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4. Sasha, Lёsha, and Kolya started a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.) | Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The museum has 16 halls, arranged as shown in the diagram. In half of them, paintings are exhibited, and in the other half, sculptures. From any hall, you can go to any adjacent one (sharing a common wall). During any tour of the museum, the halls alternate: a hall with paintings - a hall with sculptures - a hall wi... | Answer: 15.
Solution: One of the possible routes is shown in the diagram. Let's prove that if a tourist wants to visit each hall no more than once, they will not be able to see more than 15 halls. Note that the route starts in a hall with paintings (A) and ends in a hall with paintings (B). Therefore, the number of ha... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Sasha, Lёsha, and Kolya start a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.) | Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - ... | # Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of rai... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. We will call a number palindromic if it reads the same from left to right as it does from right to left. For example, the number 12321 is palindromic. How many five-digit palindromic numbers are there that are divisible by 5? | Answer: 100.
Solution: A number that is divisible by 5 must end in 5 or 0. A mirrored number cannot end in 0, as then it would have to start with 0. Therefore, the first and last digits are 5. The second and third digits can be anything from the combination 00 to the combination 99 - a total of 100 options. Since the ... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - ... | # Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of rai... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.4. 2011 Warehouses are connected by roads in such a way that from any warehouse you can drive to any other, possibly by driving along several roads. On the warehouses, there are $x_{1}, \ldots, x_{2011}$ kg of cement respectively. In one trip, you can transport an arbitrary amount of cement from any warehouse to ano... | Answer. In 2010 trips.
Solution. First, we show that the plan cannot always be completed in 2009 trips. Suppose (with any road scheme) that initially all the cement is located at one warehouse $S$, and it needs to be evenly distributed to all warehouses. Then, cement must be delivered to each warehouse, except $S$, in... | 2010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits of $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? Find all p... | Answer. The digit 6.
Solution. By the condition, $M=3 N$, so the number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.
We will show that $N$ cannot end in 1. If $N$ e... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Pentagon $A B C D E$ is inscribed in circle $\omega$. Diagonal $A C$ is the diameter of circle $\omega$. Find $\angle B E C$, if $\angle A D B=20^{\circ}$. | Answer: $70^{\circ}$.
Solution. Fig. 4. Since $\angle A D B=20^{\circ}$, the arc $A B$ is $40^{\circ}$. Since $A C$ is a diameter, the arc $A B C$ is $180^{\circ}$, so the arc $B C$ is $180^{\circ}-40^{\circ}=140^{\circ}$. The angle $B E C$ subtends the arc $B C$, which means it is equal to $140^{\circ} / 2=70^{\circ}... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the edge $A A^{\prime}$ of the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge length 2, a point $K$ is marked. In space, a point $T$ is marked such that $T B=\sqrt{11}$ and $T C=\sqrt{15}$. Find the length of the height of the tetrahedron $T B C K$, dropped from vertex $C$. | Answer: 2.
Fig. 5: to problem 5
Solution. Notice that
$$
T B^{2}+B C^{2}=11+4=15=T C^{2}
$$
From this, by the converse of the Pythagorean theorem, it follows that angle $T B C$ is a right... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Inside the magician's hat, there live 100 rabbits: white, blue, and green. It is known that if 81 rabbits are randomly pulled out of the hat, there will definitely be three of different colors among them. What is the minimum number of rabbits that need to be taken out of the hat to ensure that there are defi... | # Answer: 61.
Solution. We will prove that if 61 rabbits are randomly pulled out of the hat, then among them there will be two of different colors. Suppose the opposite: let there be $a \geqslant 61$ rabbits of some color (for example, white). Let the second color by number of rabbits be blue. Then there are at least ... | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In an $n \times n$ square, there are 1014 dominoes (each covering two adjacent cells). No two dominoes share any points (even corner points). For what smallest $n$ is this possible? | Answer. For $n=77$.
Solution. Attach four cells to the right and below each domino so that they form a $2 \times 3$ rectangle (if the domino is vertical) or a $3 \times 2$ rectangle (if it is horizontal). If the rectangles of two dominoes have at least one common cell, then the dominoes have a common point. Therefore,... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) Percival's castle had a square shape. One day, Percival decided to expand his domain and added a square extension to the castle. As a result, the perimeter of the castle increased by $10 \%$. By what percentage did the area of the castle increase? | Answer: $4 \%$.
Solution. Let the width of the castle be $a$, and the width of the extension be $b$. Then the original perimeter is $4 a$, and the final perimeter is $4 a+2 b$. Therefore:
$$
1.1 \cdot 4 a=4 a+2 b \Leftrightarrow b=0.2 a
$$
From this, the area of the castle becomes $a^{2}+(0.2 a)^{2}=1.04 a^{2}$, whi... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)$ take?
## Answer: 0. | Solution. Note that the equality $a^{2}+b=b^{2}+c$ can be written as: $a^{2}-b^{2}=c-b$. Similarly, we have $b^{2}-c^{2}=a-c, c^{2}-a^{2}=b-a$. Substituting these equalities into the desired expressions, we get that
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Lёsha did not hesitate to calculate the sum
$$
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017}
$$
and wrote it on the board. How many times is the digit 1 written in the final result? | Answer: 2013.
Solution. Transform the expression:
$$
\begin{aligned}
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017} & =(10-1)+(100-1)+\ldots+\left(10^{2017}-1\right)= \\
& =\underbrace{1 \ldots 10}_{2017}-2017=\underbrace{1 \ldots 1}_{2013} 09093 .
