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4. Alexei drew 7 lines on a plane, which divided it into several parts. Then he chose two adjacent parts (adjacent parts are those that share a common side), counted how many sides each of them contains, and added these two numbers. What is the largest number he could have obtained? Explain your answer.
. Return the line we removed. It will be a side for both small parts (+... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. The numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ form a geometric progression. Among them, there are both rational and irrational numbers. What is the maximum number of terms in this progression that can be rational numbers? | Answer: 3.
Example: let $a_{1}=1, q=\sqrt{2}$, we get the geometric progression $1, \sqrt{2}, 2, 2\sqrt{2}, 4$. Evaluation. If there are 4 rational numbers among them, then there will be two consecutive rational members of the geometric progression. This means that the common ratio of the progression (the ratio of the... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a box, there are oranges, pears, and apples, a total of 60 fruits. It is known that there are 3 times more apples than non-apples, and there are 5 times fewer pears than non-pears. How many oranges are in the box? | Answer: 5.
Solution. Since there are 3 times more apples than non-apples, apples make up $\frac{3}{4}$ of the total number of fruits, i.e., there are $\frac{3}{4} \cdot 60=45$ apples. Since there are 5 times fewer pears than non-pears, pears make up $\frac{1}{6}$ of the total number of fruits, i.e., there are $\frac{1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. In the cells of a $2 \times 35$ table (2 rows, 35 columns), non-zero real numbers are placed, and all numbers in the top row are distinct. For any two numbers in the same column, the following condition is satisfied: one number is the square of the other.
(a) (1 point) What is the maximum number of negati... | Answer: (a) 35. (b) 12.
Solution. (a) In any column, there can be no more than one negative number, so the total number of negative numbers is no more than 35. There can be exactly 35 if, for example, the top numbers are $-1, -2, -3, \ldots, -35$, and below them are the numbers $1^{2}, 2^{2}, 3^{2}, \ldots, 35^{2}$ re... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2. What is the greatest number of consecutive natural numbers, each of which has exactly four natural divisors (including 1 and the number itself)? | Answer: three numbers.
Solution. Suppose there are four consecutive numbers that satisfy the condition. Note that among four consecutive numbers, one is divisible by 4. Then, in the prime factorization of this number, there are at least two twos. If there is another prime divisor $p$ different from two, then the numbe... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A sequence of 2016 numbers is written. Each one, except the first and the last, is equal to the sum of its neighbors. Find the sum of all 2016 numbers. | Answer: the sum of all numbers is zero.
Solution. Let the first number be $a$, and the second $b$. Denote the third as $x$. Then $a+x=b$ and, therefore, $x=b-a$. Now, express the fourth in terms of $a$ and $b$: $b+y=b-a$ and, therefore, $y=-a$. Continuing, we get: $a, b, b-a, -a, -b, a-b, a, b, \ldots$ This means the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On a $5 \times 5$ board, a cross consisting of five cells (a cell and all its neighbors) is located. What is the minimum number of detectors needed to place on the board cells to accurately determine the position of the cross? (A detector indicates whether a cell belongs to the cross or not, and the detectors trigge... | Answer: 4.
Solution. There are 9 possible positions for the cross on the board, which is the same as the number of positions for the central cell of the cross. A detector has two states, so the total number of possible states for three detectors is $2^{3}=8$, and thus, they cannot distinguish between 9 positions of th... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3.7. Katya passes the time while her parents are at work. On a piece of paper, she absentmindedly drew Cheburashkas in two rows (at least one Cheburashka was drawn in each row).
Then, after some thought, between every two adjacent Cheburashkas in a row, she drew a Gena the Crocodile. And then to the left of ea... | Answer: 11.
Solution. The Krokodilofes are drawn exactly in the gaps between the other characters. In each of the two rows of gaps, there is 1 less gap than there are characters, so
There are 2 fewer Krokodilofes than all the other characters. Therefore, Cheburashkas, Crocodile Gens, and Old Ladies Shapoklyak total $... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.3. Vitya and his mother left home at the same time and walked in opposite directions at the same speed: Vitya - to school, and his mother - to work. After 10 minutes, Vitya realized he didn't have the keys to the house, and he would return from school earlier than his mother, so he started to catch up with he... | Answer: 5 minutes.
Solution. Let Vitya and his mother initially walk at a speed of $s$ meters per minute. After 10 minutes, when Vitya realized he had forgotten his keys, the distance between him and his mother became $10 s + 10 s = 20 s$. When Vitya started to catch up with his mother, he walked at a speed of $5 s$ m... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. How many real numbers $x$ exist such that the value of the expression $\sqrt{123-\sqrt{x}}$ is an integer? | Answer: 12.
Solution. From the condition, it follows that the value $s=123-\sqrt{x} \leqslant 123$ is a square of an integer. Since $11^{2}<123<12^{2}$, this value can take one of 12 values $0^{2}$, $1^{2}, 2^{2}, \ldots, 11^{2}$. And for each of these 12 values of $s$, there is a unique value of $x=(123-s)^{2}$ (obvi... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. At exactly noon, a truck left the village and headed for the city, at the same time, a car left the city and headed for the village. If the truck had left 45 minutes earlier, they would have met 18 kilometers closer to the city. And if the car had left 20 minutes earlier, they would have met $k$ kilometer... | Answer: 8.
