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1. Indicate the largest possible number, in decimal notation, in which all digits are different, and the sum of its digits is 37.
# | # Answer: 976543210.
Sketch of the solution. The sum of all ten digits is 45, so to achieve the maximum, we exclude only one digit. It will be 8. Among the nine-digit numbers, the larger one has the higher significant digits. Hence the answer. | 976543210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,283 |
2. Does there exist a natural number, in the representation of which only the digits 7 and 5 appear in equal quantities, and which is divisible by 7 and 5? Justify your answer. | Answer: exists.
Sketch of the solution. Examples are based on the fact that 1001 is divisible by 7. Therefore, a number of the form $\overline{a b c a b c}=\overline{a b c} \times 1001$ is divisible by 7. From this: $777777555555=777777 \times 1000000+555555$.
$5775=5005+770$
For example, the number: 777777555555 or... | 777777555555or5775 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,284 |
4. There is a contour of a square with a side of 20 cm, it was cut into two groups of equal segments consisting of three and four segments. What is the length of these segments? Find all possible answers. | Answer: 20 cm and 5 cm.
Sketch of the solution.
If each side of the square is cut into at least two segments, there will be 8 segments, and their $4+3=7$. This means one of the segments is equal to the side of the square, and there are three such segments, each 20 cm long. Four segments are obtained by cutting one si... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,286 |
5. There are five ancient coins, two of which are counterfeit. An expert can, for a chocolate bar, indicate how many of any two coins are counterfeit. Vasya, the collector, has four chocolate bars. Can Vasya find the counterfeit coins if the expert requires him to specify all pairs of coins he should check in advance a... | Answer: He can. Sketch of the solution. Number the coins from 1 to 5. The expert should check the following four pairs of coins: 1 and 2, 2 and 3, 3 and 4, 4 and 5 for four chocolates. Let the answers be a, b, c, d. If a+c=2, then the fifth coin is genuine, otherwise it is counterfeit. If a +d=2, then the third coin is... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,287 |
6. A rectangle $3 \times 100$ consists of 300 squares $1 \times 1$. What is the maximum number of diagonals that can be drawn in the squares so that no two diagonals share a common endpoint? (In one square, two diagonals can be drawn without sharing endpoints. Common internal points are allowed.) | Answer: 200.
Example. Let's number the rows and columns containing the squares. In each square with both odd numbers, we will draw two diagonals.
Another example. In all cells of the first and third rows, we will draw parallel diagonals.
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,291 |
4-7. Along a straight alley, 100 lamps are placed at equal intervals, numbered in order from 1 to 100. At the same time, from different ends of the alley, Petya and Vasya started walking towards each other at different constant speeds (Petya from the first lamp, Vasya from the hundredth). When Petya was at the 22nd lam... | Answer. At the 64th lamppost.
Solution. There are a total of 99 intervals between the lampposts. From the condition, it follows that while Petya walks 21 intervals, Vasya walks 12 intervals. This is exactly three times less than the length of the alley. Therefore, Petya should walk three times more to the meeting poin... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,295 |
4-8. In a chess club, 90 children attend. During the session, they divided into 30 groups of 3 people, and in each group, everyone played one game with each other. No other games were played. In total, there were 30 games of "boy+boy" and 14 games of "girl+girl". How many "mixed" groups were there, that is, groups in w... | Answer: 23.
Solution: There were a total of 90 games, so the number of games "boy+girl" was $90-30-14=46$. In each mixed group, two "boy+girl" games are played, while in non-mixed groups, there are no such games. In total, there were exactly $46 / 2=23$ mixed groups.
## Grade 5 | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,296 |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,297 |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,298 |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
We will call suc... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,299 |
5-7. On the faces of a die, the numbers $6,7,8,9,10,11$ are written. The die was rolled twice. The first time, the sum of the numbers on the four "vertical" (that is, excluding the bottom and top) faces was 33, and the second time - 35. What number can be written on the face opposite the face with the number 7? Find al... | Answer: 9 or 11.
Solution. The total sum of the numbers on the faces is $6+7+8+9+10+11=51$. Since the sum of the numbers on four faces the first time is 33, the sum of the numbers on the two remaining faces is $51-33=18$. Similarly, the sum of the numbers on two other opposite faces is $51-35=16$. Then, the sum on the... | 9or11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,300 |
6-3. The red segments in the figure have equal length. They overlap by equal segments of length $x$ cm. What is $x$ in centimeters?
 | Answer: 2.5.
Solution. Adding up the lengths of all the red segments, we get 98 cm. Why is this more than 83 cm - the distance from edge to edge? Because all overlapping parts of the red segments have been counted twice. There are 6 overlapping parts, each with a length of $x$. Therefore, the difference $98-83=15$ equ... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,302 |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? 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 | 18,305 |
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment ... | Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$. | 219 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,306 |
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 | 18,308 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,309 |
1. Let $S(n)$ denote the number of hits on the target by the shooter in $n$ shots. At the beginning of the shooting, $S(n)$ was less than $90\%$ of $n$, and by the end of the shooting, it was more than $90\%$. Was there necessarily a moment during the shooting when $S(n)$ was exactly $90\%$ of $n$? | # Solution.
Assume that such a moment did not exist. Then, for some $n$, the following inequalities must hold:
$\left\{\begin{array}{l}\frac{S(n)}{n}<\frac{9}{10} ;\end{array} \Rightarrow 9 n-1<10 \cdot S(n)<9 n\right.$.
The last chain of inequalities cannot be satisfied for any $n$ and $S(n)$, since there cannot be... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,310 |
2. Winnie-the-Pooh stocked up on chocolate bars for the winter: $60\%$ of the total number were "Snickers", $30\%$ were "Mars", and $10\%$ were "Bounty". In the spring, it turned out that the number of "Bounty" bars eaten by Winnie-the-Pooh was $120\%$ of the number of "Mars" bars eaten and $30\%$ of the number of "Sni... | # Solution.
Let there be $3 k$ "Bounty" chocolate bars in total. Then there were $9 k$ "Mars" bars and $18 k$ "Snickers" bars. Since $k$ "Bounty" bars were eaten, $\frac{k}{1.2}=\frac{5 k}{6}$ "Mars" bars were eaten. Therefore, $k$ is divisible by 6. "Snickers" bars eaten were $\frac{k}{0.3}=\frac{10 k}{3}$, and the r... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,311 |
3. Find all values of the parameter $a$, for which the interval ( $3 a ; 5 a-2)$ contains at least one integer.
# | # Solution.
The left end of the interval must be less than the right, so
$3 a < 5a - 2$. Further, if the length of the interval is greater than one, it definitely contains an integer. $5 a - 2 - 3 a > 1 \Rightarrow a > 1.5$. Thus, the interval $(1.5 ; \propto)$ is part of the solution set. Consider the values $a \in ... | (1.2;4/3)\cup(1.4;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,312 |
4. Given a circle of radius $R$. At a distance of $2 R$ from the center of the circle, a point $A$ is chosen. From this point, a tangent and a secant are drawn, and the secant is equidistant from the center of the circle and the point of tangency. Find the length of the segment of the secant enclosed within the circle. | # Solution.
