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19. Grandfather Frost has many identical dials in the form of regular 12-sided polygons, on which numbers from 1 to 12 are printed. He places these dials in a stack on top of each other (one by one, face up). In doing so, the vertices of the dials coincide, but the numbers in the coinciding vertices do not necessarily ... | 19. Answer: 12.
Let there be $k$ clock faces in the stack. Consider any two adjacent columns. The sum of the numbers in them differs by either $k$ or $k-12s$ (where $s$ is the number of clock faces for which one column has 12 and the next has 1). For $k=1,2, \ldots, 11$, all these differences are not divisible by 12, ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21. Baron Münchhausen told a story. "A whole crowd gathered. We reached a crossroads. Then half of our group turned left, a third - right, and a fifth - straight ahead." "But excuse me," the duke noticed, - the sum of a half, a third, and a fifth is not equal to one, so you are lying!" The baron objected: "I am not lyi... | 21. Answer: 37 (18 to the left, 12 to the right, 7 straight).
Evaluation: $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30}$. If there were a total of people, then $\frac{1}{30} \cdot$ is the error. Let's calculate the maximum error (in fractions of a person): $\frac{1}{2}+\frac{1}{3}+\frac{2}{5}=\frac{37}{30}$. Ther... | 37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
32. In the country, there are 1000000 people, each of whom is acquainted with at least one resident. After surveys, a paradoxical situation has emerged: exactly $90 \%$ of the population admitted that they believe in Santa Claus, however, each resident can claim that among their acquaintances, exactly $10 \%$ believe i... | 32. Each person knows at least 10 people. In total, we have 900,000 people who believe in Santa Claus, and they have at least 8,100,000 acquaintances with those who do not believe in him. In total, 100,000 people do not believe in Santa Claus, so among them, there will be someone who has at least 81 acquaintances who b... | 810 | Combinatorics | proof | Yes | Yes | olympiads | false |
35. There is a pile of 660 stones on the table. In one move, you can split any of the existing piles into two smaller ones. At the same time, the sizes of any two piles on the table at the same time must differ by less than a factor of two. What is the maximum number of piles that can result?
## 9th Grade
## Plot 1
... | 35. Answer: 30
Example: Let's go in reverse - we will combine two piles into one without violating the conditions of the problem. Let's say we initially had piles (two of each kind): $15,15,16,16, \ldots, 29,29$. It is easy to see that the condition is not violated, and the sum is 660. We will combine the two smallest... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
40.
$$
\frac{x^{2}}{x^{2}+2 y z}+\frac{y^{2}}{y^{2}+2 z x}+\frac{z^{2}}{z^{2}+2 x y} \geqslant 1
$$ | 40. Since $2 x y \leqslant x^{2}+y^{2}$, we get
$$
\begin{aligned}
\frac{x^{2}}{x^{2}+2 y z}+\frac{y^{2}}{y^{2}+2 z x} & +\frac{z^{2}}{z^{2}+2 x y} \geqslant \\
& \geqslant \frac{x^{2}}{x^{2}+y^{2}+z^{2}}+\frac{y^{2}}{x^{2}+y^{2}+z^{2}}+\frac{z^{2}}{x^{2}+y^{2}+z^{2}}=1
\end{aligned}
$$ | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
41.
$$
\frac{x^{2}+2 y^{2}+2 z^{2}}{x^{2}+y z}+\frac{y^{2}+2 z^{2}+2 x^{2}}{y^{2}+z x}+\frac{z^{2}+2 x^{2}+2 y^{2}}{z^{2}+x y}>6
$$ | 41. Transform the left side using $2 x y \leqslant x^{2}+y^{2}$, multiplying the numerator and denominator by 2:
$$
\begin{aligned}
& \frac{x^{2}+2 y^{2}+2 z^{2}}{x^{2}+y z}+\frac{y^{2}+2 z^{2}+2 x^{2}}{y^{2}+z x}+\frac{z^{2}+2 x^{2}+2 y^{2}}{z^{2}+x y} \geqslant \\
& \geqslant 2\left(\frac{x^{2}+2 y^{2}+2 z^{2}}{2 x^... | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
65. What is the smallest degree that a polynomial $P$ can have if it is known that there exists an integer $b$ such that $A_{P}$ contains elements both greater and less than $b$, but does not contain $b$? | 65. Answer: 4.
For example, consider the polynomial $P(x)=(x-3)(x-1)(x+1)(x+3)$. It is even. Therefore, if $c \neq P(0)$, then $l_{P}(c)$ is even. Therefore, in $A$ there can only be one odd number, namely, $l_{P}(P(0))$. Thus, $b=1$ or $b=3$ is the desired number, since $0,4 \in A_{P}$. Why can't the degree be less? ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
65. What is the smallest degree that a polynomial $P$ can have if it is known that there exists an integer $b$ such that $A_{P}$ contains elements both greater and smaller than $b$, but does not contain $b$? | 65. Answer: 4.
For example, consider the polynomial $P(x)=(x-3)(x-1)(x+1)(x+3)$. It is even. Therefore, if $c \neq P(0)$, then $l_{P}(c)$ is even. Therefore, in $A$ there can only be one odd number, namely, $l_{P}(P(0))$. Thus, $b=1$ or $b=3$ is the desired number, since $0,4 \in A_{P}$. Why can't the degree be less? ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The number 2015 was divided into 12 addends, after which all numbers that can be obtained by adding some of these addends (from one to nine) were written down. What is the minimum number of numbers that could have been written down? | 5.
Answer: 10
Example: Let's break down 2015 into 11 numbers of 155 and one number of 310. In this case, the sum of the addends can range from $155 \times 1$ to $155 \times 10$.
We will prove that it is impossible to get fewer than 10 different numbers: first, note that 2015 is not divisible by 12, so we will have ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. There are 1000 crows sitting on a wire. At the end of each minute, every third (third, sixth, ninth, and so on) crow flies away.
a) Which crows, by their initial count, will remain on the wire in the end?
b) How many minutes will pass before the crows stop flying away? | Solution. a) It is clear that at the end there will be no more than two ravens left, and the first and second never fly away. Therefore, it will be precisely these two. Answer: 1 and 2.
b) It is easy to see that the number of ravens flying away in the next minute is one-third of their number, rounded down. Thus, it on... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Seven natives from several tribes are sitting in a circle by the fire. Each one says to the neighbor on their left: “Among the other five, there are no members of my tribe.” It is known that the natives lie to foreigners and tell the truth to their own. How many tribes are represented around the fire? | Solution. If there are at least 4 natives from one tribe, then two of them sit next to each other, and one of them will lie to the other, although they should tell the truth. If there is only one native from a certain tribe, then he tells the truth to his left neighbor, although he should lie. Therefore, each tribe has... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. On a 5 by 5 board, a certain number of bishops and knights were placed. Then, they counted how many times each bishop is attacked, and all the obtained numbers were added together. What is the maximum sum that could have been obtained?