Solution. We will express all distances in kilometers, time in hours, and speed in kilometers per hour. Let the distance between the village and the city be $S$, the speed of the truck be $x$, and the speed of the car be $y$. The time until the first meeting is $\frac{S}{x+y}$, so the distance from the vill... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a grid table with 5 rows and 6 columns, each cell contains either a cross, a zero, or a star. It is known that:
- in each column, the number of zeros is not less than the number of crosses;
- in each column, the number of zeros is not less than the number of stars;
- in each row, the number of crosses... | Answer: 6.
Solution. Since in each column there are no fewer O's (noughts) than X's (crosses), then in the entire table there are no fewer O's than X's. Similarly, considering the rows, it follows that in the entire table there are no fewer X's than O's. Therefore, there are an equal number of X's and O's in the entir... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 3.1. Condition:
Danil took a white cube and numbered its faces with numbers from 1 to 6, writing each one exactly once. It turned out that the sum of the numbers on one pair of opposite faces is 11. What can the sum of the numbers on none of the remaining pairs of opposite faces NOT be?
## Answer options:
$\square... | # Solution.
If the sum of the numbers on opposite faces is 11, then the numbers on these faces are 5 and 6. The remaining numbers can be paired in three ways: $(1,2)$ and $(3,4); (1,3)$ and $(2,4); (1,4)$ and $(2,3)$, i.e., the sums that CAN be obtained are: $3,7,4,6$ and 5. Therefore, the sums that CANNOT be obtained... | 9 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
# 5.1. Condition:
Polina came to the cafeteria and saw that 2 puff pastries cost 3 times more than 2 pies. Polina didn't have enough money for 2 puff pastries, but she did have enough for 1 pie and 1 puff pastry. After the purchase, she wondered how many times more money she spent buying 1 puff pastry and 1 pie instea... | Answer: 2
## Solution.
2 puff pastries are 3 times more expensive than 2 pies, so one puff pastry is 3 times more expensive than one pie, which means one puff pastry costs as much as 3 pies. Therefore, 1 puff pastry and 1 pie cost as much as 4 pies. Then they are 2 times more expensive than 2 pies. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. If the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic polynomial $g(x)=(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$. | Answer: 6.
Solution: We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=$ $4 f(-2)$.
Comment: Correct answer without justification - 0 points. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Six people - liars and knights - sat around a table. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, 3 people said: "I have one coin," while the other 3 said: "I have no coins." What is the maximum ... | Answer: 4.
Solution: After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 6. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs from t... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3. On the board, there are $N$ prime numbers (not necessarily distinct). It turns out that the sum of any three numbers on the board is also a prime number. For what largest $N$ is this possible | Answer: $N=4$.
Solution. Consider the remainders when the $N$ written numbers are divided by 3. All three remainders cannot occur, because in this case, the sum of three numbers with different remainders will be divisible by 3 (and will be greater than 3), so it will not be a prime number. Therefore, there can be no m... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. Each cell of a $7 \mathrm{x} 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the n... | Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones. In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones. Thus... | 8 | Combinatorics | 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 |
3. Three motorcyclists start simultaneously from one point on a circular highway in the same direction. The first motorcyclist caught up with the second for the first time after making 4.5 laps from the start, and 30 minutes before that, he caught up with the third motorcyclist for the first time. The second motorcycli... | # Solution
Let $x, y, z$ be the speeds of the first, second, and third motorcyclists in circles per hour, respectively.
Express the time it takes for the first motorcyclist to catch up with the second, the first to catch up with the third, and the second to catch up with the third in different ways.
We obtain the sy... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. It is known that $a^{2}+2=b^{4}, b^{2}+2=c^{4}, c^{2}+2=a^{4}$. What is the value of the product $\left(a^{2}-1\right)\left(b^{2}-1\right)\left(c^{2}-1\right)$? Find all possible values and prove that there are no others. | Solution: Subtract 1 from both sides of each equation, we get:
$$
\begin{gathered}
a^{2}+1=b^{4}-1=\left(b^{2}-1\right)\left(b^{2}+1\right) \\
b^{2}+1=\left(c^{2}-1\right)\left(c^{2}+1\right), \quad c^{2}+1=\left(a^{2}-1\right)\left(a^{2}+1\right)
\end{gathered}
$$
Multiply the left and right parts of all three equat... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.5. What is the smallest value that the GCD $(x, y)$ can take, where $x$ and $y$ are natural numbers, if the LCM $(x, y)=(x-y)^{2}$? | Solution:
When $x=4$ and $y=2$, LCM $(x, y)=(x-y)^{2}=4$, and GCD $(x, y)=2$. Thus, 2 is achievable.
Assume that GCD $(x, y)=1$ (the numbers are coprime). Then LCM $(x, y)=x y$. From the condition, we get that $x y=x^{2}-2 x y+y^{2}$. We obtain $x^{2}-3 x y+y^{2}=0$. Solving the obtained quadratic equation with respe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A natural number is written on the board. Nikolai noticed that he can append a digit to the right of it in two ways so that the resulting number is divisible by 9. In how many ways can he append a digit to the right of the given number so that the resulting number is divisible by 3? | Answer: 4 ways.
