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6. Several cells on a $14 \times 14$ board are marked. It is known that no two of the marked cells are in the same row and the same column, and also that a knight can, starting from any marked cell, reach any other marked cell via marked cells. What is the maximum possible number of marked cells?
|
Answer: 14.
Sketch of the solution.
Estimate. In each row, there is no more than one marked cell (field), and therefore no more than 14 cells are marked.
Example in the image.
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Ten football teams each played one game against each of the others. As a result, each team ended up with exactly $x$ points. What is the greatest possible value of $x$? (Win - 3 points, draw - 1 point, loss - 0 points.)
|
Answer: 13.
Evaluation. Let's prove that $x$ cannot be greater than 13. Indeed, in each match, either 3 points are awarded (if one of the teams wins) or 2 points (if there is a draw). In total, $\frac{10 \cdot 9}{2}=45$ matches were played, meaning no more than 135 points were awarded, that is, the total points scored by all teams is no more than 135. Thus, $10 x \leqslant 135$, which means $x \leqslant 13.5$. Since $x$ is an integer, $x \leqslant 13$.
Example. Let's show that teams could have scored 13 points each. Arrange the teams in a circle and divide them
| | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | $\mathbf{6}$ | $\mathbf{7}$ | $\mathbf{8}$ | $\mathbf{9}$ | $\mathbf{1 0}$ | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{A}$ | | 1 | 3 | 3 | 3 | 3 | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{1 3}$ |
| $\mathbf{B}$ | 1 | | 3 | 3 | 3 | 3 | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{1 3}$ |
| $\mathbf{C}$ | $\mathbf{0}$ | 0 | | $\mathbf{1}$ | 3 | 3 | 3 | 3 | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{1 3}$ |
| $\mathbf{D}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{1}$ | | 3 | 3 | 3 | 3 | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{1 3}$ |
| $\mathbf{E}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | 0 | | $\mathbf{1}$ | 3 | 3 | 3 | 3 | $\mathbf{1 3}$ |
| $\mathbf{F}$ | 0 | 0 | 0 | 0 | 1 | | 3 | 3 | 3 | 3 | $\mathbf{1 3}$ |
| $\mathbf{G}$ | 3 | 3 | 0 | 0 | 0 | 0 | | 1 | 3 | 3 | $\mathbf{1 3}$ |
| $\mathbf{H}$ | 3 | 3 | 0 | 0 | 0 | 0 | 1 | | 3 | 3 | $\mathbf{1 3}$ |
| $\mathbf{I}$ | 3 | 3 | 3 | 3 | 0 | 0 | 0 | 0 | | $\mathbf{1}$ | $\mathbf{1 3}$ |
| $\mathbf{K}$ | 3 | 3 | 3 | 3 | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{0}$ | $\mathbf{1}$ | | $\mathbf{1 3}$ |
sequentially into 5 pairs. Let the teams of each pair play a draw against each other, each of them win against four teams following the given pair clockwise, and lose to the remaining teams. Then each team scored exactly 13 points.
A similar example can be provided in the form of a table.
+ full justified solution
干 correct answer and example provided
$\mp$ only proved that $x \leqslant 13$.
- only the answer is provided
Note: The symbol "干" in the last part seems to be a typo or a non-standard symbol. It has been left as is in the translation.
|
13
|
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
|
1. The numbers $p$ and $b$ are roots of the quadratic equation $x^{2}+2020 a x+c=0, a \neq 0$. Find the sum of the roots of the quadratic equations $a x^{2}+b x+d=0$ and $a x^{2}+p x+q=0$, if each of them has 2 distinct real roots.
|
Solution. Since $p$ and $b$ are the roots of the quadratic equation $x^{2}+2020 a x+c=0$, by Vieta's theorem, $p+b=-2020 a$. Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2}+b x+d=0$, and $x_{3}$ and $x_{4}$ be the roots of the equation $a x^{2}+p x+q=0$. Then, by Vieta's theorem, $x_{1}+x_{2}=-\frac{b}{a}$ and $x_{3}+x_{4}=-\frac{p}{a}$. Therefore, $x_{1}+x_{2}+x_{3}+x_{4}=-\frac{b}{a}-\frac{p}{a}=-\frac{b+p}{a}=-\frac{-2020 a}{a}=2020$.
Answer: 2020.
Comment. The participant of the olympiad incorrectly refers to Vieta's theorem as the inverse theorem of Vieta - no points should be deducted.
|
2020
|
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
|
1. A positive number $x$ was increased by $69 \%$. By what percentage did the number $\sqrt{\frac{x}{5}}$ increase?
|
# Solution.
The increased number will be $1.69 x$, then from the proportion
$\sqrt{\frac{x}{5}}-100 \%$, we determine $y=\frac{1.3 \sqrt{\frac{x}{5}} \cdot 100}{\sqrt{\frac{x}{5}}}=130$ (\%), so the difference will be
equal to $30 \%$.
Answer: by $30 \%$.
Instructions. Only the answer - 0 points; the answer obtained for some specific $x$ (without justification of the approach's correctness) - 2 points; the square root found approximately, for example, using a calculator - 0 points.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In each cell of a $10 \times 10$ board, there is a grasshopper. At the whistle, each grasshopper jumps over one cell diagonally (not to the adjacent diagonal cell, but to the next one). As a result, some cells may end up with more than one grasshopper, while some cells will be unoccupied. Prove that in this case, there will be at least 20 unoccupied cells.
|
# Solution.
We will paint the cells of the board black and white, as shown in the figure. As a result, 60 cells will be painted black, and 40 cells will be painted white. Notice that from a black cell, the grasshopper can only jump to a white cell, and from a white cell, it can only jump to a black cell. Consequently, after the grasshoppers have made their jump, 40 grasshoppers are on the 60 black cells. Therefore, at least 20 black cells will remain unoccupied.
Hint. Only specific cases were considered - 0 points.
|
20
|
Combinatorics
|
proof
|
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
|
2. Thirty beads (blue and green) were laid out in a circle. For 26 beads, the neighboring one was blue, and for 20 beads, the neighboring one was green. How many blue beads were there?
|
Answer: 18 blue beads.
Solution. Let's find the number of beads that have both a blue and a green bead next to them: $26+20-30=$ 16. The number of beads that have only blue beads next to them is $26-16=10$. The number of blue beads is $\frac{10 \cdot 2+16}{2}=18$. Here is an example of such an arrangement. Let's denote a blue bead as (b) and a green bead as (g). Suppose the beads are arranged in a circle in the following order clockwise:
b g bbbb g b g g b g g bbbb g g g g b b g b b g g b b g g g,
where the first and last beads are adjacent. In this arrangement, all conditions are met, and there are 18 blue beads.
Comment. If the correct answer is given, obtained by trial and error, and a valid example of the arrangement is provided, but it is not shown that there cannot be other solutions - 4 points. Points should not be deducted for the lack of an example. For arithmetic errors, deduct $1-2$ points.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 1. Clone 1
The figure is divided into 7 equal squares and several rectangles. The perimeter of rectangle A is 112 cm. What is the perimeter of rectangle B? Express your answer in centimeters.
#
|
# Answer: 168
## Solution
1st method. Rectangles A and B are composed of identical squares: Rectangle A consists of three, and Rectangle B consists of four. Let's call the side of such a square a "stick" and count the perimeter in sticks. The perimeter of Rectangle B is four sticks more than the perimeter of Rectangle A. Also, note that four sticks are half the perimeter of Rectangle A. Therefore, the perimeter of Rectangle B is $112 / 2 + 112 = 168$ cm.
2nd method. Rectangles A and B are composed of identical squares: Rectangle A consists of three, and Rectangle B consists of four. Let's call the side of such a square a "stick" and count the perimeter in sticks. The perimeter of Rectangle A is 8 sticks, and the perimeter of Rectangle B is 12 sticks. The length of one stick is $112 / 8 = 14$ cm. Therefore, the perimeter of Rectangle B is $12 \cdot 14 = 168$ cm.
## Clone 2
The figure is divided into 7 equal squares and several rectangles. The perimeter of Rectangle A is 116 cm. What is the perimeter of Rectangle B? Express your answer in centimeters.
Answer: 174
## Clone 3
The figure is divided into 7 equal squares and several rectangles. The perimeter of Rectangle A is 122 cm. What is the perimeter of Rectangle B? Express your answer in centimeters.
Answer: 183
## Clone 4
The figure is divided into 7 equal squares and several rectangles. The perimeter of Rectangle A is 126 cm. What is the perimeter of Rectangle B? Express your answer in centimeters.
Answer: 189
#
|
168
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. Clone 1
A rope was divided into 19 equal parts and arranged in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of them is 8 meters, and the shortest is 2 meters. What was the length of the rope before it was cut? Express your answer in meters.
|
Answer: 114
Solution
Let's find the length of each of the 19 equal parts into which the rope was divided. To do this, we need to add the shortest piece to half of the longest piece obtained after the cut. We get $8 / 2 + 2 = 6$. Then the total length of the rope is $19 \cdot 6 = 114$ meters.
## Clone 2
A rope was divided into 19 equal parts and laid out in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of which is 8 meters, and the shortest is 1 meter. What was the length of the rope before it was cut? Express your answer in meters.
Answer: 95
## Clone 3
A rope was divided into 19 equal parts and laid out in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of which is 10 meters, and the shortest is 2 meters. What was the length of the rope before it was cut? Express your answer in meters.
## Answer: 133
## Clone 4
A rope was divided into 19 equal parts and laid out in a snake-like pattern. After that, a cut was made along the dotted line. The rope split into 20 pieces: the longest of which is 10 meters, and the shortest is 3 meters. What was the length of the rope before it was cut? Express your answer in meters.
Answer: 152
|
152
|
Logic and Puzzles
|
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
|
5. In a chess tournament, where each participant played against every other, three chess players fell ill and dropped out of the tournament before it reached its halfway point. A total of 130 games were played in the tournament. How many chess players participated in the tournament?
|
Answer: 19 people
Solution.
If 16 chess players participated in the tournament, the number of games they played should not exceed (16 x 15):2 = 120 games. Therefore, more than 16 people played in the tournament. Let's consider the following cases.
A) The tournament started with 17 participants. Then 14 of them, who finished the tournament, played (14 x 13):2 = 91 matches among themselves, and the eliminated players played 39 matches together. Therefore, someone among them played no less than 13 matches, i.e., was eliminated in the second half of the tournament. This contradicts the condition.
B) The tournament started with 18 participants. Then 15 of them, who finished the tournament, played (15 x 14):2 = 105 matches among themselves, and the eliminated players played 25 matches together. Since half of the tournament consists of 8 rounds, the eliminated participants could not have played more than 24 matches together.
