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8.5. Snow White entered a room where 30 chairs were arranged around a round table. Some of the chairs were occupied by dwarfs. It turned out that Snow White could not sit down without having someone next to her. What is the minimum number of dwarfs that could have been at the table? Explain how the dwarfs should have b... | Answer: 10.
Solution: If there were three consecutive empty chairs at the table in some place, Snow White could sit down in such a way that no one would sit next to her. Therefore, in any set of three consecutive chairs, at least one must be occupied by a dwarf. Since there are 30 chairs in total, there cannot be fewe... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On Eeyore's Birthday, Winnie-the-Pooh, Piglet, and Owl came to visit. When Owl left, the average age in this company decreased by 2 years, and when Piglet left, the average age decreased by another 1 year. How many years older is Owl than Piglet?
Answer: Owl is 6 years older than Piglet. | Solution. Let the average age of those who remained after Piglet be x, and Piglet's age be y. Then $2x + y = 3(x + 1)$, which means $y = x + 3$. Let Owl's age be K. Then $3(x + 1) + K = 4(x + 3)$, which means $K = x + 9$. Therefore, Owl is older than Piglet by $(x + 9) - (x + 3) = 6$ years.
Criteria. If the solution i... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.3. In the park, all bicycle paths run from north to south or from west to east. Petya and Kolya simultaneously started from point $A$ and rode their bicycles at constant speeds: Petya - along the route $A-B-C$, Kolya - along the route $A-D-E-F-C$ (see fig.), and both spent 12 minutes on the road. It is known that Kol... | Answer: 1 minute.
Solution. Draw the segment $D H$, as shown in Fig. 2. Kolya travels 1.2 times faster than Petya, so it would take him $12 / 1.2=10$ minutes to travel the route $A-B-C$. The difference in time
. Then each of them said: "Among my neighbors, there is a liar." What is the maximum number of people sitting at the table who can say: "Among my neighbors, there is a k... | # Answer: 8.
Solution. Note that two liars cannot sit next to each other (otherwise, each of them would be telling the truth). Therefore, no liar can say the second phrase.
On the other hand, 3 knights also cannot sit next to each other (otherwise, the middle one would have lied by saying that he has a neighbor who i... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Sixteen boys gathered for fishing. It is known that every boy who put on boots also put on a cap. Without boots, there were 10 boys, and without a cap - two. Which boys are more and by how many: those who wore a cap but no boots, or those who put on boots? Be sure to explain your answer. | Answer. Those who were in caps but without boots were 2 more than those who were in boots.
Solution. Out of 16 boys, 10 were without boots, which means 6 were in boots. Two were without caps, so 14 were in caps. Since everyone who wore boots also wore a cap, out of the 14 who wore caps, 6 also wore boots, and the rema... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 5. Rectangle $A B C D$ was divided into four
smaller rectangles with equal perimeters (see figure). It is known that $A B=18$ cm, and $B C=16$ cm. Find the lengths of the sides of the other rectangles. Be sure to explain your answer. | Answer. 2 cm and 18 cm are the lengths of the sides of rectangle $A B L E$, 6 cm and 14 cm are the lengths of the sides of the other rectangles.
Solution. Since the perimeters of the three vertical rectangles are equal and the segments $E D, F G, K H$ and $L C$ are also equal, the segments $E F$, $F K$ and $K L$ are a... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A point in a triangle is connected to the vertices by three segments. What is the maximum number of these segments that can equal the opposite side?
# | # Answer: One.
## Solution:
Let $B M = A C$ and $A B = M C$ (see fig.). Triangles $A B M$ and $M C A$ are equal by three sides. Therefore, angle $B A M$ is equal to angle $A M C$, which means $A B \parallel M C$.
Similarly, $A C \parallel M B$.
Thus, $A B M C$ is a parallelogram, but this is not the case, because a... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Buratino buried two ingots on the Field of Wonders: a gold one and a silver one. On the days when the weather was good, the gold ingot increased by $30 \%$, and the silver one by $20 \%$. On the days when the weather was bad, the gold ingot decreased by $30 \%$, and the silver one by $20 \%$. After a week, it turned... | Solution. Increasing a number by $20 \%$ is equivalent to multiplying it by 1.2, and decreasing a number by $20 \%$ is equivalent to multiplying it by 0.8 (for $30 \%$ - by 1.3 and 0.7, respectively). Therefore, the result does not depend on the sequence of good and bad weather days, but only on the number of good and ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Find all positive roots of the equation $x^{x}+x^{1-x}=x+1$.
# | # Solution
Since $x>0$, then
$0=x^{2 x}+x-x^{x+1}-x^{x}=x^{x}\left(x^{x}-1\right)-x\left(x^{x}-1\right)=x\left(x^{x}-1\right)\left(x^{x-1}-1\right)$.
Thus, $x=1$.
## Answer $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Several numbers are written on the board. It is known that the square of any written number is greater than the product of any two other written numbers. What is the maximum number of numbers that can be on the board
# | # Answer. 3 numbers.
Solution. Suppose there are at least four numbers, and $a-$ is the number with the smallest absolute value. Among the remaining numbers, at least two have the same sign (both non-negative or both non-positive). Let these numbers be $b$ and $c$; then $bc = |bc| \geqslant |a|^2 = a^2$, which contrad... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. At one meal, Karlson can eat no more than 5 kg of jam. If he opens a new jar of jam, he must eat it completely during this meal. (Karlson will not open a new jar if he has to eat more than 5 kg of jam together with what he has just eaten.)
Little Boy has several jars of raspberry jam weighing a total of ... | Answer: 12.
Solution. We will prove that in 12 meals, Karlson will always be able to eat all the jam.