\end{aligned}
$$
Criteria. Any correct solution: 7 points.
It is sh... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (7 points) Several sages lined up in a column. All of them wore either black or white caps. It turned out that among any 10 consecutive sages, there were an equal number of sages with white and black caps, while among any 12 consecutive sages - not an equal number. What is the maximum number of sages that could be
... | Solution. We will prove that there cannot be more than 15 sages. Suppose the opposite, that there are at least 16 sages. Sequentially number all the sages. Consider nine consecutive sages. If we add one of the two neighboring sages to them, then among them there will be an equal number of sages with white and black hat... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Petya has four cards with digits $1,2,3,4$. Each digit appears exactly once. How many natural numbers greater than 2222 can Petya make from these cards? | Answer: 16.
Solution: Let's find out how many different numbers can be formed from these cards: the first digit can be chosen in 4 ways, the second can be appended in 3 ways, the third in 2 ways, and the last one is uniquely determined. That is, a total of 24 different numbers can be obtained (it is also possible to v... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Variant 1.
Nine integers from 1 to 5 are written on the board. It is known that seven of them are not less than 2, six are greater than 2, three are not less than 4, and one is not less than 5. Find the sum of all the numbers. | Answer: 26.
Solution. A number not less than 5 is 5. There is exactly one number 5. Three numbers are not less than 4, so exactly two numbers are equal to 4. Six numbers are greater than 2, meaning all of them are not less than 3. Therefore, exactly three numbers are equal to 3. Seven numbers are not less than 2, so o... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya said that from April 1 to April 20, the café was open for 17 days, and from April 10 to April 30, it was open for 18 days. It is known that he made a mistake once. What was the date of the last Tuesday in April? | Answer: 29
Solution: Since there are exactly 21 days from April 10 to April 30, each day of the week occurred exactly 3 times during this period. Therefore, this statement cannot be false. This means the first statement is false, and there were only 2 Mondays from April 1 to April 20 (there could not have been four, a... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1.
A rectangle was cut into three rectangles, two of which have dimensions 9 m $\times$ 12 m and 10 m $\times$ 15 m. What is the maximum area the original rectangle could have had? Express your answer in square meters. | Answer: 330
Solution. Since the sizes of the two rectangles are fixed, in order for the original rectangle to have the maximum area, the third rectangle must have the largest area. Since the two given rectangles do not have the same sides, the largest area will be obtained by attaching the smaller side of one rectangl... | 330 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1.
In the addition example, where the numbers were written on cards, two cards were swapped, resulting in the incorrect expression: $37541+43839=80280$. Find the error and write down the correct sum. | Answer: 80380
Solution. Let's start checking the example from right to left. There are no errors in the units and tens place, but an error appears in the hundreds place. This means that one of the digits in this place $-2, 8$ or $5-$ is transposed.
Let's consider the following cases: | 80380 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 6. Option 1.
Nезнайка named four numbers, and Ponchik wrote down all their pairwise sums on six cards. Then he lost one card, and the numbers left on the remaining cards were $270, 360, 390, 500, 620$. What number did Ponchik write on the lost card? | Answer: 530.
Solution. Let the original numbers be $a \leq b \leq c \leq d$. Suppose the card with the maximum sum is lost. Then this sum is $c+d$. Therefore, $a+b=270$ and $a+b+c+d>270+620=890$. On the other hand, the sum of all numbers on the cards is $3a+3b+2c+2d=270+360+390+500+620=2140$. We get that $2140>1780+2c... | 530 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
101 natural numbers are written in a circle. It is known that among any 3 consecutive numbers, there is at least one even number. What is the smallest number of even numbers that can be among the written numbers? | Answer: 34.
Solution: Consider any 3 consecutive numbers. Among them, there is an even number. Fix this number and its neighbor, and divide the remaining 99 into 33 sets of 3 consecutive numbers. In each such set, there is at least one even number. Thus, the total number of even numbers is no less than $1+33=34$. Such... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1.
Identical coins are laid out on a table in the shape of a hexagon. If they are laid out so that the side of the hexagon consists of 2 coins, then 7 coins are enough, and if the side consists of 3 coins, then a total of 19 coins are required. How many coins are needed to build a hexagon with a side cons... | Answer: 271.
Solution.
Method 1.
We will divide the coins into layers (contours) from the center. The first layer contains 1 coin, the second layer contains 6, and so on. Notice that each new layer contains 6 more coins than the previous one (if we remove the coins at the vertices, we get exactly as many coins as th... | 271 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. There were 10 chatterboxes sitting in a circle. At first, one of them told one joke, the next one clockwise told two jokes, the next one told three, and so on around the circle, until one of them told 100 jokes at once. At this point, the chatterboxes got tired, and the next one clockwise told 99 jokes, the next one... | Answer: 1000 jokes.
Solution. Let's number the chatterboxes from 1 to 10 clockwise, starting with the one who told the first joke. Then for any pair of chatterboxes with numbers $k$ and $k+1 (1 \leq k \leq 9)$, the $(k+1)$-th chatterbox initially tells one more joke per round than the $k$-th, for 10 rounds. After the ... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $ABC$, the bisector $BD$ was drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \| AC$. Find the angle $DEF$.
| Answer: $45^{\circ}$.