C) The tournament started with 19 participants. Then 16 of them, who finished the tournament, played (16 x 15):2 = 120 matches among themselves, and the eliminated players played 10 matches together. Each of them could have been eliminated in the first half of the tournament.
D) The tournament started with 20 participants. Then 17 of them, who finished the tournament, played (17 x 16):2 = 136 matches among themselves, which is more than all the participants in the tournament.
Therefore, 19 people participated in the tournament.
|
19
|
Combinatorics
|
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
|
10.8. On a plane, $N$ points are marked. Any three of them form a triangle, the angles of which in degrees are expressed by natural numbers. For what largest $N$ is this possible?
|
Answer: 180.
First solution. Example. First, we show that for $N=180$, the required condition is possible. Mark 180 points on a circle, dividing it into 180 equal arcs, each $2^{\circ}$. The measure of any arc with endpoints at two of the marked points is expressed as an even number of degrees, so the measure of any inscribed angle in the circle, formed by three marked points, is a natural number of degrees. Therefore, 180 marked points satisfy the condition of the problem.
Estimate. It remains to prove that $N \leqslant 180$. Any three marked points form a triangle, so they cannot lie on the same line. Assuming the marked points are located on a coordinate plane, let $A$ be any of them with the maximum ordinate. Among the remaining points, choose points $B$ and $C$ such that the angle $B A C$ is maximized.
From the condition of the problem, it follows that in triangle $A B C$, the measures of angles $A B C$ and $A C B$ are at least $1^{\circ}$, so the measure of angle $B A C$ is at most $178^{\circ}$. Due to the choice of points $B$ and $C$, the remaining $N-3$ marked points lie strictly inside angle $B A C$, and each ray starting from point $A$ contains no more than one of them. By drawing a ray through each marked point inside angle $B A C$ with the origin at point $A$, we obtain $N-3$ distinct rays, dividing $\angle B A C$ into $N-2$ angles. If $N-2>178$, then at least one of these angles has a measure less than $1^{\circ}$ and is an angle of some triangle with vertices at three marked points, which contradicts the condition of the problem. Therefore, $N-2 \leqslant 178$, that is, $N \leqslant 180$, which was to be proved.
Remark 1. The choice of the point $A$ used in the solution can also be described in the following ways.
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.9. In the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip is shifted one position clockwise relative to its initial position?
(S. Berlov)
|
# Answer: 50.
Solution. Example. The chip 50 is sequentially exchanged 99 times with the next one counterclockwise. We get the required arrangement.
There are several ways to prove the estimate, below we provide two of them.
The first way. Suppose that for some $k<50$ the required arrangement is obtained.
At any moment, we consider the arc from chip 100 to chip 1 clockwise to be painted. Since chips 100 and 1 cannot be swapped in one move, each specific chip $m$ $(2 \leqslant m \leqslant 99)$ could only enter the painted arc or leave the painted arc by swapping with one of the chips 1 or 100.
Since initially and at the end, chip $m$ was not on the painted arc, it made an equal number of entries onto the painted arc and exits from the painted arc. For $m \leqslant 50$, chip $m$ could not swap with chip 100, so it could only enter or exit by swapping with chip 1. Upon entry, chip 1 moves 1 position clockwise, and upon exit, 1 position counterclockwise. We conduct similar reasoning for chips $m \geqslant 51$, which cannot swap with chip 1.
Thus, we obtain that chips 1 and 100 will make an equal number of moves clockwise and counterclockwise, so they will remain in their positions. Contradiction.
The second way. We will count the shifts of the chips relative to their initial position, with a shift clockwise being counted as positive and a shift counterclockwise as negative. Then, when two chips are swapped, +1 is added to the shift of one of them, and -1 to the other. Therefore, the sum of the shifts after the operations will be 0.
We reason by contradiction: suppose that for $k<50$ each chip $i$ is finally shifted one position clockwise, i.e., its shift is $1+100 t_{i}$ (here $t_{i}$ is the integer "number of rotations" clockwise, in particular, if $t_{i}<0$, chip $i$ made $\left|t_{i}\right|$ rotations counterclockwise). Then the total shift of all 100 chips is $100\left(t_{1}+t_{2}+\ldots+t_{100}\right)+100$. Since it must equal 0, we have $t_{1}+t_{2}+\ldots+t_{100}=-1$.
Since $k<50$, chips with numbers $i$ and $j$, where $j \geqslant i+50$, could not swap places, so their shifts at any moment are definitely less than 100 apart, meaning the "number of rotations" $t_{i}$ and $t_{j}$ are equal for $j \geqslant i+50$. Hence, we have $t_{1}=t_{51}$, $t_{2}=t_{52}, \ldots, t_{50}=t_{100}$. Then the sum $t_{1}+t_{2}+\ldots+t_{100}=2\left(t_{1}+\right.$ $\left.+t_{2}+\ldots+t_{50}\right)$ is even, and thus cannot equal -1. Contradiction.
Remark 1. The last reasoning can be modified as follows.
From this, $t_{1}=t_{51}, t_{1}=t_{52}, \ldots, t_{1}=t_{100}, t_{2}=t_{100}, t_{3}=t_{100}$, $\ldots, t_{50}=t_{100}$, thus all $t_{i}$ are equal to $t_{1}$. Then $t_{1}+t_{2}+$ $+\ldots+t_{100}=100 t_{1} \neq-1$.
Remark 2. The proof can be completed by contradiction in another way, replacing the last paragraph of the solution with the following reasoning.
We will paint chips red for which $t_{i} \geqslant 0$, and blue for which $t_{i}<0$. Clearly, at some point, a blue and a red chip must swap, since the difference in their shifts is at least 100. Since $k<50$, the pair of chips 1 and 51 are of the same color, similarly pairs of chips 1 and $52, \ldots, 1$ and 100, 2 and 100, 3 and $100, \ldots, 50$ and $100-$ are of the same color. Thus, all chips are of the same color. We know that $t_{1}+t_{2}+\ldots+t_{100}=-1$, so among the numbers $t_{1}, \ldots, t_{100}$ there are both non-negative and negative ones, i.e., there are both red and blue chips - contradiction.
Comment. Only the correct answer - 0 points.
Points for the example and the estimate are summed.
A correct example is provided - 2 points.
There is a complete proof of the estimate - 5 points.
In the absence of a complete proof of the estimate, the following progress is evaluated (points for progress (1) are not summed with points for progress (2a), (2b), (2c), points for progress (2a), (2b), (2c) are summed):
(1) The idea of tracking the membership of chips to the arc 100-1 (or the relative position of the triplet of chips $1, m, 100)-2$ points.
(2a) shifts are considered and it is shown that the sum of shifts is $0-1$ point.
(2b) in the assumption that the chips are finally shifted 1 position clockwise, an equation for the sum of "number of rotations" is written -1 point.
(2c) it is shown that chips with different "number of rotations" or with "number of rotations" of different signs must have swapped places at some point -1 point.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. The angle formed by the bisector of angle $A B C$ with its sides is 6 times smaller than the angle adjacent to angle $A B C$. Find angle $A B C$.
|
Answer: 45 degrees. Solution. Let $x$ be the degree measure of angle $A B C$. From the condition of the problem, we get the equation $\frac{x}{2}=\frac{180-x}{6} \Leftrightarrow 8 x=360 \Leftrightarrow x=45$ (degrees).
|
45
|
Geometry
|
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
|
7.4. Does there exist a six-digit natural number that, when multiplied by 9, is written with the same digits but in reverse order?
|
Answer: it exists. Solution. Let $\overline{a b c d e f}$ be the desired number, i.e., $\overline{a b c d e f} \cdot 9=\overline{f e d c b a}$. Then it is obvious that $a=1, b=0$ (otherwise, multiplying by 9 would result in a seven-digit number). Therefore, $f=9$, and the second-to-last digit $e=8$ (which follows from column multiplication). Then the third digit $c$ can be 8 or 9. But if $c=8$, then $d=1$, since the sum of the digits is divisible by 9, but the number 108189 does not fit upon verification. If, however, $c=9$, then $d=9$ and the number 109989 is the only one that satisfies the condition, and it fits upon verification.
|
109989
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On the board, 2020 quadratic equations are written:
$$
\begin{gathered}
2020 x^{2} + b x + 2021 = 0 \\
2019 x^{2} + b x + 2020 = 0 \\
2018 x^{2} + b x + 2019 = 0 \\
\ldots \\
x^{2} + b x + 2 = 0
\end{gathered}
$$
(each subsequent equation is obtained from the previous one by decreasing the leading coefficient and the constant term by one). Find the product of the roots of all the equations written on the board, given that each of them has two real roots.
|
Solution. According to the theorem converse to Vieta's theorem, the product of the roots of the first equation is $\frac{2021}{2020}$, the product of the roots of the second equation is $\frac{2020}{2019}$, the third is $-\frac{2019}{2018}, \ldots$, the two thousand and twentieth is $\frac{2}{1}$. Therefore, the product of the roots of all equations is $\frac{2021}{2020} \cdot \frac{2020}{2019} \cdot \frac{2019}{2018} \cdot \frac{2018}{2017} \cdot \ldots \cdot \frac{3}{2} \cdot \frac{2}{1}=2021$. Answer: 2021.
Comment. The participant of the olympiad mistakenly calls the statement of the theorem converse to Vieta's theorem as Vieta's theorem - no points should be deducted.
|
2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A team consisting of boys and girls from the Rostov region went to a CS:GO esports tournament. The average number of points scored by the girls turned out to be 22, by the boys - 47, and the average number of points for the entire team - 41. What is the percentage of girls in this team?
|
Solution. Let the number of girls be $x$, the number of boys be $y$, and the total points scored by them be $S_{1}$ and $S_{2}$ respectively. From the conditions, the following equations can be derived: $\frac{S_{1}}{x}=22, \frac{S_{2}}{y}=47$ and $\frac{S_{1}+S_{2}}{x+y}=41$. Then $S_{1}=22 x$ and $S_{2}=47 y$, from which $\frac{22 x+47 y}{x+y}=41$. Consider the left part of the last equation: $\frac{22 x+47 y}{x+y}=\frac{22 x+22 y+25 y}{x+y}=22+\frac{25 y}{x+y}$. Therefore, $22+\frac{25 y}{x+y}=41$ or $\frac{y}{x+y}=\frac{19}{25}$. Since $\frac{19}{25}=\frac{76}{100}$, the proportion of boys is $76\%$, and thus the proportion of girls is $24\%$. Answer: $\mathbf{24\%}$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a computer game, one person can play as one of three factions: $T, Z$ or $P$. There is a network play mode in which 8 players are divided into two teams of 4 players each. How many different matches can there be, differing in the sets of factions? Matches are considered different if one match has a team that the other does not. The order of the teams and the order of listing the factions in the team do not matter. For example, matches $(P Z P T ; T T Z P)$ and $(P Z T T ; T Z P P)$ are considered the same, while matches $(P Z P Z ; T Z P Z)$ and $(P Z P T ; Z Z P Z)$ are different.
|
Solution. First, let's calculate the number of ways to form one team from the specified factions. Let's number the factions. The number of options is equal to the number of solutions to the equation $x_{1}+x_{2}+x_{3}=4, x_{i} \geqslant 0$, where $x_{i}$ is the number of players from faction $i$. The number of solutions can be found using the formula for combinations with repetition: $C_{4+3-1}^{3-1}=15$ or by directly listing the options. We will divide all match options into two sets based on whether the teams are the same. If the teams are the same, we get 15 matches, and if they are different, we get $C_{15}^{2}=105$. Thus, we get the answer: $15+105=120$ matches. Answer: 120.
|
120
|
Combinatorics
|
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
|
3. There are seven red cubes, three blue ones, and nine green ones. Ten cubes were placed in a gift bag. In how many different ways could this have been done?
|
Answer: 31.
