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2. In triangle $A B C$, the median $B N$ is twice as short as side $A B$ and forms an angle of $20^{\circ}$ with it. Find the angle $A B C$. | Answer: $100^{\circ}$.
Solution. Let $N$ be the midpoint of segment $B D$. Then $A B C D-$ is a parallelogram. In triangle $A B D$, we have the equality of sides $A B$ and $B D$. Therefore,
$$
\angle B D A=\frac{1}{2}\left(180^{\circ}-20^{\circ}\right)=80^{\circ} \text {. }
$$
Angles $A D B$ and $C B D$ are equal as... | 100 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,409 |
4. What is the greatest length that a closed, non-self-intersecting broken line can have, which runs along the grid lines of a $6 \times 10$ cell field?
# | # Answer: 76.
Solution. We will color the nodes of the grid in a checkerboard pattern, with black and white colors. The length of a closed non-self-intersecting broken line is equal to the number of nodes it passes through. Each segment of the broken line connects a black and a white node. When traversing the broken l... | 76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,411 |
1. Compute the area of the figure defined by the inequality
$$
x^{2}+y^{2} \leqslant 4(x+y)
$$ | Answer: $8 \pi$.
Solution. If we rewrite the inequality as
$$
(x-2)^{2}+(y-2)^{2} \leqslant 8
$$
it is easy to see that it defines a circle, the square of the radius of which is 8.
Evaluation. 10 points for the correct solution. | 8\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,412 |
4. 100 balls of the same mass move along a trough towards a metal wall with the same speed. After colliding with the wall, a ball bounces off it with the same speed. Upon collision of two balls, they scatter with the same speed. (The balls move only along the trough). Find the total number of collisions between the bal... | Answer: 4950.
Solution. We will assume that each ball has a flag. Imagine that upon collision, the balls exchange flags. Then each flag flies to the wall at a constant speed, and after hitting the wall, it flies in the opposite direction. The number of collisions between the balls is equal to the number of flag exchan... | 4950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,413 |
# Problem №1 (15 points)
A car starts moving when the traffic light turns green. As it approaches the next traffic light, the red signal lights up for it again. The graph shows the dependence of the car's speed on the distance traveled. Determine the time of the car's movement between the traffic lights.
Let's find the acceleration:
$a=\frac{v^{2}}{2 S}=0.5 \, \text{m/s}^2$
The time spent on this segment:
$t_{1}=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 \cdot 900}{0.5}}=60 \, \text{s}$
On the second segment, the motion is ... | 126.7\, | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,414 |
# Problem №3 (10 points)
An electric field is created in space, the intensity of which depends on the coordinate as shown in the figure.
This field moved a particle with a mass of 1 mg and a... | # Solution:
The intensity at the point with a coordinate of 10 cm is $200 \mathrm{~B} / \mathrm{M}$
The potential difference between the final and initial points is equal to the area under the graph:
$\Delta \varphi=\left(\frac{100+200}{2} \cdot 0.05\right)+(200 \cdot 0.05)=17.5 B$
## b) points
The work done by th... | 187\mathrm{M}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,415 |
Problem 1. Let's call a number small if it is a 10-digit number and there does not exist a smaller 10-digit number with the same sum of digits. How many small numbers exist | Solution. Answer: 90.
It is clear that the sum of the digits of a 10-digit number can take any value from 1 to 90 inclusive. For each of the 90 possible sums of the digits, there is a unique smallest 10-digit number with such a sum of digits. Therefore, there are 90 small numbers.
## Criteria
One of the largest suit... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,416 |
Task 2. For real numbers $a, b, c, x, y$, it is known that
$$
\frac{1}{a+x}=6, \quad \frac{1}{b+y}=3, \quad \frac{1}{c+x+y}=2
$$
Prove that among the numbers $a, b, c$, one is equal to the sum of the other two. | Solution. From the equalities in the condition, it follows that $a+x=\frac{1}{6}, b+y=\frac{1}{3}, c+x+y=\frac{1}{2}$. Therefore, $c+x+y=\frac{1}{2}=\frac{1}{6}+\frac{1}{3}=(a+x)+(b+y)$. Then $c+x+y=a+x+b+y$, from which we get $c=a+b$, as required.
## Criteria
One of the largest suitable criteria is used:
7 p. Any c... | +b | Algebra | proof | Yes | Yes | olympiads | false | 4,417 |
Problem 3. On the side $BC$ of an acute-angled triangle $ABC$, a point $D$ is chosen such that $AB + BD = DC$. Prove that $\angle ADC = 90^{\circ}$, given that $\angle B = 2 \angle C$. | Solution. Mark point $M$ on the extension of side $B C$ beyond point $B$ such that $A B = B M$. In the isosceles triangle $A B M$, the external angle at vertex $B$ is equal to the sum of the two equal angles at the base $A M$, so $\angle A M B = \frac{\angle B}{2} = \angle C$. Therefore, triangle $A M C$ is isosceles a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,418 |
Problem 4. On the board, there are $N$ natural numbers, where $N \geqslant 5$. It is known that the sum of all the numbers is 80, and the sum of any five of them is no more than 19. What is the smallest value that $N$ can take? | Solution. Answer: 26.
The condition of the problem is equivalent to the sum of the five largest numbers not exceeding 19, and the sum of all numbers being 80. Note that among the five largest numbers, there must be a number not greater than 3 (otherwise, if all of them are not less than 4, their sum is not less than $... | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,419 |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1010 of them such that the sum of any two does not equal 2021 or 2022. How many
ways are there to do this | Solution. Answer: $\frac{1011 \cdot 1012}{2}=511566$.
Let's write our numbers in the following order: $2021,1,2020,2,2019,3, \ldots, 1012,1010,1011$.
Notice that if two numbers sum to 2021 or 2022, they stand next to each other in this sequence. Our task is reduced to the following: 2021 objects (for convenience, let... | 511566 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,420 |
Problem 1. Petya was given several physics problems and several math problems for homework. All the problems solved by Petya constitute $5 \%$ of the total number of physics problems and $20 \%$ of the total number of math problems. What percentage of the total number of problems did Petya solve? | Solution. Answer: $4 \%$.
Let Pete solve $N$ problems. This constitutes $5 \%$ ( $\frac{1}{20}$ of) the physics problems, so the total number of physics problems was $20 \mathrm{~N}$. Similarly, the number of math problems was $5 N$, so the total number of problems assigned was $20 N + 5 N = 25 N$. The problems solved... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,421 |
Problem 2. In a two-digit number, each digit was increased by 2 or by 4 (different digits could be increased by different numbers), as a result of which the number increased fourfold. What could the original number have been? Find all possible options and prove that there are no others. | Solution. Answer: 14.
Let $x$ be the original number, then the resulting number equals $4 x$. In this case, after increasing two digits, the number itself was increased by 22, 24, 42, or 44. This results in four cases:
- $4 x-x=22$
- $4 x-x=24$
- $4 x-x=42$
- $4 x-x=44$.
