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# 8. Variant 1.
On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, 2023 natives, among whom $N$ are liars, stood in a circle and each said: "Both of my neighbors are liars." How many different values can $N$ take? | Answer: 337.
Solution: Both neighbors of a knight must be liars, and the neighbors of a liar are either two knights or a knight and a liar. Therefore, three liars cannot stand in a row (since in this case, the middle liar would tell the truth). We can divide the entire circle into groups of consecutive liars/knights. ... | 337 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. The students of a school went on a trip in six buses. The number of students in the buses was not necessarily equal, but on average, there were 28 students per bus. When the first bus arrived at the destination, the average number of students in the buses that continued moving became 26. How many students were in ... | Answer: 38.
Solution: The initial total number of schoolchildren was $28 \cdot 6=168$. After the first bus finished its trip, there were $26 \cdot 5=130$ schoolchildren left. Therefore, there were $168-130=38$ schoolchildren in the first bus.
Comment: A correct answer without justification - 0 points. | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. On weekdays (from Monday to Friday), Petya worked out in the gym five times. It is known that in total he spent 135 minutes in the gym, and the time spent in the gym on any two different days differed by at least 7 minutes. What is the maximum duration that the shortest workout could have been? | Answer: 13 minutes.
Solution. Let the minimum training time be $x$ minutes, then the second (in terms of duration) is no less than $x+7$, the third is no less than $x+14$, the fourth is no less than $x+21$, and the fifth is no less than $x+28$. Therefore, the total duration of the trainings is no less than $5 x+70$ mi... | 13 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.5. One hundred and one numbers are written in a circle. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 101 written numbers? | Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is a positive one. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there will be at least two positive numbers. Thus, the total number of positive numbers is at least $1+2 \cdot 20=41$. Such a situ... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. What is the maximum number of L-shaped pieces
| |
| :--- |
consisting of three $1 x 1$ squares, that can be placed in a 5x7 rectangle? (The L-shaped pieces can be rotated and flipped, but they cannot overlap). | Solution: The area of the corner is 3, and the area of the rectangle is 35, so 12 corners cannot fit into the rectangle. The image below shows one way to place 11 corners in the rectangle.
A... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the board, three two-digit numbers are written, one of which starts with 5, the second with 6, and the third with 7. The teacher asked three students to each choose any two of these numbers and add them. The first student got 147, and the answers of the second and third students are different three-digit numbers ... | Answer: Only 78.
Solution: Let the number starting with 7 be denoted as $a$, the number starting with 6 as $b$, and the number starting with 5 as $c$. The sum $a+b \geqslant 70+60=130$, so it must be equal to 147. The maximum sum of numbers starting with 7 and 6 is $69+79=148$. The number 147 is only 1 less than 148, ... | 78 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a deck of 52 cards, each person makes one cut. A cut consists of taking the top $N$ cards and placing them at the bottom of the deck, without changing their order.
- First, Andrey cut 28 cards,
- then Boris cut 31 cards,
- then Vanya cut 2 cards,
- then Gena cut several cards,
- then Dima cut 21 cards.
The last... | Answer: 22.
Solution: Removing $N$ cards will result in the same outcome as moving $N$ cards one by one from the top to the bottom. We will consider that each of the boys moved one card several times.
After the last move, the order of the cards returned to the initial state, meaning the total number of card moves was... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Two equilateral triangles $C E F$ and $D I H$ are positioned as shown in the diagram. The diagram indicates the measures of some angles. Find the measure of angle $x$. Provide your answer in degrees.
- $\angle C E M=60^{\circ}$: this is the angle of the equilateral triangle $C E F$;
- $\angle ... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$, the lengths of the sides are known: $A B=4, B C=5, C A=6$. Point $M$ is the midpoint of segment $B C$, and point $H$ is the foot of the perpendicular dropped from $B$ to the angle bisector of angle $A$. Find the length of segment $H M$. If necessary, round your answer to the hundredths.
# | # Answer. 1.
Solution. Let $D$ be the intersection point of line $B H$ with line $A C$. Triangle $A B D$ is isosceles because in it the bisector and the altitude from vertex $A$ coincide. Therefore, $H$ is the midpoint of segment $B D$. Then $H M$ is the midline of triangle $B C D$. Note that $C D = A C - A D = A C - ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. The school stage of the Magic and Wizardry Olympiad consists of 5 spells. Out of 100 young wizards who participated in the competition,
- 95 correctly performed the 1st spell
- 75 correctly performed the 2nd spell
- 97 correctly performed the 3rd spell
- 95 correctly performed the 4th spell
- 96 correctly performed... | # Answer: 8.
Solution. The number of students who correctly performed all spells is no more than 75, since only 75 students correctly performed the second spell. The number of students who made mistakes in the 1st, 3rd, 4th, or 5th spells is no more than $(100-95)+(100-97)+(100-95)+(100-96)=$ $5+3+5+4=17$. If a studen... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Rational numbers a, b, and c are such that $(a+b+c)(a+b-c)=4 c^{2}$. Prove that $\mathrm{a}+\mathrm{b}=0$.
