problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
8.2. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved three seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. ... | Answer: 3
Solution. Let's see how the seat number changed for everyone except Anya. Varya's seat number increased by 3, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decreased by ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.3. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved one seat to the right, Galia had moved three seats to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. ... | # Answer: 3
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 1, Galia's decreased by 3, and the sum of Diana's and El's seat numbers did not change. At the same time, the total sum of the seats did not change, so Anya's seat number must have increased by 2. ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.4. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved one seat to the right, Galia had moved two seats to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Wh... | # Answer: 4
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 1, Galia's decreased by 2, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have increase... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the pantry, Winnie-the-Pooh keeps 11 pots, seven of which contain jam, and four contain honey. All the pots are lined up, and Winnie remembers that the pots with honey are standing together. What is the minimum number of pots Winnie-the-Pooh needs to check to find a pot with honey? | Answer: one.
Solution. Let's number the pots from 1 to 11 in the order of their arrangement in a row. Exactly one of the pots numbered 4 and 8 contains honey. Therefore, it is sufficient to check one of them. It is impossible to identify the pot with honey without checking, as any pot may contain either honey or jam.
... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. To the number $A$, consisting of eight non-zero digits, a seven-digit number, consisting of identical digits, was added, and the eight-digit number $B$ was obtained. It turned out that the number $B$ can be obtained from the number $A$ by rearranging some of the digits. What digit can the number $A$ start with if ... | Answer: 5.
Solution: Since the numbers $A$ and $B$ have the same sum of digits, their difference is divisible by 9. Therefore, the added seven-digit number with identical digits is divisible by 9. This means it consists of nines. That is, we can consider that $10^7$ was added to the number $A$ and 1 was subtracted. Th... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. On a checkerboard of size $8 \times 8$, 8 checkerboard ships of size $1 \times 3$ are placed such that no two cells occupied by different ships share any points. One shot is allowed to pierce all 8 cells of one row or one column. What is the minimum number of shots needed to guarantee hitting at least one ship? | Answer: 2 shots.
Solution. We will make 2 shots as shown in Fig. 5. Suppose we did not hit any ship. Then there are no ships in area 1. In each of areas 2 and 3, there is no more than 1 ship. Therefore, there are at least 6 ships in area 4.
Area 4 is a $5 \times 5$ square. Then in this area, horizontally placed ships... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. From Central Square to the station, there is a straight street divided by 11 intersections into 12 equal blocks. At each intersection, there is a traffic light. All traffic lights simultaneously turn green for 3 minutes, then red for 1 minute. It takes the bus two minutes to travel one block (from intersection to in... | Solution. A car will travel three blocks without obstacles in the first three minutes. Upon approaching the intersection separating the third and fourth blocks, the car will stop at the traffic light for 1 minute. Thus, to travel three blocks and start moving on the fourth block, the car will need four minutes. The car... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The head of the fish weighs as much as the tail and half of the body, the body weighs as much as the head and the tail together. The tail weighs 1 kg. How much does the fish weigh? | Answer: 8 kg.
Solution 1. The body weighs as much as the head and tail, i.e., two tails and half the body. This means that half the body weighs as much as two tails, i.e., the body weighs 4 kg. Then the head weighs $1+2=3$ kg, and the whole fish weighs $4+3+1=8$ kg.
Solution 2. Let $\Gamma, T, X$ be the weight of the... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Real numbers $x, y, z$ (non-zero) satisfy the equation: $x+y=z$. Find all possible values that the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}$ can take. | 1. Answer: 3.
Solution. Transform the expression $\frac{z}{x}+\frac{y}{z}-\frac{x}{y}+\frac{x}{z}+\frac{z}{y}-\frac{y}{x}=$
$=\frac{z}{x}-\frac{y}{x}+\frac{z}{y}-\frac{x}{y}+\frac{y}{z}+\frac{x}{z}=\frac{z-y}{x}+\frac{z-x}{y}+\frac{y+x}{z}$.
Using the condition $x+y=z$ and the derived formulas $y=z-x$ and $x=$ $z-y$... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. For different numbers $a$ and $b$, it is known that $\frac{a}{b}+a=\frac{b}{a}+b$. Find $\frac{1}{a}+\frac{1}{b}$. | Answer: -1.
Solution. The given equality can be written as $\frac{a}{b}-\frac{b}{a}=b-a$, from which $\frac{a^{2}-b^{2}}{a b}=b-a$ or $\frac{(a-b)(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the requi... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. During a math test, Oleg was supposed to divide a given number by 2, and then add 6 to the result. But he hurried and instead multiplied the given number by 2, and then subtracted 6 from the result. Nevertheless, he got the correct answer. What number was given to Oleg | Answer: 8.
Solution. Since Oleg multiplied the number by 2 instead of dividing it, he got a result that is four times greater than the required one. This means that the difference between these two results is three times greater than the required result. However, according to the condition, this difference is equal to... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. In the kingdom, there live counts, dukes, and marquises. One day, each count dueled with three dukes and several marquises. Each duke dueled with two counts and six marquises. Each marquise dueled with three dukes and two counts. It is known that all counts dueled with an equal number of marquises. How many ... | Answer: With 6 marquises.