Let's put red cubes into the bag (8 ways from 0 to 7), now we place the blue cubes (4 ways from 0 to 3). Add the necessary number of green cubes (1 way). In total, $8 \times 4=32$.
One operation is impossible: 10 green. Therefore, there is one fewer way.
|
31
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Pete wants to color several cells of an $8 \times 8$ square so that for any vertex, there is a colored square to which it belongs. What is the minimum number of squares he must color?
|
Answer: 25
Let's mark 25 vertices of an $8 \times 8$ square (see the figure on the right). At each marked vertex, there must be a shaded square. Each square touches only one such vertex, so there must be at least 25. The example shown below indicates that the 25 shaded squares touch all the vertices of the grid.
squares touch all the vertices of the grid.
Criteria. Example: 4 points. Evaluation: 3 points.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. In the castle, there are 25 identical square rooms, forming a $5 \times 5$ square. In these rooms, 25 people—liars and knights (liars always lie, knights always tell the truth)—settled, one person per room. Each of these 25 people said: “At least one of the neighboring rooms to mine is occupied by a liar.” What is the maximum number of knights that could be among these 25 people? Rooms are considered neighboring if they share a wall.
|
Answer: 18 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 7 liars (thus showing that there are no more than 18 knights).
First, consider dividing the rooms into 6 groups (2 rooms, marked in gray, do not belong to any group). In each group, there must be at least one liar (otherwise, one of the knights in the group will not have any liar neighbors). Therefore, there are at least 6 liars. We will prove that there cannot be exactly 6 liars. Assume the contrary. Then each of the marked groups will have exactly one liar. This means that the "gray rooms" must definitely be occupied by knights.
If we make a similar division but rotated by $90^{\circ}$, we will find that there must be knights in two more rooms. These 4 rooms are marked in gray on the diagram. This means that a liar must live in the central room.
Since each of the 6 groups in the division must have exactly 1 liar, the rooms marked in gray on the diagram must be occupied by knights.
By rotating the diagram by $90^{\circ}$, we find that knights must live in the rooms marked in gray on the diagram.
But then, in each group of 3 "corner rooms," there must be at least 2 liars. In total, there will be at least 9 liars. This is a contradiction. Therefore, there must be at least 7 liars.
The diagram below shows how 18 knights and 7 liars could be accommodated.
| $\mathrm{P}$ | $\mathrm{P}$ | L | $\mathrm{P}$ | $\mathrm{P}$ |
| :---: | :---: | :---: | :---: | :---: |
| L | $\mathrm{P}$ | $\mathrm{P}$ | $\mathrm{P}$ | L |
| $\mathrm{P}$ | $\mathrm{P}$ | L | $\mathrm{P}$ | $\mathrm{P}$ |
| L | $\mathrm{P}$ | $\mathrm{P}$ | $\mathrm{P}$ | L |
| $\mathrm{P}$ | $\mathrm{P}$ | L | $\mathrm{P}$ | $\mathrm{P}$ |
Comment: A correct answer without justification - 0 points. An example of 7 liars and 18 knights - 2 points. Proving that there are no more than 18 knights - 5 points.
Note: In the example, liars should not be neighbors.
|
18
|
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
|
7.3. For breakfast, Karlson ate $40 \%$ of the cake, and Little One ate 150 g. For lunch, Fräulein Bock ate $30 \%$ of the remainder and another 120 g, and Matilda licked the remaining 90 g of crumbs from the cake. What was the initial mass of the cake?
|
Answer: 750 g.
Solution. First method (solving "from the end").
1) $90+120=210$ (g) of the cake remained after Fröken Bok ate $30\%$ of the remainder.
Since Fröken Bok ate $30\%$ of the remainder, 210 g is $70\%$ of the remainder.
2) $210 \div 0.7 = 300$ (g) of the cake was left before Fröken Bok started her lunch.
3) $300 + 150 = 450$ (g) of the cake was left before Little Man started eating.
Since Karlson ate $40\%$ of the cake, 450 g is $60\%$ of the cake.
4) $450 \div 0.6 = 750$ (g) is the initial mass of the cake.
Second method (formulating an equation). Let $x$ g be the initial mass of the cake, then after Karlson and Little Man's breakfast, $0.6x - 150$ (g) remained, and after Fröken Bok's lunch, $0.7(0.6x - 150) - 120 = 0.42x - 225$ (g) remained, which is 90 g licked by Matilda. We get the equation $0.42x - 225 = 90$, the solution to which is $x = 750$.
## Grading Criteria:
+ A complete and well-reasoned solution is provided
$\pm$ A solution "from the end" is provided, all actions are correctly performed and the correct answer is obtained, but explanations are missing
$\pm$ An equation is correctly formulated and solved, but explanations for formulating the equation are missing
$\mp$ An equation is correctly formulated, but it is not solved or an arithmetic error is made in solving it
$\pm$ The idea of solving "from the end" is present and at least the first action is performed correctly, but the solution is not completed or is completed with errors
$\mp$ Only the correct answer is provided (possibly with a check that it is valid)
- The problem is not solved or is solved incorrectly
|
750
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In a correspondence mathematics olympiad, out of 500 participants, exactly 30 did not like the problem conditions, exactly 40 did not like the organization of the event, and finally, exactly 50 did not like the method of determining the winners of the olympiad. We will call an olympiad participant "significantly dissatisfied" if they were dissatisfied with at least two out of the three aspects of the olympiad. What is the maximum number of "significantly dissatisfied" participants that could have been at this olympiad?
|
Solution. In total, there were $120=30+40+50$ "dissatisfactions" expressed, so there cannot be more than 60 "significantly dissatisfied" individuals, because otherwise they would have more than 120 "dissatisfactions" in total, which contradicts the condition. It remains to show that there could have been 60 of them. For example, participants 1 - 40 are dissatisfied with the organization of the event, participants 1 - 10 and $41-60-$ are dissatisfied with the problem conditions, and participants 11 - 60 are dissatisfied with the determination of the winners.
Answer: 60.
|
60
|
Combinatorics
|
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
|
9.1. A four-digit number $n$ was doubled and 1000 was added to the result. As a result, a four-digit number was obtained, written with the same digits as $n$, but in reverse order. Find all possible values of $n$.
|
Answer: 2996.
Solution. Let $a, b, c, d$ be the number of thousands, hundreds, tens, and units in the number $n$ respectively. Write the condition in the form:
| $a$ | $b$ | $c$ | $d$ |
| ---: | ---: | ---: | ---: |
| $a$ | $b$ | $c$ | $d$ |
| $+\quad 1$ | 0 | 0 | 0 |
| $d$ | $c$ | $b$ | $a$ |
Since $a$ is the last digit of $2d$, $a$ is even. Since $2a+1 \leq d \leq 9$, it follows that $a \leq 4$. Therefore, either $a=2$ or $a=4$.
If $a=4$, then from $2a+1 \leq d \leq 9$ it follows that $d=9$. But then the last digit of the number $2d$ is 8, and $a \neq 8$.
If $a=2$, then from $2a+1 \leq b \quad c \quad d$ it follows that $d \geq 5$. And since the last digit of $2d$ is 2, then $d=6$. Thus, the equation
reduces to the form:
Here, either $2c+1=b$ and $2b=10+c$, or $2c+1=10+b$ and $2b+1=10+c$. In the first case, the system has no integer solutions, in the second case we get $b=c=9$.
|
2996
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. In the basket, there are oranges and bananas. If you add as many oranges as there are currently bananas (in pieces), then the percentage of oranges
will be twice as much as it would be if you added as many bananas as there are currently oranges. What is the current percentage of oranges in the basket?
|
Answer: 50.
Solution. Let $a$ be the number of oranges, and $b$ be the number of bananas in the basket. If we add as many oranges as there are currently bananas, then there will be $a+b$ oranges out of $a+2b$ fruits, and the proportion of oranges will be $(a+b)/(a+2b)$. If we add as many bananas as there are currently oranges, then there will be $a$ oranges among $2a+b$ fruits, and the proportion of oranges will be $a/(2a+b)$. According to the problem, we have $(a+b)/(a+2b) = 2a/(2a+b)$. From this, we get $b^2 = ab$. Since there are more than zero bananas in the basket, $b = a$, meaning there are an equal number of bananas and oranges in the basket.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Find the value of the expression $x-\sqrt{2022 x}+2023$,
if $x-\sqrt{\frac{2022}{x}}=2023$.
|
Answer: 2024.
Solution. Transform the condition $x-\sqrt{\frac{2022}{x}}=2023$ given that $x>0$. We get: $x^{2}-\sqrt{2022 x}=2023 x ; x^{2}-2022 x-x-\sqrt{2022 x}=0$;
$$
(x-\sqrt{2022 x})(x+\sqrt{2022 x})-(x+\sqrt{2022 x});(x-\sqrt{2022 x}-1)(x+
$$
$\sqrt{2022 x})=0$.
The second factor is positive for all $x>0$, which implies that $x-$ $\sqrt{2022 x}-1=0$, from which $x-\sqrt{2022 x}+2023=2024$.
|
2024
|
Algebra
|
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
|
7.4 On the table, there are 2020 boxes, some of which contain candies, while the others are empty. On the first box, it is written: "All boxes are empty." On the second: "At least 2019 boxes are empty." On the third: "At least 2018 boxes are empty," and so on, up to the 2020th, which says: "At least one box is empty." It is known that the inscriptions on the empty boxes are false, while those on the boxes with candies are true. Determine how many boxes contain candies. Justify your answer.
|
Solution. Note that if the inscription "At least $A$ boxes are empty" is true for some $A$, then all subsequent inscriptions are also true. This means that the boxes with candies are all boxes starting from some number $N$. Then the number of boxes without candies is exactly $N-1$. The inscription on the $N$-th box reads: "At least $2021-N$ boxes are empty." This is true. The inscription on the previous box, numbered $N-1$, reads: "At least $2021-(N-1)$ boxes are empty." This is false. Therefore, the number of empty boxes is $2021-N$. From the equation $N-1=2021-N$, we find that $N=1011$. Thus, the number of empty boxes is 1010, and the number of boxes with candies is $2020-1010=1010$.