# | # Solution:
Answer: 80.
The example is constructed through a chessboard coloring - bishops on one color, rooks on the other.
Evaluation: First, place bishops everywhere. The total sum will be 64. Now, in each cell, write the number of points that can be gained by replacing a bishop with a knight. We will get the fol... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.3. Let $N=15$ and only the second operation is allowed. Prove that from any arrangement, less than $10^{4}$ others can be obtained. | Solution. We will paint the positions in three colors; the parity of the number of blacks on all positions changes simultaneously. With a fixed set of parities, there are exactly $2^{12}$ arrangements, and with a fixed set up to inversion $-2^{13}<10000$. | 10000 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Timur thought of a three-digit number and told it to Anton, who then wrote down three numbers that result from replacing one of the digits of the original number with 1 (for example, if the number 215 was thought of, Anton would write down 115, 215, 211). The sum of all the numbers written down by Anton turned out t... | Solution: Notice that from the number abc, we get the sum $1 b c + a 1 c + a b 1 = 111 + 2^{\star} a b c$. From the equation $1243 = 111 + 2 * a b c$, it follows that $a b c = 566$. | 566 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In each cell of a $10 \times 10$ board, there sits a rabbit. Between rabbits in adjacent cells, there are partitions that can be removed. What is the minimum number of partitions that need to be removed so that any rabbit can visit any other rabbit, traveling through no more than 17 cells (not counting the starting ... | Solution. An example for 100 partitions is given in the figure. The correctness of the example follows from the fact that, as can be easily seen, any rabbit can reach the central 4-cell square by moving no more than 8 cells. From the central square to any cell, it can also be reached in no more than 8 moves. No more th... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to triple the salary for everyone earning up to 500 dollars, and increase the salary by 1000 dollars for the rest, so the total salary will become 24000 dollars. A mean manager proposes to reduce the salary to 500... | Solution. Note that the increase proposed by the kind manager is twice as large as the salary proposed by the evil manager (this is true for both poor and rich employees). The increase according to the kind manager's proposal is 14000, so the salary according to the evil manager's proposal is 7000. | 7000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. There are 30 people in the bar. The bartender knows that among them, there are 10 knights (who always tell the truth), 10 liars (who always lie), and 10 troublemakers. The bartender can ask person $X$ about person $Y$: "Is it true that $Y$ is a troublemaker?" If $X$ is not a troublemaker, they will answer the questi... | Answer: 19. Indeed, with the first question, a peaceful client may be identified, so the bartender cannot guarantee to keep all peaceful clients.
Let's show how the bartender can leave 19 peaceful clients.
Solution 1. First, he asks everyone about client $A$ until someone throws him out of the bar. Let's say $B$ thre... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. In a $100 \times 100$ square, 10000 cuts were made along the grid lines (each cut one cell long), and it split into 2500 four-cell figures. How many of them are $2 \times 2$ squares? | Solution:
Answer: 2300.
Solution. Note that the perimeter of a $2 \times 2$ square is 8, while for the other four-cell figures (rectangle $1 \times 4$, T-shape, L-shape, or S-shape) it is 10. Let the number of squares be $x$, and the number of other figures be $2500-x$. Then their total perimeter is $8 x + 10 \cdot (... | 2300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Milla was writing the letters M and L in uppercase. At the end, she counted that the letter matched the previous one 59 times, and did not match 40 times. Determine the maximum number of letters M that Milla could have written, and prove that it is indeed the maximum.
| Solution. Answer: 80 letters.
For each letter except the first, it is known whether it matches the previous letter or not. Therefore, Milla wrote $1+59+40=100$ letters.
Divide all the written letters into groups of consecutive identical letters. Then, groups of M (M-groups) will alternate with groups of L (L-groups).... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There are 20 chocolate candies on the table. Masha and Bear are playing a game according to the following rules. Players take turns. In one move, a player can take one or several candies from the table and eat them. Masha goes first, but on this move, she cannot take all the candies. In all other moves, players cann... | Solution. Masha will win if she eats 4 candies on her first move.
Let's arrange the candies in a row, number them, and assume that the players take candies in a row from left to right.
Suppose Masha eats one candy on her first move. In this case, she will lose, as until the end of the game, the players will take one ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Petya thought of a 9-digit number obtained by rearranging the digits of the number 123456789. Vitya is trying to guess it. For this, he chooses any 9-digit number (possibly with repeated digits and zeros) and tells it to Petya, who then responds with how many digits of this number match the ones in his thought numbe... | Solution. Vitya's first move is to name the number 122222 222. Then Petya can only respond with 0, 1, or 2.
If Petya answers 0, then neither the one nor the two hit their places. This means the two must be in the first position. There's no need to ask further.
If Petya answers 2, then both the one and the two hit the... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.4. Prove that Shakti will always be able to find a strange sociable set of no more than 82 troglodytes.
# | # Solution.
Consider a communicative set $A$ of the smallest size. Note that we can choose $A$ such that for each troglodyte $a$ in $A$, there exists a troglodyte $v_{a}$ whose only friend in the set $A$ is the troglodyte $a$. Indeed, otherwise, if $a$ is friends with someone in $A$, it can simply be removed, and if i... | 82 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. The number 100 is represented as the sum of several two-digit numbers, and in each addend, the digits are swapped. What is the largest number that could result from the new sum? | Solution. If the digits in the two-digit number $\overline{b a}$ are swapped, the number increases by $9(a-b)$. Therefore, the new sum is $S=100+9 U-9 D$, where $D$ is the sum of the tens digits, and $U$ is the sum of the units digits in the original addends. Since $10 D+U=100$, then $S=1000-99 D$. Thus, we need to min... | 406 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Given a string of 2021 letters A and B. Consider the longest palindromic substring. What is its minimum possible length? A palindrome is a string that reads the same from right to left and from left to right. | Solution. The minimum possible length of the maximum palindrome is 4.