Among them, only the third one fits, correspo... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,422 |
Problem 3. In an acute-angled triangle $ABC$, the altitude $AD$ is drawn. It turns out that $AB + BD = DC$. Prove that $\angle B = 2 \angle C$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution. Mark point $M$ on the extension of side $B C$ beyond point $B$ such that $A B = B M$. Then, in the isosceles triangle $A B M$, the external angle at vertex $B$ is equal to the sum of the two equal angles at the base $A M$, so $\angle A B C = 2 \angle A M B$.
The triangle $A M C$ is also isosceles, since segm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,423 |
Problem 5. A $9 \times 9$ table is divided into nine $3 \times 3$ squares. Petya and Vasya take turns writing the numbers from 1 to 9 in the cells of the table according to the rules of Sudoku, meaning that no two identical numbers should appear in any row, any column, or any of the nine $3 \times 3$ squares. Petya sta... | Solution. Answer: Petya.
We will present a winning strategy for Petya. His first move is to place the number 5 in the center of the table (let's call this cell $C$), and then, if Vasya places a number $a$ in some cell, Petya will place the number $10-a$ in the cell symmetric to Vasya's cell relative to the center of t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,424 |
Problem 3. We will call a natural number odd-powered if all its prime divisors enter its factorization with an odd exponent. What is the maximum number of odd-powered numbers that can occur consecutively?
## Answer: 7. | Solution. Note that among any eight consecutive natural numbers, there will definitely be a number that is divisible by 4 but not by 8. Then the number 2 will appear in its factorization to the second power.
Example of seven odd-power numbers: $37,38,39,40,41,42,43$.
Problem 4/1. On side $A C$ of triangle $A B C$, a ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,425 |
Problem 1. Grandfather Frost had 120 chocolate candies and 200 jelly candies. At the morning performance, he gave candies to the children: each child received one chocolate candy and one jelly candy. Counting the candies after the performance, Grandfather Frost found that there were three times as many jelly candies le... | Answer: 80.
Solution. Let the total number of children be $x$, then after the morning party, Grandfather Frost had $120-x$ chocolate candies and $200-x$ jelly candies left. Since there were three times as many jelly candies left as chocolate candies, we get the equation $3 \cdot(120-x)=200-x$. Solving this, we get $x=... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,426 |
Task 2. On an island, there live knights who always tell the truth, and liars who always lie. One day, 30 inhabitants of this island sat around a round table. Each of them said one of two phrases: "My left neighbor is a liar" or "My right neighbor is a liar." What is the smallest number of knights that can be at the ta... | # Answer: 10.
Solution. Next to each liar, there must be at least one knight (otherwise, if liars sit on both sides of a liar, the statement made by the liar would definitely be true). Therefore, among any three consecutive residents, there is at least one knight. If we divide all the people sitting at the table into ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,427 |
Problem 3. Five different natural numbers are written in a circle on the board. Each of them, Petya divided by the next one in the clockwise direction, and then wrote down the 5 resulting numbers (not necessarily integers) on a piece of paper. Can the sum of the 5 numbers on the paper be an integer? | Answer: Yes, it can.
Solution. For example, for the numbers $1,2,4,8,16$, after division, the sum is $1 / 2+1 / 2+1 / 2+1 / 2+16=18$.
## Criteria
One suitable criterion is used:
7 p. Any complete solution to the problem.
0 p. There is only the correct answer, but the correct example is not provided. | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,428 |
Problem 4. Carlson has three boxes, each containing 10 candies. One box is labeled with the number 4, another with 7, and the third with 10. In one operation, Carlson sequentially performs the following two actions:
- he takes from any box a number of candies equal to the number written on it;
- he eats 3 of the taken... | Answer: 27.
Solution. Since at each step, Karlson eats 3 candies, the total number of candies eaten is divisible by 3. We will prove that it does not exceed 27; for this, it is sufficient to show that it cannot equal 30, that is, Karlson cannot eat all the candies. Indeed, if this were possible, then before the last o... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,429 |
Problem 5. A grid rectangle $42 \times 44$ was cut along the grid lines into rectangles $1 \times 8$, one square $2 \times 2$, and one tetromino. Prove that this tetromino is also a square. (All possible tetrominoes are shown in the figure below, they can be rotated and flipped.)
 that intersects this tetromino at exactly one cell. Now let's paint black the entire row, as well as all rows that are at a distance from it that is a multiple of four (for example, if the 11th row is chosen, the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 4,430 |
2. Quadratic trinomials $P(x)=x^{2}+\frac{x}{2}+b$ and $Q(x)=x^{2}+c x+d$ with real coefficients are such that $P(x) Q(x)=Q(P(x))$ for all $x$. Find all real roots of the equation $P(Q(x))=0$. | 2. Quadratic trinomials $P(x)=x^{2}+\frac{x}{2}+b$ and $Q(x)=x^{2}+c x+d$ with real coefficients are such that $P(x) Q(x)=Q(P(x))$ for all $x$. Find all real roots of the equation $P(Q(x))=0$.
Answer: $x \in\left\{-1, \frac{1}{2}\right\}$.
In the equation $P(x) Q(x)=Q(P(x))$, expand the brackets and equate the coeffi... | x\in{-1,\frac{1}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,432 |
3. Does there exist a set of numbers $x_{1}, x_{2}, \ldots, x_{99}$ such that each of them is equal to $\sqrt{2}+1$ or $\sqrt{2}-1$ and the equality
$$
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+\ldots+x_{98} x_{99}+x_{99} x_{1}=199 ?
$$ | 3. Does there exist a set of numbers $x_{1}, x_{2}, \ldots, x_{99}$ such that each of them is equal to $\sqrt{2}+1$ or $\sqrt{2}-1$ and the equality
$$
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+\ldots+x_{98} x_{99}+x_{99} x_{1}=199
$$
holds?
Answer: No, such a set does not exist.