We will distribute the jars into piles according to the following algorithm. In each pile (starting with the first, then the second, and so on), we will place jars one by one until the pile contains more than 5 kg of... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. The graph of the function $y=x^{2}+a x+b$ is shown in the figure. It is known that the line $A B$ is perpendicular to the line $y=x$. Find the length of the segment $O C$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: 1.
Solution. Since $y(0)=b$, then $B(0 ; b)$. Now let's find the length of the segment $O A$.
First method. Since the line $A B$ is perpendicular to the line $y=x$, it is parallel to the line $y=-x$. Moreover, this line passes through the point $B(0 ; b)$. Therefore,
. Then $\angle F A H=\angle H B F=\alpha$ (inscribed angles subtending the same arc). From the right triangle $A D C: \angle C=90^{\circ}-\alpha$, and from the right triangle $E C B: \angle E B C=90^{\circ}-\angle C=\alpha$.
Thus, $B E$ is the height an... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.5. A natural number $n$ is called good if each of its natural divisors, increased by 1, is a divisor of the number $n+1$. Find all good natural numbers.
(S. Berlov) | Answer. One and all odd prime numbers.
Solution. It is clear that $n=1$ satisfies the condition. Also, all odd primes satisfy it: if $n=p$, then its divisors increased by 1 are 2 and $p+1$; both of them divide $p+1$. On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisi... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the sum $\sin x + \sin y + \sin z$, given that $\sin x = \tan y$, $\sin y = \tan z$, $\sin z = \tan x$ | Answer: 0.
First solution. From $\sin x = \operatorname{tg} y$, we get $\sin x \cos y = \sin y$. Therefore, $|\sin x| \cdot |\cos y| = |\sin y|$. This means $|\sin x| \geq |\sin y|$, and the inequality becomes an equality only if either $\sin y = \sin x = 0$ or $|\cos y| = 1$ (which again implies $\sin y = \sin x = 0$... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. In the castle, there are 16 identical square rooms forming a $4 \times 4$ square. Sixteen people, who are either liars or knights (liars always lie, knights always tell the truth), have settled in these rooms, one person per room. Each of these 16 people said: "At least one of the rooms adjacent to mine is occupi... | Answer: 12 knights.
Solution: Note that for each knight, at least one of their neighbors must be a liar. We will show that there must be no fewer than 4 liars (thus showing that there are no more than 12 knights). Suppose there are no more than 3 liars, then there will be a "vertical row" of rooms where only knights l... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In a square grid of cells, some cells are painted black. It turned out that no black cell shares a side with more than one other black cell? What is the maximum number of cells that could be painted black?
In a square of cells, some cells are painted black. It turned out that no black cell shares a side with more t... | Solution. An example of properly coloring 8 squares is shown in Fig. 1a. Suppose more than 8 were colored, then at least one of the four $2 \times 2$ squares would have at least three cells colored (Fig. 1b). But then at least one of them would share adjacent sides with two other black cells. This contradiction complet... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the tournament, 15 volleyball teams are playing, and each team plays against all other teams only once. Since there are no draws in volleyball, there is a winner in each match. A team is considered to have performed well if it loses no more than two matches. Find the maximum possible number of teams that performe... | Answer: 5.
Solution.
Evaluation. If the number of teams that played well is not less than 6, then consider six of them. They could have lost no more than $6 \times 2=12$ matches. But the games between them amounted to $6 \times 5 / 2=15$. Thus, they lost no fewer than 15 matches in total. Contradiction.
Example. Pla... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let there be $x$ seventh-graders participating in the tournament, who together scored $n$ points. Then the number of eighth-graders participating in the tournament is $10 * x$ people, and the total points they scored is $4.5 * n$ points. Therefore, a total of $11 * x$ students participated in the tournament, and the... | Answer: 1 student from 7th grade participated in the tournament and scored 10 points. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) In Sun City, they exchange a honey cake for 6 pretzels, and for 9 pretzels, they give 4 doughnuts. How many doughnuts will they give for 3 honey cakes? Explain your answer. | Answer: 8.
Solution.
If for one cookie you get 6 pretzels, then for 3 cookies you will get $3 \times 6=18$ pretzels. 18 pretzels is 2 times 9 pretzels. Therefore, for them, you will get 2 times 4 gingerbread cookies, i.e., 8 gingerbread cookies.
## Grading Criteria.
- Any correct and justified solution - 7 points.
... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. For a natural number $a$, the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot a$ is denoted as $a$ !.
(a) (2 points) Find the smallest natural number $m$ such that $m$ ! is divisible by $23 m$.
(b) (2 points) Find the smallest natural number $n$ such that $n$ ! is divisible by $33n$. | # Answer:
(a) (2 points) 24.
(b) (2 points) 12.
Solution. (a) The condition is equivalent to $(m-1)!$ being divisible by 23. Since 23 is a prime number, at least one of the numbers $1, 2, \ldots, m-1$ must be divisible by 23, so $m-1 \geqslant 23$ and $m \geqslant 24$. Clearly, $m=24$ works, since in this case $\fra... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. In the morning, 5 foreign cars were parked along the road. By noon, 2 domestic cars were parked between each pair of foreign cars. And by evening, a motorcycle was parked between each pair of adjacent cars. How many motorcycles were parked in total $?$
Answer: 12 . | Solution. Between 5 foreign cars there are 4 gaps, so there were $4 \cdot 2=8$ domestic cars parked there; that is, a total of $5+8=13$ cars were parked. Between them, 12 motorcycles were parked. | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the table, there were cards with digits from 1 to 9 (a total of 9 cards). Katya chose four cards such that the product of the digits on two of them equals the product of the digits on the other two. Then Anton took one more card from the table. In the end, the cards with the digits $1,4,5,8$ remained on t... | # Answer: 7.