Solution. Let there be $x$ counts, $y$ dukes, and $z$ marquises in the kingdom. Each count fought with three dukes, so there were $3 x$ duels between counts and dukes. But each duke fought with two counts, meaning there were $2 y$ such duels. Therefore, $3 x=2 y$.
Each duke fought with six m... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. On an island, there live knights who always tell the truth, and liars who always lie. One day, 15 natives, among whom were both knights and liars, stood in a circle, and each said: "Of the two people standing opposite me, one is a knight, and the other is a liar." How many of them are knights? | Answer: 10 knights.
Solution. Consider any knight. He tells the truth, which means that opposite him stand a knight and a liar. One of the people opposite the found liar is the initial knight, so next to him stands another knight. Opposite this new knight stands the previously found liar and another person, who must b... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.1
## Condition:
Given a triangle $\mathrm{ABC}$, in which $\mathrm{AB}=5$. The median $\mathrm{BM}$ is perpendicular to the bisector $\mathrm{AL}$. Find $\mathrm{AC}$. | Answer: 10
Exact match of the answer -1 point
Solution.
Consider triangle ABM: since the bisector drawn from vertex A is perpendicular to side $\mathrm{BM}$, triangle ABM is isosceles.
Therefore, $\mathrm{AB}=\mathrm{AM}=\mathrm{MC}$. Hence, $\mathrm{AC}=10$.
# | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.1
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid ... | Answer: 6
Exact match of the answer -1 point
## Solution.
It is claimed that each painting by Ivan Konstantinovich was sold for $1000 \times 2^{\text {x }}$ rubles, where $\mathrm{x}$ is some natural number. Indeed, initially, the painting's price is 1000 rubles, and each participant doubles its price, meaning the p... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.2
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid ... | Answer: 4
Exact match of the answer - 1 point
Solution by analogy with task №7.1
# | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 7.3
## Condition:
Artist Ivan Konstantinovich decided to sell several of his paintings at the Broken Auction. The rules of the Broken Auction are as follows: first, Ivan Konstantinovich names a certain starting price for his painting, after which the participants who want to buy this painting begin to bid on... | Answer: 6
Exact match of the answer - 1 point
Solution by analogy with task №7.1
# | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. In triangle $ABC$, the median $BM$ was drawn. It turned out that $\angle ABM=40^{\circ}$, $\angle MBC=70^{\circ}$. Find the ratio $AB: BM$. Justify your answer.
To solve problem 9.2 Sol... | Answer: $A B: B M=2$.
| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| An example of a triangle satisfying the conditions of the problem is provided | 0 points |
| Incorrect solution or its absence | 0 points | | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. After the bankruptcy of the company "Horns and Hooves," 17 horns, 2 hooves, and one weight remained. All this wealth was divided equally by weight between Pankovskiy and Balaganov, with the weight entirely going to Balaganov. The horns and hooves were not cut into pieces. Each horn is heavier than each hoof and li... | Solution: Let one hoof weigh $k$, and one horn weigh $k+\delta$ (all weights in the same units of measurement, for example, in puds). Then, according to the condition, the weight of the weight (the iron piece) is $k+2\delta$, and the total weight of the divided property is $20k+19\delta$. Each person received $10k+9.5\... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Masha is distributing tennis balls into identical boxes. If she uses 4 boxes, there is still room for 8 more balls in the last box, and if she uses 3 boxes, 4 balls will not fit into the boxes. How many balls is one box designed to hold? | Answer: 12
Solution.
Method 1
Notice that if the balls are in 3 boxes, there are 4 tennis balls left. Take these 4 balls and put them in the fourth box. According to the condition, there will still be room for 8 more balls in this box. This means that the box can hold a total of 12 balls.
Method 2.
Let the box hol... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Variant 2.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to Apri... | Answer: 11
Option 3.
The cafe "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "the cafe was open for 18 days from April 1 to April 20" and "the cafe was also open for 18 days from April 10 to April 30." It is known that he was wrong once. How many days was the cafe open from Ap... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
Petya wrote 9 consecutive natural numbers on the board. Kolya calculated their sum and got the answer 43040102. It turned out that he made a mistake only in the first digit of the sum. What should the first digit be | Answer: 8.
Solution: Let the consecutive natural numbers be $a-4, a-3, \cdots, a+3, a+4$. The sum of nine consecutive numbers is divisible by 9, indeed, this sum is equal to $(a-4)+(a-3)+\cdots+(a+3)+(a+4)=9a$. Therefore, the desired sum must be divisible by 9. This means that the sum of the digits of the obtained ans... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 3 blue points. And on any segment with endpoints at blue points, containing 2 blue points inside, there are at least 2 red point... | # Answer. 3.
Solution. Note that on a segment with endpoints at red points, not containing other red points, there cannot be 4 blue points. Indeed, in this case, between the outermost blue points there will be 2 blue points, which means there will be at least 2 more red points. Therefore, there are no more than 3 blue... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2. Petya runs down from the fourth floor to the first floor 2 seconds faster than his mother rides the elevator. Mother rides the elevator from the fourth floor to the first floor 2 seconds faster than Petya runs down from the fifth floor to the first floor. How many seconds does it take Petya to run down from the f... | Answer: 12 seconds.