We will prove that it cannot be less than 4. Consider the 5 letters in the center of the string. If these are alternating letters, then it is a palindrome of length 5. Suppose among these five letters there are two identical letters standing next to... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the bus route, there are only four stops - "Initial", "First", "Final", and "Last". At the first two stops, passengers only got on, and at the remaining stops, they only got off. It turned out that 30 passengers got on at the "Initial" stop, and 14 passengers got off at the "Last" stop. At the "First" stop, three... | Answer: Those traveling from "First" to "Final" are six more.
Let $x$ be the number of people who got on at "First". Then, $3x$ people got off at "Final". Since the number of people getting on equals the number of people getting off, we have $30 + x = 3x + 14$, from which $x = 8$. Let $y$ be the number of people trave... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Baron Munchausen placed a horse in some cells of an $N \times N$ board. He claims that no one will find two different $4 \times 4$ squares on this board (with sides along the grid lines) with the same number of horses. For what largest $N$ can his words be true?
# | # Solution:
Answer: $N=7$.
The number of knights in a $4 \times 4$ square can range from 0 to 16, i.e., there are 17 possible variants. The number of $4 \times 4$ squares on an $N \times N$ board is $(N-3)^{2}$ (since the top-left cell of the square can occupy positions from the far left to the fourth from the right ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Pompous Vova has an iPhone XXX, and on that iPhone there is a voice-command calculator: "Multiply my number by two and subtract two from the result," "Be so kind as to multiply my number by three and then add four," and finally, "Add seven to my number!" The iPhone knows that Vova initially had the number 1. How man... | Answer: 9000 (or 18000, if negative numbers are considered).
We will prove that we can obtain all four-digit numbers. Note that the command +7 allows us to obtain from the current number all larger numbers with the same remainder when divided by seven. Therefore, it is sufficient to use the first two buttons to obtain... | 9000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Winnie-the-Pooh decided to give Eeyore a pot of honey. On the way to Eeyore, he tried the honey from the pot several times. When he tried the honey for the first time, the pot became half as heavy. And after the second time, the pot became half as heavy again. And after the third! And after the fourth! Indeed, after... | Solution. $\quad$ Answer: 3000 g. Indeed, $3000=(((200 \cdot 2) \cdot 2) \cdot 2) \cdot 2)-200$. | 3000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Petya has several 5-ruble coins and several 2-ruble coins. Vanya has as many 5-ruble coins as Petya has 2-ruble coins, and as many 2-ruble coins as Petya has 5-ruble coins. Petya has 60 rubles more than Vanya. Which coins does Petya have more of - 5-ruble or 2-ruble? By how many? | Solution. Answer: Petya has 20 more 5-ruble coins.
Suppose Petya has more 2-ruble coins than 5-ruble coins. Let Petya remove one 2-ruble coin, and Vanya remove one 5-ruble coin. Then the difference in the sums of Petya's and Vanya's coins will increase by 3. If they repeat this operation until Petya has an equal numbe... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The Elector George has 100 coins, some of which are counterfeit (possibly all or none). George can show the expert from 10 to 20 coins, and the expert will tell him how many of them are counterfeit. The problem is that the only expert in the entire region is Baron Münchhausen, and he exaggerates: the result given by... | Solution. It will work. We will submit $X$ random coins for expertise, and then the same $X$ plus 1 more. If the baron says the same number both times, the added coin is genuine; otherwise, it is counterfeit. This way, we can divide the coins into groups of 10, and check each group for counterfeits in 11 queries: first... | 110 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There are several cards. On each of them, on each of the two sides, a circle is drawn: red, blue, or yellow. Among any 30 cards, there is a card with a red circle, among any 40 cards, there is a yellow circle, and among any 50, there is a blue one. In total, there are 20 cards with circles of different colors. Prove... | # Solution:
There are no more than 29 cards without red circles, no more than 39 cards without yellow, and no more than 49 cards without blue. Adding all these numbers, we get no more than 117. In this process, single-colored cards have been counted twice, and two-colored cards have been counted once. Therefore, the d... | 48 | Combinatorics | proof | Yes | Yes | olympiads | false |
7. Kopyatych took some natural number, raised it to the 1st, 2nd, 3rd, 4th, and 5th powers. Then he encrypted the numbers by replacing the same digits with the same letters and different digits with different letters. He wrote each encrypted number on a separate piece of paper. But Nyusha left only a fragment of each p... | # Solution:
It is clear that the letters К, Ё, Ж, И are of the same parity, and among them there are no 0 or 5.
If К, Ё, Ж, И are even, then all powers except the first one must be divisible by 4, and then О, Ш are also even digits, which is not good.
The last digit of the number in the first and fifth power is the ... | 189 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. We remind you that the sum of the angles in a triangle equals 180 degrees. In triangle $A B C$, angle $A$ is a right angle. Let $B M$ be the median of the triangle, and $D$ be the midpoint of $B M$. It turns out that $\angle A B D = \angle A C D$. What are these angles? | # Solution.
Draw $A D$. Since triangle $A B M$ is a right triangle, its median $A D$ equals half the hypotenuse $B M$. Therefore, $\angle A B D = \angle B A D = \alpha$, and $\angle A D M$ equals $2 \alpha$, as the exterior angle of triangle $A B D$. However, since $D M = A D$, $\angle D A M = \angle D M A = 90^{\circ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In the Magic and Wizardry club, all first and second-year students wear red robes, third-year students wear blue, and fourth-year students wear black.
Last year, at the general assembly of students, there were 15 red, 7 blue, and several black robes, while this year - blue and black robes are equal in number, and r... | # Solution.
(a) 7 mantles. Since 7 blue mantles from last year will turn black this year and there are an equal number of blue and black mantles this year, there will also be 7 blue mantles this year, which will turn black next year.