Assume the contrary: let such a set of nu... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,433 |
5. The incircle of triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $I_{A}$ is the excenter of the excircle opposite side $BC$ of triangle $ABC$, and points $K$ and $L$ are the midpoints of segments $DI_{A}$ and $EI_{A}$ respectively. Lines $BK$ and $CL$ intersect at point $F$, which... | 5. The circle inscribed in triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$ respectively. Point $I_{A}$ is the center of the excircle opposite side $BC$ of triangle $ABC$, and points $K$ and $L$ are the midpoints of segments $DI_{A}$ and $EI_{A}$ respectively. Lines $BK$ and $CL$ intersect at point $F$,... | 130 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,435 |
6.2. Masha and the Bear ate a basket of raspberries and 40 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate both raspberries and pies 3 times faster than Masha. How many pies did Masha eat, if they ate t... | Answer: 4 pies. Solution: The bear ate his half of the raspberries three times faster than Masha. This means Masha ate pies for three times less time than the bear. Since she eats three times slower, she ate 9 times fewer pies than the bear. Dividing the pies in the ratio of $9: 1$, we see that Masha got the 10th part,... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,437 |
6.3. At a round table sit 10 elves, each with a basket of nuts in front of them. Each was asked, "How many nuts do your two neighbors have together?" and, going around the circle, the answers received were 110, 120, 130, 140, 150, 160, 170, 180, 190, and 200. How many nuts does the elf who answered 160 have? | Answer: 55. Solution. We will call the elves 1-m, 2-m, etc., in the order of receiving answers. The odd-numbered ones sit every other. In their answers, the number of nuts of each even number is counted twice, so the sum $110+130+150+170+190=750$ is equal to twice the number of nuts of all even numbers. Therefore, the ... | 55 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,438 |
6.4. A natural number is called a palindrome if it remains unchanged when its digits are written in reverse order (for example, the numbers 4, 55, 626 are palindromes, while 20, 201, 2016 are not). Represent the number 2016 as a product of two palindromes (find all options and explain why there are no others). | Answer: 8.252. Solution: The number 2016 is not divisible by 11. Therefore, none of the palindromes are two-digit (they all are divisible by 11). This means one of the palindromes is a single-digit (if both factors have at least 3 digits, the product is no less than $100 \cdot 100 > 2016$). The number 2016 has only the... | 8\cdot252 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,439 |
6.5. 30 students are walking in pairs, with students of different heights in each pair. Prove that they can stand in a circle so that the height of each differs from the height of their neighbors. | Solution. Let one of the students be called Petya, and all other students of Petya's height be called Vasya. We will build a row from left to right, adding pairs. First, we place Petya with his partner, with Petya on the left. If there is a pair with Vasya, we place it next, with Vasya also on the left. Then to the lef... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,440 |
Problem 9.1. Fisherman Vasya caught several fish. The three largest fish, accounting for $35 \%$ of the total catch weight, he put in the refrigerator. The three smallest, accounting for $5 / 13$ of the weight of all the remaining fish, the fisherman gave to the cat. Vasya ate all the rest of the fish he caught. How ma... | Answer: 10.
Solution. From the condition, it follows that the cat got $65 \% \cdot \frac{5}{13}=25 \%$ of the weight of the catch. Therefore, Vasya ate $40 \%$ of the caught fish. If we denote by $x$ the number of fish eaten by Vasya, then the average weight of the eaten fish as a percentage of the total catch weight ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,441 |
Problem 9.2. At a party, 24 people gathered. A guest is considered an introvert if they have no more than three acquaintances among the other guests. It turned out that each guest has at least three acquaintances who are introverts. How many introverts could there have been at the party? Provide all possible answers an... | # Answer: 24.
Solution. Let $a$ be the number of pairs of introverts who know each other, and $b$ be the number of pairs of acquaintances where one of the pair is an introvert. Each person who came to the party is included in at least three pairs, since they know at least three other introverts, and pairs of introvert... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,442 |
Problem 9.3. A circle $\omega$ with center at point $I$ is inscribed in a convex quadrilateral $ABCD$ and touches side $AB$ at point $M$, and side $CD$ at point $N$, with $\angle BAD + \angle ADC < 180^{\circ}$. A point $K \neq M$ is chosen on the line $MN$ such that $AK = AM$. In what ratio can the line $DI$ divide th... | Answer: $1: 1$.
Solution. Let $P$ be the intersection point of lines $A B$ and $C D$ (such a point exists by the condition, otherwise
Fig. 1: to the solution of problem 9.3
$\left.\angle B... | 1:1 | Geometry | proof | Yes | Yes | olympiads | false | 4,443 |
Problem 9.4. It is known that the number 400000001 is the product of two prime numbers $p$ and $q$. Find the sum of the natural divisors of the number $p+q-1$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: 45864.
Solution. The number $n=400000001$ can be written in the following form
\[
\begin{aligned}
n & =4 \cdot 10^{8}+1=4 \cdot 10^{8}+4 \cdot 10^{4}+1-4 \cdot 10^{4}= \\
& =\left(2 \cdot 10^{4}+1\right)^{2}-\left(2 \cdot 10^{2}\right)^{2}=\left(2 \cdot 10^{4}+2 \cdot 10^{2}+1\right)\left(2 \cdot 10^{4}-2 \cd... | 45864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,444 |
Problem 9.5. Positive numbers $a, b$, and $c$ satisfy the equations
$$
a^{2}+a b+b^{2}=1, \quad b^{2}+b c+c^{2}=3, \quad c^{2}+c a+a^{2}=4
$$
Find $a+b+c$. | Answer: $\sqrt{7}$.
## Kurchatov Mathematics Olympiad for School Students - 2019 Final Stage.
Solution. From a single point $T$, we lay out segments $T A, T B$, and $T C$ with lengths $a, b$, and $c$ respectively, such that $\angle A T B = \angle B T C = \angle C T A = 120^{\circ}$. Then, by the cosine theorem, takin... | \sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,445 |
4. Alexei wrote several consecutive natural numbers on the board. It turned out that only two of the written numbers have a digit sum divisible by 8: the smallest and the largest. What is the maximum number of numbers that could have been written on the board? | Answer: 16.
Example: numbers from 9999991 to 10000007.
Evaluation. We will prove that there cannot be more than 16 numbers. Let's call a number $x$ good if the sum of its digits is divisible by 8. Among the listed numbers, there must be a number $x$ ending in 0, otherwise there are too few numbers. The number $x$ is ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,446 |
Problem 11.1. Petya and Vasya participated in the election for the position of president of the chess club. By noon, Petya had $25 \%$ of the votes, and Vasya had $45 \%$. After noon, only Petya's friends came to vote (and, accordingly, voted only for him). In the end, Vasya was left with only $27 \%$ of the votes. Wha... | Answer: $55 \%$.
Solution. Let $x$ be the number of people who voted before noon, and $y$ be the number of people who voted after. Then, $0.45 x$ people voted for Vasya, which constitutes $27 \%$ of $x+y$. Thus, we have the equation $0.45 x = 0.27(x + y)$, from which $2 x = 3 y$. According to the condition, Petya rece... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,447 |
Task 11.2. Find all pairs of natural numbers $m$ and $n$ such that $m^{2019} + n$ is divisible by $m n$. | Answer: $m=1, n=1$ and $m=2, n=2^{2019}$.
Solution. From the condition, it follows that $m^{2019}+n$ is divisible by $m$, hence $n$ is divisible by $m$. Denoting $n=m n_{1}$, we conclude that $m^{2018}+n_{1}$ is divisible by $m n_{1}$. Proceeding similarly: we deduce that $n_{1}$ is a multiple of $m$, introduce the no... | =1,n=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,448 |
Problem 11.3. There are 100 pies arranged in a circle, 53 of which are with cabbage, and the rest are with rice. Alexei knows which ones are with what, and he wants to choose 67 consecutive pies such that among them there are exactly $k$ with cabbage. For which $k$ is he guaranteed to be able to do this regardless of t... | Answer: 35 or 36.
Solution. A set of 67 consecutive pies will be called a segment. Two segments will be called adjacent if the sets of pies in them differ by only a pair of extreme ones, that is, the segments differ by a shift of one pie. Then, in two adjacent segments, the number of pies with cabbage obviously differ... | 35or36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,449 |
Problem 11.4. For positive numbers $x$ and $y$, it is known that
$$
\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}=1
$$
What values can the product $x y$ take? List all possible options and prove that there are no others. | Answer: 1.