Solution. Between the first and fourth floors, there are 3 flights, and between the fifth and first floors, there are 4. According to the problem, Petya runs 4 flights in 2 seconds longer than it takes his mother to ride the elevator, and 3 flights in 2 seconds less than his mother. Therefore, Pety... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play? | Answer: 11 games.
Solution: According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of any two consecutive games, the first player must participate in at least one, which means there were no more than \(2 \cdot 10 + 1 = 21\) games. Therefore, a total of 21 games wer... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.4. Let x be some natural number. Among the statements:
$2 \times$ greater than 70
x less than 100
$3 x$ greater than 25
x not less than 10
x greater than 5
three are true and two are false. What is x | Answer: 9.
Solution
The first statement is equivalent to x > 35, and the third statement is equivalent to $\mathrm{x}>8$, since x is a natural number. The first statement is false, because if it were true, then the third, fourth, and fifth statements would also be true, i.e., there would be at least 4 true statements... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.4. Twelve chairs are arranged in a row. Sometimes a person sits on one of the free chairs. In this case, exactly one of his neighbors (if they were there) stands up and leaves. What is the maximum number of people that can be sitting simultaneously, if initially all the chairs were empty? | Answer: 11.
Solution. Evaluation. Note that it is impossible to occupy all chairs simultaneously, as at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.
Example. We will show how to seat 11 people. ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Consider the equation $\sin ^{3}(x)+\cos ^{3}(x)=-1$. How many solutions does it have on the interval $[0 ; 6 \pi]$?
Answer: 6. | Solution. From the main trigonometric identity, we have $\sin ^{2}(x)+$ $\cos ^{2}(x)=1$. Adding this to the given equality, we get
$$
0=\sin ^{2}(x) \cdot(1+\sin (x))+\cos ^{2}(x) \cdot(1+\cos (x))
$$
In this expression, all factors are non-negative, so both terms $\sin ^{2}(x)(1+\sin (x))$ and $\cos ^{2}(x)(1+\cos ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. At a ball, princesses and knights gathered - a total of 22 people. The first princess danced with seven knights, the second with eight knights, the third with nine knights, ..., the last danced with all the knights present. How many princesses were at the ball? | Answer: 8.
Solution. Note that the number of knights the princess danced with is 6 more than her number. Let there be $x$ princesses in total, then the last one has the number $x$ and danced with all the knights, and there are $x+6$ of them in total. We get that there were $x+$ $(x+6)=2x+6=22$ people at the ball, whic... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.3. Dima, Misha, and Yura decided to find out who among them is the most athletic. For this, they held 10 competitions. The winner received 3 points, the second place 1 point, and the third place received nothing (in each competition, there was a first, second, and third place). In total, Dima scored 22 points... | Answer: 10.
Solution. In each competition, the boys in total received $3+1+0=4$ points. For all competitions, they scored $4 \cdot 10=40$ points. Dima and Misha in total scored $22+8=30$ points, so the remaining $40-30=10$ points were scored by Yura. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.4. On her birthday, Katya treated her classmates with candies. After giving out some candies, she noticed that she had 10 more candies left than Artem received. After that, she gave everyone one more candy, and it turned out that all the children in the class (including Katya) had the same number of candies. ... | Answer: 9.
Solution. Initially, the number of candies Kati and Artyom had differed by 10. When Kati gave everyone one more candy, the number of candies Artyom had increased by 1, and the number of candies Kati had decreased by the number of her classmates, and they ended up with the same amount. This means that 10 is ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. Cat Matroskin, Uncle Fyodor, Postman Pechkin, and Sharik sat down at a round table. In front of each of them was a plate with 15 sandwiches. Every minute, three of them ate a sandwich from their own plate, while the fourth ate a sandwich from their neighbor's plate. After 5 minutes of the meal, there were ... | # Answer: 7
Solution. We will call the sandwiches eaten from a neighbor's plate stolen. Note that in 5 minutes, exactly 5 sandwiches were stolen.
At the same time, 7 sandwiches disappeared from Uncle Fyodor's plate, of which he himself ate no more than 5, meaning that at least 2 were stolen. From this, it is clear th... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | 2. Answer: 8.
Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent to each... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent? | 3. Answer. 5 cupcakes.
The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote $x, y$, and $z$ as the costs of a cake, a cupcake, and a bagel respectively, we get the equation: $(x+2 y+3 z)+(3 x+z)=12 y, \quad$ from which it follows that $\quad 4 x+4 z=10 y$, \quad t... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $6 \times 6$ square of cells so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.) | 5. Answer. 6.
Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted.
Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get that at l... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. Three pirates divided the diamonds they had obtained during the day in the evening: twelve each for Bill and Sam, and the rest went to John, who couldn't count. At night, Bill stole one diamond from Sam, Sam stole one from John, and John stole one from Bill. As a result, the average weight of Bill's diamonds decre... | Answer: 9 diamonds.
Solution. The first method (arithmetic). Note that the number of diamonds each pirate has did not change overnight. Since Bill has 12 diamonds, and their average weight decreased by 1 carat, the total weight of his diamonds decreased by 12 carats. Similarly, Sam also has 12 diamonds, and their aver... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. The numbers $x, y, z$ are such that $x \in[-3,7], y \in[-2,5], z \in[-5,3]$.