(b) 6 students. This year, there are 7 blue, 7 black, and $14=2 \times 7$ red mantle... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Each of five friends multiplied several consecutive numbers starting from 1. It turned out that one of the products is equal to the sum of the other four. Find all possible values of this product and show that there are no other values. | Solution. The product of all consecutive numbers from 1 to $x$ is called the factorial of the number $x$ and is denoted by $x!$. Thus, we need to solve the equation $x!=a!+b!+c!+d!$. Let the numbers $a, b, c, d$ be ordered in ascending order. Then $x>d$, i.e., $x! \geqslant x \cdot d!$. But we know that $x! \leqslant 4... | 24 | Number Theory | proof | Yes | Yes | olympiads | false |
6. In a $4 \times 4$ square, cells are colored in several colors such that in any $1 \times 3$ rectangle, there are two cells of the same color. What is the maximum number of colors that can be used? | Solution. Maximum 9 colors. See example in the picture. We will prove that more is not possible. Any row (row or column) gives a maximum of three colors, so the first row + first column will give a maximum of $3+3-1=5$ colors.
We will prove that the remaining $3 \times 3$ square will give a maximum of 4 colors. Indeed... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Captain Billy the pirate plundered 1010 gold doubloons and set sail on his ship to a deserted island to bury them as treasure. Every evening of the voyage, he paid each of his pirates one doubloon. On the eighth day of the voyage, the pirates plundered a Spanish caravel, and Billy's treasure doubled, while the numbe... | Solution. Answer: 30
Before the pirates looted the caravel, Billy managed to pay the pirates their daily wages 7 times. After this, his fortune doubled. This is equivalent to Billy having 2020 doubloons before the voyage, and he paid the wages 14 times. After this, Billy paid the wages 40 times to half of the remainin... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The length of the escalator is 200 steps. When Petya walks down the escalator, he manages to count 50 steps. How many steps will he count if he runs twice as fast? | Solution. 80 steps. Let's call the step of the escalator from which Petya begins his descent the first step. When Petya walks down, he passes 50 steps. During this time, the first step manages to descend $200-50=150$ steps. Therefore, the escalator moves three times faster than Petya walks. When Petya runs, the speed r... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. In the company, there are elves, fairies, and gnomes. Each elf is friends with all fairies except for three, and each fairy is friends with twice as many elves. Each elf is friends with exactly three gnomes, and each fairy is friends with all gnomes. Each gnome is friends with exactly half of the elves and fairies c... | Solution. Answer: 12.
Let $n$ be the number of elves, $m$ be the number of fairies, and $k$ be the number of gnomes. Then the number of friendly pairs "elf-fairy" is $n(m-3)$, and "fairy-elf" is $-m \cdot 2(m-3)$. But these are the same pairs, therefore,
$$
n(m-3)=2 m(m-3)
$$
From which, $n=2 m$. Counting the friend... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. A number is written on the board. In one move, you can either increase or decrease any of its digits by three (if it results in a digit), or swap two adjacent digits. Show how to transform the number 123456 into 654321 in 11 moves. | Solution. In the first 5 moves, we move 6 to the beginning: 612345. Then we swap 1 and 2, 4 and 5, getting 621354. Now we increase 2 and 1 by three, and decrease 5 and 4 by three - resulting in 654321. | 654321 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
3. A group of toddlers in a kindergarten has 90 teeth in total. Any two toddlers together do not have more than 9 teeth. What is the minimum number of toddlers that can be in the group? | Solution. If all children have fewer than 5 teeth, then there are no fewer than $90 / 4$, i.e., no fewer than 23 children.
If one child has exactly 5 teeth, then the others have no more than 4, and there are no fewer than $1+85 / 4$, i.e., also no fewer than 23 children.
If any of the children have between 6 and 9 te... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. A chessboard $(8 \times 8)$ was cut into several equal parts in such a way that all white cells remained uncut, while each black cell was cut. How many parts could have been obtained? | Solution. Note that there are 32 white cells, and each part contains an integer number of white cells, so the answer must be a divisor of 32.
Obviously, it cannot be 1. The answers $2, 4, 8, 16$, and 32 are possible. To achieve them,
 - are standing in a row. Each of them said one of two phrases: "To the left of me, there are twice as many knights as liars" or "To the left of me, there are as many knights as liars." In reality, there were more knights than liars... | Answer: 49.
Solution:
There are more knights than half, so:
1) Either they alternate like this: KRK...LK, but this option does not work: the phrase "there are twice as many knights as liars" cannot be said by any knight, but there are more than 50 such phrases.
2) Or some two knights stand next to each other. Two k... | 49 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Today's date is written as: 22.11.2015. How many other days in this year can be written with the same set of digits? | Solution. The month number cannot start with a two, so it is either 11 or 12. In the first case, it is 22.11, in the second case 12.12 and 21.12. Answer: two. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A group of toddlers in a kindergarten has 90 teeth in total. Any two toddlers together do not have more than 9 teeth. What is the minimum number of toddlers that can be in the group | Solution. If all children have fewer than 5 teeth, then there are no fewer than $90 / 4$, i.e., no fewer than 23 children.
If one child has exactly 5 teeth, then the others have no more than 4, and there are no fewer than $1 + 85 / 4$, i.e., also no fewer than 23 children.
If any of the children have between 6 and 9 ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.4. Let initially each island is inhabited by one colony, and let one of the islands have $d$ neighboring islands. What can the maximum possible number of colonies that can settle on this island be equal to? | Solution. Answer: $d+1$.
Example. We will prove that in a vertex of degree $d$, $d+1$ colonies can gather. Suspend the tree from this vertex as the root and prove that in each vertex from which $e$ edges go down, $e+1$ colonies can gather, conducting only migrations within its subtree. We will prove this by "induction... | +1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. From head to tail of the zebra Hippotigris - 360 stripes of the same width. Flea Masha and flea Dasha started crawling from the head of the zebra to its tail. At the same time, flea Sasha started crawling from the tail to the head. Flea Dasha crawls twice as fast as flea Masha. Before meeting flea Sasha, Masha overc... | Solution: 240 stripes.
Mashka crawled half of the zebra. Let her speed be $v$, then the closing speed of Mashka and Sashka is $2v$, and the closing speed of Dashka and Sashka is $3v$. Therefore, Sashka will crawl $3/2$ times fewer stripes before meeting Dashka than before meeting Mashka, i.e., 120 stripes. The remaini... | 240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In a row, 100 knights and 100 liars are standing (in some order). The first person was asked: "Are you a knight?", and the rest were asked in turn: "Is it true that the previous person answered 'Yes'?" What is the maximum number of people who could have said "Yes"? Knights always tell the truth, liars always lie. | Solution. 150, for example, if there are 100 knights followed by 100 liars. We will now prove that this is the maximum.