Answer: 1010 boxes.
| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| There is an arithmetic error in the correct solution process, possibly leading to an incorrect answer | 6 points |
| A correct example is provided (the first 2010 boxes are empty, the rest have candies), but it is not proven that there is no example with a different number of filled boxes | 3 points |
| It is proven that the numbers of empty (full) boxes are all natural numbers not exceeding (greater than) a fixed number | 2 points |
| Correct answer without justification or with incorrect justification | 0 points |
|
1010
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5 A businessman was driving to a business meeting. He calculated that if he traveled at a speed of 90 km/h, he would arrive an hour earlier, and if he traveled at 60 km/h, he would be an hour late. What is the minimum speed he should travel to arrive on time? Justify your answer.
|
Solution. Method 1. Suppose that three businessmen, each in their own car, were heading to a meeting. The first (Speedster) at a speed of 90 km/h, the second (Turtle) at a speed of 60 km/h, and the third - Punctual, whose speed we need to find. Consider the moment when Speedster arrives at the meeting place. The meeting is still one hour away, and Turtle's arrival is still two hours away. Thus, Turtle is currently $60 \cdot 2=120$ km away from the meeting place. Speedster was ahead of Turtle by $90-60=30$ km each hour, so he was on the road for $120: 30=4$ hours. The meeting place, therefore, is $90 \cdot 4=360$ km from the starting point, and the meeting should take place $4+1=5$ hours after the start of the businessmen's journey. Thus, Punctual's speed is $360: 5=72$ km/h.
Method 2. Let the distance the businessman needs to travel be $S$ km, the time until the meeting be $t$ hours, and the speed at which he should travel to arrive on time (which is the required speed) be $v$ km/h. Then we have the system
$\left\{\begin{array}{rl}v t & =S \\ 90(t-1) & =S \\ 60(t+1) & =S\end{array}\right.$. From the last two equations of the system, we find $t=5, S=360$. Then from the first equation $S=72$.
Answer: 72 km/h.
| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| There are arithmetic errors in the correct solution process, possibly leading to an incorrect answer | 6 points |
| The system of equations is correctly set up but not solved (or solved incorrectly), fully describing the problem | 4 points |
| At least one of the two quantities is correctly and justifiedly found: - the distance to be traveled; - the time the businessman should be on the road | 3 points |
| Answer without justification or with incorrect justification | 0 points |
|
72
|
Algebra
|
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
|
10.1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2016$ and $a \neq c$ (the numbers themselves are not given).
|
Solution. Expanding the brackets $\mathrm{x}^{2}-(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=\mathrm{x}^{2}-(\mathrm{c}+\mathrm{d}) \mathrm{x}+\mathrm{cd}$, i.e., $\quad(c+d-a-b) x=c d-a b$. By the condition $c - a = d - b$ - let's denote this by $r$. Then $a=c-r, d=b+r$ and $c d-a b=c(b+r)-(c-r) b=(b+c) r=2016 r$. Thus (since $c - -\mathrm{r}, \mathrm{d}-\mathrm{b}=\mathrm{r}) \quad 2 \mathrm{rx}=2016 \mathrm{r}, \quad \mathrm{x}=1008 \quad(\mathrm{r} \neq 0$ by the condition).
Remark. There are other solutions. For example: Let $f(x)=(x-a)(x-b)$, $g(x)=(x-c)(x-d)$. Then the condition $a+d=b+c$ gives the identity $g(x)=f(x+a-c)$, and the equation $f(x)=g(x)$ reduces to the equation $f(x)=f(x+a-c)$, which means the symmetry of the points $x$ and $x+a-c$ with respect to the abscissa $x_{0}=\frac{a+b}{2}$ of the vertex of the parabola $y=f(x)$, from which we obtain
$$
\frac{x+x+a-c}{2}=\frac{a+b}{2}, \quad \mathrm{x}=\frac{b+c}{2}=\frac{2016}{2}=1008
$$
Answer: 1008.
|
1008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1.
At the entrance to the amusement park, they sell children's and adult tickets. One children's ticket costs 600 rubles. Alexander bought 2 children's and 3 adult tickets, while Anna bought 3 children's and 2 adult tickets. It is known that Alexander paid 200 rubles more than Anna. How much did Alexander pay for the tickets in rubles?
|
Answer: 3600
## Solution.
Let $A$ be the cost of a children's ticket, and $B$ be the cost of an adult ticket. We calculate the difference $3B + 2A - 2B - 3A = B - A = 200$ rubles. This means the difference between the cost of an adult ticket and a children's ticket is 200 rubles. Then Alexander paid for the tickets $5 \cdot 600 + 3 \cdot 200 = 3600$ rubles.
|
3600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Variant 1.
On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, 2022 natives gathered around a round table, and each of them made the following statement:
"I am sitting next to a knight and a liar!"
It is known that three knights made a mistake (i.e., accidentally lied). What is the maximum number of knights that could have been at the table?
|
Answer: 1349
## Solution.
Let's divide all the people sitting at the table into blocks consisting of natives of the same type sitting in a row. Then, blocks of liars can only consist of 1 person; otherwise, a liar sitting at the edge of this block would tell the truth. If a block of knights consists of two people, then both knights in it told the truth (did not make a mistake). If a block of knights consists of 1 person, he made a mistake in his statement (he has two neighbors who are liars). If a block of knights consists of $k>2$ people, then $k-2$ knights sitting inside the block will make a mistake (each has two knight neighbors).
Let $n$ be the number of blocks of liars, and $m \leq 3$ be the number of blocks consisting of one knight. Blocks of liars and knights alternate, so there are an equal number of each. Then, the number of knights is $m+2(n-m)+3-m=2(n-m)+3$, and the number of liars is $-n$. Therefore, $2(n-m)+3+n=2022$, or $n=\frac{2019+2 m}{3} \geq \frac{2019}{3}=673$. Then, the number of knights is no more than $2022-673=1349$.
Example. 673 blocks of liars, with one in each block, alternate with 673 blocks of knights, where one block has 5 knights, and the rest have 2 knights each.
|
1349
|
Logic and Puzzles
|
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
|
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
|
Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides some existing part of the partition into two new parts, i.e., $n$ new parts will be added. (Accordingly, if the ray intersects fewer rays, then fewer new parts will be added to the partition).
1 ray - 1 part.
2 rays - $1+1$.
3 rays - $1+1+2$.
...
$n$ rays - $1+1+2+\ldots+(n-1)=1+n(n-1)/2=(n^2-n+2)/2$.
For $n=11$, the answer is $(121-11+2)/2=56$.
|
56
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cinema hall is the one located at the center of the hall, i.e., at the midpoint of the height drawn from the vertex of the triangle corresponding to seat number 1. In the figure, an example of such a triangular hall with 45 seats is shown, where the best seat is number 13.
CINEMA "TRIANGLE"
How many seats are in the cinema hall where the best seat has the number $265?$
|
Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats in each part will be the same, see the figure.
Thus, the problem reduces to finding the number $n$. We need to find the largest integer $n$ that satisfies the inequality: $1+2+\ldots+n<265$ or $n^{2}+n-530<0$, i.e., $n=22$. Then, in each of the four marked triangles, there will be $1+2+\ldots+22=23 \cdot 11=253$ seats, and in total, there will be $253 \cdot 4+23=1035$ seats in the hall.
2nd method.
Note that the "height" will consist of the numbers located in the center of the rows with odd numbers.
Let $a_{k}$ be the number of the seat located in the center of the row with number $k$ ($k$ - odd). To the right and left of it in the given row, there are $(k-1)/2$ seats. Therefore, when transitioning from $a_{k}$ to $a_{k+2}$, $(k-1)/2$ last seats from the row with number $k$, the $(k+1)$-th seat from the row with number $k+1$, and $(k+1)/2$ first seats from the row with number $k+2$ are added.
Thus, $a_{k+2}=a_{k}+(k-1)/2+(k+1)+(k+1)/2+1=a_{k}+2(k+1)$.
$a_{1}=1$
$a_{3}=a_{1}+2 \cdot 2=1+4$
$a_{5}=a_{3}+2 \cdot 4=1+4+8$
$a_{7}=a_{3}+2 \cdot 6=1+4+8+12$
$\cdots$
$a_{k}=1+4+8+12+\ldots+2(k-1)=1+4(1+2+\ldots+(k-1)/2)=1+(k+1)(k-1)/2$,
Using this formula, we will find the row number where the best seat is located.
$$
\begin{gathered}
1+(k+1)(k-1) / 2=265 \\
\left(k^{2}-1\right) / 2=264 \\
k^{2}=264 \cdot 2+1=529 \\
k=\sqrt{529}=23 .
\end{gathered}
$$
The total number of rows is $2 k-1$ (before and after the row with the best seat, there are $k-1$ rows each). Therefore, the total number of seats in the hall is given by the formula
$(1+2 k-1) \cdot(2 k-1) / 2=k \cdot(2 k-1)$. For $k=23$, we get $23 \cdot 45=1035$.
|
1035
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Variant 1.
On side $A B$ of parallelogram $A B C D$, a point $F$ is chosen, and on the extension of side $B C$ beyond vertex $B$, a point $H$ is chosen such that $A B / B F = B C / B H = 5$. Point $G$ is chosen so that $B F G H$ is a parallelogram. $G D$ intersects $A C$ at point $X$. Find $A X$, if $A C = 100$.
|
Answer: 40.
Solution.
In parallelogram $ABCD$, draw diagonal $BD$, and in parallelogram $BFGH$, draw diagonal $GB$. Let $BD$ intersect $AC$ at point $O$. We will prove that $AC \| GB$. Triangles $ABC$ and $GHB$ are similar by two sides and the angle between them: $AB / BF = BC / BH$ and $\angle GHB = \angle ABC$. Therefore, $\angle GBH = \angle ACB$, so $AC \| GB$. Since $O$ is the intersection point of the diagonals of the parallelogram, $DO = BO$, which means $XO$ is the midline of triangle $BDG$. From the similarity, it also follows that $AC / GB = 5$, $GB = 20$. Then $XO = GB / 2 = 10$, $AX = 100 / 2 - 10 = 40$.
|
40
|
Geometry
|
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
|
3. (7 points) The knightly tournament lasts exactly 7 days. By the end of the fourth day, Sir Lancelot had not yet faced one quarter of the total number of participants. And by this time, Sir Tristan had fought exactly one seventh of the knights that Sir Lancelot had faced. What is the minimum number of knights that could have participated in the tournament?
|
Answer: 20.