Solution. One of the cards that is not currently on the table has the number 7. Note that Katya could not have taken the 7, because then one of her products would be divisible by 7, while the other would not. Therefore, Anton took the 7. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Yulia thought of a number. Dasha added 1 to Yulia's number, and Anya added 13 to Yulia's number. It turned out that the number obtained by Anya is 4 times the number obtained by Dasha. What number did Yulia think of? | Answer: 3.
Solution. Note that since Anya's number is 4 times greater than Dasha's number, the difference between these numbers is 3 times greater than Dasha's number. Thus, Dasha's number is $(13-1): 3=4$. Therefore, Yulia's number is 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. Aся, Borya, Vasilina, and Grisha bought tickets to the cinema for one row. It is known that:
- There are a total of 9 seats in the row, numbered from 1 to 9.
- Borya did not sit in seat 4 or 6.
- Aся sat next to Vasilina and Grisha, and no one sat next to Borya.
- There were no more than two seats between A... | Solution. Note that Asey, Vasilina, and Grisha occupy three seats in a row, and Borya sits one seat away from them. Let's seat another child, Dima, in the free seat. Then, 5 children sit in a row. Thus, someone is sitting in the central seat of the row (that is, seat number 5). For any other seat, we can come up with a... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. Masha braided her dolls: half of the dolls got one braid each, a quarter of the dolls got two braids each, and the remaining quarter of the dolls got four braids each. She tied a ribbon in each braid. How many dolls does Masha have if she needed 24 ribbons in total? | Answer: 12.
Solution. Note that since a quarter of the dolls have four braids, the total number of ribbons used on them is the same as the total number of dolls. Half of the dolls have one braid, so the number of ribbons used on them is half the total number of dolls. And a quarter of the dolls have two braids, so the... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4. Variant 1.
A square piece of paper is folded as follows: the four corners are folded to the center so that they meet at one point (see figure),
resulting in a square again. After perfo... | Answer: 12.
Solution: After each operation, the thickness of the square doubles, and the area is halved. Since the thickness has become 16 sheets, the operation was applied 4 times. In this process, the area decreased by a factor of 16 and became equal to 9 square centimeters. Therefore, the area of the original squar... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the sequence of numbers $a_{1}, a_{2}, \ldots$, is an arithmetic progression, and the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression. It is known that $a_{1}=1$. Find $a_{2017}$. | Answer: $a_{2017}=1$;
Trunov K.V.
## Solution:
Since the sequence of numbers $a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4} \ldots$, is a geometric progression, then $\left(a_{n} a_{n+1}\right)^{2}=\left(a_{n-1} a_{n}\right)\left(a_{n+1} a_{n+2}\right)$ for $n \geq 2$. From this, we obtain that $a_{n} a_{n+1}=a_{n-1} a_{n+2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let's call a four-digit number accompanying the year $\overline{20 a b}$ if it also ends in $\overline{a b}$ and, in addition, is divisible by $\overline{a b}$ (a two-digit number), for example, the number 4623 accompanies the year 2023. How many numbers accompany the year $2022?$ | Solution. $\overline{m n 22}=\overline{m n} \cdot 100+22 \Rightarrow \overline{m n} \cdot 100: 22 \Rightarrow \overline{m n}: 11: 11,22,33,44,55,66,77$, 88,99 - 9 numbers.
Answer 9.
## Grading Criteria.
Correct answer with valid reasoning - 7 points.
Valid reasoning for divisibility by 11 and description of the set... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. How many solutions does the rebus $\frac{B+O+C+b+M+O+\breve{U}}{K+J+A+C+C}=\frac{22}{29}$ have, where different letters represent different digits, and the same letters represent the same digits? It is known that the digit 0 is not used. | Solution. In the rebus, 9 different letters are used, i.e., all non-zero digits are used. $1+2+3+0+0 \leq K+Л+A+C+C \leq 6+7+8+9+9$, i.e.
$$
6 \leq K+Л+A+C+C \leq 39 . \quad \text { Since the fraction } \frac{22}{29} \text { is irreducible, the number }
$$
$K+Л+A+C+C$ must be divisible by 29, the only case that satis... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. Different positive numbers $x, y, z$ satisfy the equations
$$
x y z=1 \quad \text{and} \quad x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$
Find the median of them. Justify your answer. | # Solution:
Method 1. Consider the polynomial $P(t)=(t-x)(t-y)(t-z)$. The numbers $x, y, z$ are its roots. Expanding the brackets or using Vieta's theorem for cubic polynomials, we get $P(t)=t^{3}+a t^{2}+b t+c$, where $a=-(x+y+z)$, $b=xy+xz+yz$, and $c=-xyz$. The condition of the problem shows that $a=-b$ and $c=-1$.... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. From a $7 \times 7$ square grid, an equal number of $2 \times 2$ squares and $1 \times 4$ rectangles were cut out along the grid lines. What is the maximum number of these figures that could have been cut out? | Answer: 12.
Solution: Both the square and the rectangle consist of 4 cells. Therefore, the number of cut-out figures is no more than 49/4, that is, no more than 12. There are an equal number of figures of both types, so there are no more than 6 squares $2 \times 2$ and rectangles $1 \times 4$. The diagram shows how to... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The beetle is 3 times heavier than the cat, the mouse is 10 times lighter than the cat, the turnip is 60 times heavier than the mouse. How many times heavier is the turnip than the beetle? Justify your answer. | # Answer. 2 times.
Solution. Cat $=10$ mice, turnip $=60$ mice. Therefore, the turnip is 6 times heavier than the cat. That is, the turnip $=6$ cats. According to the condition, Zhuchka $=3$ cats. Therefore, the turnip is 2 times heavier than Zhuchka.