Solution. Note that for each positive $y$ the function
$$
f_{y}(x)=\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}
$$
strictly decreases on the ray $(0 ;+\infty)$, since the denominators of all three fractions increase. Therefore, the function $f_{y}$ takes each value no more than once. Moreover, ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,450 |
Problem 11.5. Determine the number of possible values of the product $a \cdot b$, where $a, b-$ are integers satisfying the inequalities
$$
2019^{2} \leqslant a \leqslant b \leqslant 2020^{2}
$$
Answer: $\mathrm{C}_{2 \cdot 2019+2}^{2}+2 \cdot 2019+1=2 \cdot 2019^{2}+5 \cdot 2019+2=8162819$. | Solution. Suppose that for some two different pairs $a \leqslant b$ and $a_{1} \leqslant b_{1}$ from the interval $\left[2019^{2} ; 2020^{2}\right]$, the equality $a b=a_{1} b_{1}$ holds. Without loss of generality, assume that $a_{1} < a$. Then
$$
a-a_{1} \geqslant 2019
$$
and
$$
b-\frac{a_{1}+b_{1}}{2} \geqslant 2... | 8162819 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,451 |
9.1. The sum of the cube and the square of some non-integer number is known. Can the sign of the original number always be determined? | Answer. No, for example, the numbers $1 / 3$ and - $-3 / 3$ give the same sum 4/27.
Criterion. Counterexample 7 points. No justification required. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,453 |
9.3. Given an isosceles triangle \(ABC\). On the lateral side \(AB\), a point \(M\) is marked such that \(CM = AC\). Then, on the lateral side \(BC\), a point \(N\) is marked such that \(BN = MN\), and the angle bisector \(NH\) of triangle \(CNM\) is drawn. Prove that \(H\) lies on the median \(BK\) of triangle \(ABC\)... | Solution. In isosceles triangles $A C M$ and $M N B$, $\angle A M C = \angle A$ and $\angle B M N = \angle B$. Therefore, $\angle C M N = 180^{\circ} - \angle A M C - \angle B M N = 180^{\circ} - \angle A - \angle B = \angle C = \angle A = \angle A M C$. Thus, $H$ is the intersection point of the angle bisectors of $\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,454 |
9.4. Through the point with coordinates $(9,9)$, lines (including those parallel to the coordinate axes) are drawn, dividing the plane into angles of $9^{\circ}$. Find the sum of the x-coordinates of the points of intersection of these lines with the line $y=10-x$. | Answer: 190. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(9,9)$, 20 lines are drawn, the line $y=20-x$ intersects 19 of them. For each point on the line $y=20-x$, the sum of the coordinates is 20, so the total su... | 190 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,455 |
9.5. There are 64 checkers of three colors, divided into pairs such that the colors of the checkers in each pair are different. Prove that all the checkers can be arranged on a chessboard so that the checkers in each two-square rectangle are of different colors. | Solution. Let's distribute the checkers into three monochromatic piles. In each pile, there will be no more than 32 checkers. Now, let's distribute the largest pile on white squares, filling the rows starting from the top. The second largest pile will be distributed on black squares, filling the rows starting from the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,456 |
Problem 7.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, To... | Answer: 4.
Solution. Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,457 |
Task 7.2. Solve the puzzle
$$
A B C D E F \cdot 3 = B C D E F A
$$
List all the options and prove that there are no others. | Answer: $142857 \cdot 3=428571$ and $285714 \cdot 3=857142$.
Solution. Let the five-digit number $BCDEF$ be denoted by $x$. Then this puzzle can be rewritten as the relation $(100000 \cdot A + x) \cdot 3 = 10x + A$, which is equivalent to the equation $29999 \cdot A = 7x$, or $42857 \cdot A = x$.
Notice that digits $... | 142857\cdot3=428571285714\cdot3=857142 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,458 |
Problem 7.3. The miserly knight has 5 chests of gold: the first chest contains 1001 gold coins, the second - 2002, the third - 3003, the fourth - 4004, the fifth - 5005. Every day, the miserly knight chooses 4 chests, takes 1 coin from each, and places them in the remaining chest. After some time, there were no coins l... | Answer: in the fifth.
Solution. Let's look at the remainders of the number of coins in the chests when divided by 5. Each day, the number of coins in each chest either decreases by 1 or increases by 4, which means the remainder when divided by 5 always decreases by 1 (if it was 0, it will become 4). Since after some t... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,459 |
Problem 7.4. On the sides $AB$ and $AC$ of triangle $ABC$, points $X$ and $Y$ are chosen such that $\angle A Y B = \angle A X C = 134^{\circ}$. On the ray $YB$ beyond point $B$, point $M$ is marked, and on the ray $XC$ beyond point $C$, point $N$ is marked. It turns out that $MB = AC$ and $AB = CN$. Find $\angle MAN$.
... | Solution. Note that $\angle A C N=\angle C A X+\angle A X C=\angle B A Y+\angle A Y B=\angle A B M$. Adding
Fig. 1: to the solution of problem $?$ ?
## Kurchatov School Olympiad in Mathemat... | 46 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,460 |
Problem 7.5. A lame rook makes moves alternating between one and two squares, with the direction of the move being freely chosen (in any of the four directions). What is the maximum number of cells on a $6 \times 6$ board it can visit, if visiting the same cell twice is prohibited, but the starting cell and the first m... | Answer: 34.
Solution. An example for 34 cells is shown in Fig. ??. The visited cells are numbered from 1 to 34.
| 5 | 2 | 4 | 3 | 13 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 6 | 1 | 7 | 8 | 14 | 9 |
| 28 | 31 | | 34 | 12 | 11 |
| 27 | 32 | | 33 | 15 | 16 |
| 29 | 30 | 20 | 19 | 21 | 18 |
| 26 | 25... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,461 |
Task 1. Ant Boris moves on a coordinate plane, starting from point $P_{0}=(0,0)$, moving to point $P_{1}=(1,0)$, and then along a spiral counterclockwise (see figure).
The points with integer... | 1. Ant Borya moves on a coordinate plane, starting from point $P_{0}=$ $(0,0)$, moving to point $P_{1}=(1,0)$, and further along a spiral counterclockwise (fig.).
The points with integer coor... | (20,17) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,462 |
Problem 2. In a popular intellectual game "Clash of Minds," 10 people participate. By the end of the game, each player scores an integer number of points. It turned out that in the semifinal, all the quantities of points scored by the players have different last digits. Prove that in the final of the game, where the pl... | 2. In a popular intellectual game "Clash of Minds," 10 people participate. By the end of the game, each player scores an integer number of points. It turned out that in the semifinal, all the quantities of points scored by the players have different last digits. Prove that in the final of the game, where the players co... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,463 |
Problem 3. In an acute-angled triangle $A B C$, the altitude $A H$ is drawn. Let $P$ and $Q$ be the feet of the perpendiculars dropped from point $H$ to sides $A B$ and $A C$ respectively. Prove that $\angle B Q H=\angle C P H$. | 3. In an acute triangle $ABC$, the altitude $AH$ is drawn. Let $P$ and $Q$ be the feet of the perpendiculars dropped from point $H$ to sides $AB$ and $AC$, respectively. Prove that $\angle B Q H = \angle C P H$.