Let $O$ be the center of the circle, $B$ the point of tangency, $CG$ the secant, $BF$ and $OD$ perpendicular to the secant, and $E$ the intersection point of the secant with the ... | 2R\sqrt{\frac{10}{13}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,313 |
5. On each face of a regular icosahedron, a non-negative integer is written such that the sum of all 20 numbers is 39. Prove that there exist two different faces that share a vertex and on which the same number is written.
. Then, since the numbers on the faces are counted three times (once from each of the three ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,314 |
1. The desired number is equal to $(9+1)+(99+1)+\ldots+(\underbrace{99 \ldots 9}_{2016 \text { times }}+1)+1=$
$=10+10^{2}+\cdots+10^{2016}+1=\underbrace{11 \ldots 1}_{2017 \text { times }}$. | Answer: $\underbrace{11 \ldots 1}_{2017 \text { times }}$. | \underbrace{11\ldots1}_{2017\text{times}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,315 |
1. There are two buckets, one 7 liters, the second 3 liters. There is a tap and a sink, which allow you to fill any amount of water and, if necessary, pour it out. Write a sequence of actions to measure exactly 1 liter of water by pouring and emptying the buckets. Continue further and show how to get 2, 4, 5, 6 liters ... | Solution.
1 liter is obtained by pouring water from the 7-liter bucket into the 3-liter bucket twice. 1 liter will remain in the larger bucket.
2 liters are obtained by pouring water from the 3-liter bucket into the 7-liter bucket three times, filling it. 2 liters will remain in the smaller bucket.
4 liters are obta... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,317 |
2. There are 28 students in the class. One day, each student brought three markers, a red one, a green one, and a blue one. Can the students in the class exchange markers so that each ends up with three markers of the same color? | Solution 1. Suppose such an exchange is possible. Let's take all the students who received three red markers. According to our assumption, there are no other students who received red markers. Therefore, the total number of red markers is divisible by three. The number of red markers, according to the problem, is 28, s... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,318 |
3. Three argumentative friends are sitting in front of the TV. It is known about each of them that she either is always right about everything or always wrong about everything. The first one said: "None of us have seen this movie." The second one said: "I have seen this movie, but you both haven't." The third one said:... | Solution. If the first girlfriend is right, then the statements of the second and third are false. If the second girlfriend is right, then the statements of the first and third are false. If the third girlfriend is right, then the statements of the second and third are false. Two girlfriends cannot be right at the same... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,319 |
6. Do there exist three different natural (positive integer) numbers such that the sum of any two of them is a prime number? | Solution. The number 2 cannot be the sum of three positive integers. The other prime numbers are odd. Suppose the three numbers in question exist. The sum of the first and second numbers is odd, which means one of these numbers is even, and the other is odd. The sum of the second and third numbers is odd, which means o... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,321 |
Task 1. In a bag, there were cards with numbers from 1 to 20. Vlad drew 6 cards and said that all these cards can be divided into pairs so that the sums of the numbers in each pair are the same. Lena managed to peek at 5 of Vlad's cards: the numbers on them were $2, 4, 9, 17, 19$. What number was on the card that Lena ... | Answer: 12.
Solution. To calculate the answer, one needs to select four numbers out of the given five such that the sum of two of them equals the sum of the other two. By enumeration, it is not difficult to verify that these numbers are $2,4,17,19(2+19=4+17)$.
Thus, the number on the remaining card is $12(2+19=4+17=9... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,323 |
Problem 2. Anya, Borya, Vika, and Gena are on duty at school for 20 days. It is known that exactly three of them are on duty each day. Anya was on duty 15 times, Borya - 14 times, Vika - 18 times. How many times was Gena on duty | Answer: 13.
Solution. Since 3 people are on duty in school every day, a total of $3 \cdot 20=60$ people are needed for the duty. Therefore, Gena was on duty $60-15$ (Anya's duties) -14 (Borya's duties) -18 (Vika's duties) $=13$ times.
| Anya | $\checkmark$ | $\checkmark$ | $\checkmark$ | | $\checkmark$ | $\checkmark... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,324 |
Problem 3. During the physical education class, the entire class lined up by height (all children have different heights). Dima noticed that the number of people taller than him is four times the number of people shorter than him. And Lёnya noticed that the number of people taller than him is three times less than the ... | Answer: 21.
Solution. Let $x$ be the number of people who are shorter than Dima. Then, the total number of students in the class is $x$ (people who are shorter than Dima) $+4 x$ (people who are taller than Dima) +1 (Dima) $=5 x+1$ (total number of people in the class).
Let $y$ be the number of people who are taller t... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,325 |
Problem 4. A row of 11 numbers is written such that the sum of any three consecutive numbers is 18. Additionally, the sum of all the numbers is 64. Find the central number. | Answer: 8.
Solution. Number the numbers from left to right from 1 to 11.
Notice that the sum of the five central numbers (from the fourth to the eighth) is 64 (the sum of all numbers) $-2 \cdot 18$ (the sum of the numbers in the first and last triplets) $=28$.
Then the sixth (central) number is 18 (the sum of the fo... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,326 |
Problem 6. Koschei the Deathless has 11 large chests. In some of them lie 8 medium chests. And in some of the medium chests lie 8 small chests. The chests contain nothing else. In total, Koschei has 102 empty chests. How many chests does Koschei have in total? | Answer: 115.
Solution. Let $x$ be the number of non-empty chests.
Consider the process when Koschei just started placing chests inside each other. Initially, he had 11 empty large chests. Each time he placed 8 smaller chests into a single empty chest, the total number of empty chests increased by 7 ( -1 old empty che... | 115 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,328 |
Problem 7. The dragon has 40 piles of gold coins, and the number of coins in any two of them differs. After the dragon plundered a neighboring city and brought back more gold, the number of coins in each pile increased by either 2, 3, or 4 times. What is the smallest number of different piles of coins that could result... | Answer: 14.
Solution. Evaluation. Suppose that no more than 13 piles increased by the same factor. Then no more than 13 piles increased by a factor of 2, no more than 13 by a factor of 3, and no more than 13 by a factor of 4. Thus, the dragon has no more than 39 piles in total. Contradiction.