Solution. Note that the difference between the two numbers "noticed" by Nikolai is less than 10, but is divisible by 9. Therefore, this difference is 9. This is only possible if the appended digits are 0 and 9. Then it is easy to see that for divisibility by 3, in addition to these two digits, the digi... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. An archipelago consists of several small islands and one large island. It was decided to build bridges between the islands so that the large island would be connected to each small island by two bridges, and any two small islands would be connected by one bridge.
By November 1, all the bridges between the s... | Solution. Let's number the small islands of the archipelago. If a bridge connects islands with numbers $a$ and $b$, we write the smaller of these two numbers on this bridge.
Suppose the number of small islands in the archipelago is no more than six. Then there are no more than 5 bridges with the number 1, no more than... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. For what values of $a$ do the equations
$$
x^{2}+a x+1=0 \quad \text { and } \quad x^{2}+x+a=0
$$
have at least one common root? | Solution. Let $x_{0}$ be the common root of the equations. Substitute into the equations, equate the expressions for $x_{0}^{2}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{0}^{2} + a x_{0} + 1 = 0, \\
x_{0}^{2} + x_{0} + a = 0
\end{array}\right. \\
& (a-1) x_{0} = a-1 \quad a x_{0} + 1 = x_{0} + a \\
&
\end{alig... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $F(x)$ and $G(x)$ be polynomials of degree 2021. It is known that for all real $x$, $F(F(x)) = G(G(x))$ and there exists a real number $k, k \neq 0$, such that for all real $x$, $F(k F(F(x))) = G(k G(G(x)))$. Find the degree of the polynomial $F(x) - G(x)$. | Solution. Since $F(F(x))=G(G(x))$, then $F(k G(G(x)))=G(k G(G(x)))$. This equality holds for all $x$, i.e., for more than 2022 values of the variable (Since the polynomial $G(x)$ takes an infinite number of values). Therefore, the polynomials coincide, their difference has degree 0.
Answer. The degree of $F(x)-G(x)$ i... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 6. CONDITION
On an 8x8 chessboard, 64 checkers numbered from 1 to 64 are placed. 64 students take turns approaching the board and flipping only those checkers whose numbers are divisible by the ordinal number of the current student. A "Queen" is a checker that has been flipped an odd number of times. How many "Queen... | Solution. Obviously, each checker is flipped as many times as its number has divisors. Therefore, the number of "queens" will be the number of numbers from 1 to 64 that have an odd number of divisors, and this property is only possessed by perfect squares. Thus, the numbers of the "queens" remaining on the board will b... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. 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? | Solution. One example of four numbers that satisfy the condition of the problem is $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.
Suppose it was possible to choose five numbers. Consider the remainders of these numbers when divided by 3. If there are three identical remainders among... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
# | # Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get that $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Each of the equations $a x^{2}-b x+c=0$ and $c x^{2}-a x+b=0$ has two distinct real roots. The sum of the roots of the first equation is non-negative, and the product of the roots of the first equation is 9 times the sum of the roots of the second equation. Find the ratio of the sum of the roots of the first equatio... | # Solution.
From the condition, it follows that the coefficients $a, c \neq 0$.
By Vieta's theorem, from the condition it follows that $\frac{c}{a}=9 \frac{a}{c}$. Hence, $c^{2}=9 a^{2}$, which means $\left[\begin{array}{l}c=3 a, \\ c=-3 a .\end{array}\right.$
1 case.
$c=3 a$.
We get the equations $a x^{2}-b x+3 a... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A circle is inscribed in triangle $A B C$. Two points $E$ and $F$ are marked on the largest side of the triangle $A C$ such that $A E=A B$, and $C F=C B$. Segment $B E$ intersects the inscribed circle at points $P$ and $Q$, with $B P=1, P Q=8$. What is the length of segment $E F$? | Solution.
Let the sides of the triangle be $AB=c, AC=b, BC=a$. Then $EF=a+c-b$.
Let $G$ be the point of tangency of the inscribed circle with side $BC$. Then $BG^2 = BQ \cdot BP = 9$. There... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Vanya goes to the swimming pool every Wednesday and Friday. After one of his visits, he noticed that he had already gone 10 times this month. What will be the date of the third visit in the next month if he continues to go on Wednesdays and Fridays? | Answer: 12.
Solution. Vanya visited the swimming pool 10 times in a month, which means he visited it 5 times on Wednesdays and 5 times on Fridays.
If the first visit to the swimming pool in this month fell on a Friday, then from the first Friday to the last Friday he visited (including the first and last Friday), 29 ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.4. In parallelogram $A B C D$, the angle at vertex $A$ is $60^{\circ}$, $A B=73$, and $B C=88$. The bisector of angle $A B C$ intersects segment $A D$ at point $E$, and ray $C D$ at point $F$. Find the length of segment $E F$.
Answer, option 1. 9.
Answer, option 2. 13.
Answer, option 3. 12.
Answer, option 4. 15.