## Grading Criteria.
- Correct solution - 7 points.
- Correct ans... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.7. Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it, and write down all natural divisors of $a$, except for $a$ itself (the same numbers can appear on the board). After some time, it turned out that there are $N^{2}$ numbers on the board. ... | Answer. Only for $N=1$.
Solution. Lemma. For any natural $n>1$, the inequality holds
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<1
$$
Proof. Consider the function $f(t) = \frac{1}{t^{2}}$ for $t > 1$. The function $f(t)$ is decreasing for $t > 1$. Let $1 = d_{1} < d_{2} < \ldots < d_{k} < d_{k+1} = N$ ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.5. In the castle, there are 16 identical square rooms forming a $4 \times 4$ square. Sixteen people, who are either liars or knights (liars always lie, knights always tell the truth), moved into these rooms, one person per room. Each of these 16 people said: "At least one of the rooms adjacent to mine is occupied by ... | Answer: 8 liars.
Solution: Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's divide the rooms into 8 pairs of adjacent rooms. Then, in each pair, there can be no more than one liar. Therefore, there can be no more than 8 liars in total. Consider a chessboard coloring of ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. Initially, there were 10 piles of candies on the table, containing $1, 2, \ldots, 10$ candies respectively. The Child decided to redistribute the candies. On each odd minute, he chooses one pile and divides it into two piles, each containing at least one candy. On each even minute, he chooses two piles and merges ... | Answer: Yes.
Solution. We will provide an example of how Little One can achieve such a distribution. On the first minute, he divides the pile of 10 candies into two piles of 5 candies each. Then, on the 2nd, 4th, 6th, and 8th minutes, he combines the piles of 1+9, 2+8, 3+7, 4+6 respectively, and on the 3rd, 5th, 7th, ... | 11 | Combinatorics | proof | Yes | Yes | olympiads | false |
2. (7 points) Anya multiplied 20 twos, and Vanya multiplied 17 fives. Now they are going to multiply their huge numbers. What will be the sum of the digits of the product? | Answer: 8.
Solution. In total, 20 twos and 17 fives are multiplied. Let's rearrange the factors, alternating twos and fives. This results in 17 pairs of $2 \cdot 5$ and three additional twos, which multiply to 8. Thus, the number 8 needs to be multiplied by 10, 17 times. This results in a number consisting of the digi... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) The graph of a reduced quadratic trinomial is shown in the figure (the y-axis is erased, the distance between adjacent marked points is 1). What is the discriminant of this trinomial?
. From the condition, it follows that $x_{2}-x_{1}=2$. Since $x_{2}=\frac{-b+\sqrt{D}}{2}, x_{1}=\frac{-b-\sqrt{D}}{2}$, we get that $x_{2}-x_{1}=\sqrt{D}$, hence $D=4$.
## Answer. 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the x-coordinate of the point of intersection.
# | # Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=5, \mathrm{BC}=\sqrt{17}$, and $\mathrm{AC}=4$, a point $\mathrm{M}$ is taken on side $\mathrm{AC}$ such that $\mathrm{CM}=1$. Find the distance between the centers of the circumcircles of triangles $\mathrm{ABM}$ and $\mathrm{BCM}$. | Answer: 2.
Solution: Draw the height BH to side AC. Let $\mathrm{CH}=x$, then $\mathrm{BH}=$ 4 - $x$. By the Pythagorean theorem from two triangles, we have $B H^{2}=B C^{2}-C H^{2}=17-x^{2}$ and $B H^{2}=A B^{2}-A H^{2}=25-(4-x)^{2}$. Equating the right sides of both equations, we get $x=1$, which means points M and ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. Gleb placed the numbers $1,2,7,8,9,13,14$ at the vertices and the center of a regular hexagon such that in any of the 6 equilateral triangles, the sum of the numbers at the vertices is divisible by 3. What number could Gleb have written in the center? It is sufficient to provide one suitable example.
, and the centr... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. A field was partially planted with corn, oats, and millet. If the remaining part is completely planted with millet, then millet will occupy half of the entire field. If the remaining part is equally divided between oats and corn, then oats will occupy half of the entire field. By what factor will the amoun... | Answer: 3.
Solution. Let the area of the entire field be 1, and the empty part be $x$. Then, from the first condition, millet occupies $\frac{1}{2}-x$, and from the second condition, oats occupy $\frac{1}{2}-\frac{1}{2} x$.
Corn is left with $1-x-\left(\frac{1}{2}-x\right)-\left(\frac{1}{2}-\frac{1}{2} x\right)=\frac... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Semyon has 20 numbers: $1,2,3, \ldots, 19,20$. He formed 10 fractions by writing ten of these numbers in some order as numerators, and the remaining ten in some order as denominators. What is the maximum number of integers Semyon could obtain after simplifying all the written fractions? | Answer: 8 numbers.
## Solution:
For the fractions to have an integer value, the prime numbers $11, 13, 17, 19$ can only be numerators with a denominator of 1. Therefore, to form fractions equal to an integer, no more than 17 numbers can be used, meaning no more than 8 fractions can be formed. Example: 20/10, 19/1, $1... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.1. The numbers 7, 8, 9, 10, 11 are arranged in a row in some order. It turned out that the sum of the first three of them is 26, and the sum of the last three is 30. Determine the number standing in the middle.
. What is the value of the expression $4 a+d$, if $b+c=1$ ? | Answer: 1.