$, where $p$ is a prime number. This means that one of the numbers $x$ and $y$ is divisible by $p$. Let's consider both cases.
First, assume that $y=m p$. Then $x m^{3... | 14,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,465 |
Problem 5. In a physics competition, 17 schoolchildren are participating. The participants were offered 12 problems. As a result, each problem was solved correctly by more than half of the participants. Prove that there will definitely be three schoolchildren who, together, have solved all the problems.
http://olimpia... | 5. In a physics competition, 17 schoolchildren are participating. The participants were given 12 problems. As a result, each problem was solved correctly by more than half of the participants. Prove that there will definitely be three schoolchildren who, together, have solved all the problems.
Let's estimate the numbe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,466 |
Task 1. A grid table is given with a width of 300 and a height of 50 cells. A plastic snail is placed in the bottom-left corner, facing upwards along the table, and begins to move one cell in the direction it is facing. If the next cell does not exist, meaning the snail is at the edge of the board, or if the next cell ... | Solution. Answer: 25th row, 26th column.
We will denote a cell $(x, y)$ if it is in row $x$ and column $y$.
Initially, the snail is in cell $(1,1)$. After several moves, it will completely walk around the edge and end up in cell $(2,2)$. Notice that at this moment, the configuration of the problem repeats: the snail ... | 25th\,row,\,26th\,column | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,467 |
Problem 2. The numbers $d$ and $e$ are the roots of the quadratic trinomial $a x^{2}+b x+c$. Could it be that $a, b, c, d, e$ are consecutive integers in some order? | Solution. Answer: Yes.
This could have happened, for example, for the quadratic polynomial $2 x^{2}-2$ and its roots 1 and -1. The numbers $-2,-1,0,1,2$ are indeed consecutive integers.
## Criteria
The one most suitable criterion is used:
7 6. Any example satisfying the condition of the problem is given. | Yes | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,468 |
Problem 3. In a five-digit number, each digit was increased by 2 or by 4 (different digits could be increased by different numbers), as a result of which the number increased fourfold. What could the original number have been? Find all possible options and prove that there are no others. | Solution. Answer: 14074.
Let the original five-digit number be denoted by $N$, then the resulting number is equal to $4N$. Their difference is $3N$ and is a five-digit number consisting of 2 and 4. This difference $3N$ is not less than $3 \cdot 10000$, so it starts with a four.
By the divisibility rule for 3, the sum... | 14074 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,469 |
Problem 4. On the side $AC$ of triangle $ABC$, a point $M$ is marked such that $AM = AB + MC$. Prove that the perpendicular to $AC$ passing through $M$ bisects the arc $BC$ of the circumcircle of $ABC$. | Solution. Mark a point $X$ on the segment $A M$ such that $X M = M C$. Then, from the condition of the problem, it follows that $A X = A B$.
Consider the center of the circumscribed circle of triangle $B C X$ - it is located at the intersection of the perpendicular bisectors of segments $B C$ and $B X$. The perpendicu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,470 |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1010 of them such that the sum of any two does not equal 2021 or 2022. How many ways are there to do this? | Solution. Answer: $\frac{1011 \cdot 1012}{2}=511566$.
Let's write our numbers in the following order: $2021,1,2020,2,2019,3, \ldots, 1012,1010,1011$. Notice that if two numbers sum to 2021 or 2022, they stand next to each other in this sequence. Our task is reduced to the following: 2021 objects (for convenience, let'... | 511566 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,471 |
Problem 1. Petya has a $3 \times$ 3 table. He places chips in its cells according to the following rules:
- no more than one chip can be placed in each cell;
- a chip can be placed in an empty cell if the corresponding row and column already contain an even number of chips (0 is considered an even number).
What is th... | Solution. Answer: 9.
Let $a, b, c$ denote the left, middle, and right columns of the table, and $1, 2, 3$ denote the bottom, middle, and top rows of the table. Petya can fill all 9 cells of the table with chips, for example, in the following order: $a 1, b 2, c 3, a 2, b 3, c 1, a 3, b 1, c 2$. Obviously, it is not po... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,472 |
Problem 3. A sage has 11 visually identical diamonds: 10 ordinary ones and 1 magical one. The sage knows that all ordinary diamonds weigh the same, while the magical one differs in weight. The sage also has a balance scale, on which two piles of diamonds can be compared.
How can the sage determine in two weighings whe... | Solution. Let's present one of the possible algorithms of the sage.
Number the diamonds from 1 to 11. Place diamonds $1,2,3$ on one scale and $-4,5,6$ on the other.
First case. One of the scales tipped. Without loss of generality, the scale with $1,2,3$ was heavier. Then among the diamonds $1, \ldots, 6$ there is def... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,473 |
Problem 4. On the board, the expression is written
$$
7 * 6 * 5 * 4 * 3 * 2 * 1
$$
Masha replaces the asterisks with the signs «+» and «-». In one move, Vanya can change two consecutive signs to their opposites. Vanya wants to achieve, after several of his moves, an expression whose value is divisible by 7. Can Masha... | Solution. Answer: cannot.
We will show how Vanya can achieve an expression divisible by 7. The sign between 7 and 6 can be considered a plus (if it is a minus, we will change it and the next sign). Similarly, we can consider the sign between 6 and 5, then between 5 and 4, then between 4 and 3, and then between 3 and 2... | 2114 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,474 |
Problem 5. The numbers from 1 to 2021 are written on a board. Denis wants to choose 1011 of them such that the sum of any two does not equal 2021 or 2022. How many ways are there to do this? | Solution. Answer: 1.
Among the numbers from 1 to 2021, we can select 1010 non-overlapping pairs of numbers that sum to 2022. Specifically: $(1,2021),(2,2020), \ldots,(1009,1013),(1010,1012)$; only the number 1011 remains unpaired.
According to the condition, from each pair, we can choose no more than one number. Ther... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,475 |
Problem 1. In the Kurchatov School, exactly 2 people sit at each desk. It is known that exactly $70 \%$ of the boys have a boy as a desk partner, and exactly $40 \%$ of the girls have a girl as a desk partner. How many times more boys are there than girls? | Answer: 2 times.
Solution. Let the number of boys be $x$, and the number of girls be $y$. Note that $30\%$ of the boys sit at desks with girls and $60\%$ of the girls sit at desks with boys. Since exactly 2 people sit at each desk, then $0.3 x = 0.6 y$, from which $x = 2 y$. Thus, there are 2 times more boys than girl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,476 |
Task 2. Find the number of ways to color all natural numbers from 1 to 20 in blue and red such that both colors are used and the product of all red numbers is coprime with the product of all blue numbers. | Answer: $2^{6}-2=62$ ways.