Therefore, there will be... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,329 |
Problem 8. On one face of a die, one dot is drawn, on another face - two, on the third - three, and so on. Four identical dice are stacked as shown in the figure. How many dots in total are on the 6 faces where the dice touch?
^2$ is divisible by 3;
$(3k+1)^2=9k^2+6k+1$ gives a remainder of 1 when di... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,331 |
4. $f(x)=x^{2}+m x+n$ - a quadratic function ( $m$ and $n$ - integers).
It is known that $f(2014)>0$ and $f(2015)>0$. Prove that $f(x)>0$ for all $x$ in the interval $[2014 ; 2015]$. | Solution: We will prove by contradiction. Suppose there exists $X \in [2014 ; 2015]$ such that $f(x) \leq 0$. Then the x-coordinate of the vertex of the parabola $f(x) = x^2 + mx + n$ $X_{\text{v}} \in (2014 ; 2015)$, and the quadratic trinomial has roots $X_1$ and $X_2 \in (2014 ; 2015)$ (the roots may coincide). $x_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,332 |
5. Point $Q$ lies outside the circle $\omega_{1} \cdot Q A$ and $Q B$ are tangents to the circle ( $A$ and $B$ belong to $\boldsymbol{\omega}_{1}$ ). Through points $A$ and $B$, a second circle $\omega_{2}$ is drawn with its center at point $Q$. On the arc $A B$ of circle $\omega_{2}$, which lies inside circle $\omega_... | Let $\angle A Q B = A \breve{K} B = \boldsymbol{\alpha}$. Then $\angle A K B = \frac{360^{\circ} - \alpha}{2} = 180^{\circ} - \frac{\alpha}{2}$.
From $\triangle A Q B \quad \angle Q A B = \angle Q B A = 90^{\circ} - \frac{\alpha}{2}$.
On the other hand, $\angle Q A B$ is the angle between the tangent $Q A$ and the se... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,333 |
4. Prove that if $(a+b+c) c \leq 0$, then $b^{2} \geq 4 a c$. | Solution. Consider the quadratic trinomial $f(x)=a x^{2}+b x+c$. Obviously, $f(0)=c$, and $f(1)=a+b+c$. From the condition of the problem, it follows that either one of these numbers is zero, or they have different signs. This means that on the interval $[0,1]$ there is a root of the quadratic equation $a x^{2}+b x+c=0... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,334 |
5. Dima has 9 identical-looking balls numbered from 1 to 9. Dima knows that one of the balls is slightly heavier than the others, but to determine which one, he needs ultra-precise scales. Such scales are owned by his perpetually busy neighbor, a chemist, who agrees to help Dima under the following conditions:
- no mo... | Solution. It can be done. Here is one of the ways. First weighing: place balls 1, 2, and 3 on one scale, and balls 4, 5, and 6 on the other. Second weighing: place balls 1, 4, and 7 on one scale, and balls 2, 5, and 8 on the other.
We will show that the obtained information is sufficient to determine the heavy ball.
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,335 |
7.1. Is it possible to place rational numbers at the vertices of a square $A B C D$ such that the sum of the numbers at vertices $A$ and $B$ is 1, at vertices $B$ and $C$ is 2, at vertices $C$ and $D$ is 3, and at vertices $A$ and $D$ is 4? | Answer: No.
Solution. Let the number $x$ be placed at vertex $A$, then the number $1-x$ will be placed at vertex $B$, the number $1+x$ at vertex $C$, and the number $2-x$ at vertex $D$. Then, for any value of $x$, the sums of the numbers on the three sides will be $1, 2, 3$, respectively. For the sum of the numbers on... | No | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,337 |
7.2. On his birthday, Nikita decided to treat his classmates and took 4 bags of candies to school, each containing the same number of candies. He gave out 20 candies to his classmates, which was more than $60 \%$ but less than $70 \%$ of all the candies he had. How many candies did Nikita have in total? | Answer: 32.
Solution. Suppose there were $x$ candies in each bag.
Then $\frac{6}{10}<\frac{20}{4 x}<\frac{7}{10}$. Therefore, $\frac{50}{7}<x<\frac{50}{6}$, i.e., $7<x \leq 8, x=8$. Total candies - 32. | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,338 |
7.3. Masha and Sasha take turns (starting with Masha) writing 10 digits on the board from left to right to form a ten-digit number. Moreover, it is not allowed to write two consecutive identical digits. If the resulting number is divisible by 9, then Sasha wins; otherwise, Masha wins. Who will win with correct play fro... | Answer: Sasha.
Solution. A number is divisible by 9 if the sum of its digits is divisible by 9. Therefore, one of Sasha's possible strategies is to complement each of Masha's digits to 9. That is, if Masha writes 0, then Sasha writes 9; if Masha writes 1, then Sasha writes 8, and so on. Thus, after each pair of moves,... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,339 |
7.4. In a chest, there are 2021 coins. It is known that exactly one of them is counterfeit. Its mass differs from that of a genuine coin, while the masses of all genuine coins are equal. Can it be determined in two weighings on a balance scale without weights whether the counterfeit coin is lighter or heavier than a ge... | Answer: Yes.
Solution. First, set aside one coin, and divide the rest into 2 parts of 1010 coins each and weigh them. If their weights are equal, then weigh the set-aside coin with any other; if the set-aside coin is lighter, then the fake coin is lighter than the others, otherwise - heavier. If the weights of the fir... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,340 |
7.5. Given a triangle $A B C$ with angle $A$ equal to $60^{\circ}$. Points $M, N, K$ lie on sides $B C, A C, A B$ respectively, such that $B K=K M=M N=N C$. It turns out that $A N=2 A K$. Prove that segment $M N$ is perpendicular to $A C$. | Solution. Let $\mathrm{P}$ be the midpoint of segment $\mathrm{AN}$, then $\mathrm{AP}=\mathrm{AK}$, and angle $\mathrm{A}=60^{\circ}$ (by condition), hence triangle KAP is equilateral. Then, in isosceles triangle $\mathrm{KNP}$, angle $\angle \mathrm{KPN}=120^{\circ}$, from which $\angle \mathrm{PNK}=30^{\circ}$.
Let... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,341 |
2. Katya invited three kittens and a puppy to her place to solve math problems. After solving the problems, Katya counted the pastries in the kitchen and noticed that two were missing. Katya has a balance scale without weights, on which she can place pastries, kittens, and the puppy. All pastries weigh the same, and al... | Solution. Denote the kittens as K1, K2, K3. First, compare kittens K1 and K2. There are two possible outcomes.