(a) (1 point) Find the smallest possible value of the quantity $x^{2}+y^{2}$.
(b) (3 points) Find the smallest possible value of the quantity $x y z - z^{2}$. | # Answer:
(a) (1 point) 0.
(b) (3 points) -200.
Solution. (a) Note that $x^{2} \geqslant 0$ and $y^{2} \geqslant 0$, so $x^{2}+y^{2} \geqslant 0$. The value $x^{2}+y^{2}=0$ is possible when $x=0, y=0$.
(b) Note that $|x y z|=|x||y||z| \leqslant 7 \cdot 5 \cdot 5$ and $z^{2} \leqslant 5^{2}$, so $x y z-z^{2} \geqsla... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. The graph of the quadratic function $y=a x^{2}+c$ intersects the coordinate axes at the vertices of an equilateral triangle. What is the value of ac? | Answer: -3.
Solution. Since the graph intersects the OX axis at two points, the numbers a and c have different signs. The points of intersection are: $A\left(\sqrt{-\frac{c}{a}} ; 0\right) ; B\left(-\sqrt{-\frac{c}{a}} ; 0\right)$. The graph intersects the OY axis at point $C(0 ; c)$. Then the side $AB$ of the equilat... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given $n>2$ natural numbers, among which there are no three equal, and the sum of any two of them is a prime number. What is the largest possible value of $n$? | 1. Answer: 3. Note that the triplet $1,1,2$ satisfies the condition. Suppose there are more than 3 numbers. Consider any 4 of them a, b, c, d. Among these four numbers, there cannot be two even numbers, otherwise their sum would be an even prime greater than two. Therefore, at least 3 of them must be odd. Their pairwis... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) The kids were given the task to convert the turtle's speed from centimeters per second to meters per minute. Masha got an answer of 25 m/min, but she thought there were 60 cm in a meter and 100 seconds in a minute. Help Masha find the correct answer. | Answer: 9 m/min.
Solution. The turtle covers a distance of 25 Machine "meters" in one Machine "minute," meaning it crawls $25 \cdot 60$ centimeters in 100 seconds. Therefore, the speed of the turtle is $\frac{25 \cdot 60}{100}=15$ cm/sec. Thus, in 60 seconds, the turtle will crawl $15 \cdot 60$ centimeters, which is $... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon (i.e., exactly at 12 o'clock) and, without stopping, arrived: one at $B$ at 4 PM, and the other at $A$ at 9 PM. At what time was dawn that day? | Answer: at 6 AM.
Solution. Let's denote the meeting point as $C$. Let $x$ be the number of hours from dawn to noon.
The speed of the first pedestrian on segment $A C$ is $A C / x$, and on segment $B C$ it is $B C / 4$. Since his speed is constant, we have $\frac{A C}{x}=\frac{B C}{4}$, which can be rewritten as $\fra... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. There are three acrobat brothers. Their average height is 1 meter 74 centimeters. The average height of two of these brothers: the tallest and the shortest - 1 meter 75 centimeters. What is the height of the middle brother? Justify your answer. | Answer: 1 meter 72 centimeters.
Solution. Since the average height of all three is 1 meter 74 centimeters, the total height of all three is 5 meters 22 centimeters. The average height of the two brothers is 1 meter 75 centimeters, so their combined height is 3 meters 50 centimeters. Therefore, the height of the middle... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On the island, there live knights who always tell the truth and liars who always lie. In the island's football team, there are 11 people. Player number 1 said: "In our team, the number of knights and the number of liars differ by one." Player number 2 said: "In our team, the number of knights and the number of liars... | 3. Answer: the only knight plays under number 9. The solution is that the number of knights cannot be equal to the number of liars, so one of the answers must be correct. Two answers cannot be true, as they contradict each other. Therefore, there is exactly 1 knight and 10 liars in the team, which is stated by the play... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A student passed 31 exams over 5 years of study. Each subsequent year, he passed more exams than the previous year, and in the fifth year, he passed three times as many exams as in the first year. How many exams did he pass in the fourth year? | # Answer: 8.
Solution: Let $a, b, c, d, e$ be the number of exams taken in each year of study. According to the problem, $a+b+c+d+e=31, a<b<c<d<e$.
Replace the numbers $b, c, d, e$ in the equation with definitely not larger values:
$a+(a+1)+(a+2)+(a+3)+3a \leq 31$. We get $7a \leq 25$. Replacing these numbers with d... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A circle with center at point $O$ is circumscribed around quadrilateral $A B C D$. The diagonals of the quadrilateral are perpendicular. Find the length of side $B C$, if the distance from point $O$ to side $A D$ is 1. | Answer: $B C=2$
Solution. Let $O E \perp A D$, then $O E=1$. Draw a line through point $A$ perpendicular to $A D$, which intersects the circle at point $M$. Then $D M$ is a diameter and $D O=O M$. Since $\angle D B A=\angle D M A$ (as inscribed angles subtending the same arc), $\angle M D A=90^{\circ}-\angle D M A$ an... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | Answer: 8.