Solution. Note that all even numbers must be of the same color. Since among them are the numbers 6, 10, and 14, the numbers divisible by 3, 5, and 7 must be of the same color. The remaining numbers are $1, 11, 13, 17$, and 19. Note that they can be distributed in any way among the two colors... | 62 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,477 |
Problem 3. On the board, the numbers $2,3,5, \ldots, 2003,2011,2017$ are written, i.e., all prime numbers not exceeding 2020. In one operation, two numbers $a, b$ can be replaced by the largest prime number not exceeding $\sqrt{a^{2}-a b+b^{2}}$. After several operations, only one number remains on the board. What is t... | Answer: The maximum value is 2011.
Solution. Note that if $a$ $2011^{2}$, therefore, as a result of this operation, the number 2011 will appear on the board.
Remark. The inequality $a<\sqrt{a^{2}-a b+b^{2}}<b$ for $a<b$ can be proven geometrically. The matter is that in a triangle with sides $a$ and $b$ and an angle ... | 2011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,478 |
Problem 4. Let $S A B C D$ be a regular quadrilateral pyramid with base $A B C D$. On the segment $A C$, there is a point $M$ such that $S M = M B$ and the planes $S B M$ and $S A B$ are perpendicular. Find the ratio $A M: A C$. | Answer: $3: 4$.
Solution. Let the center of the base $A B C D$ be $O$. Then $S O \perp A C$ and $B D \perp A C$, which means $S B \perp A C$. This implies that the perpendicular bisector of the segment $S B$ (let's denote this plane as $\alpha$) is parallel to $A C$. On the other hand, from $S M = M B$, it follows tha... | 3:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,479 |
Problem 5. Prove that for a natural number $n>2$, the numbers from 1 to $n$ can be divided into two sets such that the products of the numbers in the sets differ by no more than $\frac{n-1}{n-2}$ times. | Solution. We will prove this statement by induction on $n$.
Base for $n=3,4,5$.
For $n=3$, we divide into sets $\{1,2\}$ and $\{3\}$, the ratio is $\frac{3}{2}$, which is less than $\frac{2}{1}$.
For $n=4$, we divide into sets $\{1,2,3\}$ and $\{4\}$. The ratio will be $\frac{6}{4}$, which is exactly $\frac{3}{2}$.
... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,480 |
Problem 6. Prove that there exist such sequences of natural numbers $a_{n}$ and $b_{n}$ that the following conditions are simultaneously satisfied:
- sequences $a_{n}$ and $b_{n}$ are non-decreasing;
- sequences $A_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}$ and $B_{n}=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\... | Solution. Consider the sequence $c_{k}=2^{k}$. It is clear that all sums
$$
C_{n}=\frac{1}{c_{1}}+\frac{1}{c_{2}}+\ldots+\frac{1}{c_{n}}
$$
are bounded. We will construct the original sequences $a_{n}$ and $b_{n}$ such that $\max \left(a_{n}, b_{n}\right)=c_{n}$. We will sequentially divide the natural number sequenc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,481 |
1. On a plane, an overlapping square and a circle are drawn. Together they occupy an area of 2018 cm². The area of intersection is 137 cm². The area of the circle is 1371 cm². What is the perimeter of the square? | Answer: 112 cm.
The area of the part of the circle outside the square is $1371-137=1234$ cm $^{2}$, therefore, the area of the square can be expressed by the formula $2018-1234=784 \mathrm{~cm}^{2}$. In conclusion, the length of the side of the square is $\sqrt{784}=28$ cm, and its perimeter is 112 cm.
## Criteria
+... | 112 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,482 |
2. In front of Vasya lies a stack of 15 red, 15 blue, and 15 yellow cards. Vasya needs to choose 15 out of all 45 cards to earn the maximum number of points. Points are awarded as follows: for each red card, Vasya gets one point. For each blue card, Vasya gets a number of points equal to twice the number of selected re... | Answer: 168.
Let the number of red, blue, and yellow cards taken by Vasya be denoted by $\mathrm{K}$, C, and Y, respectively. The number of points Vasya will receive is given by the formula
$$
\mathrm{K}+2 \cdot \mathrm{K} \cdot \mathrm{C}+3 \cdot \mathrm{C} \cdot \mathrm{K}=\mathrm{K} \cdot(2 \mathrm{C}+1)+\mathrm{K... | 168 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,483 |
3. Find all pairs of prime numbers $p$ and $q$, for which $p^{2}+p q+q^{2}$ is a perfect square. | Answer: $(3,5)$ and $(5,3)$.
The equality $p^{2}+p q+q^{2}=a^{2}$, implied in the problem, can be rewritten as follows:
$$
p q=(p+q)^{2}-a^{2}=(p+q-a)(p+q+a) \text {. }
$$
Since the factor $p+q+a$ on the right side is greater than both $p$ and $q$, and using the fact that $p$ and $q$ are prime numbers, we conclude t... | (3,5)(5,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,484 |
Problem 1. In 20 boxes, 60 black and 60 white balls were distributed - 6 balls in each. Vanya noticed that in each of the first 14 boxes, there were more black balls than white ones. Is it true that among the last 6 boxes, there will definitely be one in which all the balls are white? | Answer: Yes, it is correct.
Solution. From the condition, it follows that each of the first 14 boxes contains at least 4 black balls. Therefore, there are at least $14 \cdot 4=56$ black balls in the first 14 boxes. If there were at least one black ball in each of the remaining 6 boxes, then the total number of black b... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,485 |
Problem 2. Gosha entered a natural number into the calculator. Then he performed the following operation, consisting of two actions, three times: first, he extracted the square root, and then took the integer part of the obtained number. In the end, he got the number 1. What is the largest number that Gosha could have ... | Answer: 255.
Solution. Suppose he entered a number not less than 256. Then after the first operation, he would get a number not less than $[\sqrt{256}]=16$, after the second - not less than $[\sqrt{16}]=4$, after the third - not less than $[\sqrt{4}]=2$, which is a contradiction.
Let's assume Gosha entered the number... | 255 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,486 |
Problem 3. Let's call small all natural numbers not exceeding 150. Does there exist a natural number $N$ that does not divide by some 2 consecutive small numbers, but divides by 148 of the remaining small numbers? | Answer: Yes, it exists.
Solution. Let $150!=2^{k} \cdot a$, where $a$ is odd. We will prove that the number $N=2^{6} \cdot \frac{a}{127}$ satisfies the condition of the problem, specifically that it is not divisible by 127 and 128, but is divisible by all other small numbers.
Obviously, the number 127 is prime, and $... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,487 |
Problem 4. Given a triangle $ABC$ such that $\angle BAC = 2 \angle BCA$. Point $L$ on side $BC$ is such that $\angle BAL = \angle CAL$. Point $M$ is the midpoint of side $AC$. Point $H$ on segment $AL$ is such that $MH \perp AL$. On side $BC$, there is a point $K$ such that triangle $KMH$ is equilateral. Prove that poi... | Solution. From the condition, it follows that $\angle B A L=\angle C A L=\angle B C A$. Therefore, triangle $C A L$ is isosceles with base $A C$, and its median $L M$ is the axis of symmetry of this triangle (as well as the altitude and bisector).