Solution: Since the graphs pass through the point with coordinates (2; 4), then $4=4a+2b+1$ and $4=4+2c+d$. Therefore, $4a+2b=3$, and $2c+d=0$, or $4a=3-2b, d=-2c$. Summing up the obtained expressions: $4a+d=3-2b-2c=3-2(b+c)=3-2=1$. Note: The condition is satisfied by the functions $y=ax^2+bx+1$ and $y=x^2+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two candles of different thickness, each 24 cm long, are lit simultaneously. The thin candle burns out in 4 hours, the thick one in 6 hours. After what time will one candle be half the length of the other? It is assumed that each candle burns uniformly, meaning it decreases by the same height over equal time interva... | Solution. Let $x$ hours $(x<4)$ have passed since the candles were lit. Since the thin candle decreases by $24: 4=6$ cm every hour, it will decrease by a length of $6 x$ in this time, leaving an unburned stub of length $24-6 x$. Similarly, the stub of the second candle will have a length of $24-4 x$. For one candle to ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. On a sheet of paper, three intersecting circles are drawn, forming 7 regions. We will call two regions adjacent if they share a common boundary. Regions that border at exactly one point are not considered adjacent.
Two regions already have numbers written in them. Write integers in the remaining 5 region... | Answer: -8.
Solution. Let the unknown number be $x$. Then we can place the numbers in the other regions as follows:
The sum of the numbers in the regions surrounding $x$ should be equal to $... | -8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. Petya and Daniil are playing the following game. Petya has 36 candies. He lays out these candies in the cells of a $3 \times 3$ square (some cells may remain empty). After this, Daniil chooses four cells forming a $2 \times 2$ square and takes all the candies from there. What is the maximum number of cand... | # Answer: 9.
Solution. If Petya places 9 candies in each corner cell (and does not place any candies in the other cells), then in any $2 \times 2$ square there will be exactly 9 candies. After this, Daniil will be able to take only 9 candies.
Let's prove that Daniil can get at least 9 candies. Suppose the opposite: l... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.2. During a physical education class, 25 students from 5B class lined up. Each of the students is either an excellent student who always tells the truth, or a troublemaker who always lies.
Excellent student Vlad stood in the 13th place. Everyone except Vlad stated: "There are exactly 6 troublemakers between ... | Answer: 12.
Solution. Note that students in places $7-12$ are troublemakers, since there are fewer than 6 people between each of them and Vlad. Therefore, the student with number 6 is an excellent student. The same can be said about the student in the 5th place, then about the 4th, the 3rd, the 2nd, and the 1st.
Thus... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 5.4. Masha drew two little people in her notebook. The area of each cell is 1.
Which of the little people has a larger area?
What is the difference? If the areas are the same, write "0" in the answer.
,
- 2 large triangl... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.1. 20 schoolchildren came to the mathematics Olympiad. Everyone who brought a pencil also brought a pen. 12 people forgot their pencils at home, and 2 schoolchildren forgot their pen. By how many fewer schoolchildren brought a pencil than those who brought a pen but forgot a pencil? | Answer: 2.
Solution: 8 students brought a pencil, which means they also brought a pen. 18 students brought a pen. Therefore, 10 students brought a pen without a pencil. Then 10-8=2. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. 5.1. An isosceles trapezoid \(ABCD\) is inscribed in a circle with diameter \(AD\) and center at point \(O\). A circle is inscribed in triangle \(BOC\) with center at point \(I\). Find the ratio of the areas of triangles \(AID\) and \(BIC\), given that \(AD=15, BC=5\). | Answer: 9.
## Solution.
From the fact that $A D$ is the diameter of the circle circumscribed around the trapezoid, it follows that $A O = B O = C O = D O$ as radii. Therefore, triangle $B O D$ is isosceles. Hence, $\angle O B D = \angle O D B = \angle D B C$ (the last equality is due to $A D \parallel B C$). Thus, $B... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.1. Petya writes on the board such different three-digit natural numbers that each of them is divisible by 3, and the first two digits differ by 2. What is the maximum number of such numbers he can write if they end in 6 or 7? | Answer: 9
Solution. A number is divisible by 3 if the sum of its digits is a multiple of 3. If the number ends in 6, then the sum of the other two digits leaves a remainder of 0 when divided by 3. Such numbers are: 2,4 and 4,2; 5,7 and 7,5. If the number ends in 7, then the sum of the other two digits leaves a remaind... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3.1. Every hour, between two adjacent nettle bushes in a row, two more of the same grow. How many bushes do you need to plant initially so that after three hours, the total number of bushes together is 190? | Answer: 8
Solution. If at the moment there are $\mathrm{n}$ bushes, then on the next move their number increases by $2(n-1)$. Thus, if after 3 hours the total number of bushes should be 190, then one hour before that, there should be $\frac{190-1}{3}+1=64$. One more hour back, $\frac{64-1}{3}+1=22$. And on the next mo... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.1. 40 people came into a room where there were 40 chairs, black and white, and sat on them. All of them said they were sitting on black chairs. Then they somehow resat, and exactly 16 claimed they were sitting on white chairs. Each of those sitting either lied both times or told the truth both times. How many of them... | Answer: 8
Solution. Initially, everyone who told the truth sat on black chairs, and everyone who lied sat on white ones. After some of them switched chairs, 16 claimed they were sitting on white chairs. Obviously, this group includes those who told the truth and were sitting on white chairs, and those who switched wit... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. On Fyodor's bookshelf, there are volumes of works by various authors. When a friend borrowed two volumes of Pushkin, Fyodor noticed that among the remaining books, he had read at least half of them in full. After the friend returned the two volumes of Pushkin and borrowed two volumes of Lermontov, Fyodor realized ... | Answer: 12
Solution. Let's number all of Fyodor's books from 1 to $\mathrm{n}$.
Let the first two volumes be Pushkin's, $s_{1}$ - the number of them read by Fyodor; the last two - Lermontov... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. There are nuts in three boxes. In the first box, there are six fewer nuts than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer. | Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is given by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.