Solution: Note that it is impossible to arrange all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor, the card with the number 1. Therefore, both cards 5 and 7 must be at the ends, and the card with the number 1 must be adjacent ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Rational numbers $a, b$ and $с$ are such that $(a+b+c)(a+b-c)=2 c^{2}$. Prove that $c=0$.
---
Translation:
4. Rational numbers $a, b$ and $c$ are such that $(a+b+c)(a+b-c)=2 c^{2}$. Prove that $c=0$. | Solution. The initial equality is equivalent to the following $(a+b)^{2}-c^{2}=2 c^{2}$, or $(a+b)^{2}=3 c^{2}$. If $c \neq 0$, we get $((a+b) / c)^{2}=3 .|(a+b) / c|={ }^{-}$. On the left, we have a rational number, since the sum, quotient, and absolute value of rational numbers are rational, while on the right, we ha... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
10.3. Given three quadratic trinomials $f(x)=a x^{2}+b x+c, g(x)=b x^{2}+c x+a, h(x)=c x^{2}+$ $a x+b$, where $a, b, c$ are distinct non-zero real numbers. From them, three equations were formed: $f(x)=g(x), f(x)=h(x), g(x)=h(x)$. Find the product of all roots of these three equations, given that each of them has two d... | Answer. 1.
Solution. Since it is known that all equations have roots, we can use Vieta's theorem. Then the product of all roots will be equal to $\frac{c-a}{a-b} \cdot \frac{a-b}{b-c} \cdot \frac{b-c}{c-a}=1$.
Comment. Correct answer without justification - 0 points. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. In a $4 \times 4$ square, each of the 16 cells was painted either black or white. Then, in each of the nine $2 \times 2$ squares that can be identified within this square, the number of black cells was counted. The resulting numbers were 0, 2, 2, 3, 3, 4, 4, 4, 4. How many black cells can there be in the large squ... | Answer: 11
Solution. Note that squares with 0 white cells and 4 white cells cannot intersect. If a square with 0 white cells is not in a corner, then it does not intersect with more than three other squares (the square can be in the center or adjacent to the middle of a side). Let the white square be in the bottom-lef... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. What is the largest number of different natural numbers that can be chosen so that the sum of any three of them is a prime number? | Answer: 4 numbers. Example: $1,3,7,9$. Indeed, the numbers $1+3+7=11, 1+3+9=13$, $1+7+9=17, 3+7+9=19$ are prime. Evaluation. Note that among five natural numbers, it is always possible to choose three whose sum is a composite number. Consider the remainders of these five numbers when divided by 3. If there are three id... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Solve the inequality: $\sqrt{(x-2)^{2}\left(x-x^{2}\right)}<\sqrt{4 x-1-\left(x^{2}-3 x\right)^{2}}$. | 11.1. Answer: $x=2$.
The left side of the original inequality is defined at $x=2$ and. At the point $x=2$, the inequality is true. We will prove that there are no solutions on the interval $[0 ; 1]$. For this, we will square both sides and bring the inequality to the form $x^{2}(1-x)<-1$. The last inequality is not sa... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. All horizontal and vertical distances between adjacent points are equal to 1. What is the area of the triangle with vertices at the black points? | Answer. 1.
Solution. The area of the triangle can be found, for example, by subtracting from half the area of the square the area of the square and the areas of two right triangles. We get $S=10-4-2-3=1$. The figure can be divided in other ways.
Comment. An answer without justification - 0 points. A partition that al... | 1 | Geometry | 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?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle AB... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some number (not necessarily an integer). Then the first said: “My number is greater than 1”, the second said: “My number is greater than $2”, \ldots$, the tenth said: “My number is greater th... | Answer: 9 knights.
Solution. Estimation. Note that none of the knights could have said the phrase "My number is greater than 10," otherwise the number they thought of would indeed be greater than 10. But then he could not have said any of the phrases "My number is less than 1," "My number is less than 2," ..., "My num... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. One day Uncle Fyodor weighed Sharik and Matroskin. It turned out that Sharik is 6 kg heavier than Matroskin, and Matroskin is three times lighter than Sharik. How much did Matroskin weigh? | Answer: 3 kg.
Solution: Since Matroskin is three times lighter than Sharik, Matroskin is lighter than Sharik by two of his own weights. According to the condition, this is equal to 6 kg, i.e., Matroskin weighs $6: 2=3$ kg. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Mom bought a box of lump sugar (sugar in cubes). The children first ate the top layer - 77 cubes, then the side layer - 55 cubes, and finally, the front layer. How many sugar cubes are left in the box? | Answer: 300 or 0.
Solution. A box has three dimensions: height, width, and depth. To find out how many cubes are in the top layer, you need to multiply the width by the depth, and for the side layer, multiply the height by the depth. After the top layer is eaten, the height decreases by 1, while the depth remains the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 10.5. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars al... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10-2-1. Petya writes down a sequence of numbers: if the current number is equal to $x$, then the next one is $\frac{1}{1-x}$. The first number in the sequence is 2. What is the five hundredth number? | Answer: -1.