... | Solution 1. Lines $A D$ and $B C$ are parallel, therefore
$$
\angle A B C=180^{\circ}-\angle B A D=120^{\circ}
$$
(angles $A B C$ and $B A D$ are consecutive interior angles). Then $\angle A B E=60^{\circ}$ and $\angle C B E=60^{\circ}$, since $B E$ is the bisector of angle $A B C$. In triangle $A B E$, the angles at... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
3. Given a triangle $A B C$, where $\angle B A C=60^{\circ}$. Point $S$ is the midpoint of the angle bisector $A D$. It is known that $\angle S B A=30^{\circ}$. Find DC/BS. | 3. Answer: 2. Solution: We have $\mathrm{BS}=\mathrm{AS}=\mathrm{SD}$, therefore, triangle $\mathrm{ABD}$ is a right triangle (in it, the median is equal to half the side), then $\angle A C D=30^{\circ}$, hence $A D=D C$, from which we get that the required ratio is 2.
Criteria: correct solution - 7 points, proved tha... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.1. For what least natural value of \( b \) does the equation
$$
x^{2}+b x+25=0
$$
have at least one root? | Answer: 10
Solution. The equation has at least one root if and only if the discriminant $D=b^{2}-4 \cdot 25=$ $=b^{2}-100$ is greater than or equal to 0. For positive $b$, this condition is equivalent to the condition $b \geqslant 10$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. It is known that the area of the shaded region of the figure is $\frac{32}{\pi}$, and the radius of the smaller circle is 3 times smaller than the radius of the larger circle. What is the length of the smaller circle?
^{2}=9\pi R^{2}$. Therefore, the area of the shaded part is $S_{2}-S_{1}=8\pi R^{2}$. We have $8\pi R^... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. How many four-digit numbers exist that are divisible by 17 and end in 17? | 1. Let's denote the four-digit number as $\overline{x y 17}$. Then the number $\overline{x y 17}-17$ is also divisible by 17. But $\overline{x y 17}-17=100 \cdot \overline{x y}+17-17=100 \cdot \overline{x y}$. Since the numbers 100 and 17 are coprime, the two-digit number $\overline{x y}$ is divisible by 17. By enumera... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. The teacher drew the graph of the function $y=\frac{k}{x}$ and three lines with a slope of $k$ ( $k$ is not equal to zero). Petya wrote down the abscissas of all six points of intersection and multiplied them. Prove that the result does not depend on the choice of the number $k$. | 10.1. A line parallel to the line $y=k x$ has the equation $y=k x+b$ The abscissas of its intersection points with the hyperbola are both roots of the equation $\frac{k}{x}=k x+b, \quad$ equivalent to the equation $k x^{2}+b x-k=0$. The product of the roots of this equation is -1. Multiplying three such products, we ge... | -1 | Algebra | proof | Yes | Yes | olympiads | false |
2. The plane departed from Perm on September 28 at noon and arrived in Kirov at 11:00 AM (all departure and arrival times mentioned in the problem are local). At 7:00 PM the same day, the plane departed from Kirov to Yakutsk and arrived there at 7:00 AM. Three hours later, it departed from Yakutsk to Perm and returned ... | # Solution.
The plane was absent in Perm for 23 hours. Out of these, it was stationed in Kirov for 8 hours (from 11 to 19) and for 3 hours in Yakutsk. In total, out of these 23 hours, it was stationed $8+3=11$ (hours), i.e., the plane was in the air for $23-11=12$ (hours).
## Grading Criteria.
- Correct solution -7 ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. On a glade, 25 gnomes gathered. It is known that 1) every gnome who put on a hat also put on shoes; 2) 12 gnomes came without a hat; 3) 5 gnomes came barefoot. Which gnomes are more and by how many: those who came in shoes but without a hat, or those who put on a hat? | Answer. There are 6 more gnomes who put on a cap.
## Solution.
From condition 2, it follows that $25-12=13$ gnomes came in a cap.
From condition 1, we get that exactly 13 gnomes came both in a cap and in shoes.
From condition 3, it follows that a total of $25-5=20$ gnomes came in shoes.
Thus, $20-13=7$ gnomes came... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The difference of the squares of two numbers is 6, and if each of these numbers is decreased by 2, then the difference of their squares becomes equal to 18. What is the sum of these numbers? | Answer: -2.
## Solution.
Given:
\[
\begin{aligned}
& a^{2}-b^{2}=6 \\
& (a-2)^{2}-(b-2)^{2}=18
\end{aligned}
\]
There are different ways to proceed.