---
The translation maintains the original text's line breaks and formatting. | Solution. The initial equality is equivalent to the following $(a+b)^{2}-c^{2}=4 c^{2}$, or $(a+b)^{2}=5 c^{2}$. If $c \neq 0$, we get $((a+b) / c)^{2}=5 .|(a+b) / c|={ }^{-}$. On the left, we have a rational number, since the sum, quotient, and absolute value of rational numbers are rational, while on the right, we ha... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find $AB$, if $CA=9, CB=4$. | Answer: 6.
Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. 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 |
9.2. Solve the system of equations
$$
\left\{\begin{aligned}
10 x^{2}+5 y^{2}-2 x y-38 x-6 y+41 & =0 \\
3 x^{2}-2 y^{2}+5 x y-17 x-6 y+20 & =0
\end{aligned}\right.
$$ | Solution: We will eliminate the product $xy$. For example, multiply the first equation by 5, the second by 2, and add the left and right parts of the obtained equations. We get $56x^2 + 21y^2 - 224x - 42y + 245 = 0$. Divide the equation by 7, and then complete the squares for $x$ and $y$. We have $8(x-2)^2 + 3(y-1)^2 =... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. The distance between the midpoints of sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ is equal to the distance between the midpoints of its diagonals. Find the angle formed by the lines $AD$ and $BC$ at their intersection. Justify your answer.
 $E M=A D / 2=F N ; 2) F M=B C / 2=E N$ (points $M, N, E, F-$ are the midpoints of segments $A B, C D, D B, A C$, respectively). Therefore, quadrilateral $E M F N$ is a parallelogram. According to the condition, its diagonals ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. The older brother took identical uncolored cubes from Misha and used them to build a large cube. After that, he completely painted some (not all) faces of the large cube red. When the paint dried, Misha disassembled the large cube and found that exactly 343 small cubes had no red faces. How many faces of the large... | Solution: We will call a small cube that has a red face painted. The size of the large cube is greater than 7 (since only the unpainted cubes amount to $343=7^{3}$, and there are also painted ones), but less than 9 (since all "internal" cubes are unpainted - no more than $7^{3}$). Therefore, it is equal to 8. Out of $8... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the class, there are 30 students: excellent students, average students, and poor students. Excellent students always answer questions correctly, poor students always make mistakes, and average students answer the questions given to them strictly in turn, alternating between correct and incorrect answers. All stud... | Answer: 20 C-students
Solution. Let $a$ be the number of excellent students, $b$ be the number of poor students, $c$ be the number of C-students who answered the first question incorrectly, answered the second question correctly, and answered the third question incorrectly (we will call these C-students of the first t... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. What is the maximum number of natural numbers not exceeding 2016 that can be marked so that the product of any two marked numbers is a perfect square? | Answer: 44.
Solution. Let's find the number of natural numbers whose squares are no greater than 2016. There are 44 such numbers, since $44^{2}=1936 < 2016$. Since the product of two perfect squares is a perfect square, the numbers $1=1^{2}, 4=2^{2}, \ldots, 1936=44^{2}$ can be marked.
We will prove that it is imposs... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. If you add the last digit of a number to the number itself, you get 5574, and if you add the second-to-last digit, you get 557. What is this number? | Answer: 5567
Solution. Note that the guessed number differs from 5574 by one digit, so the guessed number is not less than $5574-9=5565$ and not more than 5574. Then the second to last digit of the number is 6 or 7, so our number is 5566 or 5567. The first one does not fit, but the second one does. | 5567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. In the table, there are numbers. It turned out that six sums: three sums by rows and three sums by columns are the same. The minimum of the numbers $a, b$, and $c$ is 101. Find the sum $a+b+c$.
| $\mathrm{a}$ | 1 | 4 |
| :---: | :---: | :---: |
| 2 | $\mathrm{~b}$ | 5 |
| 3 | 6 | $\mathrm{c}$ | | Answer: 309
Solution. Note that $a+5=b+7=c+9$, therefore, the number $c$ is the smallest, and the others are 2 and 4 more, that is, 103 and 105. The sum of all three numbers $101+103+105=309$ | 309 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Alina and Masha wanted to create an interesting version of the school tour olympiad. Masha proposed several problems and rejected every second problem of Alina's (exactly half), Alina also proposed several problems and did not reject only every third problem of Masha's (exactly a third). In the end, there were 10 ... | Answer: 15
Solution. In total, 17 tasks were rejected. Note that among the tasks proposed by Alina, an equal number were included and rejected, and among the tasks proposed by Masha, the number of rejected tasks is twice the number of tasks that remained in the variant. Since the number of rejected tasks is 7 more tha... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.1. On a grid sheet, a $1 \times 5$ rectangle was painted. Each minute, all those uncolored cells that have at least one side-adjacent cell already colored are colored. For example, after one minute, 17 cells will be colored. How many cells will be colored after 5 minutes? | Answer: 105
Solution 1. Note that after 5 minutes, we will get a stepped figure, in which the rows will have $5,7,9,11,13,15,13,11,9,7$ and 5 cells. In total, this is 105.