Solution. Let Lancelot not have fought with $x$ knights. Then the total number of knights is $4 x$, and Lancelot fought with $3 x-1$ knights (the total number minus $x$ and Lancelot himself). Then Tristan fought with $\frac{3 x-1}{7}$ knights. To find the smallest possible number of knights, we need to find the smallest $x$ such that $3 x-1$ is divisible by 7. The values $x=1,2,3,4$ do not work, but $x=5$ does. Thus, the smallest possible number of knights is 20.
Criteria. Any correct solution: 7 points.
For the absence of justification that there could indeed be exactly 20 knights, points are not deducted.
Only the correct answer is provided: 2 points.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (7 points) In triangle $A B C$, the median $A M$ is drawn. Find the angle $A M C$, if angles $B A C$ and $B C A$ are equal to $45^{\circ}$ and $30^{\circ}$ respectively.
Answer: $135^{\circ}$.
|
Solution. Let $B H$ be the height of triangle $A B C$. According to the problem, angle $B A C$ is $45^{\circ}$, so $B H = A H$. In triangle $C B H$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B C = 2 B H$. The median $H M$ of the right triangle $B H C$ is equal to half the hypotenuse $B C$.
Combining all the equalities of the segments, we get $A H = B H = H M = M B = M C$. Therefore, triangle $M B H$ is equilateral, and angle $C M H$ is $120^{\circ}$. Additionally, triangle $A H M$ is isosceles, its angle $A H M$ is $90^{\circ} + 60^{\circ} = 150^{\circ}$, so angle $A M H$ is $15^{\circ}$. Thus,
$$
\angle A M C = \angle A M H + \angle H M C = 120^{\circ} + 15^{\circ} = 135^{\circ}.
$$
Criteria. Any correct solution: 7 points.
Proved that triangle $A H M$ is isosceles, but the correct answer was not obtained: 5 points.
Only the correct answer was provided: 1 point.
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
|
Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ rectangles. At the same time, one one-cell rectangle at the intersection of the strips is counted twice (and only it is). Therefore, the answer to the problem is the number $15+10-1=24$.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advanced the clock by 7 hours. How many times did "cuckoo" sound during this time?
|
Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+12+1+2+3+4=43
$$
|
43
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
|
Answer: 27.
Solution. The area of the entire square is $6 \cdot 6=36$ sq. cm.
Divide the triangle located in the middle of the square into two smaller triangles, as shown in the picture on the left. Then the dark gray triangles can be combined into a rectangle $1 \times 3$, and the light gray triangles into a rectangle $2 \times 3$. Therefore, the area of the figure that remains after cutting out all the triangles is 36-3-6 $=27$ sq. cm.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.6. On the board, there is one three-digit number and two two-digit numbers. The sum of the numbers that have a seven in their notation is 208. The sum of the numbers that have a three in their notation is 76. Find the sum of all three numbers.
|
Answer: 247.
Solution. Let the three-digit number be $A$, the two-digit numbers be $-B$ and $C$. Among the numbers whose sum is 76, there cannot be a three-digit number. The sum cannot consist of a single number either, because otherwise that number would be 76, but it does not contain a three. Therefore, 76 is the sum of two two-digit numbers, each of which contains a three:
$$
B+C=76
$$
Among the numbers whose sum is 208, there must be a three-digit number (since the sum of two-digit numbers is only 76). It cannot be the only one, because the number 208 does not contain a seven. Therefore, there is at least one two-digit number in this sum - but not both at the same time (otherwise the three-digit number would be $208-76=132$, but it does not contain a seven).
Without loss of generality, let's assume that the two-digit number in this sum is $B$. Then
$$
A+B=208
$$
and both of these numbers contain a seven.
Therefore, the number $B$ contains both a three and a seven, i.e., $B=37$ or $B=73$.
If $B=73$, then $C=76-B=3$ - not a two-digit number. Therefore, $B=37, C=76-37=39$, $A=208-37=171$.
The sum of all the numbers is $171+37+39=247$.
|
247
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.5. In the park, paths are laid out as shown in the figure. Two workers started to asphalt them, starting simultaneously from point $A$. They lay asphalt at constant speeds: the first on the section $A-B-C$, the second on the section $A-D-E-F-C$. In the end, they finished the work simultaneously, spending 9 hours on it. It is known that the second works 1.2 times faster than the first. How many minutes did the second spend laying asphalt on the section $D E$?
|
Answer: 45.
Solution. Let the line $A D$ intersect the line $C F$ at point $G$, as shown in the figure below. Since $A B C G$ and $D E F G$ are rectangles, we have $A B=C G, B C=A G, E F=D G$ and $D E=F G$.
The second worker works 1.2 times faster than the first, and the working time was the same, so the second worker laid 1.2 times more asphalt than the first. Let the first worker lay $x=A B+B C$ asphalt on the section $A-B-C$, then the second worker on the section $A-D-E-F-C$ laid
$$
\begin{aligned}
1.2 x & =A D+D E+E F+F G+G C=(A D+E F+C G)+(D E+F G)= \\
& =(B C+A B)+(D E+D E)=x+2 D E
\end{aligned}
$$
From this, we find that $D E=0.1 x$, which is 12 times less than the total amount of asphalt on the second section. Therefore, the second worker spent 12 times less time on the section $D E$ than on his entire path. He worked a total of $9 \cdot 60=540$ minutes, so he spent $\frac{540}{12}=45$ minutes on $D E$.
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.8. There are exactly 120 ways to color five cells in a $5 \times 5$ table so that each column and each row contains exactly one colored cell.
There are exactly 96 ways to color five cells in a $5 \times 5$ table without a corner cell so that each column and each row contains exactly one colored cell.
How many ways are there to color five cells in a $5 \times 5$ table without two corner cells so that each column and each row contains exactly one colored cell?
|
Answer: 78.
Solution. Consider the colorings of a $5 \times 5$ table described in the problem (i.e., such that in each column and each row exactly one cell is colored).
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner cell be called $B$. Suppose that among the colorings of the first table, exactly $a$ have the cell $A$ colored, and exactly $b$ have the cell $B$ colored. Clearly, $a=b$ due to symmetry.
Notice that the number of colorings of the second table is equal to the number of colorings of the first table where the cell $A$ is not colored. The number of colorings of the third table is equal to the number of colorings of the first table where neither $A$ nor $B$ is colored. To find this, we subtract from 120 the number of colorings of the first table where $A$ or $B$ is colored. To count the number of such colorings, we add $a$ and $b$, and then subtract what has been counted twice - the ways where both $A$ and $B$ are colored.
All colorings of the first table can be divided into two types: those in which the cell $A$ is colored, and those in which it is not. This leads to the equation $120=a+96$, i.e., $a=24$. Then $b=24$ as well.
The number of colorings where both $A$ and $B$ are colored is the same as the number of ways to color the central $3 \times 3$ square. It is easy to see that there are exactly $3!=6$ such ways (to choose the colored cell in its top row, there are 3 ways, in the middle row - 2 ways, and in the bottom row - 1 way).
Thus, the answer to the problem is the number $120-(24+24-6)=78$.
## 7th grade
|
78
|
Combinatorics
|
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 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
|
Answer: 16.
Solution. An equal number of black and white cells can only be in squares $2 \times 2$ or $4 \times 4$ (in all other squares, there is an odd number of cells in total, so there cannot be an equal number of black and white cells). There are only two non-fitting $2 \times 2$ squares (both of which contain the center of the table, but do not contain any cells above the center), so there are exactly $16-2=14$ fitting $2 \times 2$ squares. And among the $4 \times 4$ squares, only the two lower ones fit.
Thus, the total number of squares with an equal number of black and white cells is exactly $14+2=16$.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
|
Answer: 21.
Solution. Since $AC$ is the diameter of the circle, point $O$ is the midpoint of $AC$, and $\angle AKC=90^{\circ}$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA + \angle CAK = \angle BKA = 90^{\circ}.
$$
The area of the right triangle $ABC$ can be found as $\frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot 6 \cdot 14 = 42$. Since the median $BO$ divides its area in half, the area of triangle $BOC$ is $\frac{42}{2} = 21$.
|
21
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angle $B A M$ measure?
|
Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does not exceed the angle $PAK$, and these angles are equal only if points $M$ and $P$ coincide. Therefore, $M$ is this point of tangency.
The radius $KM$ of the circle is perpendicular to the tangent $AM$. Also, in the right triangle $AMK$, the leg $MK$ is half the hypotenuse $AK$, so $\angle MAK=30^{\circ}$. Additionally, from the condition, we get that $\angle BAC=180^{\circ}-2 \cdot 53^{\circ}=74^{\circ}$. Therefore,
$$
\angle BAM=\angle BAC-\angle MAK=74^{\circ}-30^{\circ}=44^{\circ}
$$
|
44
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
|
Answer: 84.
Solution. Rhombuses consisting of eight triangles can be of one of three types:
It is clear that the number of rhombuses of each orientation will be the same, so let's consider only the vertical ones. Each of them is uniquely determined by its top triangle. Now it is easy to count the number of such triangles.
In the first row, there is 1 such triangle, in the second row - 2, in the third row - 3, ..., in the seventh row - 7. In total, there are \(1+2+3+\ldots+7=28\) vertical rhombuses, and therefore, the answer to the problem is \(28 \cdot 3 = 84\).
|
84
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
|
Answer: 972.
Solution. Let's mark the center of the circle $I$, as well as the points of tangency $P, Q, K$ with the sides $B C$, $A D, A B$ respectively. Note that $I P \perp B C, I Q \perp A D$, i.e., points $P, I, Q$ lie on the same line, and $P Q$ is the height of the given trapezoid, equal to the diameter of its inscribed circle. Also, the segments of tangents $A K=A Q, B K=B P, C P=C L, D L=D Q$ are equal.
From the condition, it follows that $C P=C L=\frac{1}{5} \cdot 30=6, D Q=D L=\frac{4}{5} \cdot 30=24$. Then $B K=B P=$ $B C-C P=9-6=3$.
Since the lines $C I$ and $D I$ are the angle bisectors of angles $C$ and $D$ of the trapezoid, we have $\angle I C D+$ $\angle I D C=\frac{1}{2}(\angle C+\angle D)=\frac{180^{\circ}}{2}=90^{\circ}$, i.e., triangle $C I D$ is a right triangle with a right angle at vertex $I$. Since $I L$ is its height dropped to the hypotenuse, $I L=$ $\sqrt{C L \cdot D L}=12$. This is the radius of the circle.