Solution. Let segments $B D$ and $E F$ intersect at point $G$. From the condition, we have $\angle E D G = \angle E D A = \angle D E G$, hence $G E = G D$. Similarly, $G F = G D$. Therefore, $G E = G F$, which means $B G$ is the bisector and median, and thus also the altitude in triangle $B E F$.... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a football tournament where each team played against each other once, teams A, B, C, D, and E participated. For a win, a team received 3 points, for a draw 1 point, and for a loss 0 points. In the end, it turned out that teams A, B, C, D, and E each had 7 points. What is the maximum number of points that team $\m... | Answer: 7 points.
Solution: In a match where one of the teams won, the teams together score 3 points, in a match that ended in a draw - 2 points. Since 7 is not divisible by 3, the team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament.... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. How many different triangles can be formed from: a) 40 matches; b) 43 matches? Solution. We need to find the number of triples of natural numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ such that $\mathrm{x} \leq \mathrm{y} \leq \mathrm{z}$, $\mathrm{x}+\mathrm{y}+\mathrm{z}=40$ and $\mathrm{x}+\mathrm{y}>\mathrm{z}$. ... | Answer: a) $33 ;$ b) 44. | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 2. Solve the equation
$$
\sqrt{3 x-2-x^{2}}+\sqrt{x^{2}-4 x+3}=\sqrt{2}(1-\sqrt{x})
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{c}
3 x-2-x^{2} \geq 0 \\
x^{2}-4 x+3 \geq 0
\end{array}\right.
$$
we obtain that the domain of the function on the left side of the equation is $\{1\}$. The domain of the function on the right side of the equation is the numerical ray $[0 ;+\infty)$... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. First solution. Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, t... | The second solution. Divide all the coins into five equal piles, each containing 8 coins, and number them. Place the 1st and 2nd piles on one side of the scales, and the 3rd and 4th piles on the other.
Consider the first case - the scales balance. Then either there is one fake coin on each side, or all the coins being... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Postman Pechkin is riding a bicycle along a highway. He noticed that every 4.5 kilometers, a suburban bus overtakes him, and every 9 minutes, a suburban bus passes him in the opposite direction. The interval of bus movement in both directions is 12 minutes. At what speed is Pechkin riding? | Answer: 15 km/h
Solution: Let x (km/h) and y (km/h) be the speeds of the cyclist and the bus, respectively. Since the interval between bus movements is 12 minutes (1/5 hour), the distance between two consecutive buses is y/5 km. Therefore, at the moment the cyclist meets a bus, the distance between the cyclist and the... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given a square $A B C D$. Point $N$ lies on side $A D$, such that $A N$ : $N D=2: 3$, point $F$ lies on side $C D$ and $D F: F C=1: 4$, point $K$ lies on side $A B$, such that $A K: K B=1: 4$. Find the angle $K N F$.
 $m \cdot x = 94 = 2 \cdot 47$. From this, it is clear that the numbers $m$ and $x$ are exactly the numbers 2 and 47, i.e., $m = 2$, $x = 47$. (since the number... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 9.1
All three-digit numbers are written in a row: $100101102 \ldots 998$ 999. How many times in this row does a zero follow a two?
## Number of points 7 Answer: 19
# | # Solution
Since a three-digit number cannot start with zero, the two followed by a zero cannot be in the units place of any three-digit number in the sequence. Let's assume the two is in the tens place of a three-digit number. Then the zero following it is in the units place of the same number, i.e., the number ends ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. Harry, Ron, and Hermione wanted to buy identical waterproof cloaks. However, they lacked the money: Ron was short by a third of the cloak's price, Hermione by a quarter, and Harry by one fifth of the cloak's price. When the price of the cloak dropped by 9.4 sickles during a sale, the friends pooled their savings a... | Answer: 36 sickles.
Solution. Let the initial cost of the cloak be $x$ sickles, then Ron had $\frac{2}{3} x$ sickles, Hermione had $\frac{3}{4} x$ sickles, and Harry had $\frac{4}{5} x$ sickles. During the sale, the cloak cost $(x-9.4)$ sickles, and three cloaks cost $3(x-9.4)$ sickles. Since the friends bought three ... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3. Each of the thirteen dwarfs is either a knight, who always tells the truth, or a liar, who always lies. One day, all the dwarfs in turn made the statement: “Among the statements made previously, there are exactly two more false ones than true ones.” How many knights could there have been among the dwarfs? | Answer: 6
Solution. The first two statements are obviously false, as there were fewer than two statements made before each of them. The third statement is true, as there were 2 false statements and zero true statements made before it. The fourth statement is false, as it adds one true statement to the two false ones, ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. At each vertex of a cube lives a number, not necessarily positive. All eight numbers are distinct. If a number is equal to the sum of the three numbers living in the adjacent vertices, then it is happy. What is the maximum number of happy numbers that can live at the vertices of the cube? | Answer: 8.
Solution. See, for example, Fig. 7.5a. It is easy to verify that each vertex of the cube contains a lucky number.
There are other examples as well. Let's understand how they are structured (this was not required of the olympiad participants). Let's denote the numbers at the vertices of the cube (see Fig. 7... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.1. In the numerical example АБВ $+9=$ ГДЕ, the letters А, Б, В, Г, Д, and Е represent six different digits. What digit is represented by the letter Д? | Answer: 0.