## Method 1.
\((a-2)^{2}-(b-2)^{2}=a^{2}-4a+4-b^{2}+4b-4=a^{2}-b^{2}-4(a-b)\). Since from the first condition \(a^{2}-b^{2}=6\), we get \(6-4(a-b)=18\). Hence, \(a-b... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. On the board, the numbers $\sqrt{2}$ and $\sqrt{5}$ are written. It is allowed to add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that it is possible to write the number 1 on the board. | Solution: The simplest way is to provide a sequence of numbers that will lead to the number 1. For example, the following sequence works:
$$
\begin{gathered}
\sqrt{2}+\sqrt{5}, \quad 2 \sqrt{2}+\sqrt{5}, \quad 3 \sqrt{2}+\sqrt{5}, \quad \sqrt{2}-\sqrt{5}, \quad 2 \sqrt{2}-\sqrt{5}, \quad 3 \sqrt{2}-\sqrt{5} \\
(\sqrt{... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
9.6. Let \(a\) and \(b\) be positive numbers. Find the minimum value of the fraction \(\frac{(a+b)(a+2)(b+2)}{16ab}\). Justify your answer. | Solution: By the inequality between the arithmetic mean and the geometric mean of two numbers, the following three inequalities hold:
$$
a+b \geqslant 2 \sqrt{a b}, \quad a+2 \geqslant 2 \sqrt{2 a}, \quad b+2 \geqslant 2 \sqrt{b}
$$
Multiplying the left and right sides of these three inequalities, we get
$$
(a+b)(a+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Matvey decided to start eating properly and every day he ate one bun less and one pear more than the previous day. In total, during the time of proper nutrition, he ate 264 buns and 187 pears. How many days was Matvey on a proper diet? | Answer: 11 days.
Solution: If we "reverse" the sequence of the number of buns, while leaving the pears unchanged, the total number of buns and pears will not change, and the difference between the number of buns and pears eaten each day will become constant. Since $264-187=77=7 \cdot 11$, the correct diet lasted eithe... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
Along a road, 10 lampposts were placed at equal distances, and the distance between the outermost posts was $k$ meters. Along another road, 100 lampposts were placed at the same distances, and the distance between the outermost posts was $m$ meters. Find the ratio $m: k$. | Answer: 11.
Solution: Let the distance between adjacent posts be $x$ meters. Then, in the first case, the distance between the outermost posts is $(10-1) x=9 x=k$ meters. And in the second case, $(100-1) x=99 x=m$ meters. Therefore, the desired ratio is $m: k=(99 x):(9 x)=11$. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Sasha had 47 sticks. Using them all, he formed several letters "B" and "V" as shown in the figure. What is the maximum number of letters "B" that Sasha could have formed?
 | Answer: 8.
Solution. To form the letter "Б", 4 sticks are needed, and to form the letter "В", 5 sticks are needed.
- Sasha could not have formed at least 12 letters "Б" because it would require no less than 48 sticks.
- If Sasha had formed 11 letters "Б", he would have $47-11 \cdot 4=3$ sticks left. This would not be... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.6. Zhenya took a $3 \times 3$ board and placed a column of blue and red cubes on each cell. Then he drew a diagram of the resulting arrangement: he labeled the number of cubes of both colors in each column (the order of the cubes is unknown).
What is the maximum number of blue cubes Zhenya can see if he look... | Answer: 12.
Solution. Let's understand the maximum number of blue cubes Zhenya can see in each of the three rows: left, middle, and right.
Left row. The first column consists of 5 cubes (2 red and 3 blue), so it completely blocks the second column, as well as 5 out of 7 cubes in the last column.
Thus, Zhenya sees al... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, it is allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal? | Answer: 11.
Solution. Suppose $N$ moves were made, after which the number of coins in all stacks became equal.
Let's slightly change the rules. Suppose initially there were not 9, 7, 5, and 10 coins in the stacks, but $N+9, N+7, N+5$, and $N+10$ respectively; and we will perform the moves as follows: instead of addin... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 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 |
4. Dad is preparing gifts. He distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest gifts, there are 50 candies. How many bags are the candies distributed into? How many candies are in the smallest gift? | Answer: 10 packages, 5 candies.
Solution. Let's number the gifts from the smallest to the largest, from 1 to $n$. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.4 Let $n$ - be a natural number greater than 10. What digit can stand immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$? Provide all possible answers and prove that there are no others. | Solution. Method 1. $n^{2}+n=(n+0.5)^{2}-0.2510$ ), the digit immediately after the decimal point is no less than 4. That is, it is 4.
Method 2. The required digit is the last digit in the number $\left[10 \sqrt{n^{2}+n}\right]=$ $\left[\sqrt{100 n^{2}+100 n}\right]$. Note that $100 n^{2}+100 n100 n^{2}+80 n+16=(10 n+... | 4 | 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 = \tan 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$). Similarly... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." That child said to their right neigh... | Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. A group of boys sitting next to each other is followed by a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so their number... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, you are allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal?
# | # Answer: in 11 moves.
Solution 1. Consider the differences between the number of coins in each of the other stacks and the number of coins in stack No. 3. In one move, either each of these differences increases by 1 (if all stacks except 3 are chosen for adding coins), or exactly 1 of these differences decreases by 1... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. 4.1. In a right triangle $ABC$ (right angle at $C$), the bisector $BK$ is drawn. Point $L$ on side $BC$ is such that $\angle CKL = \angle ABC / 2$. Find $KB$, if $AB = 18, BL = 8$. | Answer: 12.
## Solution.
Note that $\angle L K B=\angle C K B-\angle C K L=\angle C A B+\angle A B K-\angle C K L$ (the last equality holds because $\angle C K B$ is an exterior angle of triangle $A B K$). Since $\angle C K L=\angle A B C / 2=\angle A B K$, we have that $\angle L K B=\angle C A B$. From the fact that... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the rebus
$$
\mathbf{K}\mathbf{O}>\mathbf{H}>\mathbf{A}>\mathbf{B}>\mathbf{U}>\mathbf{P}>\mathbf{y}>\mathbf{C}
$$
different letters represent different digits. How many solutions does the rebus have? | Answer: 0.