Similarly, considering the height $I K$ in the right triangle $A I B$, we get $12=$ $I K=\sqrt{A K \cdot B K}$. Using $B K=3$, we extract $A K=A Q=48$.
Thus, the area of the trapezoid is
$$
S_{A B C D}=\frac{B C+A D}{2} \cdot C H=\frac{9+(48+24)}{2} \cdot 24=972
$$
|
972
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.6. Thirty girls - 13 in red dresses and 17 in blue dresses - were dancing in a circle around a Christmas tree. Later, each of them was asked if their right neighbor was in a blue dress. It turned out that those who answered correctly were only the girls standing between girls in dresses of the same color. How many girls could have answered affirmatively?
(R. Zhenodarov)
|
Answer: 17.
Solution: Consider any girl. The colors of the dresses of her left and right neighbors could have been: blue-blue, blue-red, red-blue, red-red. The girl answered "yes" in exactly the first two cases; therefore, she said "yes" exactly when her left neighbor was wearing a blue dress.
Thus, since exactly 17 girls had a left neighbor wearing a blue dress, the answer "yes" was given 17 times.
Remark: There are other (more complex) justifications for why exactly 17 girls in the circle answered "yes."
Comment: An answer alone (without justification, or obtained by considering specific cases of arrangement) -1 point.
For attempts at justification that do not consider all possible arrangements of the girls, no additional points are awarded.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.8. In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of the figure $\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). We will call such a placement unsuccessful if the sum of the numbers in the four cells of the figure is not equal to 0. Find the smallest possible number of unsuccessful placements.
|
Answer: 36.
Solution: We will show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Suppose the opposite; let the numbers in the outer cells of the cross be \(a, b, c, d\), and the number in the central cell be \(e\); denote by \(S\) (M. Antipov) the sum of all these five numbers. Then, according to our assumption, \(S-a = S-b = S-c = S-d = 0\), from which it follows that \(a = b = c = d\). Therefore, \(S-a = e + 3a = 0\), which means \(e = -3a = \pm 3\), which is impossible.
Thus, in each of the 36 "crosses" (with centers in all non-edge cells), there is an unsuccessful placement. Clearly, each placement is contained in no more than one cross; therefore, there are at least 36 such placements.
On the other hand, Figure 2 shows an example of a placement where the number of unsuccessful placements is 36 (in each cell, the sign of the corresponding number is indicated). Indeed, in any cross, there is exactly one unsuccessful placement, and all placements adjacent to the long side of the board are successful.
Comment: Only the answer without justification - 0 points.
Only an example with exactly 36 unsuccessful placements is provided - 2 points.
It is only proven that there must be at least 36 placements, but the corresponding example is missing (or incorrect) - 4 points.
It is only proven that at least one of the four placements of the figure in the "cross" is unsuccessful - 2 points. (These 2 points can be combined with points for a correct example.)
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar?
|
Answer: In the sixth jar, there are -16 beetles.
Solution. Let there be $x$ beetles in the first jar, then in the second jar there are no fewer than $x+1$ beetles, in the third jar no fewer than $x+2$ beetles, and so on. Thus, in the tenth jar there are no fewer than $x+9$ beetles. Therefore, the total number of beetles is no less than $10 x+45$. Considering that a total of 150 beetles were distributed, we get: $x \leqslant 10$.
On the other hand, in the tenth jar there should be no more than $2 x$ beetles, in the ninth jar no more than $2 x-1$ beetles, and so on. This means that in the first jar there are no more than $2 x-9$ beetles, and the total number of beetles is no more than $20 x-45$. Since a total of 150 beetles were distributed, then $x \geqslant 10$.
Thus, there are exactly 10 beetles in the first jar, and 19 or 20 in the last jar. Let's find the sum of eleven consecutive numbers starting from ten: $10+11+\ldots+19+20=165$. Since there should be a total of 150 beetles, the jar with 15 beetles is missing. Therefore, the distribution is uniquely determined: 10, 11, 12, 13, 14, 16, 17, 18, 19, and 20 beetles from the first to the tenth jar, respectively. Thus, in the sixth jar, there are 16 beetles.
Having proved that $x \leqslant 10$, we can continue the reasoning differently. Since in the tenth jar there are no fewer than $x+9$ beetles, and $x+9 \leqslant 2 x$, then $x \geqslant 9$. Then consider two cases: $x=9$ and $x=10$, estimating the number of beetles in the tenth jar.
## Evaluation Criteria:
+ The correct answer and a fully justified solution are provided
$\pm$ The correct answer and generally correct estimates of the number of beetles both "from above" and "from below" are provided, but they contain some gaps
$\pm$ Correct and justified estimates of the number of beetles both "from above" and "from below" are provided, the distribution of beetles in jars is correctly found, but the answer to the question of the problem is incorrect or missing
干 Only one of the two required estimates is correctly performed
干 The distribution of beetles in jars is correctly indicated, but it is not justified
- Only the answer is provided
- The problem is not solved or is solved incorrectly
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1 In the glass, there was a solution in which water made up $99 \%$. The glass with the solution was weighed, and the weight turned out to be 500 gr. After that, some of the water evaporated, so that in the end, the proportion of water was 98\%. What will be the weight of the glass with the resulting solution, if the weight of the empty glass is 300 gr.?
|
Answer: 400 g.
Indication: Initially, the water was $0.99 \cdot(500-300)=198$ (g), and the substance was $200-198=2$ (g). After the water evaporated, 2 g of the substance make up $100 \%-98 \%=2 \%$ of the solution, so the entire solution weighs 100 g, and together with the glass, it weighs 400 g.
|
400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Usain runs one lap around the school stadium at a constant speed, while photographers Arina and Marina are positioned around the track. For the first 4 seconds after the start, Usain was closer to Arina, then for 21 seconds he was closer to Marina, and then until the finish, he was closer to Arina again. How long does it take for Usain to run a full lap?
|
# Solution:
It is not hard to see that regardless of Arina and Marina's positions, the entire circle of the school stadium is divided into two equal parts - half of the circle is closer to Arina and the other half is closer to Marina (this is half of the shorter arc between Arina and Marina and half of the longer arc between Arina and Marina - highlighted with a thick line in the diagram on the right). Therefore, the distance Usain covers while being closer to Marina is half of the entire circle. He covers this distance in 21 seconds, so he will cover the entire circle in 42 seconds.
## Criteria:
Correct answer without justification or with incorrect justification - 2 points.
Incorrect justification includes justification based on a specific case of the positions of Arina, Marina, and Usain at the start (for example, if the author places them at opposite points of the circle, with Usain at the midpoint of the arc between them).
|
42
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. Thirty girls -13 in red dresses and 17 in blue dresses - were dancing in a circle around a Christmas tree. Later, each of them was asked if their right neighbor was in a blue dress. It turned out that those who answered correctly were only the girls standing between girls in dresses of the same color. How many girls could have answered affirmatively?
(R. Zhenodarov)
|
# Answer: 17.
Solution. Consider any girl. The colors of the dresses of her left and right neighbors could have been: blue-blue, blue-red, red-blue, red-red. The girl answered "yes" in exactly the first two cases; therefore, she said "yes" exactly when her left neighbor was wearing a blue dress.
Thus, since exactly 17 girls had a left neighbor wearing a blue dress, the answer "yes" was given 17 times.
Remark. There are other (more complex) justifications for why exactly 17 girls in the circle answered "yes."
Comment. An answer alone (without justification, or obtained by considering specific cases of arrangement) -1 point.
For attempts at justification that do not consider all possible arrangements of the girls, no additional points are awarded.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.8. On a circle of length 2013, 2013 points are marked, dividing it into equal arcs. A chip is placed at each marked point. We define the distance between two points as the length of the shorter arc between them. For what largest $n$ can the chips be rearranged so that there is again one chip at each marked point, and the distance between any two chips that were initially no more than $n$ apart has increased?
( D. Khramov)
|
Answer. $n=670$.
Solution. Let's number the points and the chips placed on them in a clockwise direction with consecutive non-negative integers from 0 to 2012. Consider an arbitrary permutation and the chips with numbers 0, 671, and 1342, initially located at the vertices of an equilateral triangle. The pairwise distances between them are 671. After the permutation, the sum of the pairwise distances between these chips will not exceed the length of the circumference, which means that the distance between some two of them will not exceed $2013 / 3 = 671$; hence, the distance between these two chips will not increase. Therefore, for $n \geqslant 671$, the required permutation is impossible.
Now, let's provide an example of the desired permutation for $n=670$. Each chip with a number $i \leqslant 1006$ will be moved to the point with number $a_{i}=2i$, and each chip with a number $i \geqslant 1007$ will be moved to the point with number $a_{i}=2i-2013$. In other words, $a_{i}$ is the remainder of the division of $2i$ by 2013. It is easy to see that each point will have one chip. It remains to show that the distances between pairs of chips, initially separated by no more than 670, will increase.
Consider any two chips with numbers $i$ and $j$; let the distance between them be $d \leqslant 670$. Then one of the arcs between points $a_{i}$ and $a_{j}$ will have a length of $2d$, meaning the distance between these points is $d'=\min\{2d, 2013-2d\}$. Note that $2d > d$ and $2013-2d > d$ (the latter because $3d < 2013$). Therefore, $d' > d$, which is what we needed to prove.
Comment. Only the correct answer without justification - 0 points.
Proving that for $n \geqslant 671$ the required arrangement does not exist (making an estimate) - 3 points.
Providing an example showing that $n=670$ works, with justification that the example meets the condition - 4 points.
Providing a correct example showing that $n=670$ works, without sufficient justification that the example meets the condition - 3 points.
Points for progress in proving the estimate and in constructing the example are cumulative.
|
670
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. A paper rectangle $3 \times 7$ was cut into squares $1 \times 1$. Each square, except those that stood at the corners of the rectangle, was cut along both diagonals. How many small triangles were obtained?
|
Answer: 68.
Solution. Note that the total number of squares cut along the diagonals is $3 \cdot 7-4=17$. Each of them is cut into 4 small triangles. Therefore, there will be $4 \cdot 17$ - 68 small triangles.
|
68
|
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 3. Masha has 4 pieces of red plasticine, 3 pieces of blue plasticine, and 5 pieces of yellow plasticine. First, she divided each non-red piece of plasticine in half, and then she divided each non-yellow piece of plasticine in half. How many pieces of plasticine did Masha get
|
Answer: 30 pieces of plasticine.
Solution. After Masha's first action, the number of blue and yellow pieces of plasticine doubles. They become 6 and 10, respectively. After Masha's second action, the number of red and blue pieces of plasticine doubles. They become 8 and 12, respectively. Then the total number of pieces of plasticine is $8+12+10=30$.