1) One of the kittens turned out to be heavier, for example, $\mathrm{K} 1 > \mathrm{K} 2$. Then, for the second weighing, compare K1 and K3 + pastry. If $\mathrm{K} 1 > \mathrm{K} 3 +$ pastry, then both past... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,343 |
3. Find the numerical value of the expression
$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}+\frac{2}{x y+1}
$$
if it is known that $x$ is not equal to $y$ and the sum of the first two terms is equal to the third. | Answer: 2.
Solution. Let's bring the condition to a common denominator
$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}=\frac{2}{x y+1}
$$
we get
$$
\frac{\left(x^{2}+y^{2}+2\right)(x y+1)-2\left(x^{2}+1\right)\left(y^{2}+1\right)}{\left(x^{2}+1\right)\left(y^{2}+1\right)(x y+1)}=0
$$
expand all brackets in the numerator, c... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,344 |
4. In the multi-story fraction
$$
P+\frac{1}{A+\frac{1}{B+\frac{1}{H+\frac{1}{0}}}}
$$
digits other than zero were substituted for the letters. What is the smallest result that can be obtained in this case? (As usual, different letters correspond to different digits). | Answer. $1+\frac{1}{9+\frac{1}{2+\frac{1}{8+\frac{1}{3}}}}=1 \frac{53}{502}$.
Solution. Let the entire fraction be T and write
$$
\mathrm{T}=\mathrm{P}+\frac{1}{X}, \quad X=\mathrm{A}+\frac{1}{Y}, \quad Y=\mathrm{B}+\frac{1}{\mathrm{Z}}, \quad Z=\mathrm{H}+\frac{1}{\mathrm{O}}
$$
Notice that all numbers X, Y, Z are ... | 1\frac{53}{502} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,345 |
5. After the topic "triangles," all students were given paper triangles and scissors to solve the next problem. Mischievous Kolya, not listening to the teacher, did the following:
he folded his triangle along a straight line, made a straight cut with scissors, and got three pieces, unfolded the folded parts. Kolya ran ... | Answer: Yes, they can.
Solution. Consider a triangle $ABC$ where $\angle C=90^{\circ}, \angle A=60^{\circ}$. Draw its angle bisector $AD$ and the perpendicular $DE$ to the leg $BC$ (see the left figure). Fold the triangle along the line $DE$, then point $A$ will take the position $A'$, after which we make a cut along ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,346 |
9.1 Karlson eats three jars of jam and one jar of honey in 25 minutes, while Little Man takes 55 minutes. One jar of jam and three jars of honey Karlson eats in 35 minutes, while Little Man takes 1 hour 25 minutes. How long will it take them to eat six jars of jam together? | 9.1 20 minutes.
From the condition, it follows that if Karlson eats three jars of jam and one jar of honey, and then immediately eats one jar of jam and three jars of honey, he will spend $25+35=60$ minutes. For Little Man, this time will be 140 minutes. Therefore, Karlson will spend 15 minutes on one jar of jam and o... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,347 |
9.4. On the diagonal $BD$ of parallelogram $ABCD$, a point $K$ is taken. Line $AK$ intersects lines $CD$ and $BC$ at points $L$ and $M$ respectively. Prove that $AK^2 = KL \cdot KM$. | 9.4. On the diagonal $BD$ of parallelogram $ABCD$, a point $K$ is taken. Line $AK$ intersects lines $CD$ and $BC$ at points $L$ and $M$ respectively. Prove that $AK^2 = KL \cdot KM$.
The relation $AK^2 = KL \cdot KM$ is equivalent to $\frac{AK}{KL} = \frac{KM}{AK}$. From the similarity of triangles $ABK$ and $LDK$, it... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,349 |
9.5. A football team coach loves to experiment with the lineup. During training sessions, he divides 20 available field players into two teams of 10 players each, adds goalkeepers, and arranges a game between the teams. He wants any two field players to end up on different teams at some training session. What is the mi... | # 9.5. 5 Training Sessions.
Note that four training sessions are insufficient. Let's take 10 football players who played on the same team during the first training session. During the second training session, at least five of them will be on the same team again. During the third training session, at least three of the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,350 |
1. In the addition example: $\square+\triangle+\square=\square \square$, fill in the same digit in each square and a different digit in the triangle so that the example becomes correct. | Answer: $1+9+1=11$.
Comment (how the answer could be found): On the right, there is a two-digit number with two identical digits. Since the largest sum of three digits is $9+9+9=27$, the number on the right is less than 30, i.e., it is 11 or 22. If it is 11, we get: $1+\triangle+1=11$, from which $\triangle=9$. If it ... | 1+9+1=11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,351 |
2. At the first stop, 18 passengers entered an empty bus. Then at each stop, 4 people got off and 6 people got on. How many passengers were in the bus between the fourth and fifth stops? | Answer: 24 people.
## Solution.
## Method 1.
After each stop, except the first one, the number of passengers in the bus increases by 2 people. Therefore, from the second to the fourth stop, the number of people increased by 6 people. That is, it became $18+6=24$ people.
Method 2.
From the second to the fourth stop... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,352 |
3. Three foxes: Alice, Larisa, and Inessa were talking on a meadow. Larisa: "Alice is not the most cunning." Alice: "I am more cunning than Larisa." Inessa: "Alice is more cunning than me." It is known that the most cunning fox lied, and the others told the truth.
a) Can Alice be the most cunning fox? Why?
b) Which f... | Answer. a) Cannot, b) Inessa.
## Solution.
a) Alice cannot be the smartest, because if she is the smartest, then she is smarter than Larisa, i.e., Alice told the truth. But the smartest fox should have lied.
b) The smartest fox is Inessa. Let's prove this. We have already established in part a) that Alice cannot be t... | Inessa | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,353 |
4. How to form a rectangle, with all sides greater than 1, from 13 rectangles of sizes $1 \times 1, 2 \times 1, 3 \times 1, \ldots, 13 \times 1$? | Answer. One of the possible examples is shown in the figure:
Comment 1 (how to guess the example). If we group the rectangles: the first with the last, the second with the second-to-last, and... | 7\times13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,354 |
5. A sign engraver makes signs with letters. He engraves identical letters in the same amount of time, and different letters possibly in different times. For two signs “ДОМ МОДЫ” and “ВХОД” together, he spent 50 minutes, and one sign “В ДЫМОХОД” he made in 35 minutes. How long will it take him to make the sign “ВЫХОД”? | Answer: 20 minutes.