Solution 2. Note that if we shorten the strip to $1 \times 1$, 44 cells will disappear, which is $4 \cdot 11=44$. If there is only one cell at the... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. A cube lies on a plane. Each face is marked with 1 to 6 points such that the sum of points on opposite faces is always 7. The cube is rolled over the plane as shown in the picture. How many points will be on the top face of the cube when it lands on the last cell?
![](https://cdn.mathpix.com/cropped/2024_05_06_ff... | Answer: 5
Solution. To find out the answer, you need to imagine rolling the die in the opposite direction and each time keep track of where the face that will end up on top will be. It turns out that this face will be at the bottom in the initial position of the die. Since the top face has 2 dots, the bottom face now ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. A four-digit number is called "beautiful" if it is impossible to append a digit to the right so that the resulting five-digit number is divisible by 11. How many beautiful numbers are there that are greater than 3100 and less than 3600? | Answer: 46
Solution. Note that the number 3101 is beautiful because 31009 and 31020 are divisible by 11, so 3101* cannot be divisible by 11. Suppose the number is beautiful, then we have found 10 consecutive numbers, none of which are divisible by 11, but among 11 consecutive numbers, at least one is divisible by 11, ... | 46 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the minimum value of the expression $\frac{25 x^{2} \sin ^{2} x+16}{x \sin x}$ for $0<x<\pi$. | Answer: 40.
Solution. Let $y = x \sin x$. Then the expression can be written as
$$
\frac{25 y^{2} + 16}{y} = 25 y + \frac{16}{y}.
$$
Note that $y > 0$ (since $x > 0$ and $\sin x > 0$), so we can apply the inequality between the means:
$$
25 y + \frac{16}{y} \geq 2 \sqrt{25 y \cdot \frac{16}{y}} = 40.
$$
Equality h... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a right-angled triangle, the lengths of the leg $a$ and the hypotenuse $c$ are expressed as two-digit integers, and both numbers use the same set of digits. The length of the leg $b$ is a positive rational number. What can the length of the hypotenuse be? (Provide all answers and prove that there are no others.) | Answer: 65.
Solution. Since $a \neq c$, in two-digit numbers, two different digits must be used, and they will stand in different places. Let $a=10 e+d, c=10 d+e$. Then $b=\sqrt{(10 d+e)^{2}-(10 e+d)^{2}}=3 \sqrt{11(d-e)(d+e)}$.
By the condition, $b$ is rational, so it cannot contain square roots. The number 11 is pr... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number $N$, which is divisible by $r$, ends in $r$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and $2p+1$ is the cube of a natural number. | Answer: 11713.
Solution. Let $2 p+1=n^{3}$. Then $(n-1)\left(n^{2}+n+1\right)=2 p$. The number $2 p$ can only have the following positive divisors: $1,2, p, 2 p$. The number $n$ is obviously odd, so $n-1$ is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=p$. From this, $n=3, p=13$.
The... | 11713 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A paper triangle with sides $34, 30, 8 \sqrt{13}$ was folded along the midlines and formed into a triangular pyramid. Through the opposite edges of the pyramid, parallel planes were drawn (a total of 6 planes). Prove that the parallelepiped formed by the intersection of these planes is a rectangular parallelepiped, ... | Answer: 1224.
Solution: All four faces of the pyramid are equal, so the opposite edges of the pyramid are pairwise equal. On opposite faces of the parallelepiped, different diagonals are equal, as they are the opposite edges of the pyramid (see figure). Thus, in the parallelogram, which is a face of the parallelepiped... | 1224 | Geometry | proof | Yes | Yes | olympiads | false |
Problem 6. On the board, there are five "equations" of the form $x^{2}+\ldots x+\ldots=0$. Two players take turns filling in the dots with natural numbers from 1 to 10, with each number being used only once. The game ends when all numbers are filled in. The player who makes the first move wants to have as many equation... | Answer. 3. Solution. To obtain three equations, each having two distinct roots, the first player needs to use the three initial moves to write the largest of the numbers not yet written in the place of the coefficient of $x$ in the "equation" where no number has yet been written. This is possible because if no more tha... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the store, apples were sold. On the second day, they sold a quarter of the amount of apples sold on the first day, and an additional eight kilograms. On the third day, they sold a quarter of the amount of apples sold on the second day, and an additional eight kilograms. How many kilograms of apples were sold on t... | Solution. Let's reason from the end. On the third day, 18 kg of apples were sold. If we subtract 8 kilograms, the remaining 10 kg will be a quarter of the amount sold on the second day. Therefore, 40 kilograms of apples were sold on the second day. Of these, 32 kilograms are a quarter of the amount sold on the first da... | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 10.5
The altitudes AD and $\mathrm{BE}$ of an acute-angled triangle $\mathrm{ABC}$ intersect at point $\mathrm{H}$.
The circumcircle of triangle $\mathrm{ABH}$ intersects sides $\mathrm{AC}$ and $\mathrm{BC}$ at points $\mathrm{F}$ and $\mathrm{G}$, respectively. Find FG, if $\mathrm{DE}=5$ cm.
