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1. The lengths of the sides of a right-angled triangle are expressed by natural numbers. Prove that at least one of the sides is divisible by 2, and at least one of the sides is divisible by 3. | Solution.
Let $a$ and $b$ be the legs, and $c$ be the hypotenuse. Then $b^{2}=c^{2}-a^{2}=(c-a)(c+a)$.
Consider the "worst" of the possible cases.
If $\boldsymbol{c}$ and $\boldsymbol{a}$ are both odd, then their difference and sum are even, hence $b^{2}$ is even, and $\boldsymbol{b}$ is even.
If $a$ and $b$ are bo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,719 |
2. Based on $AC$ of the isosceles triangle $ABC$, a point $E$ is taken, and on the lateral sides $AB$ and $BC$ - points $K$ and $M$ such that $KE$ is parallel to $BC$ and $EM$ is parallel to $AB$. What part of the area of triangle $\mathrm{ABC}$ is occupied by the area of triangle $KEM$, if $\mathbf{BM}: \mathbf{EM}=2:... | Answer: $\frac{6}{25}$.
Solution.
Quadrilateral BKEM is a parallelogram. Triangles AKE and EMC are isosceles. Each of them is similar to triangle $\mathrm{ABC}$ with similarity coefficients $\frac{\mathrm{KE}}{\mathrm{BC}}=\frac{2 x}{2 x+3 x}=\frac{2}{5} \quad$ and $\quad \frac{\mathrm{MC}}{\mathrm{BC}}=\frac{3 x}{2 ... | \frac{6}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,720 |
3. Three brigades, working together, must complete a certain job. It is known that the first and second brigades together can complete it 36 minutes faster than the third brigade. In the time it takes for the first and third brigades to complete the job together, the second brigade can complete only half of the job. In... | Answer: 1 hour 20 minutes.
Solution.
Let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ be the productivity of the first, second, and third teams, respectively, that is, the part of the work that a team completes in 1 hour. Using the first condition of the problem, we form an equation. Since the productivity of the first and s... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,721 |
4. Solve the equation: $3 x^{2}+14 y^{2}-12 x y+6 x-20 y+11=0$. | Answer: $\boldsymbol{x}=3, \boldsymbol{y}=2$
Solution.
Consider the equation as a quadratic in $\boldsymbol{x}$, with $\boldsymbol{y}$ treated as a parameter. The equation $3 x^{2}-6 x(2 y-1)+14 y^{2}-20 y+11=0$ has real roots if and only if its discriminant is non-negative, i.e., $D=36(2 y-1)^{2}-12\left(14 y^{2}-20... | 3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,722 |
5. In a chess tournament, where each participant played against every other, three chess players fell ill and dropped out of the tournament before it reached its halfway point. A total of 130 games were played in the tournament. How many chess players participated in the tournament? | Answer: 19 people
Solution.
If 16 chess players participated in the tournament, the number of games they played should not exceed (16 x 15):2 = 120 games. Therefore, more than 16 people played in the tournament. Let's consider the following cases.
A) The tournament started with 17 participants. Then 14 of them, who ... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,723 |
1. All gnomes are divided into liars and knights. Liars always lie, and knights always tell the truth. On each cell of a $4 \times 4$ board, there is a gnome. It is known that among them, there are both liars and knights. Each gnome stated: “Among my neighbors (by side) there are an equal number of liars and knights.” ... | Answer: 12 liars.
Solution: Any dwarf standing on the side of the square but not in a corner cannot be telling the truth, because they have three neighbors, and among them, there cannot be an equal number of knights and liars. Therefore, these eight dwarfs are liars. Thus, the dwarfs standing in the corners are also l... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,724 |
4. Does there exist a natural number $n$ such that the number $n^{2}+3$ has exactly ten digits, and all of them are different? | Answer. No, it does not exist.
Solution. If in a ten-digit number $n^{2}+3$ all digits are different, then it is divisible by 9 (by the divisibility rule for 9: the sum of its digits is 45). But then it is divisible by 3, and thus the number $n$ itself is divisible by 3, since 3 is a prime. Then $n^{2}+3$ gives a rema... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,725 |
10.6. Three consecutive odd numbers are written on the board. Can the sum of the remainders of dividing these three numbers by 2022 be a prime number?
( | Answer: It cannot.
Solution: Let $r$ be the remainder of the smaller of the given odd numbers when divided by 2022, so that $r$ is some odd number from the set $\{1,3,5, \ldots, 2021\}$. If $r \leqslant 2017$, then the other two remainders are $-r+2$ and $r+4$, so the sum of the remainders is $r+(r+2)+(r+4)=3(r+2)$ - ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,727 |
10.7. Given a cyclic quadrilateral $A B C D$, in which $\angle A=2 \angle B$. The bisector of angle $C$ intersects side $A B$ at point $E$. Prove that $A D+A E=B E$.
(A. Kuznetsov) | Solution. Let $\angle A B C=\alpha$, then by the condition $\angle D A B=2 \alpha$. On the extension of the segment $A B$ beyond point $A$, mark point $F$ such that $A D=A F$. Then triangle $A F D$ is isosceles, and its base angles are equal. Since $\angle F A D=180^{\circ}-2 \alpha$, then $\angle A F D=\angle A D F=\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,728 |
10.8. On a plane, $N$ points are marked. Any three of them form a triangle, the angles of which in degrees are expressed by natural numbers. For what largest $N$ is this possible? | Answer: 180.