## Solution.
In the signs FASHION HOUSE ENTRANCE and IN CHIMNEY, we separate the letters that form the word EXIT, then from the first sign, D, O, M, M, O, D will remain, and from the second - D, M, O. Note that FASHION HOUSE ENTRANCE differs from IN CHIMNEY by the letters D, O, M, and in time - by... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,355 |
11.1. The polynomial $P(x)$ is such that the polynomials $P(P(x))$ and $P(P(P(x)))$ are strictly monotonic on the entire real line. Prove that $P(x)$ is also strictly monotonic on the entire real line. (K. Sukhov) | First solution. Since the polynomial $P(P(x))$ is monotonic, it must have an odd degree, and thus it takes all real values.
Let $a>b$, then there exist numbers $x_{a}$ and $x_{b}$ such that $P\left(P\left(x_{a}\right)\right)=a, P\left(P\left(x_{b}\right)\right)=b$. Since the leading coefficient of the polynomial $P(P(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,356 |
11.2. Given positive numbers $x_{1}, x_{2}, \ldots, x_{n}$, where $n \geqslant 2$. Prove that
$$
\frac{1+x_{1}^{2}}{1+x_{1} x_{2}}+\frac{1+x_{2}^{2}}{1+x_{2} x_{3}}+\ldots+\frac{1+x_{n-1}^{2}}{1+x_{n-1} x_{n}}+\frac{1+x_{n}^{2}}{1+x_{n} x_{1}} \geqslant n
$$
(F. Petrov) | Solution. In all solutions, we assume that the numbering of variables is cyclic, that is, $x_{n+1}=x_{1}, x_{n+2}=x_{2}$, and so on.
First Solution. Notice that for all $i=1,2, \ldots, n$, the inequality $\left(1+x_{i}^{2}\right)\left(1+x_{i+1}^{2}\right) \geqslant\left(1+x_{i} x_{i+1}\right)^{2}$ holds, since
$$
\le... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,357 |
11.3. Given a natural number $k$. On a grid plane, $N$ cells are initially marked. We will call the cross of cell $A$ the set of all cells that are in the same vertical or horizontal line with $A$. If the cross of an unmarked cell $A$ contains at least $k$ other marked cells, then cell $A$ can also be marked. It turned... | Answer. $N=\left[\frac{k+1}{2}\right] \cdot\left[\frac{k+2}{2}\right]=\left\{\begin{array}{ll}m(m+1), & \text { if } k=2 m ; \\ m^{2}, & \text { if } k=2 m-1\end{array}\right.$.
Solution. Let $N(k)$ be the answer to the problem; set $f(k)=\left[\frac{k+1}{2}\right] \cdot\left[\frac{k+2}{2}\right]$. First, we prove tha... | N=[\frac{k+1}{2}]\cdot[\frac{k+2}{2}]={\begin{pmatrix}(+1),&\text{if}k=2;\\^{2},&\text{if}k=2-1\end{pmatrix}.} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,358 |
11.4. On the sides $AB$ and $AC$ of triangle $ABC$, points $P$ and $Q$ are chosen respectively such that $PQ \parallel BC$. Segments $BQ$ and $CP$ intersect at point $O$. Point $A'$ is symmetric to point $A$ with respect to the line $BC$. Segment $A'O$ intersects the circumcircle $\omega$ of triangle $APQ$ at point $S$... | The first solution. The case $A B=A C$ follows from symmetry; without loss of generality, we will assume that $A C>A B$.
Fig. 17
Choose a point $X$ on $\omega$ such that $P A X Q$ is an iso... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,359 |
11.5. On a table, 1000 cards are laid out in a circle, each with a different natural number written on it. First, Vasya chooses one of the cards and removes it from the table. Then he repeats the following operation. If the number on the last removed card is $k$, Vasya counts $k$ cards clockwise from it and removes tha... | # Answer. It could.
Solution. Temporarily abandon the condition of the distinctness of the numbers on the cards. Let $A$ and $B$ be two adjacent cards ($A$ lies after $B$ clockwise). Write an arbitrary number on card $A$, the number 2 on card $B$, and ones on all the other cards. If Vasya removes any card first, excep... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,360 |
11.6. Three diagonals of a regular $n$-sided prism intersect at one internal point $O$. Prove that point $O$ is the center of the prism. (A diagonal of a prism is a segment connecting two of its vertices that are not in the same face.) | The first solution. Let the diagonals $A A_{1}, B B_{1}$, and $C C_{1}$ intersect at point $O$, with vertices $A, B$, and $C$ lying on one base of the prism, and vertices $A_{1}, B_{1}$, and $C_{1}$ on the opposite base (see Fig. 19). Then points $A, A_{1}, B$, and $B_{1}$ lie in the same plane $\alpha$, which intersec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,361 |
11.7. Let's define the sequence $a_{1}, a_{2}, a_{3}, \ldots$ by the formula $a_{n} = \left[n^{\frac{2018}{2017}}\right]$. Prove that there exists a natural number $N$ such that among any $N$ consecutive terms of the sequence, there is one whose decimal representation contains the digit 5. (As usual, $[x]$ denotes the ... | Solution. Let $\beta=\frac{1}{2017}$. Recall that a particular case of Bernoulli's inequality $(1+x)^{2017} \geqslant 1+2017 x$ (for $x \geqslant-\beta$) can be rewritten as $1+\beta y \geqslant(1+y)^{\beta}$ (for $y=2017 x \geqslant-1$).
Lemma 1. For any natural number $n$, the inequalities $\frac{n+1+\beta}{n+1} \le... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,362 |
11.8. Initially, in the lower left and lower right corners of a $2018 \times 2018$ board, there are two knights - a red one and a blue one, respectively. Kolya and Sasha take turns; Kolya starts. On a turn, a player moves their knight (Kolya - the red one, and Sasha - the blue one), shifting it simultaneously 20 cells ... | # Answer. Sasha.