Consider any liar, except possibly the first one. Either they said "no," or the previous person said "no." Thus, to each of the 99 liars, we can associate at least one "no" response (either theirs or... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The gnomes went to work, and Snow White is feeling lonely. She laid out a pile of 36 stones on the table. Every minute, Snow White splits one of the existing piles into two, and then adds a new stone to one of them. After some time, she had seven piles, each with an equal number of stones. How many stones ended up i... | Solution. There will be seven piles after six moves. After six moves, there will be $36+6=42$ stones - meaning, 6 stones in each pile. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain city, the fare scheme for traveling by metro with a card is as follows: the first trip costs 50 rubles, and each subsequent trip costs either the same as the previous one or one ruble less. Petya spent 345 rubles on several trips, and then on several subsequent trips - another 365 rubles. How many trips... | Solution. A total of 710 rubles was spent. The total number of trips could not have been 14 or less (for 14 trips, a maximum of 700 rubles could be spent), but it also could not have been 17 or more $(50+49+48+\cdots+35+$ $34=714$, and this is the minimum that could be spent). Therefore, the choice is between two optio... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Seven people stood in a circle, each of whom is either a knight, who always tells the truth, or a liar, who always lies, or a traveler, who alternates between truth and lies.
The first and second said in unison: "Among us there is exactly 1 liar," the second and third: "Among us there are exactly 2 knights," the th... | Solution. If all 7 statements are lies, then everyone lied, meaning everyone is a liar, and then 7 and 1 told the truth. This means that at least one statement is true. On the other hand, there are no more than 2 true statements (one about the liars, the other about the knights). That is, there are either 0 or 1 knight... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. When the child was born, their parents were not yet 40 years old, but they were already adults. When the child turned 2 years old, the age of exactly one of the parents was divisible by 2; when the child turned 3 years old, the age of exactly one of the parents was divisible by 3, and so on. How long could such a pa... | Solution. The condition means that at the moment of the child's birth, the age of exactly one of the parents was divisible by 2, the age of exactly one of the parents - by 3, and so on (as long as this pattern continued).
Let one of the parents be 24 years old, and the other 35 years old. Then this pattern could conti... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. In a company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to triple the salary for those earning up to 500 dollars, and increase the salary by 1000 dollars for the rest, so the total salary will become 24000 dollars. A mean manager proposes to reduce the salary to 500 doll... | Solution. Note that the increase proposed by the kind manager is twice as large as the salary proposed by the evil manager (this is true for both poor and rich employees). The increase according to the kind manager's proposal is 14000, so the salary according to the evil manager's proposal is 7000. | 7000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. 31 cars started simultaneously from one point on a circular track: the first car at a speed of 61 km/h, the second at 62 km/h, and so on (the 31st at 91 km/h). The track is narrow, and if one car overtakes another by a full lap, they crash into each other, both fly off the track, and are eliminated from the race. In... | Solution. First, the fastest car collides with the slowest, then the second fastest collides with the second slowest, and so on. In the end, the car with the median speed remains, i.e., the 16th. It travels at a speed of $76 \mathrm{Km} /$ h.
Criteria. Full solution - 3 points.
1 point if only the answer 76 is writte... | 76 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Five brothers were dividing their father's inheritance equally. The inheritance included three houses. Since the houses couldn't be divided, the three older brothers took them, and the younger brothers were given money: each of the three older brothers paid $2000. How much did one house cost?
## Solution: | Answer: $5000.
Since each of the brothers who received money got $3000, the total inheritance was estimated at $15000. Therefore, each house was worth a third of this amount. | 5000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the Parliament of the Emerald City, 5 parties are represented, which together developed 100 laws over the year (each law was developed by exactly one of the parties). It is known that any three parties together developed no fewer than 50 laws. What is the maximum number of laws that the Green Lenses party could h... | Solution:
Answer: 33.
Evaluation. Let this quantity be $x$. Since any three parties together have developed no less than 50, any party together with the Green Lenses party has developed no more than 50. Therefore, each of the four other parties has developed no more than $50-x$, and all of them together have develope... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Andrey placed chips of a hundred different colors in the cells of a $10 \times 10$ board. Each minute, one of the chips changes color, and only a chip that was unique (i.e., differed in color from all others) in its row or column before this operation can change color. After $N$ minutes, it turned out that no chip c... | Solution. Let's look at the moment when there are no moves left. Suppose we have $k$ colors left, then we have repainted at least $100-k$ cells. Note that if nothing can be repainted, then there are at least 4 cells of each color. In total, there are at least 25 colors left after this process. Therefore, we will repain... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$, angle $A$ is equal to $50^{\circ}, B H$ is the altitude. Point $M$ on $B C$ is such that $B M=B H$. The perpendicular bisector of segment $M C$ intersects $A C$ at point $K$. It turns out that $A C=2 \cdot H K$. Find the angles of triangle $A B C$. | Solution. Draw a perpendicular to $BC$ from point $M$. Let $K^{\prime}$ be the point of its intersection with $AC$. Then the segment $MK$ is the midline in $\triangle K^{\prime}MC$, that is, $K^{\prime}K = KC$. From the fact that $HK = KC + AH$, we get $AH = HK^{\prime}$. Right triangles $BHA$ and $BHK^{\prime}$ are eq... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Masha bought 2021 weights of pairwise distinct masses. Now Masha places one weight on each pan of a two-pan balance (weights placed on the balance previously are not removed). Each time the balance is in equilibrium, Masha rejoices. What is the maximum number of times she can find a reason to be happy | Solution. Answer: 673 times.