Solution. From the rebus, it follows that $\mathbf{P}>\mathbf{O}>\mathbf{P}$. This cannot be! Evaluation. 7 points for the correct solution. | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Seryozha decided to start eating properly and every day he ate one fewer chocolate cookie and one more sugar-free cookie than the previous day. In total, during the time of proper nutrition, he ate 264 chocolate cookies and 187 sugar-free cookies. How many days was Seryozha on a proper diet? | Answer: 11 days.
Solution 1. The total number of cookies eaten each day is the same. In total, Seryozha ate $264+187=451=11 \cdot 41$ cookies. Since the proper diet lasted more than one day and he ate more than one cookie each day, either he ate for 11 days, 41 cookies each day, or for 41 days, 11 cookies each day. Bu... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Three bear cubs were dividing three pieces of cheese weighing 10 g, 12 g, and 15 g. A fox came to help them. She can simultaneously bite and eat 1 g of cheese from any two pieces. Can the fox leave the bear cubs equal pieces of cheese? | Answer: She can.
Solution. Let's provide one of the possible examples of how the fox could do it. For convenience, let's record the results of the fox's "work" in a table.
| 10 | 12 | 15 |
| :---: | :---: | :---: |
| 9 | 12 | 14 |
| 8 | 12 | 13 |
| 7 | 12 | 12 |
| 7 | 11 | 11 |
| 7 | 10 | 10 |
| 7 | 9 | 9 |
| 7 | 8 |... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In the underwater kingdom, there live octopuses with seven and eight legs. Those with 7 legs always lie, while those with 8 legs always tell the truth. One day, a conversation took place between three octopuses.
Green octopus: "We have 21 legs together." Blue octopus (to the green one): "You're lying!" Red octopus:... | Answer. 1) Could not.
2) The green octopus has 7 legs, the blue one has 8 legs, and the red one has 7 legs.
Solution.
1) If the green octopus had told the truth, then each octopus would have 7 legs. This means that the green octopus, according to the condition, should have lied. We get a contradiction, so the green ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 11.5. (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 |
# 3. CONDITION
In what minimum number of colors should natural numbers be painted so that any two numbers, the difference between which is 3, 4, or 6, are of different colors? | Solution. Let the number $n=1$ be color $A$, then the numbers 4, 5, and 7 must be painted in another color. Let $n=4$ be color $B$, then from $7-4=3$, it follows that the number $n=7$ is of the third color $C$. Therefore, at least 3 colors are required. The coloring $A A A B B B C C C A A A B B B \ldots$ is the desired... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by o... | Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. Thus, the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y = x - n ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by ... | # Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. This means the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y =... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were wor... | 1. Answer: by 8 percent. Solution. Let's take 1 for the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Triangle $A B C$ is similar to the triangle formed by its altitudes. Two sides of triangle $A B C$ are 4 cm and 9 cm. Find the third side. | 4. Answer. 6 cm. Solution. Let $A B=9, B C=4, A C=x, S$ - the area of triangle $A B C$. Then the heights of the triangle are $2 S / 4, 2 S / 9, 2 S / x$. By the triangle inequality, side $A C$ can be either the largest or the middle in length. Applying the triangle inequality to the triangle formed by the heights, we s... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. Six different natural numbers from 6 to 11 are placed on the faces of a cube. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 36, and the second time it was 33. What number is written on the face opposite the one with the number 10? Justify your answer. | Solution: The sum of the numbers on all 6 faces of the die is 51. Therefore, the sum of the numbers on the top and bottom faces during the first roll is 15, and during the second roll, it is 18. The number 15 can be obtained (as the sum of two different integers from the interval [6; 11]) in two ways: 15 = 9 + 6 = 8 + ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2. 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? | Answer: 8.
Solution: Note that it is impossible to lay out 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 card with the number 1. Therefore, both cards 5 and 7 must be at the edges, and the card with the number 1 must be adjacent to e... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. 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? | Answer: 5 cupcakes.
Solution: The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote P, K, and B as the costs of a cake, a cupcake, and a bagel, respectively, we get the equation: $(P + 2K + 3L) + (3P + 5) = 12K$, from which it follows that $4P + 4B = 10K$, or $2P +... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. What is the minimum number of 3-cell corners that need to be painted in a $6 \times 6$ square so that no more corners can be painted? (Painted corners must not overlap.) | Answer: 6.
Solution. 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 tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. From points $A$ and $B$, two cars set off towards each other simultaneously with constant speeds. One hour before the first car arrived at $B$ and four hours before the second car arrived at $A$, they met. Find the ratio of the speeds of the cars. | 3. Answer: the speed of the first car is twice as high.