Solution. We will present a strategy that allows Sasha to win. We will color the entire board in a chessboard pattern. Let $K$ and $S$ be the squares where the red ($k$) and blue ($s$) knights are initially placed, respectively; these squares are of different colors. It is not difficult to understand ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,363 |
7.1. The younger brother takes 25 minutes to reach school, while the older brother takes 15 minutes to walk the same route. How many minutes after the younger brother leaves home will the older brother catch up to him if he leaves 8 minutes later? | Answer: in 17 minutes. Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 25 minutes, in 8 minutes he will cover the distance $\frac{8 S}{25}$. After the older brother leaves the house, the rate at which they are closing the distance between them will be $\frac{S}{1... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,364 |
7.2. One side of the rectangle (width) was increased by $10 \%$, and the other (length) - by $20 \%$. a) By what percentage is the area of the new rectangle larger than the area of the original? b) Find the ratio of the sides of the original rectangle, if it is known that the perimeter of the new rectangle is $18 \%$ l... | Answer. a) by $32 \%$; b) $1: 4$. Solution. a) Let $a$ and $b$ be the sides of the original rectangle. Then the sides of the new rectangle are $1.1 a$ and $1.2 b$, and its area $S=1.1 a \cdot 1.2 b=1.32 a b$, which is $132 \%$ of the area of the original rectangle. Thus, the area has increased by $32 \%$. b) From the c... | )32;b)1:4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,365 |
7.3. On a plane, 100 lines were drawn. Could the number of all intersection points be a) 100? b) 99? | Answer. a) yes, it could; b) yes, it could. Solution. We can provide such examples (see fig.). a) Construct a bundle of 99 lines passing through one point, and intersect all 99 lines with the hundredth line. b) In this case, draw 99 parallel lines and intersect them with the hundredth line. We can
$ denote the sum of the digits of a natural number $n$. Does there exist an $n$ such that $n \cdot s(n)=100200300$? | Answer. Does not exist. Solution. By the divisibility rule for 3 (and for 9), the numbers $n$ and $s(n)$ either both are divisible by 3 (by 9), or both are not divisible by 3 (respectively, by 9). First, consider the case when $n$ and $s(n)$ are divisible by 3. Then the product $n \cdot s(n)$ is divisible by 9. In the ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,367 |
7.5. Petya tells his neighbor Vova: «In our class, there are 30 people, and there is an interesting situation: any two boys have a different number of girlfriends in the class, and any two girls have a different number of friends among the boys in the class. Can you determine how many boys and how many girls we have in... | Answer. a) Vova is wrong; b) 15 boys and 15 girls. Solution. a) To show that Vova is wrong, we will provide an example of a class that meets the conditions of the problem. For a class with 15 girls and 15 boys, we will number them and present a "friendship" table. In the cells of the table, there is a "+", if the corre... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,368 |
1. Ruslan and Kolya collected coins of 1, 2, and 5 rubles, and it turned out that Ruslan's piggy bank does not have coins of the same denomination as Kolya's. Can the boys pay 2006 rubles from their piggy banks with the same number of coins? Explain your answer. | 1. They cannot. Suppose the boys can pay 2006 rubles from their piggy banks with the same number of coins. Then, one of the boys must have coins of only one denomination (1, 2, or 5 rubles), otherwise, it would contradict the condition. If someone has 5-ruble coins in their piggy bank, they must also have coins of anot... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,369 |
2. In the USA, the date is typically written as the month number, followed by the day number, and then the year. In Europe, however, the day comes first, followed by the month and the year. How many days in a year cannot be read unambiguously without knowing which format it is written in? | 2. It is clear that these are the days whose date can be the number of the month, i.e., takes values from 1 to 12. There are such days $12 \cdot 12=144$. However, the days where the number matches the month are unambiguous. There are 12 such days. Therefore, the number of days sought is $144-12=132$. | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,370 |
3. Four spacecraft are moving around a planet along the same orbit. They started moving simultaneously from one point on the orbit (let's call it the starting point) and moved at constant speeds. It is known that for any three spacecraft, there was a moment when they met. Prove that there will be a moment when all four... | 3. It is sufficient to solve the problem for the case when the slowest spacecraft is at the starting point, and the others (let's call them $1,2,3$) are flying at speeds reduced by the speed of the slowest one; under such a change in conditions, the moments of spacecraft encounters will remain unchanged. Let the 1st sp... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 18,371 |
5. A traveler arrived on an island inhabited by liars (L) and truth-tellers (P). Each L, when asked a question "How many..?", gives a number that is 2 more or 2 less than the correct answer, while each P answers correctly. The traveler met two residents of the island and asked each how many L and P live on the island. ... | 5. I - L, II - P. On the island, there are 1000 L and 1000 P. The answers of the first and second are different, so the option P and P is impossible. The option L and L is also impossible, because the numbers 1001 and 1000 differ by 1, while the answers of the liars regarding the number of L should differ by 4, or coin... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,373 |
11.1. All equations of the form $(x-a)(x-b)=(x-c)(x-d)$ are considered, where $a, b, c$, and $d$ are some real numbers that satisfy the condition $a+d=b+c=2017$. Prove that all considered equations have a common root and find it. | Solution: $(x-a)(x-b)=(x-c)(x-d) \Leftrightarrow x(c+d-a-b)=cd-ab$. From the condition $a+d=b+c=2017$, it follows that $d=2017-a$ and $b=2017-c$. Then the equation takes the form $x(c+(2017-a)-(2017-c)-a)=2017c-ac-(2017a-ac)$. After equivalent transformations, we get the equality $(c-a)(2x-2017)=0$. This becomes an ide... | \frac{2017}{2} | Algebra | proof | Yes | Yes | olympiads | false | 18,374 |
11.2. Fishermen caught several carp and pike. Each caught as many carp as all the others caught pike. How many fishermen were there if the total number of carp caught was 10 times the number of pike? Justify your answer. | # Solution:
Method 1. Each fisherman caught as many carp and pike together as the total number of pike caught. Summing the catches of all fishermen, we get that the total catch of all fishermen (in terms of the number of fish) is equal to the total number of pike caught, multiplied by the number of fishermen. On the o... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,375 |
11.3. Does there exist a polynomial $p(x)$ of degree 13 with the coefficient of the leading term equal to $\frac{1}{1001}$, which takes integer values at all integer points? Justify your answer. | Solution: For example, the polynomial $p(x)=\frac{x(x-1)(x-2) \ldots(x-12)}{1001}$ works. Let's show this. When expanding the brackets, the highest power of $x$ will be 13, and the coefficient of this term will be $\frac{1}{1001}$. Since $1001=7 \cdot 11 \cdot 13$, it remains to show that for any integer $x$, the numbe... | exists | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,376 |
11.4. Can a regular hexagon with a side length of $\frac{2}{3}$ fit inside a unit cube? Justify your answer. | Solution: Consider the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ and its section by a plane passing through the midpoints of the edges $A B, B C$, and $C C_{1}$ - see the figure. In the section, a hexagon $E F G H I J$ is obtained - see the figure. We will show that it is regular. Indeed, each of its sides is equal to hal... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,377 |
11.5. In triangle $ABC$, the median $BK$ and the bisector $CL$ are drawn. Let $P$ be their point of intersection. Prove that $\frac{PC}{PL} - \frac{AC}{BC} = 1$. | # Solution:
Method 1. Draw a line through point $l$ parallel to line $B K$; let it intersect side $A C$ at point $F$. By Thales' theorem, $\frac{C P}{P L}=\frac{C K}{K F}$. By the same theorem, $\frac{K F}{F A}=\frac{B L}{L A}$. By the property of the angle bisector of a triangle, $\frac{B L}{L A}=\frac{B C}{C A}$. Th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,378 |
11.6. Solve the system of equations
\[
\left\{\begin{aligned}
x+2 y+3 z & =2 \\
\frac{1}{x}+\frac{1}{2 y}+\frac{1}{3 z} & =\frac{5}{6} \\
x y z & =-1
\end{aligned}\right.