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. A square area was paved with square tiles (all tiles are the same). A total of 20 tiles adjoin the four sides of the area. How many tiles were used in total?
|
Answer: 36 tiles.
Solution. Let's call the tiles adjacent to the top side of the platform "top tiles". Similarly, we define "right", "left", and "bottom" tiles. Notice that if we add the number of top tiles, the number of bottom tiles, the number of right tiles, and the number of left tiles, we will count the corner tiles twice, so the result will be 4 more than the total number of boundary tiles, i.e., we will get $20+4=24$. Then the number of tiles adjacent to one side of the square is $24: 4=6$. Therefore, the total number of tiles is $6 \cdot 6=36$.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. Merlin decided to weigh King Arthur on enchanted scales that always err by the same weight in the same direction. When Merlin weighed Arthur, they showed a weight of 19 stones. Then Merlin weighed the royal horse and got a weight of 101 stones. Finally, Merlin weighed Arthur on the horse, and the scales showed 114 stones. How many stones does King Arthur actually weigh?
|
Answer: 13 stones.
Solution. Note that if we add 19 stones and 101 stones, we get the combined weight of Arthur and the horse, to which the scale error has been added (or subtracted) twice. Meanwhile, 114 stones is the combined weight of Arthur and the horse, to which the scale error has been added only once. Therefore, the scale error is $19+101-114=6$. This means King Arthur weighs $19-6=13$ stones.
|
13
|
Logic and Puzzles
|
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 7. In the queue for the school cafeteria, 16 schoolchildren are standing in such a way that boys and girls alternate. (The first is a boy, followed by a girl, then a boy again, and so on.) Any boy who is followed by a girl in the queue can swap places with her. After some time, it turned out that all the girls are at the beginning of the queue, and all the boys are at the end. How many swaps were made
|
Answer: 36 exchanges.
Solution. Note that in the end, each boy will make one exchange with each girl who is in line after him. That is, the first boy will make 8 exchanges, the second - 7, the third - 6, and so on. Then the total number of exchanges is
$$
8+7+6+5+4+3+2+1=36.
$$
It is not hard to understand that if each boy strives to give his place to the next girl in line, then sooner or later all the girls will end up at the beginning of the line. Indeed, suppose that at some point no boy can give his place, but not all girls are at the beginning of the line. Then there is a girl who is standing among the boys. Then there is a boy who can give her his place. Contradiction.
|
36
|
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
|
7.4. How to cut a $5 \times 5$ square with straight lines so that the resulting pieces can be used to form 50 equal squares? It is not allowed to leave unused pieces or overlap them.
|
Solution. First, cut the $5 \times 5$ square into 25 squares of $1 \times 1$, then cut each of the resulting squares along the diagonals into 4 triangles, from which, by attaching the longer sides of two triangles to each other, 2 squares can be formed:
|
50
|
Geometry
|
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
|
11.3. Let's call the distance between two cells on a checkerboard the minimum number of moves a chess king can make to get from one to the other. Find the maximum number of cells that can be marked on a $100 \times 100$ board such that there are no two marked cells with a distance of 15 between them.
(I. Bogdanov)
|
Answer. $55^{2}=3025$ cells.
Solution. Divide the board into 9 squares $30 \times 30$, 6 rectangles $10 \times 30$, and one square $10 \times 10$ (see Fig. 5). In each $30 \times 30$ square, the cells are divided into $15^{2}$ groups of four such that the distance between any two cells in the same group is 15 (each group consists of cells with coordinates $(a, b)$, $(a, b+15)$, $(a+15, b)$, and $(a+15, b+15)$). Therefore, in any such group, no more than one cell can be marked, meaning the total number of marked cells in such a square does not exceed $15^{2}$.
Fig. 5
Fig. 6
Similarly, each $10 \times 30$ rectangle (with the longer horizontal side) is divided into pairs of cells, each pair being 15 units apart (with coordinates $(a, b)$ and $(a+15, b)$) - therefore, no more than $15 \cdot 10$ cells can be marked in such a rectangle. Finally, the $10 \times 10$ square has $10^{2}$ cells. In total, the number of marked cells does not exceed $9 \cdot 15^{2} + 6 \cdot 15 \cdot 10 + 10^{2} = (3 \cdot 15 + 10)^{2} = 55^{2}$.
An example with this number of marked cells is shown in Fig. 6.
Comment. Only the answer - 0 points.
Only the correct example of marked cells provided - 1 point.
Proved only that the total number of marked cells is no more than $55^{2} - 5$ points.
|
3025
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the equation $1-(2-(3-(\ldots 2010-(2011-(2012-x)) \ldots)))=1006$.
|
# Answer. $x=2012$
Solution. Opening the brackets, we get $1-2+3-4+\ldots+2011-2012+x=$ 1006; $-1006+x=1006 ; x=2012$.
|
2012
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part.
|
Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In five 15-liter buckets, there are 1, 2, 3, 4, and 5 liters of water respectively. It is allowed to triple the amount of water in any container by pouring water from one other container (if there is not enough water to triple the amount, then it is not allowed to pour from this bucket). What is the maximum amount of water that can be collected in one bucket using such actions?
|
# Answer
The answer depends on the interpretation of the condition -- whether it is allowed to pour NOT all the contents of the bucket (essentially -- whether it is possible to measure OUT ONE LITER of water)
A) If it is not allowed, then the answer is 9 liters
B) If it is allowed, then the answer is 12 liters
## Solution:
Variant $A$ )
We will show how to collect 9 liters in one of the buckets:
$1,2,3,4,5=>1,6,3,0,5=>1,0,9,0,5$.
## Variant B)
We will show how to collect 12 liters in one of the buckets:
$1,2,3,4,5=>1,6,3,4,1=>1,0,9,4,1=>1,0,1,12,1$
In any case, it is necessary to prove that this is the maximum number (that it is impossible to get MORE).
Let the maximum number of liters be $n>=9$.
Consider the last operation with this bucket (at least one operation was performed
--- otherwise $n<=5$ ).
Since $n$ is the maximum number of liters, the last operation could not have involved pouring from this bucket
(otherwise, there would have been more before), i.e., it was filled, so $n$ is a multiple of 3.
There are 15 liters in all the buckets combined.
Note that after each step, there is a non-empty bucket, the number of liters in which is a multiple of three. (solution)
Assume that $n=15$.
Then, on the previous step, there were exactly two non-empty buckets, one with 5 and the other with 10 liters.
But neither of these numbers is a multiple of 3. Contradiction with (*)
## Variant A)
Assume that $n=12$.
Then, on the previous step, there should have been two non-empty buckets: 4 and 8 liters.
Then, due to condition (*), the remaining 3 liters should have been in one bucket.
But then, one step earlier, the amount of water in the buckets should have been 1, 2, 4, 8, which contradicts the condition (*).
Thus, $n=9$, an example of how to get 9 liters is provided above.
Variant B) Thus, $n=12$, an example of how to get 12 liters is provided above.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. While walking in the park, Seryozha and Misha stumbled upon a meadow surrounded by lindens. Seryozha walked around the meadow, counting the trees. Misha did the same but started from a different tree (although he went in the same direction). The tree that was the $20-\mathrm{th}$ for Seryozha was the $7-\mathrm{th}$ for Misha, and the tree that was the $7-\mathrm{th}$ for Seryozha was the 94-th for Misha. How many trees grew around the meadow?
|
# Answer: 100.
Solution. First method. Let there be $n$ trees growing around the glade. We will calculate in two ways the number of intervals between the two trees mentioned in the problem's condition. During Sergei's walk: $20-7=13$. During Misha's walk: $7+(n-94)=n-87$. Therefore, $n-87=13$, which means $n=100$.
Second method. Between the first and second mentioned trees, Misha counted $94-7-1=86$ more trees. And Sergei counted $20-7-1=12$ trees between the second tree and the first tree. Thus, around the glade, there are $86+12=98$ trees plus the two mentioned trees, which makes exactly 100 trees.
## Grading criteria:
+ the correct answer and a fully justified solution are provided
partially correct reasoning is provided, but there is an error of one tree in the "jump"
- only the answer is provided
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. In triangle $A B C$, angle $C$ is $75^{\circ}$, and angle $B$ is $60^{\circ}$. The vertex $M$ of the isosceles right triangle $B C M$ with hypotenuse $B C$ is located inside triangle $A B C$. Find angle $M A C$.
|
Answer: $30^{\circ}$.
First method. From the problem statement, it follows that $\angle B A C=45^{\circ}$. Draw a circle with center $M$ and radius $M B=M C$ (see Fig. 11.2). Since $\angle B M C=90^{\circ}$, the larger arc $B C$ of this circle is the locus of points from which the chord $B C$ is seen at an angle of $45^{\circ}$. Therefore, vertex $A$ lies on this circle. Hence, triangle $A M C$ is isosceles, so $\angle M A C=\angle M C A=\angle B C A-\angle M C B=75^{\circ}-45^{\circ}=30^{\circ}$.
Second method. Let $B C=a$, then from triangle $B M C: M C=\frac{a}{\sqrt{2}}$. From triangle $A B C$ using the Law of Sines, we get $\frac{B C}{\sin 45^{\circ}}=\frac{A C}{\sin 60^{\circ}}$, which means $A C=a \sqrt{\frac{3}{2}}$.
Next, from triangle $C M A$ using the Law of Cosines: $A M^{2}=C M^{2}+C A^{2}-2 C M \cdot \operatorname{CA} \cdot \cos \angle M C A=\frac{a^{2}}{2}+\frac{3 a^{2}}{2}-2 \frac{a}{\sqrt{2}} \cdot a \times$ $\times \sqrt{\frac{3}{2}} \cdot \frac{\sqrt{3}}{2}=\frac{a^{2}}{2}$, so $A M=\frac{a}{\sqrt{2}}$. Therefore, triangle $A M C$ is isosceles.
Further calculations are outlined above.
## Grading Criteria:
+ correct answer and a fully justified solution
$\pm$ it is proven that triangle $A M C$ is isosceles, but an arithmetic error is made in further calculations
- only the answer is provided
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.6. On a circle, 20 points are marked. How many triples of chords with endpoints at these points exist such that each chord intersects each other (possibly at the endpoints)?
|
Answer: 156180.
Solution. The ends of the sought chords can be 3, 4, 5, or 6 points. Let's consider these cases.
1) The ends of the chords are 3 points (see Fig. 11.6a). They can be chosen in $C_{20}^{3}$ ways. Each triplet of points can be connected by chords in a unique way.
Fig. 11.6a
Fig. 11.6b
Fig. 11.6c
Fig. 11.6d
Fig. 11.6e
2) The ends of the chords are 4 points. There are two possible configurations of the chords (see Fig. 11.6b, c). Four points can be chosen in $C_{20}^{4}$ ways. For each set of four points, there are 8 ways to connect them with chords.