## Number of p... | # Answer:
$\mathrm{FG}=10 \mathrm{~cm}$
## Solution
We have only one numerical given.
So, the length of FG will either be equal to ED or a multiple of it. It doesn't seem to be equal. It should be larger (judging by the diagram), by some factor. What do we know about such relationships? There is the midline of a tr... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On the swamp, there are 64 bumps, they are arranged in an $8 \times 8$ square. On each bump sits a frog or a toad. Frogs never lie, while toads only tell lies. Each of them, both frogs and toads, croaked: “At least one of the neighboring bumps has a toad.” What is the maximum number of toads that could sit on these ... | Solution. Note that frogs cannot sit on adjacent lily pads, otherwise they would be telling the truth. If we divide the lily pads into pairs, there will be no more than one frog in each pair, otherwise they would be telling the truth. Therefore, there can be no more than 32 frogs, as the lily pads can be divided into 3... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Vanya thought of a seven-digit number, and then subtracted from it the sum of all the digits except one. He got 9875352. What number would Vanya have gotten if he had subtracted the sum of all the digits except the second from the left? | Solution. Since the number is a seven-digit number, the sum of its digits is no more than 63. Adding 63 to 9875352, we get an upper estimate for the original number: 9875415. We see that in the intended number, after subtracting the sum of all digits, the first four digits do not change (the hundreds place initially ha... | 9875357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the largest natural number without repeating digits, in which the product of any two consecutive digits is divisible by 6. | 3. Answer: 894326705.
Among two adjacent digits, there is certainly one divisible by 3, and there are only four such digits. Therefore, a 10-digit number is impossible. Let's try to construct a 9-digit number. The digits $0,3,6,9$, which are multiples of three, must be in even positions. Moreover, next to 3 and 9, the... | 894326705 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. How many four-digit numbers are there for which the sum of the digits is a multiple of ten? | 4. Answer: 900.
To any three-digit number, you can append exactly one decimal digit on the right so that the sum of the digits becomes a multiple of 10. For example, to 780, we append 5; to 202, we append 6; to 334, we append 0, and so on. Since three-digit numbers range from 100 to 999 inclusive, there are exactly $9... | 900 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Cut a $3 \times 9$ rectangle into 8 squares.
7 points are awarded for a complete solution to each problem
The maximum total score is 35 | 5. First, cut the rectangle into three squares of size $3 \times 3$. Leave two of them, and from the third, cut out a square of size $2 \times 2$. Cut the remaining part into 5 squares of size $1 \times 1$. In total, you will have 8 squares. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. The cells of the pyramid are filled according to the following rule: above every two adjacent numbers, their arithmetic mean is written. Some numbers were erased, and the structure shown in the figure was obtained. What number was in the bottom right cell? (The arithmetic mean of two numbers is their sum d... | Answer: 6.
Solution. Let's restore the numbers in the table by going through it from top to bottom. For example, if the numbers 21 and $x$ are in the second row, then from $18=\frac{1}{2}(21+x)$ we get $x=15$. Similarly, in the third row, we get that next to the number 14 is 16, and next to it is -26; in the last row,... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. On the side $B C$ of rectangle $A B C D$, a point $K$ is marked. Point $H$ on segment $A K$ is such that $\angle A H D=90^{\circ}$. It turns out that $A K=B C$. How many degrees does angle $A D H$ measure if $\angle C K D=71^{\circ}$?
Fig. 5: to the solution of problem 8.3
Solution. Since $B C \| A D$, we get $\angle A D K=\angle C K D=71^{\circ}$ (Fig. 5).
Since $A K=B C=A D$, triangle $A K D$ is isosceles a... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Thirty-six children are standing in a circle, each dressed in a red or blue sweater. It is known that next to each boy stands a girl, and next to each girl stands a person in a blue sweater. Find the maximum possible number of girls in red sweaters.
# | # Answer: 24.
Solution. Note that there will not be 3 girls in red sweaters standing in a row (otherwise, the condition would not be satisfied for the middle one). By dividing 36 children into 12 triplets, we get that in each of them there are no more than 2 girls in red sweaters, and the total number of girls in red ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. A car left the city for the village at the same time a cyclist left the village for the city. When the car and the cyclist met, the car immediately turned around and headed back to the city. As a result, the cyclist arrived in the city 35 minutes after the car. How many minutes did the cyclist spend on the... | Answer: 55.
Fig. 7: to the solution of problem 8.5
Solution. In Fig. 7, let's mark village $A$, city $B$, point $P$ where the car and the cyclist meet, and point $Q$, where the cyclist was ... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. In an acute-angled triangle $A B C$, the altitude $B H$ is drawn. It turns out that $C H=A B+A H$. How many degrees does the angle $B A C$ measure if $\angle A B C=84^{\circ}$ ?
 | Answer: 64.