Let the speed of one car be $k$ times the speed of the other. Since they started at the same time, one car will travel a distance $k$ times greater. After this, the segments will switch, and the slower car, whose speed is $k$ times less, will have to cover a dist... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On his birthday, Piglet baked a large cake weighing 10 kg and invited 100 guests. Among them was Winnie-the-Pooh, who is fond of sweets. The birthday boy announced the rule for dividing the cake: the first guest cuts a piece of the cake that is $1 \%$, the second guest cuts a piece of the cake that is $2 \%$ of the ... | Solution. The first guests in the queue receive increasingly larger pieces of the pie because the remaining part of the pie is large at the initial stages of division. However, since the remaining part of the pie decreases, there will come a point when guests start receiving smaller pieces of the pie. At which guest wi... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. On the website of the football club "Rostov," a poll is being conducted to determine which of the $m$ football players the website visitors consider the best at the end of the season. Each visitor votes once for one player. The website displays the rating of each player, which is the percentage of votes cast for the... | Solution. Let $a$ be the greatest loss of a percentage point due to rounding when determining a footballer's rating. Then, according to rounding rules, $aa m \geqslant 5$, which means $0.5 m>5$ or $m>10$.
We will show that a solution exists when $m=11$. For example, let $m=11$ and 73 visitors voted, with 33 of them vo... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 10.1
For all real $x$ and $y$, the equality $f\left(x^{2}+y\right)=f(x)+f\left(y^{2}\right)$ holds. Find $f(-1)$.
## Number of points 7 | Answer 0.
Solution
Substituting $x=0, y=0$, we get $f(0)=f(0)+f(0)$, that is, $f(0)=0$.
Substituting $x=0, y=-1$, we get $f(-1)=f(0)+f(1)$, that is, $f(-1)=f(1)$.
Substituting $x=-1, y=-1$, we get $f(0)=f(-1)+f(1)$. Therefore, $2 f(-1)=0$.
# | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board?
(S. Berlov)
... | # Answer. Two.
Solution. Note that no two squares of natural numbers differ by 1, because $x^{2}-y^{2}=(x-y)(x+y)$, where the second bracket is greater than one. Therefore, the numbers $a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation of the num... | # Answer. $k=6$.
Solution. By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
We will use the following known fact: t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Three brigades, working together, must complete a certain job. It is known that the first and second brigades together can complete it 36 minutes faster than the third brigade. In the time it takes for the first and third brigades to complete the job together, the second brigade can complete only half of the job. In... | Answer: 1 hour 20 minutes.
Solution.
Let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ be the productivity of the first, second, and third teams, respectively, that is, the part of the work that a team completes in 1 hour. Using the first condition of the problem, we form an equation. Since the productivity of the first and s... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. All gnomes are divided into liars and knights. Liars always lie, and knights always tell the truth. On each cell of a $4 \times 4$ board, there is a gnome. It is known that among them, there are both liars and knights. Each gnome stated: “Among my neighbors (by side) there are an equal number of liars and knights.” ... | Answer: 12 liars.
Solution: Any dwarf standing on the side of the square but not in a corner cannot be telling the truth, because they have three neighbors, and among them, there cannot be an equal number of knights and liars. Therefore, these eight dwarfs are liars. Thus, the dwarfs standing in the corners are also l... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2. A car, moving at a constant speed, traveled from point A to point B in 3 hours. To reduce the travel time on the return trip, the driver left point B at a speed 25% higher, and upon reaching the midpoint of the journey between A and B, increased the speed by another 20%. How long will the return trip take? | Answer: 2 hours 12 minutes. Solution. Let the distance from A to B be $a$ (km), and the speed of movement from A to B be $v$ (km/h). Then $\frac{a}{v}=3$. Let $\mathrm{C}$ be the midpoint of the path between A and B. Then the time of movement on the return trip from B to C is $\frac{a / 2}{v \cdot 1.25}=\frac{2}{5} \fr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. There are candies in five bags. The first has 2, the second has 12, the third has 12, the fourth has 12, and the fifth has 12. Any number of candies can be moved from any bag to any other bag. What is the minimum number of moves required to ensure that all bags have an equal number of candies? | Answer: 4.
There are 50 candies in total and they should be 10 each. In four bags, there are 12 each, which is more than 10, and these bags participate in the redistributions to reduce to 10. Therefore, there are no fewer than 4 redistributions.
From the second to the fifth bag, two candies each go to the first.
Cri... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.4 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$ if it is known that $\mathrm{a}_{1}=1,... | Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The sides of the quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ have the following lengths: $A B=9, B C=2$, $C D=14, D A=5$. Find the length of the diagonal $\boldsymbol{A} \boldsymbol{C}$, if it is known that it is an integer. | Solution. Apply the triangle inequality to $\triangle ABC$ and to $\triangle ACD$: for the first, we will have that $AB + BC > AC$, that is, $AC < AB + BC = 11$; for the second, we will have that $AC + CD > AD$, that is, $AC > CD - DA = 9$. Therefore, $9 < AC < 11$, from which $AC = 10$.
Answer: $AC = 10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Unlucky Emelya was given several metal balls, from which he broke 3 of the largest ones (their mass was $35 \%$ of the total mass of all the balls), then lost 3 of the smallest ones, and brought the remaining balls (their mass was $8 / 13$ of the unbroken ones) home. How many balls were given to Emelya | Answer: 10 balls.