3) The ends of the chords are 5 points. In this case, exactly two chords share a common vertex, and the third chord connects the remaining two points (see Fig. 11.6d). Five points can be chosen in $C_{20}^{5}$ ways. For each set of five points, there are five ways to draw the chords (one for each point where two chords meet).
4) The ends of the chords are 6 points (see Fig. 11.6e). Six points can be chosen in $C_{20}^{6}$ ways. For each set of six points, there is a unique way to draw the chords, as the chords must intersect pairwise at internal points.
Thus, the total number of ways to draw the chords is
$$
\begin{aligned}
C_{20}^{3}+C_{20}^{4} \cdot 8 & +C_{20}^{5} \cdot 5+C_{20}^{6}=\frac{20 \cdot 19 \cdot 18}{2 \cdot 3}+\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 8}{2 \cdot 3 \cdot 4}+\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 5}{2 \cdot 3 \cdot 4 \cdot 5}+\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}= \\
& =\frac{20 \cdot 19 \cdot 18}{2 \cdot 3}\left(1+34+\frac{17 \cdot 16}{4}+\frac{17 \cdot 16 \cdot 15}{4 \cdot 5 \cdot 6}\right)=10 \cdot 19 \cdot 6 \cdot(35+68+34)=10 \cdot 19 \cdot 6 \cdot 137=156180
\end{aligned}
$$
The answer can be left in the form $C_{20}^{3}+C_{20}^{4} \cdot 8+C_{20}^{5} \cdot 5+C_{20}^{6}$.
## Grading Criteria:
+ Correct answer and a fully justified solution
$\pm$ All cases are correctly analyzed, but there is an arithmetic error in the final calculation
干 Some individual cases are correctly analyzed
- Only the answer is provided
|
156180
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. One hundred athletes are lined up. Each of them is wearing a red or blue sports suit. If an athlete is wearing a red suit, then the athlete standing nine people away from him is wearing a blue suit. Prove that no more than 50 athletes are wearing red suits.
|
8.5. Solution. Let's number the athletes in the row and consider the first 20
According to the condition, in each such pair, there is no more than one athlete in a red costume. Therefore, among the first 20, there are no more than 10 athletes dressed in red costumes. Reasoning similarly, we get the same statement for the other groups of 20 athletes: 21-40, 41-60, 61-80, 81-100. In total, there are no more than 50 athletes.
|
50
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
6. (7 points) If the number $A$ is written on the board, you can add any of its divisors, except 1 and $A$ itself. Can you get 1234321 from $A=4$? Answer: Yes.
|
Solution. Adding a divisor $n$ to a number means adding $n$ to a number of the form $k n$. The result will be a number of the form $(k+1)n$. Note that the number 1234321 is divisible by 11. Then to the number $A=4=2 \cdot 2$, we will add 2 until we get the number $2 \cdot 11$: $2 \cdot 2 \rightarrow 2 \cdot 3 \rightarrow 2 \cdot 4 \rightarrow 2 \cdot 5 \rightarrow \ldots \rightarrow 2 \cdot 11$. Then we will add 11:
$$
2 \cdot 11 \rightarrow 3 \cdot 11 \rightarrow 4 \cdot 11 \rightarrow 5 \cdot 11 \rightarrow \ldots \rightarrow 112211 \cdot 11=1234321
$$
## Grading Criteria.
- Any correct algorithm for obtaining the number - 7 points.
- There is an idea of how to obtain a number that is a multiple of a proper divisor of the number 1234321, -3 points.
- Answer "yes" without justification - 0 points.
## Maximum score for all completed tasks - 42.
|
1234321
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Point $\mathbf{E}$ is the midpoint of side AB of parallelogram ABCD. On segment DE, there is a point F such that $\mathrm{AD}=\mathbf{B F}$. Find the measure of angle CFD.
|
Solution: Extend $\mathrm{DE}$ to intersect line $\mathrm{BC}$ at point $\mathrm{K}$ (see the figure). Since $\mathrm{BK} \| \mathrm{AD}$, we have $\angle \mathrm{KBE} = \angle \mathrm{DAE}$. Additionally, $\angle \mathrm{KEB} = \angle \mathrm{DEA}$ and $\mathrm{AE} = \mathrm{BE}$, so triangles $\mathrm{BKE}$ and $\mathrm{ADE}$ are congruent. Therefore, $\mathrm{BK} = \mathrm{AD} = \mathrm{BC}$.
Thus, in triangle $\mathrm{CFK}$, the median $\mathrm{FB}$ is equal to half the side to which it is drawn, so this triangle is a right triangle with a right angle at $\mathrm{F}$. Therefore, angle $\mathrm{CFD}$ is also a right angle.
Answer: $90^{\circ}$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter Г on the board, at least one colored cell is found?
|
Solution: Consider a $2 \times 3$ rectangle. In it, obviously, a minimum number of cells need to be colored. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown in the figure.
Answer: 12.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Two painters are painting a 15-meter corridor. Each of them moves from the beginning of the corridor to its end and starts painting at some point until the paint runs out. The first painter has red paint, which is enough to paint 9 meters of the corridor; the second has yellow paint, which is enough for 10 meters.
The first painter starts painting when he is two meters from the beginning of the corridor; and the second finishes painting when he is one meter from the end of the corridor. How many meters of the corridor are painted with exactly one layer?
|
Answer: 5 meters.
Solution. The first painter starts painting 2 meters from the beginning of the corridor and finishes at $2+9=11$ meters from the beginning of the corridor.
The second painter finishes painting 1 meter from the end of the corridor, which is 14 meters from the beginning of the corridor, and starts at $14-10=4$ meters from the beginning of the corridor (Fig. 1).
Fig. 1: to the solution of problem 2
Thus, there are two sections of the corridor that are painted with one layer:
- the section from 2 meters (from the beginning of the corridor) to 4 meters (from the beginning of the corridor) was painted only by the first painter (length of the section 2 meters);
- the section from 11 meters (from the beginning of the corridor) to 14 meters (from the beginning of the corridor) was painted only by the second painter (length of the section 3 meters).
This results in $3 \mathrm{m}+2 \mathrm{m}=5 \mathrm{m}$.
## Criteria
3 p. The correct answer is obtained.
4 6. The correct answer and justification are present.
The justification is considered to be a "picture" showing the sections painted by the painters.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. Lёsha bought a chocolate bar in the shape of a heart (see the picture on the right). Each whole small square of the bar weighs 6 g. How much does the entire bar weigh?
Answer: 240 g.
|
Solution. The tile consists of 32 whole squares and 16 triangles. Each triangle is half a square, meaning it weighs $6: 2=3$ g. Therefore, the weight of the chocolate tile is calculated as follows:
$$
32 \cdot 6+16 \cdot 3=240_{\Gamma} \text {. }
$$
## Criteria
4 p. The correct answer is obtained.
|
240
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. As is known, balance scales come to equilibrium when the weight on both pans is the same. On one pan, there are 9 identical diamonds, and on the other, 4 identical emeralds. If one more such emerald is added to the diamonds, the scales will be balanced. How many diamonds will balance one emerald? The answer needs to be justified.
|
Answer: 3 diamonds.
Solution. From the condition of the problem, it follows that 9 diamonds and 1 emerald weigh as much as 4 emeralds. Thus, if we remove one emerald from each side of the scales, the equality will be preserved, that is, 9 diamonds weigh as much as 3 emeralds. This means that 3 diamonds weigh as much as 1 emerald.
## Criteria
1 p. The correct answer is obtained.
4 p. The correct answer and justification are present.
If the work mentions (or illustrates) three cases of balance on the scales:
- 9 diamonds and 1 emerald balance 4 emeralds,
- 9 diamonds balance 3 emeralds,
- 3 diamonds balance 1 emerald,
then this is considered a correct justification.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Once, a team of Knights and a team of Liars met in the park and decided to ride the circular carousel, which can accommodate 40 people (the "Chain" carousel, where everyone sits one behind the other). When they sat down, each person saw two others, one in front of them and one behind them, and said: "At least one of the people sitting in front of me or behind me belongs to my team." One seat remained empty, and they invited another Liar. This Liar said: "Together with me, we can arrange ourselves on the carousel so that this rule is satisfied again." How many people were in the team of Knights? (A Knight always tells the truth, a Liar always lies.)
|
Solution. Let's consider the initial seating arrangement. From the statement of each Liar, it follows that none of those sitting in front of or behind him is a Liar, i.e., each Liar is surrounded by two Knights. Knights and Liars cannot alternate, as in front of or behind each Knight, there must be at least one Knight. Therefore, Liars and Knights in the seating arrangement form sequences ...KKLKKLKK... Between two Knights, whose neighbors are Liars, any number of Knights can be added in the initial seating, for example, ...KKLKKKKKKLKK... Therefore, the minimum number of Knights will only be in the case where the seating is in triplets ...KKLKK.L... Initially, 39 seats are occupied, this number is divisible by 3, so the maximum number of Liars is 13, and the minimum number of Knights is 26. Note that if one Liar is replaced by a Knight, then we get 12 Liars, for the formation of triplets, 24 Knights are needed, and 3 more Knights can be placed arbitrarily between other Knights.
The problem states that the seating is redone when another Liar appears. Let's see how 40 people, including the additional Liar, can be seated. From his statement, it follows that it is impossible to seat them so that at least one of those sitting in front of or behind each belongs to his team. If we had an extra triplet of Knights, which we placed freely during the first seating, then the additional Liar together with two of them could form a triplet ..KKL.. Therefore, there could not have been more than 26 Knights in the initial seating.
Answer. The team of Knights consisted of 26 people.
|
26
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1.
Find the smallest natural number whose sum of digits is 47.
|
Answer: 299999.
Solution. To find the smallest number, you need to get by with as few digits as possible. The largest digit is 9, so you can't do with fewer than 6 digits ( $5 \cdot 9<47$ ). We cannot put less than 2 in the first place, and by taking 2, the other five digits are uniquely determined: they are nines.
|
299999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Variant 1.
In the apartment, there are four square rooms, which are marked as room №1, №2, №3, №4, and a corridor (№5). The perimeter of room №1 is 16 m, and the perimeter of room №2 is 24 m. What is the perimeter of the corridor (№5)? Give your answer in meters.
|
Answer: 40.
Solution. The side of room No.1 is $16: 4=4$ meters (we divide by 4 because a square has 4 equal sides), and the side of room No.2 is $24: 4=6$ meters. Then, the side of room No.3 is $6+4=10$ meters, so the side of room No.4 is $10+4=14$ meters. Therefore, the longer side of the corridor is $14+4=18$ meters, and the shorter side is $6-4=2$ meters. The perimeter of the corridor is $(2+18) \cdot 2=40$ meters.
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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