Let point $H$ under reflection about line $L M$ map to some point $H_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,488 |
Problem 5. A buyer came to an antique shop. The trader laid out 2022 coins on the table, among which there are genuine and counterfeit coins, and warned the buyer that the genuine coins among them are more than half. To the buyer, all the coins look indistinguishable, while the trader knows exactly which coins are genu... | Solution. Let's outline the buyer's strategy. The first move is to ask the trader about any two coins. In subsequent moves, we will always ask about the coin that remained on the table from the previous move and any other coin that has not been involved yet.
We will show the rule for removing coins from the table. The... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,489 |
# 1.3 10-11 grades
1/1. For what value of $k$ do the numbers $28+k, 300+k$ and $604+k$ in the given order serve as the squares of three consecutive terms of some increasing arithmetic progression | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,490 | |
Task 1. Three travelers approached a wide river, where they managed to find an old raft at the shore. An inscription on the raft states that it can carry no more than 7 puds at a time. How can the travelers cross the river if two of them weigh 3 puds each, and the third one weighs 5 puds? | Solution. First, the two "light" travelers cross to the opposite bank. Then one of them returns with the raft. Next, the "heavy" traveler crosses alone on the raft. After this, the "light" traveler who remained on the opposite bank can return to the starting bank with the raft; this will allow both light travelers to f... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,491 |
Problem 2. A train consists of 20 cars, numbered from 1 to 20, starting from the beginning of the train. Some of the cars are postal cars. It is known that
- the total number of postal cars is even;
- the number of the nearest postal car to the beginning of the train is equal to the total number of postal cars
- the n... | Answer: $4,5,15,16$.
Solution. The number of the last postal car cannot exceed 20; the number of postal cars is four times less than this number and therefore does not exceed 5. Since the number of postal cars is even, there are 2 or 4 of them.
Suppose there are two. Then the first one has the number 2, and the last ... | 4,5,15,16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,492 |
Task 3. Lёnya has cards with digits from 1 to 7. How many ways are there to glue them into two three-digit numbers (one card will not be used) so that each of them is divisible by 9? | Answer: 36.
Solution. The sum of the digits in each number is divisible by 9, which means the overall sum of the used digits is also divisible by 9. The sum of all given digits $1+2+\ldots+7$ is 28. If we remove the digit 1, the remaining sum is 27, which is divisible by 9; removing any other digit cannot result in a ... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,493 |
Problem 4. Five numbers $2,0,1,9,0$ are written in a circle on the board in the given order clockwise (the last zero is written next to the first two). In one move, the sum of each pair of adjacent numbers is written between them. For example, such an arrangement of numbers (on the right) will be after the first move:
... | Answer: $8 \cdot 3^{5}=1944$.
Solution. Let's see how the numbers between the first and second zero (which Polina will count at the end) change. On each move, each "old" number is included as an addend in two "new" numbers. This means that if all the "new" numbers are added together, each "old" number will be summed t... | 1944 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,494 |
Problem 5. Ivan the Tsarevich has 10 gold coins. He knows that among them there are 5 genuine and 5 counterfeit coins, but he cannot distinguish them. As is known, Baba Yaga can distinguish counterfeit coins from genuine ones. Ivan agreed with Baba Yaga that he would show her any three coins, and she would choose two o... | Answer: Yes.
Solution. Let Baba Yaga choose one counterfeit coin and one real coin, and call them secret. Then she can ensure that Ivan does not find out which is which in the following way.
If Ivan asks about three coins, one of which is secret, Baba Yaga will choose two non-secret coins and say how many of them are... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,495 |
Problem 1. For two non-zero numbers $a$ and $b$, it is known that
$$
a^{2}+\frac{b^{3}}{a}=b^{2}+\frac{a^{3}}{b} .
$$
Is it true that the numbers $a$ and $b$ are equal? | Answer: No
Solution. The equality holds when $a=-b$, for example, when $a=1$ and $b=-1$.
Remark. In fact, the equality holds only when $a=b$ or $a=-b$. Indeed, multiplying it by $a / b^{3}$ and letting $a / b=\lambda$, we get $\lambda^{3}+1=\lambda+\lambda^{4}$. This is equivalent to the equation $\lambda^{4}-\lambda... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,496 |
Problem 2. In a $5 \times 5$ square, there are 5 columns, 5 rows, and 18 diagonals, including diagonals of length one. In each cell of this square, Vova wrote the number $1, 3, 5$ or 7, and Lesha calculated the sum of the numbers in each column, row, and diagonal. Prove that among the sums obtained by Lesha, there are ... | Solution. We will call a row, column, or diagonal along which Andrey summed the numbers a line. Note that there are a total of 20 lines consisting of an odd number of cells (5 lines in each direction). Since all numbers in the table are odd, the sums in these lines are also odd. At the same time, they cannot exceed \(5... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,497 |
Problem 3. Dodson, Williams, and their horse Bolivar want to get from city A to city B as quickly as possible. Along the road, there are 27 telegraph poles that divide the entire journey into 28 equal segments. Dodson takes 9 minutes to walk one segment, Williams takes 11 minutes, and either of them can ride Bolivar to... | Solution. Let the distance from A to B be taken as a unit, and time will be measured in minutes. Then Dodson's speed is $1 / 9$, Williams' speed is $1 / 11$, and Bolivar's speed is $-1 / 3$.
Let the desired post have the number $k$ (i.e., the distance from city A is $k / 28$). Then Dodson will arrive in time
$$
\frac... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,498 |
Problem 4. In a right triangle $ABC$ with a right angle at $A$, the altitude $AH$ is drawn. On the extension of the hypotenuse $BC$ beyond point $C$, a point $X$ is found such that
$$
H X=\frac{B H+C X}{3} .
$$
Prove that $2 \angle A B C=\angle A X C$.
$, or $H X=\frac{1}{2}(B H-C H)$.
It is clear that the value $H X=\frac{1}{2}(B H-C H)$ is positive. It turns out that it is equal to the distance from $H$ to the midpoint of $B C$, which we will denote by $M$. Indeed,
$$
... | 2\angleABC=\angleAXC | Geometry | proof | Yes | Yes | olympiads | false | 4,499 |
Problem 5. In the cells of an $8 \times 8$ chessboard, there are 8 white and 8 black chips such that no two chips are in the same cell. Additionally, no column or row contains chips of the same color. For each white chip, the distance to the black chip in the same column is calculated. What is the maximum value that th... | Answer: 32
Solution. An example of chip placement where the sum of distances is 32 is shown in the figure:
We will prove that the sum of distances cannot be greater than 32. For this, consid... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,500 |
4. Vasya remembers that his friend Petya lives on Kurchatovskaya Street, house number 8, but he forgot the apartment number. In response to a request to clarify the address, Petya replied: “My apartment number is a three-digit number. If you rearrange the digits, you can get five other three-digit numbers. So, the sum ... | Answer: 425.
Let the first, second, and third digits of Petya's apartment number be denoted by $a, b, c$ respectively. The condition of the problem can be algebraically written as:
$$
\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}=2017
$$
Adding $\overline{a b c}$ to both sides ... | 425 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,501 |
5. In each cell of a $15 \times 15$ table, a number $-1, 0$, or $+1$ is written such that the sum of the numbers in any row is non-positive, and the sum of the numbers in any column is non-negative. What is the smallest number of zeros that can be written in the cells of the table? | Answer: 15.