Solution: In the addition, the second digit of the first addend АБВ has changed (Д instead of Б). This could only happen if 1 was carried over from the units place to the tens place during the addition. However, the first digit also changed (Г instead of А). This means that 1 was also carried over from the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.2. Lines parallel to the sides of a square form a smaller square, the center of which coincides with the center of the original square. It is known that the area of the cross formed by the smaller square (see the figure on the right) is 17 times the area of the smaller square. How many times is the area of the origin... | Answer. 81 times.
Solution. Let the square have dimensions of 1 cm $\times 1$ cm, and the larger square have dimensions of $n$ cm $\times n$ cm. Then the area of the cross is $(2 n-1)$ cm $^{2}$ (the vertical column has dimensions $n \times 1$, the horizontal row has dimensions $1 \times n$, and the area of the square... | 81 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.5. In the room, there are 10 people - liars and knights (liars always lie, and knights always tell the truth). The first said: "In this room, there is at least 1 liar." The second said: "In this room, there are at least 2 liars." The third said: "In this room, there are at least 3 liars." And so on,
up to the tenth,... | Answer: 5.
Solution: Let there be $k$ liars in the room. Then the first $k$ people told the truth (and thus were knights), while the remaining $(10-k)$ lied (and were liars). Therefore, $k=10-k$, from which $k=5$.
Comment: The answer is obtained by considering an example -3 points. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, there are as many knights as there are liars." Player number 2 said: "In our team, the number of knights and the number of liars differ by one,... | 4. Answer: There are either no knights at all, or there is only one and he plays under number 10. The two answers cannot both be true, as they contradict each other. This means there can be no more than one true answer. If there is no true answer, then the team consists entirely of liars, in which case indeed none of t... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2 Out of 24 matches, a figure in the form of a $3 \times 3$ square is laid out (see figure), the side length of each small square is equal to the length of a match. What is the smallest number of matches that can be removed so that there are no whole $1 \times 1$ squares left, formed from matches.
![](https://cdn.ma... | Answer: 5 matches.
Solution: Estimation: It is impossible to manage with four matches, as by removing a match, we "ruin" no more than two squares (each match is a side of one or two adjacent squares), but initially, we have 9 small squares. Example for 5 matches:
. In the desired group of numbers, there cannot be a number from $A_{0... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$. | Answer: $30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
First method. We use the "folding" method. Reflect triangle $P Q R$ symmetrically relative to line $P R$, and triangle $T S R$ - relative to line $T R$ (see Fig. 11.5a). From the equality (*... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.6. A stack consists of 300 cards: 100 white, 100 black, and 100 red. For each white card, the number of black cards lying below it is counted; for each black card, the number of red cards lying below it is counted; and for each red card, the number of white cards lying below it is counted. Find the maximum possible ... | Answer: 20000.
Solution. First method. The number of different permutations of cards is finite. Therefore, their arrangement with the largest indicated sum exists (possibly not unique).
Let the cards lie in such a way that this sum is maximal. Without loss of generality, we can assume that the top card is white. Then... | 20000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In a right triangle $ABC$ with angle $\angle A=60^{\circ}$, a point $N$ is marked on the hypotenuse $AB$, and the midpoint of segment $CN$ is marked as point $K$. It turns out that $AK=AC$. The medians of triangle $BCN$ intersect at point $M$. Find the angle between the lines $AM$ and $CN$. | Solution. By the property of a right triangle with an angle of $30^{\circ}$, we get that $A B: A C=2: 1$. Therefore, $A B: A K=2: 1$. By the property of medians in a triangle, $\mathrm{BM}: \mathrm{MK}=2: 1$. Then, by the converse of the angle bisector theorem, we get that $A M$ is the bisector of angle $\angle \mathrm... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Vitya Pereperepkin always calculates percentages incorrectly in surveys: he divides the number of people who answered in a certain way by the number of all the others. For example, in the survey "What is your name?", conducted among 7 Anyas, 9 Ols, 8 Yuls, Vitya counted $50 \%$ Yuls.
Vitya conducted a survey in his... | Answer: $a=110$ (in the $2-nd$ variant $a=104$). When Vitya calculates that some part of the students constitutes $5 \%$ of the entire class, in reality, it constitutes $5 / 100=1 / 20$ of the remaining students, $1 / 21$ of the entire class. Similarly, Vitya's $50 \%$ is one third of the class. Therefore, all those wh... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In a checkers tournament, students from 10th and 11th grades participated. Each player played against every other player once. A player received 2 points for a win, 1 point for a draw, and 0 points for a loss. There were 10 times more 11th graders than 10th graders, and together they scored 4.5 times more points tha... | Answer: 20.
Solution: 1. Let $a$ be the number of tenth graders who participated in the tournament, earning $b$ points. Then, $10a$ eleventh graders played, earning $4.5b$ points. In each match, 2 points are played for, and a total of $11a$ players play $\frac{11 a(11 a-1)}{2}$ matches. Therefore, from the condition o... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. Two given quadratic trinomials $f(x)$ and $g(x)$ each have two roots, and the equalities $f(1)=g(2)$ and $g(1)=f(2)$ hold. Find the sum of all four roots of these trinomials. | Answer: 6.
First solution. Let $f(x)=x^{2}+a x+b, g(x)=x^{2}+$ $+c x+d$. Then the conditions of the problem can be written as
$$
1+a+b=4+2 c+d \quad \text { and } \quad 4+2 a+b=1+c+d
$$
Subtracting the second equation from the first, we get $-3-a=3+c$, which means $a+c=-6$. By Vieta's theorem, $-a$ is the sum of the... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than 2”, \ldots, the tenth said: “My number is greater than 10”. After that, all ten, sp... | Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My num... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.5. Each face of a cube $1000 \times$ $\times 1000 \times 1000$ is divided into $1000^{2}$ square cells with side 1. What is the maximum number of these cells that can be painted so that no two painted cells share a side?
$$
\text { (S. Dolgikh) }
$$ | Answer. $3 \cdot 1000^{2}-2000=$ $=2998000$ cells.