Answer: 8 nuts.
... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. All values of the quadratic trinomial $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed 1 in absolute value. What is the greatest value that the quantity $|a|+|b|+|c|$ can have under these conditions? For which function $f(x)$ is this value achieved? | Answer: the maximum value is 7; for example, it is achieved for $f(x)=2 x^{2}-4 x+1$.
Solution. By the condition, the values of $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed one in absolute value. In particular, $|f(0)| \leqslant 1,|f(1)| \leqslant 1,|f(2)| \leqslant 1$, which is equivalent to the syste... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. It is known that the quadratic equations in $x$, $2017 x^{2} + p x + q = 0$ and $u p x^{2} + q x + 2017 = 0$ (where $p$ and $q$ are given real numbers) have one common root. Find all possible values of this common root and prove that there are no others. | Solution: Let $x_{0}$ be the common root of these equations, that is, the equalities $2017 x_{0}^{2} + p x_{0} + q = 0$ and $p x_{0}^{2} + q x_{0} + 2017 = 0$ are satisfied. Multiply both sides of the first equation by the number $x_{0}$ and subtract the second equation. We get the consequential equation: $2017 x_{0}^{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2. On a plane, a semicircle with diameter $A B=36$ cm was constructed; inside it, a semicircle with diameter $O B=18$ cm was constructed ($O-$ the center of the larger semicircle). Then, a circle was constructed that touches both semicircles and the segment АО. Find the radius of this circle. Justify your answer. | Solution: Let the desired radius be $x$. Let points $C$ and $Q$ be the centers of the smaller semicircle and the inscribed circle, respectively, and $D$ be the point of tangency of the inscribed circle with the diameter $AB$. Due to the tangency of the circle and the larger semicircle, the ray $OQ$ passes through the p... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. We will call a natural number semi-prime if it is greater than 25 and is the sum of two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? Justify your answer. | Solution: There are many sets of five consecutive semiprime numbers, for example $30(=13+17), 31(=2+29), 32(=3+29), 33(=2+31), 34$ $(=5+29)$ or $102(=5+97), 103(=2+101), 104(=31+73), 105(=2+103)$, $106(=47+59)$. We will show that there are no six consecutive semiprime numbers.
Indeed, among any six consecutive numbers... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Inside a convex pentagon, a point is marked and connected to all vertices. What is the maximum number of the ten segments drawn (five sides and five segments connecting the marked point to the vertices of the pentagon) that can have a length of 1? (A. Kuznetsov) | Answer: 9 segments.
Solution. First, we prove that all 10 segments cannot have a length of 1. Assume the opposite. Let $A B C D E$ be a pentagon, $O$ be a point inside it, and all 10 drawn segments have a length of 1 (see Fig. 6). Then triangles $O A B, O B C$, $O C D, O D E$, and $O E A$ are equilateral, so $\angle A... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. Masha lives in apartment No. 290, which is located in the 4th entrance of a 17-story building. On which floor does Masha live? (The number of apartments is the same in all entrances of the building on all 17 floors; apartment numbers start from 1.$)$ | # Answer: 7.
Solution. Let $x$ be the number of apartments per floor, then there are $17 x$ apartments in each entrance. Thus, in the first three entrances, there are $51 x$ apartments, and in the first four $68 x$.
If $x \geqslant 6$, then in the first three entrances there are at least 306 apartments, so apartment ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. The product of positive numbers $a$ and $b$ is 1. It is known that
$$
(3 a+2 b)(3 b+2 a)=295
$$
Find $a+b$. | Answer: 7.
Solution. Expanding the brackets, we get
$$
295=6 a^{2}+6 b^{2}+13 a b=6\left(a^{2}+b^{2}\right)+13
$$
from which $a^{2}+b^{2}=47$. Then
$$
(a+b)^{2}=a^{2}+b^{2}+2 a b=47+2=49=7^{2}
$$
which gives $a+b=7$ (note that $a+b>0$, since $a>0$ and $b>0$). | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Real numbers $a$ and $b$ are such that $a^{5}+b^{5}=3, a^{15}+b^{15}=9$. Find the value of the expression $a^{10}+b^{10}$. | Answer: 5.
Solution. Let $x=a^{5}, y=b^{5}$. Then $x+y=3, x^{3}+y^{3}=9 ;(x+y)\left(x^{2}-x y+y^{2}\right)=9$; $3\left(x^{2}-x y+y^{2}\right)=9 ; x^{2}-x y+y^{2}=3 ; x^{2}+2 x y+y^{2}=9 ; 3 x y=6 ; x y=2 ; x^{2}+y^{2}=9-2 x y=9-$ $2 \cdot 2=5 ; a^{10}+b^{10}=x^{2}+y^{2}=5$
Criteria. Simplification using substitution w... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. In each cell of a $6 \times 6$ table, numbers are written. All the numbers in the top row and all the numbers in the left column are the same. Each of the other numbers in the table is equal to the sum of the numbers written in the two adjacent cells - the cell to the left and the cell above. What number can be w... | Answer: 8. Let $a$ be the number written in the corner cell at the top left. Then the table is filled in sequentially. For the bottom right corner, we get the relation $252 a = 2016$, from which it follows that $a = 8$.
Comment. An answer with a chain of calculations - 7 points. An answer obtained using combinatorial ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A $5 \times 5$ square is to be cut into two types of rectangles: $1 \times 4$ and $1 \times 3$. How many rectangles can result from the cutting? Justify your answer. | 5. Answer: 7.
Solution. Example:
If there are no more than 6 rectangles, then they occupy no more than $6 \cdot 4 = 24$ cells, while there are 25 cells. Therefore, there must be at least 7 r... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. For what value of the parameter m is the sum of the squares of the roots of the equation
$$
x^{2}-(m+1) x+m-1=0
$$
the smallest? | 3. $\mathrm{m}=3$.