Fig. 8: to the solution of problem 8.7
Solution. Mark a point \( K \) on the segment \( CH \) such that \( AH = HK \). Then, from the condition, it follows that \( AB = CK \) (Fig... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) The king decided to test his hundred sages and announced that the next day he would line them up with their eyes blindfolded and put a black or white hat on each of them. After their eyes are uncovered, each, starting from the last in line, will name the supposed color of their hat. If he fails to guess c... | Solution. Let's describe the strategy that the sages should adhere to. The last in line looks ahead, counts the number of black hats, and says "black" if this number is even. In doing so, he cannot save himself for sure. However, the 99th, 98th, ..., 1st in line receive very important information. Thus, the 99th counts... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The parabola $y=x^{2}-20 x+c$, where $c \neq 0$, intersects the $O x$ axis at points $A$ and $B$, and the $O y$ axis at point $C$. It is known that points $A$ and $C$ are symmetric with respect to the line $y=-x$. Find the area of triangle $A B C$. | Answer: 231.
Solution. Since $y(0)=c$, we have $C(0, c)$ and $A(-c, 0)$. Therefore, one of the roots of the equation $x^{2}-20 x+c=0$ is $-c$, i.e., $c^{2}+20 c+c=0$, from which $c=-21, x_{1}=21$. By Vieta's theorem, $x_{1}+x_{2}=20$. Thus, $x_{2}=-1$. The length of the base of the triangle is $x_{1}-x_{2}=22$, and th... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find $x^{2}+y^{2}+z^{2}$, if $x+y+z=2, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$. | # 9.1. Answer: 4.
## 9th grade
By getting rid of the denominators in the second equation, we get $x y + y z + z x = 0$. Squaring the first equation, we get $x^{2} + y^{2} + z^{2} + 2(x y + y z + z x) = 4$. From this, we obtain the answer. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Petya had a large wooden cube with dimensions $30 \times 30 \times 30$ cm $^{3}$. Petya decided to paint the entire cube, which took 100 grams of paint. Later, Petya needed smaller cubes, so he cut the large cube with 6 cuts, parallel to the faces of the cube (2 cuts parallel to each pair of faces), into 27 cubes ... | Answer: 200 grams.
First solution. Each cut increases the unpainted surface area by an amount equal to twice the area of the cut. It is clear that the area of the cut is the same as the area of a face. Therefore, it is necessary to paint an additional surface area equal to \(6 \cdot 2 = 12\) faces, which is twice the ... | 200 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. For four consecutive natural numbers, it is known that the largest of them is a divisor of the product of the other three. Find all values that the largest of these numbers can take. | Answer: 6.
First solution. Let our numbers be $n-3, n-2, n-1, n$, where $n-3 \geqslant 1, n \geqslant 4$. Since the numbers $n-1$ and $n$ are coprime, $(n-3)(n-2)$ is divisible by $n$. Note that $(n-3)(n-2) = n^2 - 5n + 6 = n(n-5) + 6$, and since both numbers $(n-3)(n-2)$ and $n(n-5)$ are divisible by $n$, 6 is also d... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4. Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the water. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have caught on the three days - Mon... | Answer: 50.
Solution. If the old man caught 25 fish each of the first four days and caught nothing on Friday, the conditions of the problem are met, and exactly 50 fish were caught over the specified three days.
Let's prove that in the specified days, fewer than 50 fish could not have been caught. Indeed, suppose few... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$. | Answer: $45^{\circ}$.
Solution. Let $H$ be the orthocenter (the point of intersection of the altitudes) of triangle $ABC$. Then the altitude $AT$ of triangle $ABC$ contains the median of triangle $AMN$, that is, it intersects segment $MN$ at its midpoint - point $E$ (see Fig. 10.5a, b). We can reason in different ways... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the drawer, there are 23 socks: 8 white and 15 black. Every minute, Marina approaches the drawer and pulls out a sock. If at any moment Marina pulls out more black socks than white ones, she exclaims: "Finally!" - and ends the process.
What is the maximum number of socks Marina can pull out before she exclaims: ... | Answer: 17.
Solution: If Marina takes out 17 socks, then among them there will be no more than 8 white ones, which means there will be at least 9 black ones. Therefore, at this moment, she will definitely exclaim, "Finally!"
On the other hand, if she takes out only 16 socks, she might be unlucky: if she sequentially ... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Circle $k_{1}$ with a radius of 8 cm lies inside circle $k$. Both circles intersect circle $k_{2}$ with a radius of 15 cm, as shown in the figure. What is the radius of $k$ if the shaded area inside $k$ but outside $k_{1}$ is equal to the total shaded area inside $k_{2}$?
 \Leftrightarrow\left(2^{x}-2^{-x}\right)^{2}=4 \cos ^{2 a x} / 2\right) \Leftrightarrow$
$$
\left[\begin{array} { l }
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = 2 \operatorname { cos } \frac { a x } { 2 } ... | 4042 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. For the angles of triangle $ABC$, it is known that $\sin \angle A + \cos \angle B = \sqrt{2}$ and $\cos \angle A + \sin \angle B = \sqrt{2}$. Find the measure of angle $C$. | Answer: $90^{\circ}$.