Solution. Consider an arbitrary coloring that satisfies the condition. Divide all the cells on the surface into "edges" as shown in Fig. 2 -... | 2998000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial
$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
has at least one integer root?
(P. Kozlov) | Answer. For $n=6$.
Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.
Suppose the numbers $a_{1}, a_{2}, \ldots, ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
 | Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. In a notebook, a triangular grid is drawn (see figure). Tanya placed integers at the nodes of the grid. We will call two numbers close if they are in adjacent nodes of the grid. It is known that
- the sum of all ten numbers is 43;
- the sum of any three numbers such that any two of them are close is 11.
... | Answer: 10.
Solution. Let's denote the numbers by variables as shown in the figure.
Then
\[
\begin{gathered}
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}=c_{1}+c_{2}+c_{3}=11 \\
\left(a_{1}+a_{2}+a... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. The least common multiple of four pairwise distinct numbers is 165. What is the maximum value that the sum of these numbers can take? | Answer: 268.
Solution. Since 165 is the least common multiple of four numbers, these numbers are divisors of 165. To maximize the sum of these numbers, it is sufficient to take the four largest divisors of 165. If one of them is the number 165 itself, then the LCM will definitely be equal to it.
Then the maximum sum ... | 268 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. The teacher wrote a fraction on the board, where the numerator and the denominator are natural numbers. Misha added 30 to the numerator of the given fraction and wrote the resulting fraction in his notebook, while Lesha subtracted 6 from the denominator of the fraction written on the board and also wrote t... | Answer: 5.
Solution. Let $\frac{a}{b}$ be the original fraction. Then Misha wrote down the fraction $\frac{a+30}{b}$ in his notebook, and Lёsha wrote down $-\frac{a}{b-6}$.
Let's write the equation
$$
\frac{a+30}{b}=\frac{a}{b-6}
$$
Transforming it, we get
$$
\begin{gathered}
(a+30)(b-6)=a b \\
a b+30 b-6 a-180=a ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. Given a cyclic quadrilateral $A B C D$. It is known that $\angle A D B=48^{\circ}, \angle B D C=$ $56^{\circ}$. Inside triangle $A B C$, a point $X$ is marked such that $\angle B C X=24^{\circ}$, and ray $A X$ is the angle bisector of $\angle B A C$. Find the angle $C B X$.
$.
The length of segment $A B$ is equal to the absolute value of the difference of the roots of the equation $x^{2}+a x+b=s$. We can express the difference of the... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. On a line, two red points and several blue points are marked. It turned out that one of the red points is contained in exactly 56 segments with blue endpoints, and the other - in 50 segments with blue endpoints. How many blue points are marked? | Answer: 15.
Solution. Let there be $a$ blue points to the left of the first red point, and $b$ blue points to the right; $c$ blue points to the left of the second red point, and $d$ blue points to the right. Then $a b=56, c d=50$. Additionally, $a+b=c+d$ - the number of blue points.
Notice that among the numbers $c$ ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Pantelej and Gerasim received 20 grades each in November, and Pantelej received as many fives as Gerasim received fours, as many fours as Gerasim received threes, as many threes as Gerasim received twos, and as many twos as Gerasim received fives. At the same time, their average grade for November is the same. How m... | Solution. Let's add one to each of Gerasim's grades. His total score will increase by 20. On the other hand, it will become greater than Pantelej's total score by four times the number of Pantelej's twos.
Answer: 5 twos. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Each pair of numbers $A$ and $B$ is assigned a number $A * B$. Find $2021 * 1999$, if it is known that for any three numbers $A, B, C$ the identities are satisfied: $A * A=0$ and $A *(B * C)=(A * B)+C$. | Solution.
$A *(A * A)=A * 0=A *(B * B)=A * B+B$,
$A *(A * A)=(A * A)+A=0+A=A$, then $A * B+B=A$.
Therefore, $A * B=A-B$. Hence, $2021 * 1999=2021-1999=22$.
Answer: 22. | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total? | 8.1. Let a pen cost $r$ rubles, and the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bought is $\frac{357}{2... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. In 8th grade class "G", there are enough underachievers, but Vovochka studies the worst of all. The pedagogical council decided that either Vovochka must correct his twos by the end of the quarter, or he will be expelled. If Vovochka corrects his twos, then the class will have $24 \%$ of underachievers, and if he ... | 8.2. Let there be $n$ students in the class now. According to the condition,
$$
0.24 n = 0.25(n-1)
$$
i.e., $0.01 n = 0.25$. Therefore, $n = 25$. One person constitutes $4\%$ of 25, so there are now $24 + 4 = 28\%$ of underachievers.