Note that the discriminant $\mathrm{D}$ of the equation is equal to
$$
\mathrm{D}=(\mathrm{m}+1)^{2}-4(\mathrm{~m}-1)=(\mathrm{m}-1)^{2}+4>0
$$
therefore, the equation has two roots for any $\mathrm{m}$. Let's represent the sum of the squares of the roots $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ as
... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.1. Arina wrote down a number and the number of the month of her birthday, multiplied them and got 248. In which month was Arina born? Write the number of the month in your answer.
 | Answer: 8
Solution. $248=2 * 2 * 2 * 31$. Obviously, 31 is the number, then 8 is the month number. | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. It is known that in the past chess tournament, in each round all players were paired, the loser was eliminated (there were no draws). It is known that the winner played 6 games. How many participants in the tournament won at least 2 games more than they lost? | Answer: 8
Solution. Since the winner played 6 games, there were a total of $2^{6}=64$ players. The win-loss ratio for those eliminated in the 1st round is -1; for those who lost in the 2nd round, it is 0; for those who lost in the 3rd round, it is +1; for those who lost in the 4th round and beyond, it is at least +2. ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. In a basketball team, there is a certain number of players. The coach added up all their heights and divided by the number of players (let's call this the average height), getting 190 cm. After the first game, the coach removed Nikolai, who is 197 cm tall, and replaced him with Petr, who is 181 cm tall, after whic... | Answer: 8
Solution. The change in the total height of the players by $197-181=16$ cm leads to a change in the average height of the team by $190-188=2$ cm, so the team has $\frac{16}{2}=8$ people. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural $\boldsymbol{n}$ such that the value of the expression $\frac{2 n^{2}-2}{n^{3}-n}$ is a natural number. | Solution. $\frac{2 n^{2}-2}{n^{3}-n}=\frac{2(n-1)(n+1)}{n(n-1)(n+1)}$. For $n \neq \pm 1$, this expression is identically equal to $\frac{2}{n}$. Since $n \neq 1$, the value of the last expression is a natural number only when $n=2$.
Answer: $n=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. While tidying up the children's room before the guests arrived, mom found 9 socks. Among any four socks, at least two belong to the same owner. And among any five socks, no more than three belong to the same owner. How many children scattered the socks, and how many socks does each child own?
Answer. There are thre... | Solution. No child owned more than three socks, as otherwise the condition "among any five socks, no more than three belonged to one owner" would not be met. There are a total of 9 socks, so there are no fewer than three children. On the other hand, among any four socks, there are two socks belonging to one child, so t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. How many natural numbers $n$ exist for which $4^{n}-15$ is a square of an integer? | Answer. Two.
Solution. Let $4^{n}-15=x^{2}$, where $x$ is an integer. It is obvious that $x \neq 0$. If $x$ is negative, then $(-x)^{2}$ is also equal to $4^{n}-15$; therefore, we will assume that $4^{n}-15=x^{2}$, where $x$ is a natural number. From the equation $2^{2 n}-15=x^{2}$, we get: $2^{2 n}-x^{2}=15$, and usi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.1
## Condition:
Grandma is embroidering her grandchildren's names on their towels. She embroidered the name "ANNA" in 20 minutes, and the name "LINA" in 16 minutes. She spends the same amount of time on the same letters, and possibly different times on different letters. How long will it take her to embroi... | Express the answer in minutes.
Answer: 12
Exact match of the answer - 1 point
Solution.
Since the grandmother spends the same amount of time on identical letters, she will spend $20: 2=10$ minutes on the syllable "NA". Then she will spend
$16-10=6$ minutes on the syllable "LI". Therefore, she will spend $6+6=12$ m... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 4.1
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a five-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars above me than knights below me!"
How many liars can live in this building? | # Answer: 3
## Exact match of the answer -1 point
## Solution.
Notice that the person living on the 5th floor is definitely lying, as there is no one living above them, including liars. Therefore, there are 0 liars above them, and 0 or more knights below. Now consider the resident on the 1st floor. They definitely t... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 4.3
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a six-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars above me than knights below me!" How many liars can live in this building? | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task No. 4.1.
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a five-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars b... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.1
## Condition:
In a correct equation, identical digits were replaced with the same letters, and different digits with different letters. The result is
$$
P+\mathrm{P}+\mathrm{A}+3+Д+\mathrm{H}+\mathrm{U}+\mathrm{K}=\mathrm{U} \mathrm{U}
$$
What can U be equal to? | # Answer:
$\circ 1$
$\circ 2$
$\checkmark 3$
$\circ 4$
० 5
० 6
○ 7
○ 8
$\circ 9$
$\circ 0$
Exact match of the answer - 1 point
## Solution.
Notice that the expression involves 9 different letters, meaning all digits except one are used. On the left, we have the sum of eight different single-digit numbers. ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.1
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
Now it is easy to count the resulting pieces. For convenience, they are highl... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.1
## Condition:
On the Misty Planet, santiks, kubriks, and tugriks are in circulation. One santik can be exchanged for 1 kubrik or 1 tugrik, 1 kubrik can be exchanged for 3 santiks, and 1 tugrik can be exchanged for 4 santiks. No other exchanges are allowed. Jolly U, initially having 1 santik, made 20 ex... | # Answer: 6
## Exact match of the answer -1 point
## Solution.
To increase the number of santiks, we need to exchange them for kubriks or tugriks. In essence, we need to perform a double exchange, that is, first convert a santik into a tugrik or kubrik at a 1:1 ratio, and then increase the number of coins. Since the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.2
Condition:
In the city of Abracodabra, funtics, tubrics, and santics are in circulation. One funtic can be exchanged for 1 tubric or 1 santic, 1 tubric for 5 funtics, and 1 santic for 2 funtics. No other exchanges are allowed. Lunatic, initially having 1 funtic, made 24 exchanges and now has 40 funtics... | Answer: 9
Exact match of the answer -1 point
Solution by analogy with task No. 7.1.