## Solution:
Let the total mass of all the balls be M. Of these, Emelya broke balls with a total mass of $\frac{35}{100} M = \frac{7}{20} M$, did not break balls with a total mass of $\frac{13}{20} M$, brought home balls with a total mass of $\frac{8}{13} \cdot \frac{13}{20} M = \frac{8}{20} M$, and... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. On the road between cities A and B, there are some poles in certain places, and two numbers are written on each pole: how many kilometers from the pole to city A, and how many kilometers from the pole to city B. A tourist walking from one city to the other saw a pole where one of the numbers was three times the othe... | Answer: 7.
Solution: Without loss of generality, we can assume that the tourist saw the first post when he was closer to city A. Let the distance from the first post to A be $x$ kilometers, then the distance from it to B is $3x$. If the numbers on the second post are $y$ and $3y$, then $x + 3x = y + 3y$, since the sum... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. At competitions, an athlete's performance is evaluated by 7 judges, each of whom gives a score (an integer from 0 to 10). To obtain the final score, the best and worst scores from the judges are discarded, and the arithmetic mean is calculated. If the average score were calculated based on all seven scores, the at... | Answer: 5.
Solution: Suppose there are no fewer than six dancers. Let $A, a, S_{A}$ be the best score, the worst score, and the sum of all non-discarded scores of the winner, respectively, and $B, b, S_{B}$ be the same for the last athlete. Instead of averages, the dancers can be ranked by the sum of all scores or the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3 In a pond with carp, 30 pikes were released, which began to gradually eat each other. A pike is considered full if it has eaten three other pikes (hungry or full), and each full pike ate exactly one carp for dessert (hungry pikes did not eat carp). What is the maximum number of carp that could have been eaten? Just... | Solution. The maximum number of pikes eaten is 29, so no more than $29: 3=9.6666 \ldots$ pikes can be satiated, which means no more than 9 pikes (and, accordingly, no more than 9 roaches can be eaten). 9 roaches can be eaten as follows (the method is not unique). Choose 9 pikes and denote them as $A_{1}, A_{2}, \ldots,... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.6 On a horizontal line, points $A$ and $B$ are marked, the distance between which is 4. Above the line, two semicircles with a radius of 2 are constructed, centered at points
A and B. Additionally, one circle, also with a radius of 2, is constructed, for which the point of intersection of these semicircles is the lo... | Solution. Method 1. Let the midpoint of segment $A B$ be $E$, the center of the circle be $O$, and the second points of intersection of the circle and the semicircles be $C$ and $D$ (see figure). Since the radii of the semicircles and the radius of the circle are equal, segments $A D, E O$, and $B C$ will be equal to e... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1.
On the coordinate plane, the graphs of three reduced quadratic trinomials are drawn, intersecting the y-axis at points $-15, -6, -27$ respectively. For each trinomial, the coefficient of $x$ is a natural number, and the larger root is a prime number. Find the sum of all roots of these trinomials. | Answer: -9.
## Solution.
The parabola intersects the $O y$ axis at a point whose ordinate is equal to the free term. Therefore, our trinomials have the form $x^{2}+a x-15, x^{2}+b x-6, x^{2}+c x-27$. Let's denote their larger roots by $p, q, r$ respectively. By Vieta's theorem, the second root of the first trinomial ... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) In rectangle $A B C D$, side $A B$ is equal to 6, and side $B C$ is equal to 11. Bisectors of the angles from vertices $B$ and $C$ intersect side $A D$ at points $X$ and $Y$ respectively. Find the length of segment $X Y$.
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Alla thought of a three-digit number, in which there is no digit 0, and all digits are different. Bella wrote down the number in which the same digits are in reverse order. Galia subtracted the smaller number from the larger one. What digit stands in the tens place of the resulting difference?
# | # Answer: 9
Solution. Since we are subtracting a smaller number from a larger one, the digit in the hundreds place of the first number is greater than that of the second. Then, in the units place, conversely, the digit of the first number is smaller than that of the second. Therefore, when subtracting, we will have to... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Peter has 5 rabbit cages (the cages are in a row). It is known that there is at least one rabbit in each cage. We will call two rabbits neighbors if they sit either in the same cage or in adjacent ones. It turned out that each rabbit has either 3 or 7 neighbors. How many rabbits are sitting in the central ca... | Answer: 4 rabbits.
Solution. Let's number the cells from 1 to 5 from left to right.
Notice that the neighbors of the rabbit in the first cell are all the rabbits living in the first two cells. The neighbors of the rabbit in the second cell are all the rabbits living in the first three cells. The third cell cannot be ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. Seryozha placed numbers from 1 to 8 in the circles so that each number, except one, was used exactly once. It turned out that the sums of the numbers on each of the five lines are equal. Which number did Seryozha not use?
![](https://cdn.mathpix.com/cropped/2024_05_06_03b105aa15b6031ee02fg-3.jpg?height=326&... | Answer: 6.
Solution. Note that each number is contained in exactly two lines. Therefore, the doubled sum of all used numbers is five times greater than the sum of the numbers on one line. Thus, the sum of all numbers is divisible by 5.
Note that $1+2+3+4+5+6+7+8=36$. Therefore, for the sum of the used numbers to be d... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.1. Each of 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than $2 ”, \ldots$, the tenth said: “My number is greater than 10”. After that, all ten, sp... | Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My num... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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