Answer: $28\%$. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $\mathrm{ABC}$, $\mathrm{AC}=1$, $\mathrm{AB}=2$, $\mathrm{O}$ is the point of intersection of the angle bisectors. A segment passing through point O and parallel to side $\mathrm{BC}$ intersects sides $\mathrm{AC}$ and $\mathrm{AB}$ at points K and M, respectively. Find the perimeter of triangle $\mat... | 8.3. $\angle \mathrm{KCO}=\angle \mathrm{BCO}=\angle \mathrm{KOC}$ (alternate interior angles). Therefore, OK = KC, and similarly BM = OM. Then
$$
A K+A M+K M=A K+K C+A M+B M=3 .
$$
Answer: 3. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. How many natural numbers less than 1000 are divisible by 4 and do not contain the digits $1,3,4,5,7,9$ in their notation? | 8.4. The desired numbers are written only with the digits $0,2,6,8$.
There is exactly one single-digit number that satisfies the condition, which is the number 8.
There are six two-digit numbers, which are $20,28,60,68,80,88$.
The desired three-digit numbers can end with the following 8 combinations of digits: $00,0... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. At a round table, 2015 people are sitting, each of them is either a knight or a liar. Knights always tell the truth, and liars always lie. Each person was given a card with a number on it; all the numbers on the cards are different. After looking at their neighbors' cards, each person said: "My number is greater t... | Answer: 2013.
Solution. Let $A$ and $B$ be the people who received the cards with the largest and smallest numbers, respectively. Since both of them said the first phrase, $A$ is a knight, and $B$ is a liar. However, if they had said the second phrase, $A$ would have lied, and $B$ would have told the truth; this is im... | 2013 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. We will call a natural number interesting if the sum of its digits is a prime number. What is the maximum number of interesting numbers that can be among five consecutive natural numbers?
(
#
| # Answer: 4.
Solution. Among five consecutive natural numbers, there can be 4 interesting numbers. For example, the numbers 199, 200, 201, 202, 203 (with digit sums 19, 2, 3, 4, and 5) will work.
Now, let's prove that all 5 numbers cannot be interesting. Among our five numbers, there are three that lie within the sam... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the necessary trio of people. How ma... | Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. At the New Year's celebration, schoolchildren organized an exchange game: if they were given five tangerines, they would exchange them for three crackers and a candy, and if they were given two crackers, they would exchange them for three tangerines and a candy. Father Frost played this game with them several times ... | Solution. Ded Moroz conducted 50 exchanges, as he was given 50 candies. At the end of the game, he had no crackers left, meaning he exchanged all of them back. For every two exchanges of tangerines for crackers (ten given - six received), he conducted three exchanges of crackers for tangerines (six given - nine receive... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. How many six-digit numbers exist in which four consecutive digits form the number $2021?$ | Solution. Consider three types of six-digit numbers: $\overline{a b 2021}, \overline{a 2021 b}, \overline{2021 a b}$ (the bar denotes the decimal representation of the number, where $a, b$ - are digits).
In the first case, $\overline{a b 2021}$, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. T... | 280 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.5. This figure contains $9^{2}-4 \cdot 3=69$ cells. Figure 2 shows how a gardener can plant 60 apple trees. We will prove that it is impossible to plant more than 60 apple trees.
Figure 2
... | Answer: the maximum number of apple trees is 60. | 60 | Combinatorics | proof | Yes | Yes | olympiads | false |
10-1. Piglet has balloons of five colors. He managed to arrange them in a row in such a way that for any two different colors in the row, there will always be two adjacent balloons of these colors. What is the minimum number of balloons Piglet could have? | Answer: 11 balls.
Solution. Consider the balls of color $a$. Their neighbors must be balls of all 4 other colors. But one ball can have no more than two neighbors, so there must be at least 2 balls of color $a$. This is true for each of the 5 colors, so there must be at least 10 balls in total.
Notice that some ball ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-5. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest fig... | Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leq 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14$ - 43 $=23$ parts, and each will have between 7 and 12 parts. One figurine is clearly... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.1 A courtyard table tennis tournament among 15 players is held according to certain rules. In each round, two players are randomly selected to compete against each other. After the round, the loser receives a black card. The player who receives two black cards is eliminated from the competition. The last remaining pl... | Answer: 29
Solution. In each match, there is always exactly one loser. Since 14 players were eliminated, there were a total of $14 \cdot 2+1=29$ losses. | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1 Two ants are running along a circle towards each other at constant speeds. While one of them runs 9 laps, the other runs 6. At the point where the ants meet, a red dot appears. How many red dots will there be on the circle?
. Find the smallest natural number that hides the numbers 2021, 2120, 1220, and 1202. | # Answer: 1201201
Solution. Notice that the number contains at least two twos and one zero. If there are exactly two twos, then the zero must stand both between them and after them, but then there must be at least two zeros. Therefore, only on twos and zeros, we need 4 digits (either two twos and two zeros, or three t... | 1201201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1 In the example of addition and subtraction, the student replaced the digits with letters according to the rule: identical letters are replaced by identical digits, different letters are replaced by different digits. From how many different examples could the record $0<\overline{\overline{Б A}}+\overline{\text { БА ... | # Answer: 31
Solution. The sum of two two-digit numbers is no more than 199, so $\overline{\text { YAG }}$ is a three-digit number starting with 1, Y $=1$. Let's look at the last digit in each number, A. It is added twice and subtracted once, so the value of the expression is $\mathrm{A}$, and $\mathrm{A} \neq 0$. $\m... | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. Palm oil has increased in price by $10 \%$. Due to this, the cheese of one of the manufacturers has increased in price by $3 \%$. What is the percentage of palm oil in the cheese of this manufacturer? | Answer: $30 \%$.
