in the second cell are all the rabbits living in the first three cells. The third cell cannot be empty, so the rabbit in the second cell has more neighbors than the rabbit in the first cell. Therefore, the rabbit in the first cell has three neighbors, and the rabbit in the second cell has seven neighbors. The difference between the number of neighbors of the rabbit in the second cell and the number of neighbors of the rabbit in the first cell is equal to the number of rabbits in the central cell. Therefore, there are $7-3=4$ rabbits in the central cell.
Such an arrangement of rabbits exists. It is sufficient to place 2 rabbits in the first, second, fourth, and fifth cells, and 4 rabbits in the third cell.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Seryozha placed numbers from 1 to 8 in the circles so that each number, except one, was used exactly once. It turned out that the sums of the numbers on each of the five lines are equal. Which number did Seryozha not use?
|
Answer: 6.
Solution. Note that each number is contained in exactly two lines. Therefore, the doubled sum of all used numbers is five times greater than the sum of the numbers on one line. Thus, the sum of all numbers is divisible by 5.
Note that $1+2+3+4+5+6+7+8=36$. Therefore, for the sum of the used numbers to be divisible by 5, Seryozha must refrain from using the number one or six.
If Seryozha does not use the number one, then the sum of the numbers in each row will be equal to $(2+3+4+5+6+7+8) \cdot 2: 5=14$. But we have two rows, each containing two numbers, and 14 can be represented as the sum of two considered digits in only one way: $14=6+8$.
Thus, Seryozha does not use the digit 6. For example, he can accomplish the given task as follows:
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Each of 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than $2 ”, \ldots$, the tenth said: “My number is greater than 10”. After that, all ten, speaking in some order, said: “My number is less than 1”, “My number is less than $2 ”, \ldots$, “My number is less than 10” (each said exactly one of these ten phrases). What is the maximum number of knights that could have been among these 10 people?
|
Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My number is less than 1," "My number is less than 2," ..., "My number is less than 10." Therefore, there could not have been more than eight knights.
We will show that there could have been 8 knights. Suppose the first knight thought of the number 2, the second -3, ..., the eighth -9, and the liars thought of the numbers 5 and 6. Then the $k$-th knight could have said the phrases "My number is greater than $k$" and "My number is less than $k+2$," while the liars could have said the phrases: one - "My number is greater than 9" and "My number is less than 1," and the other - "My number is greater than 10" and "My number is less than 2."
Remark. The example given above ceases to be valid if the liars think of numbers outside the interval $[1 ; 10]$, as then some of their statements become true.
Comment. Proved that there are no more than $9-$ 0 points.
Proved that there are no more than 8 knights (or, equivalently, at least two liars -3 points.
Provided an example showing that there could have been 8 knights, with a correct indication of which person said which phrase - 4 points.
If the example provided does not fully describe the situation (for example, it is not specified what numbers the liars thought of, or it is not clearly indicated who said which phrase) - out of 4 points for the example, no more than 2 points are given.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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