Solution. Square both equalities and add them. Then use the fundamental trigonometric identity, and after transformations, we get $2 \sin \angle A \cos \angle \mathrm{B}+2 \sin \angle B \cos \angle \mathrm{A}=2 \sin (\angle A+\angle \mathrm{B})=2 \quad, \quad$ that is, $\quad \sin (\angle A+\angl... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.4 The Oddball marked the centers of 17 cells in an $N \times N$ grid such that the distance between any two marked points is greater than 2. What is the smallest value that $N$ can take? | Solution: We will show that in an $8 \times 8$ square (and then in any smaller size), it is impossible to mark the cells in this way. Indeed, let's divide the square into $2 \times 2$ squares. In each of them, the pairwise distances between the centers of the cells do not exceed $\sqrt{2}$, so no more than one cell is ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.6. Points $M$ and $N$ are the midpoints of sides $B C$ and $A D$ of quadrilateral $A B C D$. It is known that $\angle B=$ $150^{\circ}, \angle C=90^{\circ}$ and $A B=C D$. Find the angle between the lines $M N$ and $B C$. | Answer: $60^{\circ}$.
Solution. First method. Construct parallelogram $A B M K$ and rectangle $C D L M$ (see Fig. 8.6a). Since $A K\|B C\| L D$ and $A K=B M=$ $M C=L D$, then $A K D L$ is also a parallelogram. Therefore, the midpoint $N$ of its diagonal $A D$ is also the midpoint of diagonal $K L$.
. | Answer. 01.
Solution. The combination 01 occurs 16 times in the tetrad. After it, $7+8=15$ times there is a 0 or 1, and one time there is not. Therefore, one of the combinations 01 stands at the end of the line.
Criteria. Full solution - 7 points. Partial examples of sequences with the correct answer - 1. Only answer... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. There are 21 different applications installed on the phone. In how many ways can six applications be selected for deletion so that among them are three applications from the following six $T V F T' V' F'$, but none of the pairs $T T', V V', F F'$ are included? | Answer. $8 \cdot C_{15}^{3}=3640$.
Solution. Choose one application from each pair $T T^{\prime}, V V^{\prime}, F F^{\prime}-8$ ways. Thus, out of 21 applications, 6 have already been excluded, leaving 15. Choose 3 applications from the remaining 15, which can be done in $\frac{15 \cdot 14 \cdot 13}{3!}$ (or $C_{15}^{... | 3640 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Petya ran down the escalator, counting the steps. Exactly halfway down, he stumbled and the rest of the way he tumbled down (Petya flies three times faster than he runs). How many steps are there on the escalator if Petya counted 20 steps with his feet (i.e., before falling) and 30 steps with his sides (after fallin... | Answer: 80
Solution. Let the escalator have a length of $2 \mathrm{~L}$ (steps), the speed of the escalator be $u$, and Petya runs with a speed of $\mathrm{x}$, and flies with a speed of $3 \mathrm{x}$. Then the time until the fall is $\frac{L}{u+x}$ and during this time Petya will count $\frac{L x}{u+x}$ steps. From ... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. We will call a right-angled triangle elegant if one of its legs is 10 times longer than the other. Is it possible to cut a square into 2020 identical elegant triangles? | Answer: Yes
Solution. An elegant triangle with legs of 10 and 100 can be cut into 100 triangles with legs of 1 and 10 (the cuts are straight lines parallel to its sides). From four such large triangles, we can form a square (its sides will be the hypotenuses of these triangles) with a hole (in the shape of a square wi... | 2020 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 5. CONDITION
A tourist goes on a hike from $A$ to $B$ and back, and completes the entire journey in 3 hours and 41 minutes. The route from $A$ to $B$ first goes uphill, then on flat ground, and finally downhill. Over what distance does the road pass on flat ground, if the tourist's speed is 4 km/h when climbing uphi... | Solution. Let $x$ km of the path be on flat ground, then $9-x$ km of the path (uphill and downhill) the tourist travels twice, once (each of the ascent or descent) at a speed of 4 km/h, the other at a speed of 6 km/h, and spends $(9-x) / 4+(9-x) / 6$ hours on this part. Since the tourist walks $2 x / 5$ hours on flat g... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A three-digit number, all digits of which are different and non-zero, will be called balanced if it is equal to the sum of all possible two-digit numbers formed from the different digits of this number. Find the smallest balanced number. | Answer: 132.
Solution: Let the desired number be of the form $\overline{a b c}$. Then it can be represented as the sum of six different two-digit numbers: $\overline{a b c}=\overline{a b}+\overline{b a}+\overline{a c}+\overline{c a}+\overline{b c}+\overline{c b}$.
From the last equality, we get $100 a+10 b+c=22 a+22 ... | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Three circles with radii 1, 2, 3 touch each other externally at three points. Find the radius of the circle passing through these three points. | Answer: 1.