## Condition:
On the planet Mon Calamari, dataries, flans, and pegats are in circulation. One datary can be exchanged for 1 flan or 1 pegat, 1 flan - for 2 dataries, and 1 pegat - for 4 dataries. No other exchanges are allowed. Mera... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. $\left(7\right.$ points) Calculate $\frac{(2009 \cdot 2029+100) \cdot(1999 \cdot 2039+400)}{2019^{4}}$. | # Solution.
$2009 \cdot 2029+100=(2019-10) \cdot(2019+10)+100=2019^{2}-10^{2}+100=2019^{2}$.
$1999 \cdot 2039+400=(2019-20) \cdot(2019+20)+400=2019^{2}-20^{2}+400=2019^{2}$.
Then $\frac{2019^{2} \cdot 2019^{2}}{2019^{4}}=1$
Answer. 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) Find all natural solutions to the equation $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}$.
# | # Solution.
1 method. $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}, 2 n-3=\frac{1}{n^{5}}-\frac{2}{n}, 2 n-3=\frac{1-2 n^{4}}{n^{5}}$.
For $n=1$ the equality is true, for $n>1 \quad 2 n-3>0, \frac{1-2 n^{4}}{n^{5}}<0$.
Answer. $n=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2 In Pokémon hunting, 11 adults and $n$ children participated. Together, they caught $n^{2}+3 n-2$ Pokémon, with all adults catching the same number, and all children catching the same number, but each child catching 6 fewer than an adult. How many children participated in the game? | Solution: Let each child catch $m$ pokemons. Then $n m+11(m+6)=n^{2}+3 n-2$. From this, $(n+11) m=n^{2}+3 n-68$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+3 n-68=$ $n(n+11)-8(n+11)+20$, so 20 is divisible by $n+11$. The only divisor of 20 greater than 10 is 20 itself, so $n+11=20, n=9$.
## Crite... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.3. One side of a rectangle was increased by 3 times, and the other was reduced by 2 times, resulting in a square. What is the side of the square if the area of the rectangle is $54 \mathrm{m}^{2} ?$ | Answer: 9 m.
Let's reduce the side of the given rectangle by half (see Fig. 5.3a). Then the area of the resulting rectangle will be 27 m $^{2}$ (see Fig. 5.3b). Next, we will increase the other side by three times, that is, "add" two more rectangles (see Fig. 5.3c). The area of the resulting figure will become $27 \cd... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem №3
The base and the lateral side of an isosceles triangle are 34 and 49, respectively.
a) Prove that the midline of the triangle, parallel to the base, intersects the inscribed circle of the triangle.
b) Find the length of the segment of this midline that is contained within the circle. | # Answer: 8.
## Solution
a) Let $\mathrm{O}$ be the center of the inscribed circle in triangle $ABC$ with sides $AB = AC = 49$, $BC = 34$, and $AH$ be the height of the triangle. Points $M$ and $N$ are the midpoints of sides $AB$ and $AC$, respectively, and $K$ is the intersection point of $AH$ and $MN$. Since $MN$ i... | 8 | Geometry | proof | Yes | Yes | olympiads | false |
1. Dima wrote a sequence of 0s and 1s in his notebook. Then he noticed that a 1 follows a 0 sixteen times, a 0 follows a 1 fifteen times, and a 0 follows 01 eight times. How many times does a 0 follow 11? | # Answer. 7.
Solution. The combination 01 appears 16 times in the tetrad, while the combination 10 appears 15 times. Therefore, the string starts with 0, meaning that before each combination 10 there is either 0 or 1. According to the condition, eight times it is 0, so the combination 110 appears $15-8=7$ times.
Crit... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.2. How much greater is one of two positive numbers than the other if their arithmetic mean is $2 \sqrt{3}$ and the geometric mean is $\sqrt{3}$?
(Hint: the geometric mean of two numbers $m$ and $n$ is the number $p=\sqrt{\mathrm{mn}}$). | # Solution.
Let the unknown numbers be denoted by $x$ and $y$. Then, from the problem statement, we get:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \frac { x + y } { 2 } = 2 \sqrt { 3 } , } \\
{ \sqrt { x y } = \sqrt { 3 } , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3 .
\end... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the largest natural number $\mathrm{n}$, for which the system of inequalities
$1<x<2, 2<x^{2}<3, \ldots, n<x^{n}<n+1$
has a solution. (6 points) | Solution. From the condition $: n=4$.
Rewrite the inequalities as:
$\left\{\begin{aligned} & 16^{3} \text{, then for } n=5 \text{ the given system is already inconsistent: the intervals } [\sqrt[3]{3}, \sqrt[4]{4}] \text{ and } [\sqrt[5]{2}, \sqrt[5]{6}] \text{ do not intersect.} \\ & \text{For } n=4 \text{, it is not... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. There is a bag with 16 letters: А, А, А, А, В, В, Д, И, И, М, М, Н, Н, Н, Я, Я. Anna, Vanya, Danya, and Dima each took 4 letters from it, after which the bag was empty. How many of them could have successfully spelled out their names? Explain your answer. | 2. Answer: 3.
Solution. All of them could not have formed their names, as there were not enough letters D. Anna, Vanya, and Dima could have drawn cards from which they could form their names, while Danya would have been left with the cards: V, I, M, Ya. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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