Example. Let Masha have bought 673 triples of the form $x, y, x+y$ (we will choose the weights of the new triples so that they do not duplicate the old ones). The last two weights are any. Masha puts $x$ on the left pan, then $y$ on the same left pan, and finally $x+y$ on the right pan and... | 673 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Seven fishermen stand in a circle. The fishermen have a professional habit of exaggerating numbers. Each fisherman has a measure of lying (each has their own, an integer) - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catch... | Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cd... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The gnomes have gone to work, and Snow White is feeling lonely. She laid out a pile of fifteen stones on the table. Every minute, Snow White splits one pile into two non-empty piles and adds a stone to one of them. How can Snow White use these actions to get seven identical piles? | Solution. First, let's understand how many stones are in the piles. With each action, the number of piles increases by one, as does the number of stones. Therefore, 7 piles will arise after 6 actions, with the total number of stones being $15+6=21$. This means each pile should have three stones. From these consideratio... | 3333333 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In the company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to double the salary for everyone earning up to 500 dollars, and increase the salary by 500 dollars for the rest, so the total salary will become 17000 dollars. A mean manager proposes to reduce the salary to 500 ... | Solution. Answer: 7000
Let's look at the difference between the total salary from the kind manager and the current one, and we will understand that this is the total salary from the evil manager:
up to 500 - from the doubled we subtract the current, we get the current
over 500 - from the increased by 500 we subtract... | 7000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Konstantin pronounced the names of all natural numbers from 180 to 220 inclusive, while Mikhail - from 191 to 231 inclusive. Who pronounced more words and by how many? | Solution. Let's remove the numbers that both have: 191-220. Then each will have 11 numbers left: Konstantin has 180-190, and Mikhail has 221-231. Note that in the names of the numbers 181-189, 221-229, and 231, there are three words each, while in 180, 190, and 230, there are two words each. This means that Mikhail sai... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. At the festival "Hobbits - for cultural diversity!", more than 20 participants arrived. A correspondent found out that among any 15 participants of the festival, there are at least 4 humans and at least 5 elves. How many hobbits participated in the festival? Provide all possible answers and prove that there are no o... | Solution. Suppose there is at least one hobbit. If there are 10 people among the participants, then in their company with the hobbit and 4 other participants, there will not be 5 elves, which contradicts the condition. Therefore, there are no more than 9 people. Since the total number of participants is more than 20, t... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Cheburashka bought as many mirrors from Galina in the store as Gen bought from Shapoklyak. If Gen had bought from Galina, he would have 27 mirrors, and if Cheburashka had bought from Shapoklyak, he would have 3 mirrors. How many mirrors would Gen and Cheburashka buy together if Galina and Shapoklyak agreed and set t... | Solution. Let Gena have $x$ times more money than Cheburashka. If we swap Gena's and Cheburashka's money, then with the second method of purchase, the number of mirrors should be equal. Therefore, $3 x=\frac{27}{x}$, from which $3 x^{2}=27$, and $x=3$. So, Gena initially had three times more money than Cheburashka, and... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Let $N=400$. What is the maximum number of balls that the magician can guarantee are not in their correct vessels?
# | # Solution:
Answer: 533.
Let's imagine that inside the vessels, the balls are divided into cells, and each cell contains one ball. Then we can imagine that (regardless of whether they are in the same vessel or different ones) balls $i$ and $j$ simply swap cells upon the command ( $i j$ ). Thus, any sequence of comman... | 533 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. 31 cars started simultaneously from one point on a circular track: the first car at a speed of 61 km/h, the second at 62 km/h, and so on (the 31st at 91 km/h). The track is narrow, and if one car overtakes another by a full lap, they crash into each other, both fly off the track, and are eliminated from the race. In... | Solution. First, the fastest car collides with the slowest, then the second fastest collides with the second slowest, and so on. In the end, the car with the median speed remains, i.e., the 16th. It travels at a speed of $76 \mathrm{Km} /$ h.
Criteria. Full solution - 7 points.
Solution that correctly describes the o... | 76 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Each cell of a $10 \times 10$ board is painted black or white. A cell is said to be out of place if it has at least seven neighbors of a different color than itself. (Neighbors are cells that share a common side or corner.) What is the maximum number of cells on the board that can be out of place at the same time? | # Solution.
Answer 26. The example consists of 13 dominoes, which do not share any cells and do not touch the border of the board (see right).
Estimation. Let's call a cell that is not in its place an NVCT-cell. Obviously, NVCT-cells cannot be adjacent to the border. Let's examine what form a connected component of N... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Each cell of a $10 \times 10$ board is painted black or white. A cell is said to be out of place if it has at least seven neighbors of a different color than itself. (Neighbors are cells that share a common side or corner.) What is the maximum number of white cells on the board that can be out of place at the same t... | Solution. Answer: 26
Example: 13 dominoes, not touching each other or the edge of the board.
Estimate. Let's call a cell that is not in its own place an OOP cell. For each OOP cell, we draw ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Masha bought 2021 weights of pairwise distinct masses. Now Masha places one weight on each pan of a two-pan balance (weights placed on the balance previously are not removed). Each time the balance is in equilibrium, Masha rejoices. What is the maximum number of times she can find a reason to be happy | Solution. Answer: 673 times.
Example. Let Masha have bought 673 triples of the form $x, y, x+y$ (we will choose the weights of the new triples so that they do not duplicate the old ones). The last two weights are any. Masha puts $x$ on the left pan, then $-y$ on the same left pan, and finally $x+y$ on the right pan an... | 673 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Seven fishermen stand in a circle. The fishermen have a professional habit of exaggerating numbers. Each fisherman has a measure of lying (each has their own, an integer) - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catch... | Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cd... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. The lye trader Matti wrote several three-digit palindromic numbers on the board. It turned out that no two of them add up to a palindrome. Could he have written more than half of all three-digit palindromes? Recall that a palindrome is a number that remains the same when its digits are reversed. | Solution. No. There are $9 \cdot 10$ three-digit palindromes in total (9 options for the first digit, 10 for the second, and the third is uniquely determined by the first). We will prove that at least 45 palindromes are missing from the board.
If $a, b, c, d$ are such digits that $a+c=b+d=11$, then
$$
\overline{a b a... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Andrey named all natural numbers from 180 to 220 inclusive ("one hundred eighty", "one hundred eighty one", etc.). How many words did he say? | Solution. A total of 41 numbers are named. In each, the hundreds place is named (a total of 41 words). In addition to the hundreds place, the numbers 180, 190, 201-220 each contain one more word (this is 22 words). The number 200 does not contribute any more words, while the remaining 18 numbers each contain two more w... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Yesterday at the market, with one hundred tugriks you could buy 9 gingerbreads and 7 pastries (and even get some change), but today this amount is no longer enough. However, with the same one hundred tugriks today, you can buy two gingerbreads and 11 pastries (also with some change), but yesterday this amount would ... | Solution. Note that something has become more expensive (otherwise, the situation where we could buy a set yesterday but not today could not have arisen), and another sweet has become cheaper for similar reasons. Nine cookies and seven cakes together have become more expensive, so the cookie has become more expensive, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The set contained weights of 43, 70, and 57 grams, with an equal number of each type. Maly lost several weights (less than half), weighed the remaining on the scales, and got 20172 grams. How many and which weights were lost? | Solution (1). If the weights had not been lost, the total weight would end in 0. Therefore, the weight of the lost weights ends in 8. This can only be the case if 4 weights of 57 g are lost (losing 70 g or the pair $43+57$ does not affect the last digit of the total weight).