Solution variant 1. Let's list the first few terms of the obtained sequence:
$$
2,-1,1 / 2,2, \ldots
$$
We see that the sequence has entered a cycle with a period of 3. Since the number 500 when divided by 3 gives a remainder of 2, the 500th term will be the same as the 2nd. That is, the five hundredth n... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On a coordinate plane with the origin at point $O$, a parabola $y=x^{2}$ is drawn. Points $A, B$ are marked on the parabola such that $\angle A O B$ is a right angle. Find the smallest possible value of the area of triangle $A O B$. | 5. Answer: 1. Solution. Let $A\left(a, a^{2}\right), B\left(b, b^{2}\right)$ be arbitrary points on the parabola. Then, by the Pythagorean theorem, $a^{2}+a^{4}+b^{2}+b^{4}=(a-b)^{2}+\left(a^{2}-b^{2}\right)^{2} \Leftrightarrow a b=-1$. From this, we find the expression for the area of the triangle: $2 S=\sqrt{a^{2}+a^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. In the notebook, all irreducible fractions with the numerator 15 are written down, which are greater than $\frac{1}{16}$ and less than $\frac{1}{15}$. How many such fractions are written down in the notebook? | Answer: 8 fractions.
Solution. We look for all suitable irreducible fractions of the form $\frac{n}{15}$. Since $\frac{1}{16}<\frac{15}{n}$, then $\frac{15}{225}>\frac{15}{n}$, and $n>225$. Therefore, $225<n<240$. The fraction $\frac{n}{15}$ is irreducible, meaning $n$ is not divisible by 3 or 5. It is not difficult t... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. There are 2001 coins on the table. Two players play the following game: they take turns, on each turn the first player can take any odd number of coins from 1 to 99, the second player can take any even number of coins from 2 to 100, and so on. The player who cannot make a move loses. Who wins with correct play? | 5. The first one wins. Winning strategy: the first player should take 81 coins from the table on the first move. On each subsequent move, if the second player takes x coins, then the first player should take (101 - x) coins. He can always do this because if x is an even number from 2 to 100, then (101 - x) is an odd nu... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. In a class, there are 18 children. The parents decided to gift the children from this class a cake. To do this, they first asked each child the area of the piece they wanted to receive. After that, they ordered a square cake, the area of which is exactly equal to the sum of the 18 named numbers. However, upon seei... | Answer. $k=12$.
Solution. We always assume that the area of the cake is 1. We will show that for some children's requests, the parents will not be able to cut out more than 12 required pieces. Choose a number $1 / 15 > x > 1 / 16$. Suppose that 15 main children ordered a piece of cake with an area of $x$ each (and the... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Grandfather is 31 times older than his grandson. In how many years will he be 7 times older than his grandson, given that the grandfather is more than 50 but less than 90 years old? | Answer: In 8 years.
Solution: Grandfather's age is divisible by 31. But the only such number greater than 50 and less than 90 is 62. Therefore, the grandfather is 62 years old, and the grandson is 2 years old. In x years, the grandfather will be x+62 years old, and the grandson will be x+2 years old. If at that time h... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(x^{2}-|x|-2\right)^{2}+\sqrt{|2 x+1|-3}=0$. | # Solution
Since each term of the original equation is non-negative, the equation can have solutions if and only if both terms are equal to zero. We obtain the equivalent system of equations
$$
\left\{\begin{array}{l}
x^{2}-|x|-2=0 \\
|2 x+1|-3=0
\end{array}\right.
$$
The solution to this system is $x=-2$.
## Crite... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A $3 \times 3$ table was filled with prime numbers. It turned out that the sum of the numbers in any two adjacent cells is a prime number. What is the maximum number of different numbers that can be in the table? Provide an example and prove that there cannot be more different numbers.
Fill the $3 \times 3$ table w... | 3. Answer: 6.
Example:
| 5 | 2 | 29 | |
| :---: | :---: | :---: | :---: |
| 2 | 3 | 2 | |
| 11 | 2 | 17 | |
| | -2 | | 0 |
We will prove that it cannot be more. Divide the cells of th... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. What is the greatest length of an arithmetic progression of natural numbers
$a_{1}, a_{2}, \ldots, a_{n}$ with a difference of 2, in which for all $k=1,2 \ldots, n$ all numbers $a_{k}^{2}+1$ are prime | Solution.
All members of the progression must be even numbers. Consider the remainders when divided by 5. Since the common difference of the arithmetic progression is 2, the remainders in the progression will appear in the order
$\ldots 2,4,1,3,0,2,4,1,3,0, \ldots$ This means that among the terms of the progression, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A $10 \times 10$ square was cut into 17 rectangles, each with both side lengths greater than 1. What is the smallest number of squares that could be among these rectangles? Provide an example of such a cutting. | Answer: one square.
Solution. Suppose that among the rectangles, there is not a single square. Let $a$ and $b$ be the sides of an arbitrary rectangle, with $a > b$. Since the integer $b$ is greater than 1, then $b \geqslant 2$, and thus $a \geqslant 3$. Therefore, the area of each such rectangle $a \times b$ is at lea... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The diagonals of quadrilateral $A B C D$ intersect at point $O$. It is known that $A B=B C=$ $=C D, A O=8$ and $\angle B O C=120^{\circ}$. What is $D O ?$ | Answer: $D O=8$.