\] | # Solution:
Method 1. Substituting $2 y=p, 3 z=s$ leads (after transformations) to the equivalent system $\left\{\begin{aligned} x+p+s & =2 \\ x p s & =-6 . \text { By Vieta's theorem, the numbers } x, p, s-\text { are the roots } \\ x p+p s+s x & =-5\end{aligned}\right.$ of the cubic equation $t^{3}-2 t^{2}-5 t+6=0$,... | (1,-1,1),(1,\frac{3}{2},-\frac{2}{3}),(-2,\frac{1}{2},1),(-2,\frac{3}{2},\frac{1}{3}),(3,-1,\frac{1}{3}),(3,\frac{1}{2},-\frac{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,379 |
10.1. We consider all possible pairs of quadratic equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ such that each equation has two distinct roots. Is it true that the expression $\frac{1}{x_{1} x_{3}} + \frac{1}{x_{1} x_{4}} + \frac{1}{x_{2} x_{3}} + \frac{1}{x_{2} x_{4}}$, where the numbers $x_{1}, x_{2}$ are... | Solution: According to Vieta's theorem
$$
x_{1} x_{2}=-\left(x_{3}+x_{4}\right)=q \text { and } x_{3} x_{4}=-\left(x_{1}+x_{2}\right)=p
$$
Then
$$
\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}=\frac{x_{3}+x_{4}}{x_{1} x_{3} x_{4}}=-\frac{q}{p x_{1}}
$$
Similarly
$$
\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}=-\frac{... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 18,380 |
10.2. In rectangle $A B C D$, point $E$ lies on side $B C$ and does not coincide with points $B$ and $C$. Point $M$ is an arbitrary point on segment $A D$, and point $N$ is an arbitrary point on segment $M D$. Point $K$ lies on segment $A B$, and point $T$ lies on segment $C D$. Prove that the sum of the areas of trian... | # Solution:
Method 1. Notice that when point $E$ moves along side $BC$, the area of triangle $MEN$ (and thus the sum of the areas of triangles $AKM$, $MEN$, and $NDT$) does not change; therefore, we can assume that the projection of point $E$ lies on segment $MN$. Draw lines through points $M$, $E$, and $N$ parallel t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,381 |
10.3. In the distant times of stagnation in the Soviet Union, 15 and 20 kopeck coins were in circulation. Schoolboy Valera had a certain amount of money only in such coins. Moreover, the number of 20 kopeck coins was greater than the number of 15 kopeck coins. Valera spent one-fifth of all his money, paying two coins f... | Solution: One fifth of Valera's capital could be either 30, 35, or 40 kopecks. Then, after buying the ticket, he should have had 120, 140, or 160 kopecks left, and the cost of the lunch was either 60, 70, or 80 kopecks. The maximum value of three coins is 60 kopecks, so the last two scenarios are impossible. Therefore,... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,382 |
10.4. A circle with center $O$ is inscribed in angle $BAC$. A tangent to the circle, parallel to line $AO$, intersects ray $AB$ at point $P$. Prove that the equality $AP = AO$ holds. | Solution: Let $M$ and $N$ be the points of tangency of the circle with the lines $AB$ and $NP$ respectively, as shown in the figure. Denote $\angle BAO = \alpha$. Then, from the right triangle $AMO$, we find that $\angle AOM = 90^\circ - \alpha$. The angle $\angle AON$ is a right angle, so $\angle MON = \alpha$. Next, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,383 |
10.5. An increasing geometric progression consists of four different positive numbers, three of which form an arithmetic progression. What can the denominator of this progression be? Provide all possible answers and prove that there are no others. | Solution: Consider an arbitrary increasing geometric progression of four numbers $b, b q, b q^{2}, b q^{3} (b>0, q>1)$. If the first three terms (or the last three) form an arithmetic progression, then we have the equation $2 b q=b+b q^{2}$, from which $q^{2}-2 q+1=0 \Leftrightarrow q=1$. This is a contradiction.
Supp... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,384 |
10.6. There are 300 apples, any two of which differ in weight by no more than three times. Prove that they can be placed into 150 bags, with two apples in each, such that any two bags differ in weight by no more than two times. | Solution: We will sort all the apples by weight, for example, in descending order. We will form pairs as follows: the first apple with the three hundredth, the second with the two hundred and ninety-ninth, the third with the two hundred and ninety-eighth, ..., the one hundred and fiftieth with the one hundred and fifty... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,385 |
10.1. Let $f(x)=x^{2}+2 a x+b$. It is known that the equation $f(x)=0$ has two roots. Prove that then for any positive $k$ the equation $f(x)+k(x+a)^{2}=0$ also has two roots. | First solution. By the condition $a^{2}-b>0$. Let $F(x)=f(x)+k(x+a)^{2}$. We will compute the discriminant of the new quadratic polynomial $F(x)$. We have:
$$
F(x)=x^{2}+2 a x+b+k(x+a)^{2}=(k+1) x^{2}+2 a(k+1) x+\left(b+k a^{2}\right)
$$
Thus,
$$
\frac{1}{4} D_{1}=a^{2}(k+1)^{2}-(k+1)\left(b+k a^{2}\right)=(k+1)\lef... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,386 |
10.2. The circle passing through the vertices $A, B, D$ of trapezoid $A B C D$ intersects its lateral side $C D$ at point $K$. Prove that the circle circumscribed around triangle $B C K$ is tangent to the line $A B$. | Solution. Let $P$ be a point lying on the extension of segment $AB$ beyond point $B$ (see Fig. 6). Then the statement of the problem is equivalent to the angle $PBC$ being equal to the inscribed angle $CKB$. However, angles $CKB$ and $DKB$ are adjacent, so $\angle CKB = 180^\circ - \angle DKB$. On the other hand, quadr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,387 |
10.4. Is it true that any even number greater than 1000 can be represented as
$$
n(n+1)(n+2)-m(m+1)
$$
where \( m \) and \( n \) are natural numbers? | Answer: Incorrect.
Solution. Note that the product of three consecutive natural numbers $n(n+1)(n+$ $+2)$ is divisible by 3. Let's consider several cases.
If $m$ has a remainder of 0 or 2 when divided by 3, then $m(m+1)$ is divisible by 3. Therefore, the number $n(n+1)(n+2)-m(m+1)$ is divisible by 3.