Solution (2). Let's divide the initial set ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Each boy is friends with 5 girls, while all girls are friends with a different number of boys. What is the minimum number of children that can be in this group? | Solution. If there is a girl who does not have any friends, we can exclude her (the number of children will decrease, but the condition of the problem will remain).
Let us have $m$ boys and $d$ girls. Draw a segment between a boy and a girl if they are friends. Then $5m$ segments are drawn from the boys, and at least ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a volleyball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game against each other team. The ratio of the number of wins by teams from city $B$ to the number of wins by teams from city $A$ is $3: 4$. Find $n$, given that there were no draws in the tou... | (12 points) Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ games. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. What is the smallest area that a figure on the plane $Oxy$ can have, located between the lines $x=-26$ and $x=2$, bounded below by the $x$-axis, and above by the tangent to the graph of the function $y=5+\sqrt{4-x}$ at the point of tangency with abscissa $x_{0}$, where $x_{0}$ lies in the interval $-26 \leq x_{0} \l... | Solution. Solution. We will write the equation of the tangent line to the graph of the function $y=5+\sqrt{4-x}$ at the point with abscissa $x_{0}$
$$
y_{\text {tan }}=-\frac{1}{2 \sqrt{4-x_{0}}}\left(x-x_{0}\right)+5+\sqrt{4-x_{0}}
$$
 \cdot n, y_{n}=1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}$.
(20 points) | Solution: Let $z_{n}=1+2+\cdots+n=\frac{(n+1) n}{2}=\frac{n^{2}+n}{2}$. We have
$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\c... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Compute the coefficient of $x^{80}$ in the polynomial $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}$ after combining like terms. | Solution: $\left(1+x+x^{2}+\cdots+x^{80}\right)^{3}=\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)\left(1+x+x^{2}+\cdots+x^{80}\right)$
Let after expanding the brackets, the term $x^{p} \cdot x^{q} \cdot x^{r}$ is obtained if the factor $x^{p}$ is chosen from the first bracket, $x^{q}$ - from... | 3321 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Shooting at a target, with each shot the athlete scored only eight, nine, or ten points (all these points were scored at least once). Making more than 11 shots, in total he scored 100 points. How many 8-point shots did the athlete make? | Solution. $8 x+9 y+10 z=100, x, y, z \in \mathbb{N} \Rightarrow 8(x+y+z)<100$
$$
\begin{gathered}
\Rightarrow 11<x+y+z<\frac{100}{8}=12.5 \Rightarrow x+y+z=12 \\
\Rightarrow\left\{\begin{array} { c }
{ x + y + z = 1 2 } \\
{ 8 x + 9 y + 1 0 z = 1 0 0 }
\end{array} \Rightarrow \left\{\begin{array} { c }
{ x + y + z =... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The Fibonacci sequence is defined recursively: $a_{1}=a_{2}=1$, $a_{n+2}=a_{n+1}+a_{n}$ for all natural $n$. What is the last digit of $a_{2020}$? | Solution. By induction, it is proved that
1) the parity of the numbers alternates as: OOEEOOEEO...
2) the number with an index divisible by 5 is divisible by 5.
Since 2020 is divisible by 5 and leaves a remainder of 1 when divided by 3, the number $a_{2020}$ is odd and divisible by 5.
Answer: 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1 ; 0) \), and its legs lying on the lines \( y = -2 \) and \( x = 0 \)?
(12 points) | Solution. $A B: \quad y=k x+b, \quad M \in A B \Rightarrow b=-k$
$$
\begin{aligned}
& A(a ;-2) \in A B \Rightarrow-2=k a+b \Rightarrow a=1-\frac{2}{k} \\
& S_{A B C}=\frac{1}{2} B C \cdot C A=\frac{1}{2}(b+2) a=\frac{1}{2}(2-k)(1-2 / k) \\
& S^{\prime}=\frac{(2-k)(2+k)}{2 k^{2}}=0, k_{\min }=-2, \quad S_{\min }=4
\end... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Specify the greatest value of the parameter $a$ for which there is a unique solution to the system
$$
\left\{\begin{array}{l}
y=1-\sqrt{x} \\
a-2(a-y)^{2}=\sqrt{x}
\end{array}\right.
$$ | # Solution:
Equation: $\left\{\begin{array}{l}\sqrt{x}=1-y \\ a-2(a-y)^{2}=1-y\end{array}\right.$, considering that $1-y \geq 0$, we solve the resulting quadratic equation:
$$
\begin{align... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The resistance to bending of a beam with a rectangular cross-section is proportional to the product of the width of this section and the square of its height. What should be the width of the cross-section of a beam cut from a round log with a diameter of $15 \sqrt{3}$ so that its resistance to bending is maximized?
... | # Solution.
$F(a)=k a b^{2}, a-$ width of the beam, $b$ - height of the beam. Since $b^{2}=225 \cdot 3-a^{2}$, then
$F(a)=k a\left(225 \cdot 3-a^{2}\right)=k\left(225 \cdot 3 a-a^{3}\right), F^{\prime}(a)=k\left(225 \cdot 3-3 a^{2}\right)=3 k\left(225-a^{2}\right)$.