Solution. Mark a point $E$ on the line $A C$ such that triangle $B O E$ is equilateral (Fig. 2). We will prove the equality of triangles $B A E$ and $B C O$. Indeed, since $A B=B C$, triangle $A B C$ is isosceles, and thus $\angle B A C=\angle B C A$. Additionally, note another pair of equal angles $\... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A seagull is being fed from a moving boat. A piece of bread is thrown down, the seagull takes 3 seconds to pick up the piece from the sea surface, and then it takes 12 seconds to catch up with the boat. Upon entering the bay, the boat reduces its speed by half. How much time will it now take the seagull to catch up ... | Solution. Let the speed of the seagull relative to the boat be $x$ (m/s). Then in 12 seconds, it flies $12 x$ (m). The boat covers this distance in 3 seconds, so its speed is $4 x$ (m/s), and the speed of the seagull is $5 x$ (m/s). In the bay, the boat's speed becomes $2 x$ (m/s). In 3 seconds, it travels $6 x$ (m). T... | 2 | Logic and Puzzles | 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 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.5. The lateral faces of the pentagonal pyramid $S A B C D E$ are acute-angled triangles. We will call a lateral edge of the pyramid good if it is equal to the height of the opposite lateral face, drawn from the vertex of the pyramid (for example, the edge $S A$ is good if it is equal to the height of the triangle $S... | # Answer. 2.
Solution. We will show that two non-adjacent lateral edges of the pyramid cannot both be good. Suppose this is not the case, and, for example, edges $S A$ and $S C$ are good, meaning $S A = S P$ is the height of face $S C D$, and $S C = S Q$ is the height of face $S A E$. However, by the property of the h... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.1 Two couples want to sit on a bench. In how many ways can they sit so that each boy sits next to his girlfriend, if the left and right sides of the bench are distinguishable? | # Answer: 8
Solution. Let's number the seats from 1 to 4. If one of the couple sits in seat number 1, then the second must sit in seat 2. Thus, each couple either sits on the left (seats 1 and 2) or on the right (seats 3 and 4). Each couple can sit in two ways (boy on the left or right), and the couples themselves can... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4.1.2. A sports team consists of two boys and two girls. In the relay, the stages run by girls must alternate with the stages run by boys. In how many ways can the stages be distributed
# | # Answer: 8
Solution. Boys can run stages 1 and 3, or stages 2 and 4. If they run stages 1 and 3, they can be distributed in two ways, and girls can be distributed in two ways, making a total of 4 ways. In the second case, there are another 4 ways, making a total of 8 ways. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4.1.3 Two mothers with their children want to sit on a bench with 4 seats. In how many ways can they sit so that each mother sits next to her child? Each mother is walking with one child.
# | # Answer: 8
Solution. Let's number the seats from 1 to 4. If someone from a family sits in seat number 1, then the second person must sit in seat 2. Thus, each family either sits on the left (seats 1 and 2) or on the right (seats 3 and 4). Each family can sit in two ways (child on the left or right), and the families ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.1 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared three times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a hurry, stumbled, and some of the balloons burst. Eeyore received a ... | # Answer: 1
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 6 times more balloons than Piglet prepared. The next number after 31 that is a multiple of 6 is $36, 36: 6=6$, which means Piglet prepared no less than 6 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.2 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared three times as many balloons as Piglet, and Owl prepared four times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a great hurry, stumbled, and some of the balloons burst. Eeyore ... | # Answer: 4
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 8 times more balloons than Piglet prepared. The next number after 60 that is a multiple of 8 is $64, 64: 8=8$, which means Piglet prepared no less than 8 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.3 For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared four times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a great hurry, stumbled, and some of the balloons burst. Eeyore receiv... | # Answer: 2
Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 7 times more balloons than Piglet prepared. The next number after 44 that is a multiple of 7 is 49, $49: 7=7$, which means Piglet prepared no less than 6 balloons.
If all of Piglet's balloons had burst, then Eeyore woul... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1 Six children stood in a circle and whispered their favorite number to the neighbors on their right and left. Each child added the two numbers they heard and said them out loud. The picture shows what three of the children said. What number did the girl think of, next to whom there is a question mark?
 for a game. Each had to say one of the following phrases: "There is a liar below me" or "There is a k... | # Answer: 11
Solution. The formation cannot consist of only liars (in such a case, 4-8 would be telling the truth). Consider the tallest knight, let's call him R. He could only say the first phrase, so the positions 1, 2, and 3 must be occupied by liars, and R stands from 4 to 8. In addition, there is a liar who is sh... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. How many seven-digit natural numbers exist, for which the product of the first three digits is 30, the product of the three digits in the center is 7, and the product of the last three digits is 15? | Answer: 4.
Solution: Let the number be $\overline{\operatorname{abcdefg}}$. According to the condition, $c d e=7$, which means one of these digits is 7, and the other two are 1. Since 30 and 15 are not divisible by $7, d=7, c=e=1$. The number $30=5 \cdot 6 \cdot 1$, so we get two three-digit numbers $\overline{a b c}$... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. At a round table, 10 people are sitting, each of whom is either a knight, who always tells the truth, or a liar, who always lies. Each of them was given a token. Then each of them passed their token to one of their two neighbors. After that, 5 people said: "I have one token," while the other 5 said: "I have no tok... | # Answer. 7.