Evaluation. Since the sum of the numbers in all rows is non-positive, the sum of all numbers in the table is also non-positive. On the other hand, if the sum of the numbers in any column is non-negative, then the sum of all numbers in the table is non-negative. Therefore, the sum of all numbers in the tabl... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,502 |
Task 1. On an island, there live knights who always tell the truth, and liars who always lie. Some of the islanders are friends with each other (friendship is mutual).
In the morning, every islander claimed that they are friends with an odd number of knights. In the evening, every islander claimed that they are friend... | Solution. Answer: no.
We will use a well-known auxiliary statement (equivalent to the Handshaking Lemma): in any company, the number of people who are friends with an odd number of others is even.
From the morning statements, it follows that each knight is friends with an odd number of knights, and each liar, in turn... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,503 |
Problem 2. A traveler was walking from a village to a city. At 14:00, when the traveler had covered a quarter of the distance, a motorcyclist set off from the village to the city, and a truck set off from the city to the village. At 15:00, the motorcyclist caught up with the traveler, and at 15:30, the motorcyclist met... | Solution. Answer: 15:48.
Let the entire distance be denoted as $S$, and the speeds of the pedestrian, motorcycle, and truck as $V_{p}, V_{m}$, and $V_{g}$, respectively (distance is measured in kilometers, and speed in kilometers per hour). According to the problem, the motorcyclist caught up with the pedestrian in on... | 15:48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,504 |
Problem 3. In the first quadrant of the coordinate plane, two points $A$ and $B$ with integer coordinates are marked. It turns out that $\angle A O B=45^{\circ}$, where $O-$ is the origin. Prove that at least one of the four coordinates of points $A$ and $B$ is an even number. | Solution. Let point $A$ have integer coordinates $(a ; b)$, and point $B$ have coordinates $(c ; d)$. We write the scalar product of vectors $\overrightarrow{O A}(a ; b)$ and $\overrightarrow{O B}(c ; d)$ in two ways: through coordinates and through the angle between them.
$a c+b d=\overrightarrow{O A} \cdot \overrigh... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,505 |
Problem 4. The diagonals of trapezoid $ABCD (AD \| BC)$ intersect at point $O$. On $AB$, a point $E$ is marked such that line $EO$ is parallel to the bases of the trapezoid. It turns out that $EO$ is the bisector of angle $CED$. Prove that the trapezoid is a right trapezoid. | Solution. Let the line $D E$ intersect the line $B C$ at point $K$. Note that $\angle B C E=\angle C E O=\angle D E O=\angle D K C$, so triangle $C E K$ is isosceles and $C E=E K$. We will prove that segment $E B$ is its median, from which it follows that it is also an altitude, and the trapezoid will be a right trapez... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,506 |
Problem 5. There is a deck of 1024 cards, each with a different set of digits from 0 to 9, and all sets are distinct (including an empty card). We will call a set of cards complete if each digit from 0 to 9 appears exactly once on them.
Find all natural $k$ for which there exists a set of $k$ cards with the following ... | Solution. Answer: 512.
For each card, consider another card that complements it to a complete set (for example, for the card 3679, such a card would be 012458). It is clear that all 1024 cards can be divided into 512 non-overlapping pairs of cards that complement each other to a complete set. Next, we will prove that ... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,507 |
Problem 6. Given positive real numbers $a, b, c$. It is known that
$$
(a-b) \ln c+(b-c) \ln a+(c-a) \ln b=0
$$
Prove that $(a-b)(b-c)(c-a)=0$.
Here $\ln x$ is the natural logarithm (logarithm of $x$ to the base $e$). | Solution. We will use the following well-known statement.
If on the coordinate plane there are points with coordinates $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$, then for any real $\alpha$ the point with coordinates $\left(\alpha x_{1}+(1-\alpha) x_{2} ; \alpha y_{1}+\right.$ $\left.(1-\alpha) y_{2... | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,508 |
Problem 3. In a certain company, there are 100 shareholders, and any 66 of them collectively own at least $50\%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? (The percentage of shares in the company owned by a shareholder can be non-integer.) | Answer: $25 \%$.
Solution. Consider any shareholder $A$. Divide the other 99 shareholders into three groups $B, C, D$ of 33 shareholders each. By the condition, $B$ and $C$ together have at least $50 \%$ of the company's shares, similarly for $C$ and $D$, and for $B$ and $D$. Adding all this up and dividing by two, we... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,510 |
Problem 4. Let's call small all natural numbers not exceeding 150. Does there exist a natural number $N$ that is not divisible by some 2 consecutive small numbers, but is divisible by the remaining 148 small numbers? | Answer: Yes, it exists.
Solution. Let $150!=2^{k} \cdot a$, where $a$ is odd. We will prove that the number $N=2^{6} \cdot \frac{a}{127}$ satisfies the condition of the problem, specifically that it is not divisible by 127 and 128, but is divisible by all other small numbers.
Obviously, the number 127 is prime, and $... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,511 |
Problem 5. In triangle $A B C$, the angle at vertex $B$ is $120^{\circ}$, point $M$ is the midpoint of side $A C$. Points $E$ and $F$ are chosen on sides $A B$ and $B C$ respectively such that $A E=E F=F C$. Find $\angle E M F$. | Answer: $90^{\circ}$.
Solution. Note that the sum of angles $A$ and $C$ is $60^{\circ}$. Let $A E=E F=F C=u$.
On the line $E M$, place a point $G$ such that point $M$ is the midpoint of segment $E G$. Triangles $A M E$ and $C M G$ are equal by two sides and the angle between them, so $\angle M C G=\angle M A E=\angle... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,512 |
Problem 8.2. Determine all pairs of natural numbers $n$ and $m$, for which the numbers $n^{2}+4 m$ and $m^{2}+5 n$ are squares.
Answer: $(m, n)=(2,1),(22,9),(9,8)$. | Solution. Let's consider two cases. First, assume that $m \leqslant n$. Then $n^{2}+4 m \leqslant$ $n^{2}+4 n<(n+2)^{2}$. Therefore, $n^{2}+4 m=(n+1)^{2}$, from which $4 m=2 n+1$, which is impossible due to parity considerations.
Now assume that $n<m$. Then $m^{2}+5 n<m^{2}+6 m<(m+3)^{2}$, from which either $5 n=2 m+1... | (,n)=(2,1),(22,9),(9,8) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,514 |
Problem 8.3. How many lines exist that pass through the point $(0,2019)$ and intersect the parabola $y=x^{2}$ at two points with integer coordinates on the $y$-axis?
Answer: 9. | Solution. A vertical line obviously does not fit. All lines, different from the vertical one and passing through the point $(0,2019)$, are given by the equation $y=k x+2019$ for some $k$. Let such a line intersect the parabola at points $\left(a, a^{2}\right)$ and $\left(b, b^{2}\right)$, where $a^{2}$ and $b^{2}$ are ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,515 |
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