If $m$ has a re... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,389 |
10.5. Can one choose a number $n \geqslant 3$ and fill an $n \times n$ table with distinct natural numbers from 1 to $n^{2}$ in such a way that in each row there are three numbers, one of which is equal to the product of the other two | Answer: No.
Solution: Suppose the table could be filled in the required manner. Consider in each row the three numbers: two multipliers and their product. Mark the smallest multiplier in each row. Since there are $n$ rows, there are a total of $n$ smallest multipliers. And since they are all different, there will be o... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,390 |
1. A mathematician left point A for point B. After some time, a physicist also left point A for point B. Catching up with the mathematician after 20 km, the physicist, without stopping, continued to point B and turned back. They met again 20 km from B. Then each continued in their respective directions. Upon reaching p... | # Answer: 45.
Solution. From the first meeting to the second meeting, the mathematician walked a total of 100 - 20 - 20 = 60 km, while the physicist walked -100 - 20 + 20 = 100 km. It is clear from this that the ratio of their speeds is 6:10 or 3:5. From the second to the third meeting, they will walk together 100 + 1... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,391 |
4. In triangle $ABC$, the bisector $AL$ and the altitude $BH$ are drawn. It turns out that the perpendicular bisector of segment $LH$ intersects side $AB$ at its midpoint. Prove that triangle $ABC$ is isosceles. | Solution. Let $M$ be the midpoint of side $AB$ (see fig.). Then, from the right triangle $ABH$, we get $HM=\frac{1}{2} AB$. Since point $M$ lies on the perpendicular bisector of segment $LH$, we have $LM=HM=\frac{1}{2} AB$. Therefore, triangle $ABL$ is a right triangle and $AL \perp BC$. Thus, $AL$ is both the bisector... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,392 |
5. The city center is a rectangle measuring $5 \times 8$ km, consisting of 40 blocks, each $1 \times 1$ km, with boundaries formed by streets that create 54 intersections. What is the minimum number of police officers needed to be placed at the intersections so that any intersection can be reached by at least one polic... | Solution. Evaluation. Consider the intersections on the boundary. There are 26 in total. Each police officer can control no more than 5 intersections (if he is on the boundary, then exactly 5, if he is inside the city, then no more than 3 on each side and no more than 5 in the corner). Therefore, at least 6 police offi... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,393 |
6.1. A cross-country race was held in the forest. Discussing its results, one squirrel said: "The first place was taken by the hare, and the second was the fox." Another squirrel objected: "The hare took second place, and the elk was first." To which the owl remarked that in each squirrel's statement, one part was true... | Solution: If the hare took first place, then there is no true statement in the second squirrel's statements. Therefore, the first part of the first squirrel's statement is false, which means the second part of her statement is true, that is, the fox was second. Then, in the second squirrel's statements, the first part ... | Themoosetookfirstplace,thefoxwas | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,394 |
6.3. The hunter told a friend that he saw a wolf with a one-meter tail in the forest. That friend told another friend that a wolf with a two-meter tail had been seen in the forest. Passing on the news further, ordinary people doubled the length of the tail, while cowards tripled it. As a result, the 10th channel report... | Solution: Note that when information is transmitted by ordinary people, the length of the tail is multiplied by 2, and when transmitted by cowardly people, it is multiplied by 3. Therefore, the number of twos in the product equals the number of ordinary people (and the number of threes equals the number of cowardly peo... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,396 |
6.4. Three pirates were dividing a bag of coins. The first took 3/7 of all the coins; the second took 51 percent of the remainder. After this, the third received 8 fewer coins than the second. How many coins were in the bag? Justify your answer. | Solution: The third pirate received $49\%$ of the remainder, which is $2\%$ less than the third. These two percent of the remainder amount to 8 coins, so one percent is
4 coins, and the entire remainder is 400 coins. These 400 coins make up $1-\frac{3}{7}=\frac{4}{7}$ of the total. Therefore, the total number of coins... | 700 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,397 |
6.5. On cards, all two-digit numbers from 10 to 99 are written (one number per card). All these cards are lying on the table face down. What is the minimum number of cards that need to be flipped to guarantee that at least one of the revealed numbers is divisible by 7? Justify your answer. | Solution: There are exactly 13 two-digit numbers divisible by 7 (14 = 7 * 2, 21 = 7 * 3, ..., 98 = 7 * 14). If at least 13 cards are not flipped, it is impossible to exclude the situation where all these numbers remain covered. Therefore, no more than 12 cards can be left in their original position, and the rest must b... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,398 |
# Task 11.1
$P(x)$ and $Q(x)$ are reduced quadratic trinomials, each having two distinct roots.
It turned out that the sum of two numbers obtained by substituting the roots of the trinomial $P(x)$ into the trinomial $Q(x)$ is equal to the sum of two numbers obtained by substituting the roots of the trinomial $Q(x)$ i... | # Solution
Let $a_{1}$ and $a_{2}$ be the roots of the quadratic polynomial $P(x)$, and $b_{1}$ and $b_{2}$ be the roots of the quadratic polynomial $Q(x)$.
First method. $P(x)=\left(x-a_{1}\right)\left(x-a_{2}\right), Q(x)=\left(x-b_{1}\right)\left(x-b_{2}\right)$.
Therefore, $\left(b_{1}-a_{1}\right)\left(b_{1}-a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,399 |
# Task 11.2
Let $\mathrm{S}(\mathrm{n})$ be the sum of the digits of the number $\mathrm{n}$. Find all $\mathrm{n}$ for which $\mathrm{n}+\mathrm{S}(\mathrm{n})+\mathrm{S}(\mathrm{S}(\mathrm{n}))+\ldots \mathrm{S}(\mathrm{S}(\ldots \mathrm{S}(\mathrm{n}) \ldots)=2000000$ (here there are $n$ terms in the sum, and in ea... | # Solution
Proof. All $n$ terms on the left side give the same remainder when divided by 3, which matches the remainder of the number $n$ when divided by 3. By considering the three different cases, we find that the remainder of the left side when divided by 3 is either 0 or 1. However, the right side, the number 2000... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,400 |
# Problem 11.3
Prove that for any natural number $n$ the following inequality holds:
$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}>2 \sqrt{n}-\frac{3}{2}$
## Number of points 7
# | # Solution
We will prove the inequality by mathematical induction:
$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}} \geq 2 \sqrt{n}-\frac{3}{2}+\frac{1}{2 \sqrt{n}}$
1) It is true for $n=1$
2) Suppose it is true for $n=k:$
$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{k}} \geq 2 \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,401 |
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