Solution. Evaluation. After passing the chips, each person sitting at the table can have 0, 1, or 2 chips. The total number of chips will be 10. Note that if a person lies, they will state a number of chips that differs from the actual number by 1 or 2. Since the total number of chips according to the ans... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Having walked $4 / 9$ of the length of the bridge, the traveler noticed that a car was catching up to him, but it had not yet entered the bridge. Then he turned back and met the car at the beginning of the bridge. If he had continued his movement, the car would have caught up with him at the end of the bridge. Find ... | # Solution
From the condition, it follows that the time it takes for the car to approach the bridge is equal to the time it takes for the traveler to walk $4 / 9$ of the bridge. Therefore, if the traveler continues moving, by the time the car reaches the bridge, he will have walked ${ }^{8} / 9$ of the bridge. This me... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given various real numbers $p$ and $q$. It is known that there exists a number $x$ such that the equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ are satisfied. What values can the sum $p + q$ take? | Solution. Let $x$ be such that both equalities from the problem statement are satisfied. Then $x^{2}+p x+q=x^{2}+q x+p$, from which $(p-q) x=p-q$. Since $p \neq q$, it follows that $x=1$. Substituting this value into any of the equations, we find that $p+q+1=0$, that is, $p+q=-1$. Answer: -1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a football tournament, seven teams played: each team played once with each other. Teams that scored thirteen or more points advance to the next round. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round? | Solution. Evaluation. Note that the total number of games will be $\frac{7 \cdot 6}{2}=21$. Then the total number of points does not exceed $21 \cdot 3=63$. This means that no more than $\left[\frac{63}{13}\right]=4$ teams can advance to the next round.
Example. We will show that 4 teams can advance to the next round.... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. For real numbers $x$ and $y$, define the operation $\star$ as follows: $x \star y=x y+4 y-3 x$.
Calculate the value of the expression
$$
((\ldots)(((2022 \star 2021) \star 2020) \star 2019) \star \ldots) \star 2) \star 1
$$ | Answer: 12.
Solution. Let $t=(\ldots(((2022 \star 2021) \star 2020) \star 2019) \star \ldots) \star 4$. Then the value of the expression from the problem condition is
$$
\begin{aligned}
((t \star 3) \star 2) \star 1 & =((3 t+12-3 t) \star 2) \star 1=12 \star 2 \star 1= \\
& =(24+8-36) \star 1=(-4) \star 1=-4+4+12=12
... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. The numbers $a$ and $b$ are such that each of the two quadratic trinomials $x^{2} + a x + b$ and $x^{2} + b x + a$ has two distinct roots, and the product of these trinomials has exactly three distinct roots. Find all possible values of the sum of these three roots.
(S. Berlov) | Answer: 0.
Solution: From the condition, it follows that the quadratic polynomials $x^{2}+a x+b$ and $x^{2}+b x+a$ have a common root $x_{0}$, as well as roots $x_{1}$ and $x_{2}$ different from it; in particular, $a \neq b$. Then $0=\left(x_{0}^{2}+a x_{0}+b\right)-\left(x_{0}^{2}+b x_{0}+a\right)=(a-b)\left(x_{0}-1\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Natural numbers $a, x$ and $y$, greater than 100, are such that $y^{2}-1=$ $=a^{2}\left(x^{2}-1\right)$. What is the smallest value that the fraction $a / x$ can take? | # Answer. 2.
First solution. Rewrite the condition of the problem as $y^{2}=a^{2} x^{2}-a^{2}+1$. Notice that $y100$, if we set $a=2 x, y=a x-1=2 x^{2}-1$.
Second solution. We provide another proof of the estimate $a / x \geqslant 2$. Rewrite the equality from the condition as
$$
(a x-y)(a x+y)=a^{2} x^{2}-y^{2}=a^{... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. $B$ of trapezoid $A B C D: \angle A=\angle B=90^{\circ}, A D=2 \sqrt{7}, A B=\sqrt{21}, B C=2$. What is the minimum value that the sum of the lengths $X A+X B+X C+X D$ can take, where $X-$ is an arbitrary point in the plane?
, B\left(x_{b}, y_{b}\right), C\left(x_{c}, y_{c}\right)$ are taken. It is known that $x_{c}=5$ and $y_{a}=y_{b}$. Find the abscissa of the point of intersection of the medians of triangle $A B C$.
![](https://cdn.mathpix.com/cropp... | # Answer: 3
Solution. Since $y_{a}=y_{b}$, points $A$ and $B$ are symmetric with respect to the axis of the parabola, which means the midpoint $K$ of segment $A B$ has an abscissa equal to the abscissa of the vertex of the parabola $x_{k}=2$. Next, by the property of the median, the point of intersection of the median... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Little One gave Karlson 111 candies. They immediately ate some of them together, $45\%$ of the remaining candies went to Karlson for lunch, and a third of the candies left after lunch were found by Fräulein Bok during cleaning. How many candies did she find? | Answer: 11.
First method. Let Carlson have had $n$ candies before lunch. Then, after lunch, there were $\frac{55}{100} n$ left, and Fräulein Bock found $\frac{1}{3} \cdot \frac{55}{100} n=\frac{11 n}{60}$ candies. Since the number of candies must be an integer, $11 n$ is divisible by 60. Since the numbers 11 and 60 ar... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.