Solution: Let $\mathrm{O}_{1}, \mathrm{O}_{2}$ and $\mathrm{O}_{3}$ be the centers of the given circles, K, M, N the points of tangency, such that $\mathrm{O}_{1} \mathrm{~K}=\mathrm{O}_{1} \mathrm{~N}=1, \mathrm{O}_{2} \mathrm{~K}=\mathrm{O}_{2} \mathrm{M}=2$ and $\mathrm{O}_{3} \mathrm{~N}=$ $\mathrm{O}_{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1 In a competition of meaningless activity, a participant recorded 2022 numbers in a circle such that each number is equal to the product of its two neighbors. What is the maximum number of different numbers that could have been used? | Solution: If there is a zero among the numbers, then its neighbors are also zeros, so all the recorded numbers will be equal to zero. We will assume that there are no zeros among the numbers. Let $a$ and $b$ be two adjacent numbers. Then on the other side of $a$ stands $a / b$, and on the other side of $b$ stands $b / ... | 337 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Each student in class 7B ate the same number of chocolate bars during math lessons over the week. Nine of them together ate fewer than 288 chocolate bars in a week, while ten of them together ate more than 300 chocolate bars. How many chocolate bars did each student in class 7B eat? Explain your answer. | Answer: 31 chocolates each
Solution 1: Since ten students ate more than 300 chocolates, nine students ate more than (310:10)$\cdot$9 = 270 chocolates. It is also known that nine students together ate less than 288 chocolates. The only number divisible by 9 in the range from 271 to 287 is 279. Therefore, nine students ... | 31 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the largest natural number in which each digit, starting from the third, is equal to the sum of all the previous digits of the number. | Answer: 101248.
Solution: Let the first digit of the number be $a$, the second digit be $b$. Then the third digit is $(a+b)$, the fourth digit is $(2a+2b)$, the fifth digit is $(4a+4b)$, and the sixth digit is $(8a+8b)$. There cannot be a seventh digit, because if it exists, it would be equal to $(16a+16b)$, but this ... | 101248 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. Given an equilateral triangle $\mathrm{ABC}$. On the sides $\mathrm{AB}$ and $\mathrm{BC}$, isosceles right triangles ABP and BCQ are constructed externally with right angles $\angle \mathrm{ABP}$ and $\angle \mathrm{BCQ}$. Find the angle $\angle \mathrm{PAQ}$. | Answer: 90.
Solution: Consider triangle ACQ. It is isosceles because $\mathrm{AC}=\mathrm{BC}$ (by the condition of the equilateral triangle $\mathrm{ABC}$), and $\mathrm{BC}=\mathrm{CQ}$ (by the condition of the isosceles triangle BCQ). The angle at the vertex $\angle A C Q=60 \circ+90 \circ=150$ . Then $\angle \math... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.5. What is the minimum number of cells that need to be marked on a 5 by 5 board so that among the marked cells there are no adjacent ones (having a common side or a common vertex), and adding any one cell to these would violate the first condition? | Answer: 4 cells.
Solution: Estimation. Divide the board into four parts (see fig.). In each of them, a cell must be marked, otherwise the black cell contained in it can be added.
Example. The four black cells in the figure satisfy both conditions.
Criteria: Answer only - 0 points. Estimation only - 5 points. Example... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. How many five-digit numbers exist that are not divisible by 1000, and have the first, third, and last digits even?
Otvet: 9960. | Solution. The first digit of the number can be any of the four (2, 4, 6, or 8), the second and fourth can be any of ten each, and the third and fifth, if we abandon the condition "not divisible by a thousand," can be any of five (0, 2, 4, 6, or 8). Therefore, there are $4 \times 10 \times 5 \times 10 \times 5=10000$ fi... | 9960 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In a white $10 \times 10$ square, on the first move, a $1 \times 1$ cell rectangle is painted, on the second move - a $1 \times 2$ cell rectangle, on the third - $1 \times 3$ and so on, as long as it is possible to do so. After what minimum number of moves could this process end? (Cells cannot be painted over again.... | # Answer: after 6 moves.
Solution. Evaluation. On the board, 16 rectangles of size $1 \times 6$ can be highlighted, of which a maximum of $1+2+3+4+5=15$ will contain colored cells, meaning that a sixth move is always possible. Example. The process is illustrated in the diagram, showing a scenario where a seventh move ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.1
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
Now it is easy to count the resulting pieces. For convenience, they are highl... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.3
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
 balls. Each is painted in some color. If you take out any three balls from the box, there will definitely be at least one red and at least one blue among them. How many balls can be in the box | Answer: 4. Solution. In the box, there are no more than two red and blue balls (otherwise, it would be possible to draw three red or three blue balls) and no more than one ball of other colors (otherwise, it would be possible to draw one blue or red ball and two balls of other colors). Therefore, there are no more than... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.4. On each of 2013 cards, a number is written, and all these 2013 numbers are distinct. The cards are face down. In one move, it is allowed to point to ten cards, and in response, one of the numbers written on them will be reported (it is unknown which one). For what largest $t$ can it be guaranteed to find $t$ card... | Answer. $t=1986=2013-27$.
Solution. 1. First, we will show that it is impossible to guess 1987 cards. Number the cards $A_{1}, \ldots, A_{2013}$; we will demonstrate how to arrange the answers so that none of the numbers on the cards $A_{1}, \ldots, A_{27}$ can be determined.