Solution. Let's assume that the cheese cost 100 conditional rubles per kilogram. Then it increased by 3 rubles. Since this happened due to the increase in the price of palm oil, 3 rubles is $10 \%$ of the cost of palm oil in the cheese, meaning the cost of palm oil in a kilogram of cheese is 30 rubles... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. Three cyclists leave the city. The speed of the first one is 12 km/h, the second one is 16 km/h, and the third one is 24 km/h. It is known that the first one was cycling exactly when the second and third were resting (standing still), and that at no time did two cyclists cycle simultaneously. It is also known that... | Answer: 16 km
Solution: From the condition, it follows that each of the cyclists was riding at the time when the other two were standing, and, moreover, at any given moment, one of the cyclists was riding (the first one rode when the second and third were standing). Since they all traveled the same distance, and the r... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. 16 travelers, each of whom is either a liar or a knight (liars always lie, knights always tell the truth), settled into 3 rooms of a hotel. When everyone gathered in their rooms, Basil, who was staying in the first room, said: "There are more liars than knights in this room right now. Although no - there are more ... | Answer: 9 knights.
Solution: Since Vasily's statements contradict each other, Vasily is a liar. Therefore, both of Vasily's statements (about each room) are false, and in each room (when he was there) the number of liars and knights was equal. This means that in each room, without Vasily, there was one more knight tha... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. It is known that the numbers EGGPLANT and FROG are divisible by 3. What is the remainder when the number CLAN is divided by 3? (Letters represent digits, the same letters represent the same digits, different letters represent different digits).
Answer: 0 | Solution. By the divisibility rule for 3, the sums B+A+K+L+A+Z+A+N and Z+A+B+A are divisible by 3, and therefore the difference of these sums $K+N+A+H$ is also divisible by 3, which by the rule means that the number KLAN is divisible by 3, hence the remainder is 0.
Criteria. Only the answer - 0 points | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. Usually, we write the date in the format of day, month, and year (for example, 17.12.2021). In the USA, however, it is customary to write the month number, day number, and year in sequence (for example, 12.17.2021). How many days in a year cannot be determined unequivocally by its writing? | Answer: 132.
Solution. Obviously, these are the days where the date can be the number of the month, that is, it takes values from 1 to 12. There are such days $12 \times 12=144$. But the days where the number matches the month number are unambiguous. There are 12 such days. Therefore, the number of days sought is $144... | 132 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find the value of the expression $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}$, if $a+b+c=0$. | Answer: 3.
Solution. $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+(a+b+c)(3 a b+3 b c+3 c a)-3 a b c$. Using the condition $a+b+c=0$, we get: $a^{3}+b^{3}+c^{3}=3 a b c$. Then $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}=\frac{a^{3}+b^{3}+c^{3}}{a b c}=\frac{3 a b c}{a b c}=3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Does there exist a pair of unequal integers $a, b$, for which the equality
$$
\frac{a}{2015}+\frac{b}{2016}=\frac{2015+2016}{2015 \cdot 2016}
$$
holds? If such a pair does not exist, justify it. If such a pair does exist, provide an example. | # Solution.
Multiply the equation by the number $2015 \cdot 2016$. We get the following equation
$$
2016 a + 2015 b = 2015 + 2016
$$
Rewrite the equation in the following form
$$
2016(a-1) = 2015(1-b)
$$
Since the numbers 2016 and 2015 are coprime, then $(a-1)$ is divisible by 2015, i.e., there exists an integer $... | 4031 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In the parliament of a certain state, there are 2016 deputies, who are divided into 3 factions: "blues," "reds," and "greens." Each deputy either always tells the truth or always lies. Each deputy was asked the following three questions:
1) Are you a member of the "blues" faction?
2) Are you a member of the "reds" f... | # Solution.
Let the number of deputies telling the truth in the "blue," "red," and "green" factions be $r_{1}, r_{2},$ and $r_{3}$ respectively, and the number of deputies lying in the "blue," "red," and "green" factions be $l_{1}, l_{2},$ and $l_{3}$ respectively.
According to the problem:
$\left\{\begin{array}{l}r... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. At present, the exchange rates of the US dollar and the euro are as follows: $D=6$ yuan and $E=7$ yuan. The People's Bank of China determines the yuan exchange rate independently of market conditions and adheres to a policy of approximate equality of currencies. One bank employee proposed the following scheme for ch... | # Solution.
Note that if $D$ and $E$ have different parity, their sum is odd, and if they have the same parity, their sum is even. The second number in the pair is always odd. Initially, we have a pair $О Н$ (Odd, Odd).
After one year
After two years $\quad \mathrm{H}+\mathrm{H}=$ О, Н. And so on.
That is, after an... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Given an acute-angled triangle $A B C$. The feet of the altitudes $B M$ and $C N$ have perpendiculars $M L$ to $N C$ and $N K$ to $B M$. Find the angle at vertex $A$, if the ratio $K L: B C=3: 4$. | # Solution.
We will prove that triangles $N A M$ and $C A B$ are similar with a similarity coefficient of $\cos A$. The ratio of the leg $A M$ to the hypotenuse $A B$ is $\cos A$. Similarly, $A M / A C=\cos A$.
Since angle $A$ is common to both triangles
 so that the product of the remaining factors ends in 2? | Answer: 20.
Solution. From the number 99! all factors that are multiples of 5 must be removed, otherwise the product will end in 0. There are 19 such factors in total.