The derivative equals zero and changes sign from p... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find all pairs of integers $(x, y)$ that satisfy the equation $x y=20-3 x+y$. For each found pair $(x, y)$, calculate the product $x y$. In the answer, write the sum of these products. | Solution. $\quad x y=20-3 x+y, \quad x y=17+3(1-x)+y, \quad(x-1) y+3(x-1)=17$, $(x-1)(y+3)=17$. Since x and y are integers, we have four cases:
1) $\left\{\begin{array}{c}y+3=17, \\ x-1=1,\end{array} \Leftrightarrow\left\{\begin{array}{c}y=14 \\ x=2\end{array}\right.\right.$
2) $\left\{\begin{array}{c}y+3=-17, \\ x-1=... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural solutions to the inequality
$$
\frac{4}{5}+\frac{4}{45}+\frac{4}{117}+\cdots+\frac{4}{16 n^{2}-8 n-3}>n-5
$$
In your answer, write the sum of all found solutions. | Solution. We have $16 n^{2}-8 n-3=(4 n-3)(4 n+1), \quad \frac{1}{4 n-3}-\frac{1}{4 n+1}=\frac{4}{16 n^{2}-8 n-3}$,
$\frac{4}{1 \cdot 5}+\frac{4}{5 \cdot 9}+\frac{4}{9 \cdot 13}+\cdots+\frac{4}{(4 n-3)(4 n+1)}>n-5$,
$1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\cdots+\frac{1}{4 n-3}-\frac{1}{4 n+1}>n-5,1-\frac{1}{4 n+1}>n-5... | 15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. There is an unlimited supply of square glasses in 10 colors. In how many ways can 4 glasses be inserted into a $2 \times 2$ window frame so that some color appears in both the upper and lower halves of the window. | Solution. If the upper half of the glass is of one color (one of 10), then the colors of the lower glasses can be chosen in $10^{2}-9^{2}=19$ ways. If the upper glasses are different $\left(10 * 9=90\right.$ ways to choose), then the colors of the lower ones can be chosen in $10^{2}-8^{2}=36$ ways.
In total $10 * 19+9... | 3430 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x, y, z$ be the roots of the equation $t^{3}-5 t-3=0$. Find $x^{3} y^{3}+x^{3} z^{3}+y^{3} z^{3}$. | Solution. The polynomial has 3 different real roots, since $\mathrm{P}(-100)0, \mathrm{P}(0)0$. By Vieta's theorem, $\mathrm{x}+\mathrm{y}+\mathrm{z}=0, \mathrm{xy}+\mathrm{xz}+\mathrm{yz}=-5, \mathrm{xyz}=3$.
$$
\begin{gathered}
x^{3} y^{3}+x^{3} z^{3}+y^{3} z^{3}=(5 x+3)(5 y+3)+(5 x+3)(5 z+3)+(5 y+3)(5 z+3) \\
=25(x... | -98 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1; 3) \), and its legs lying on the lines \( y = x \) and \( y = -x \)?
(12 points) | Solution.
$A B: \quad y=k x+d, \quad M \in A B \Rightarrow d=3-k$
$A(a ;-a) \in A B \Rightarrow-a=k a+3-k \Rightarrow a=\frac{3-k}{k+1}$,
$B(b ; b) \in A B \Rightarrow b=k b+3-k \Rightarrow b=\frac{k-3}{k-1}$,
 ^ { 2 } } \\
{ ( y + 1 ) ^ { 2 } + ( x - 0.5 a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x-1=(y+1)^{2} \\
x-2+(x-0.5 a)^{2}=0
\end{array}\right.\right.
$$
Consider the second equation of the system
$x^{2}+x(1-a)+\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. There are 5 pieces of transparent glass of the same square shape and size. Each piece of glass is conditionally divided into 4 equal parts (right triangles) by its diagonals, and one of these triangles is painted with an opaque paint of its individual color, different from the colors of the painted parts of the othe... | Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then... | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A student wrote a program to recolor a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 19, green - 23, blue - 53, blue - 55, purple - 83, black - 128. If ... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)-k$ is the $k$-fold composition of the function $f(n)$, which is equal to $3 n-2$ for $n \leq 17$, and equal to $|129-2 n|$ for $n \geq 18$. Let's compute and write down the fi... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The wolf saw a roe deer several meters away from him and chased after her along a straight forest path. The wolf's jump is $22\%$ shorter than the roe deer's jump. Both animals jump at a constant speed. All the roe deer's jumps are of the same length, and the wolf's jumps are also equal to each other. There is a per... | Solution: Let x be the length of the roe deer's jump, then $0.78x$ is the length of the wolf's jump; y - the number of jumps the roe deer makes in a unit of time specified in the condition, $y\left(1+\frac{t}{100}\right)$ - the number of jumps the wolf makes in the same unit of time. The wolf will not be able to catch ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=75, d=0.25$.
Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$. | Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cd... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.
Segment $A_{1} B_{1}$ intersects the bisector $C C_{1}$ at point M. Find the angle $\Delta B_{1} C_{1} M$.
# | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and thus is equidistant from its sides, we get that $A_{1}$ is equid... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. What is the minimum value that the function $F(x ; y)=6 y+8 x-9$ can take, given that $x^{2}+y^{2}+25=10(x+y)$.
# | # Solution:
$x^{2}+y^{2}+25=10(x+y) \Leftrightarrow (x-5)^{2}+(y-5)^{2}=5^{2}$ - this is a circle with center $(5 ; 5)$ and radius 5. Let $F(x ; y)=\mathrm{M}$, then $\mathrm{M}=6 y+8 x-9$ - this is a straight line.
The condition for the minimum of the function is equivalent to the condition for the minimum of $\math... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (65x65 - there are a total of 4225 cells in the square), so that among any four of its cells forming a corner (an "L" shape), there is at least one painted cell.
# | # Solution.
Shading should be done diagonally every 3rd (see fig.). Thus, $\left[\frac{N^{2}}{3}\right]$ cells will be shaded. This is the minimum possible number, as within any $3 \times 3$ square, at least three cells need to be shaded. $\left[\frac{65^{2}}{3}\right]=1408$.
=10|\mathrm{p}-3|+|2 \mathrm{p}-| \mathrm{p}+\mathrm{c}||-6 \mathrm{p}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{p}$. $10>2+1+6$. Therefore, on all intervals up to $\mathrm{p}=3$, the c... | -147 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Ivan Ivanovich approached a source with two empty cans; one held 10 liters, and the other held 8 liters. Water from the source flowed in two streams - one stronger, the other weaker. Ivan Ivanovich simultaneously placed the cans under the streams and, when half of the smaller can was filled, he switched the cans. To... | Solution: Let x liters of water fill the larger can while 4 liters fill the smaller can. After the switch, (10-x) liters fill the larger can while 4 liters again fill the smaller can. Since the flow rates are constant, the ratio of the volumes of water filled in the same time is also constant. We can set up the equatio... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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