For each $i=1, \ldots, 9$, combine the ca... | 1986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 1. Option 1.
A confectionery factory received 5 rolls of ribbon, each 50 m long, for packaging cakes. How many cuts need to be made to get pieces of ribbon 2 m long? | Answer: 120.
Solution. From one roll, 25 pieces of ribbon, each 2 m long, can be obtained. For this, 24 cuts are needed. Therefore, a total of $5 \cdot 24=120$ cuts are required. | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
In a box, there are chips. Tolya and Kolya were asked how many chips are in the box. Tolya answered: “Less than 7”, and Kolya answered: “Less than 5”. How many chips can be in the box if it is known that one of the answers is correct? Find all the options. In the answer, write their sum. | Answer: 11.
Solution: If there are 7 or more chips in the box, then both boys are lying. If there are 4 or fewer chips in the box, then both boys are telling the truth. If there are 5 or 6 chips in the box, then Tolya is telling the truth, and Kolya is lying. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4. Option 1.
In a box, there are red, blue, and green pencils. It is known that if you remove two red pencils, the number of pencils of all colors will be equal. And if after that you add 10 red pencils to the box, the number of red pencils will be exactly half of all the pencils. How many pencils were in the box in... | Answer: 32.
Solution. Before adding pencils, it could be considered that there were as many red ones as green ones. Then, the 10 added pencils are the number of blue ones. So, at the moment of adding, there were 10 red, blue, and green pencils each.
# | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
The chocolate bar has the shape of a square with a side of 100 cm, divided into pieces with a side of 1 cm. The sweet tooth ate all the pieces along each of the four sides, as shown in the figure. How many pieces did the sweet tooth eat in total?
. The chocolate bar consists of $100 \times 100=10000$ pieces in total. If 2 rows of pieces are removed from each side, a square with a side length of 96 cm remains. Thus, the sweet tooth will eat $10000-96 \cdot 96=784$ pieces.
. Find the area of this flower bed (in square meters).
 | Answer: 10.
Solution. The area of the flower bed can be calculated as the difference between the area of the plot and the sum of the areas of the two triangular parts not occupied by the flower bed:
: four $3 \times 4$ and a $1 \times 1$ square. If only three cells are shaded, there will be a white rectangle of 12 cells. How to shade 4 cells is shown in the following figure:
 | # 3. Answer. 4.
Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked on the diagram. Since none of these corners can be painted, at least one cell in each of them must be painted. Note that one corner cannot paint cells of two marked corners... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. We will call a number greater than 25 semi-prime if it is the sum of some two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? | 4. Answer. 5.
Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers \(2n+1, 2n+3\), and \(2n+5\), greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers \(2n-1, 2n+1\), and \(2n+3\) are prim... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. One degree on the Celsius scale is equal to 1.8 degrees on the Fahrenheit scale, while $0^{\circ}$ Celsius corresponds to $32^{\circ}$ Fahrenheit. Can a temperature be expressed by the same number of degrees on both the Celsius and Fahrenheit scales? | Answer: Yes, it can.
From the condition, it follows that the temperature in Fahrenheit is expressed through the temperature in Celsius as follows: $T_{F}=1.8 T_{C}+32^{\circ}$. If $T_{F}=T_{C}$, then $0.8 T_{C}+32=0$, that is, $T_{C}=-40$.
Remarks. An answer without explanations - 0 points; an answer with explanation... | -40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. The monthly pension of football fan Ivan Ivanovich is
$$
\frac{149^{6}-199^{3}}{149^{4}+199^{2}+199 \cdot 149^{2}}
$$
rubles, and the cost of a ticket to a World Cup match is 22000 rubles. Will Ivan Ivanovich's pension for one month be enough to buy one ticket? Justify your answer. | Solution: Let $149^{2}=a, 199=b$. Then Ivan Ivanovich's monthly pension (in rubles) is
$$
\frac{a^{3}-b^{3}}{a^{2}+b^{2}+a \cdot b}=\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{a^{2}+a b+b^{2}}=a-b=149^{2}-199=22002
$$
Thus, his pension will be enough for one ticket.
Answer: It will be enough.
Recommendations for check... | 22002 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. The first two digits of a natural four-digit number are either each less than 5, or each greater than 5. The same can be said about the last two digits. How many such numbers are there? Justify your answer. | Solution: There are $4^{2}=16$ two-digit numbers where both digits are greater than 5, and there are $4 \cdot 5=20$ (the first digit is not zero) where both digits are less than 5. In total, there are $16+20=36$ such numbers. This is the conclusion about the first two digits of the number. The last two digits will prov... | 1476 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There is an unlimited number of chips in six colors. What is the smallest number of chips that need to be arranged in a row so that for any two different colors, there are two adjacent chips of these colors in the row. | 5. It follows from the condition that for each fixed color A, a chip of this color must be paired with a chip of each of the other 5 colors. In a row, a chip has no more than two neighbors, so a chip of color A must appear at least 3 times. Similarly for each other color. Thus, there should be no fewer than 3*6=18 chip... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=$ $90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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