The product of the remaining factors ends in 6. Indeed, the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9$ ends in 6, and similar... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Sasha marked several cells in an $8 \times 13$ table such that in any $2 \times 2$ square, there was an odd number of marked cells. Then he marked a few more cells, as a result of which in each $2 \times 2$ square, there became an even number of marked cells. What is the smallest total number of cells that Sasha cou... | Answer: 48.
Solution. See example in the figure (the digit 1 is in the cells marked the first time, the digit 2 - the second time)
Estimate. In the table, 24 independent 2x2 squares can be placed. In the first round, at least one cell in each of them was marked. Since each of them ended up with an odd number of marke... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. It is known that $a^{2}+b=b^{2}+c=c^{2}+a$. What values can the expression
$$
a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right) ?
$$
take? | Answer: 0.
Solution. From the condition, it follows that $a^{2}-b^{2}=c-b$, $b^{2}-c^{2}=a-c$, and $c^{2}-a^{2}=b-a$. Therefore, $a\left(a^{2}-b^{2}\right)+b\left(b^{2}-c^{2}\right)+c\left(c^{2}-a^{2}\right)=a(c-b)+b(a-c)+c(b-a)=0$.
## Grading Criteria:
+ a complete and justified solution is provided
the correct an... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. From Zlatoust to Miass, a "GAZ", a "MAZ", and a "KAMAZ" set off simultaneously. The "KAMAZ", having reached Miass, immediately turned back and met the "MAZ" 18 km from Miass, and the "GAZ" - 25 km from Miass. The "MAZ", having reached Miass, also immediately turned back and met the "GAZ" 8 km from Miass. What is t... | Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" $-k$ km/h. For each pair of vehicles, we equate their travel time until they meet. We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=\f... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. Square $A B C D$ and isosceles right triangle $A E F$ $\left(\angle A E F=90^{\circ}\right)$ are positioned such that point $E$ lies on segment $B C$ (see figure). Find the angle $D C F$.
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The square $A B C D$ and the isosceles right triangle $A E F$ with $\angle A E F = 90^{\circ}$ are arranged so that point... | Answer: $45^{\circ}$.
Solution. First method. Let $P$ be the foot of the perpendicular dropped from point $F$ to line $B C$ (see Fig. 9.5a). Since $\angle F E P=90^{\circ}-\angle B E A=\angl... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.6. Waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). In the process, the seller wrote down on a piece of paper the weights he used. What is the... | Answer: 6 numbers.
Solution. Let's check that with weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, one can weigh any of the given watermelons. Indeed, $2=1+1 ; 4=3+1 ; 6=5+1 ; 8=7+1 ; 11=10+1$; $12=9+3 ; 13=10+3 ; 14=9+5 ; 15=10+5 ; 16=9+7 ; 17=10+7 ; 18=9+9 ; 19=10+9 ;$ $20=10+10$. Thus, 6 different numbers could... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were wor... | 2. Answer: by 8 percent. Solution. Let's take 1 as the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In pentagon $A B C D E \quad A B=A E=C D=1, \quad B C+D E=1$, $\angle A B C=\angle A E D=90^{\circ}$. Find the area of the pentagon. | 6. Answer: 1. Solution. Extend the segment $C B$ beyond point $B$ to segment $B F=D E$, and then draw segments $C A, C E$ and $A F$. It is easy to see that triangles $C D E$ and $A B F$ are equal. Therefore, $C E=A F$ and the areas of the pentagon and quadrilateral $A E C F$ coincide. But triangles $A E C$ and $A F C$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) In a building, on all floors in all entrances there is an equal number of apartments (more than one). Also, in all entrances there are an equal number of floors. The number of floors is greater than the number of apartments per floor, but less than the number of entrances. How many floors are there in the... | Answer: 11.
Solution. Let the number of apartments per floor be K, the number of floors be F, and the number of entrances be E. According to the condition, $1<K<F<E$. The number 715 can be factored into numbers greater than one in only one way: $715=5 \cdot 11 \cdot 13$. Therefore, $K=5, F=11, E=13$.
Criterion. Corre... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Anya left the house, and after some time Vanya left from the same place and soon caught up with Anya. If Vanya had walked twice as fast, he would have caught up with Anya three times faster. How many times faster would Vanya have caught up with Anya (compared to the actual time) if, in addition, Anya walked twice... | Answer: 7.
Solution: Let Ane's speed be $v$, and Vanya's speed be $V$. The distance Ane has walked is proportional to $v$, while Vanya catches up with her at a speed of $u=V-v$. When Vanya doubles his speed, this difference triples, i.e., $u+V=2V-v=3u$. From this, we get $V=2u=2v$. If Vanya's speed doubles and Ane's s... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. In a box, there are white and blue balls, with the number of white balls being 8 times the number of blue balls. It is known that if you pull out 100 balls, there will definitely be at least one blue ball among them. How many balls are there in the box? | Answer: 108.
Solution: Since the number of white balls in the box is 8 times the number of blue balls, the total number of balls in the box is divisible by 9. Since 100 balls are drawn from it, the smallest possible number of balls in the box $n=108$, of which 12 are blue and 96 are white. Since there are fewer than a... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. Solve the equation $\sqrt{2 x-1}+\sqrt[3]{x}=\sqrt[4]{17-x}$. | Answer: 1.
Solution. It is easy to verify that $x=1$ is a root of the equation.
Since the left side of the equation is an increasing function (as the sum of two increasing functions), and the right side is a decreasing function, there are no other roots.
Comment. The correct answer without proof of its uniqueness - ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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