First solution. Example. First, we show that for $N=180$, the required condition is possible. Mark 180 points on a circle, dividing it into 180 equal arcs, each $2^{\circ}$. The measure of any arc with endpoints at two of the marked points is expressed as an even number of degrees, so the measure of any i... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,729 |
10.9. In the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip i... | # Answer: 50.
Solution. Example. The chip 50 is sequentially exchanged 99 times with the next one counterclockwise. We get the required arrangement.
There are several ways to prove the estimate, below we provide two of them.
The first way. Suppose that for some $k<50$ the required arrangement is obtained.
At any mo... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,731 |
10.10. Prove that there exists a natural number $b$ such that for any natural $n>b$ the sum of the digits of the number $n!$ is not less than $10^{100}$. | Solution. Let $a=10^{100}$. Denote by $s(m)$ the sum of the digits of the number $m$. Note the simple property $s(\ell)+s(m) \geqslant s(\ell + m)$, which is immediately clear if the numbers $\ell$ and $m$ are added in a column.
Lemma. Let $k$ be a natural number, and let the natural number $m$ be divisible by $10^{k}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,732 |
7.1. The angle formed by the bisector of angle $A B C$ with its sides is 6 times smaller than the angle adjacent to angle $A B C$. Find angle $A B C$. | Answer: 45 degrees. Solution. Let $x$ be the degree measure of angle $A B C$. From the condition of the problem, we get the equation $\frac{x}{2}=\frac{180-x}{6} \Leftrightarrow 8 x=360 \Leftrightarrow x=45$ (degrees). | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,733 |
7.2. A car, moving at a constant speed, traveled from point A to point B in 3 hours. To reduce the travel time on the return trip, the driver left point B at a speed 25% higher, and upon reaching the midpoint of the journey between A and B, increased the speed by another 20%. How long will the return trip take? | Answer: 2 hours 12 minutes. Solution. Let the distance from A to B be $a$ (km), and the speed of movement from A to B be $v$ (km/h). Then $\frac{a}{v}=3$. Let $\mathrm{C}$ be the midpoint of the path between A and B. Then the time of movement on the return trip from B to C is $\frac{a / 2}{v \cdot 1.25}=\frac{2}{5} \fr... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,734 |
7.3. We have 10 sticks of lengths $1,2,4, \ldots, 2^{9}$ (cm). Can a triangle be formed using these sticks, not necessarily all? | Answer: No. Solution. Suppose, for the sake of contradiction, that the triangle can be formed, and let $2^{n}$ be the length of the longest stick involved in the construction ($n \leq 9$). However, the side of the triangle containing this stick will then be greater than the sum of the other two sides, since $2^{n}>1+2+... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,735 |
7.4. Does there exist a six-digit natural number that, when multiplied by 9, is written with the same digits but in reverse order? | Answer: it exists. Solution. Let $\overline{a b c d e f}$ be the desired number, i.e., $\overline{a b c d e f} \cdot 9=\overline{f e d c b a}$. Then it is obvious that $a=1, b=0$ (otherwise, multiplying by 9 would result in a seven-digit number). Therefore, $f=9$, and the second-to-last digit $e=8$ (which follows from ... | 109989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,736 |
7.5. At the vertices of a cube, 8 numbers: $1, 2, \ldots, 8$ were arranged in some order, and then for each of the 12 edges of the cube, the sum of the two numbers at its ends was calculated. Prove that among these sums there are coinciding ones. | Solution. Suppose the opposite, then there will be 12 different numbers on the edges among the possible sums: from the minimum, equal to $3=1+2$, to the maximum, equal to $15=8+7$. Thus, among these 13 possible numbers, there is only one gap (not occupied by sums on the edges). If this gap is not among the numbers $\{3... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,737 |
# 3. A quadrilateral ABCD is circumscribed around a circle, with AD||BC. Prove that $A B+C D \geq 2 \sqrt{S}$, where S is the area of quadrilateral ABCD. | From the condition of the problem, it follows that the quadrilateral $ABCD$ is a trapezoid or a rhombus, in particular a square. In both cases, we have $S=\frac{AD+BC}{2} h$. It is obvious that $h \leq \frac{AB+CD}{2}$.
Quadrilateral $ABCD$ is circumscribed around a circle, so the equality $\mathrm{AD}+\mathrm{BC}=\ma... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,738 |
# 5. Write the given table of numbers so that in each column and each row there is an arithmetic progression:
| $\mathbf{1}$ | | | |
| :--- | :--- | :--- | :--- |
| | | | $\mathbf{6}$ |
| | | $\mathbf{6}$ | |
| | $\mathbf{9}$ | | | | Let $a_{1}, a_{2}, a_{3}, a_{4}$ be the differences of the arithmetic progressions written in the rows of the table. Then its cells can be filled as shown in the figure
| 1 | $1+a_{1}$ | $1+2 a_{1}$ | $1+3 a_{1}$ |
| :---: | :---: | :---: | :---: |
| $6-3 a_{2}$ | $6-2 a_{2}$ | $6-a_{2}$ | 6 |
| $6-2 a_{3}$ | $6-a_{3}... | \begin{pmatrix}\hline1&3&5&7\\\hline4.5&5&5.5&6\\\hline8&7&6&5\\\hline11.5&9&6.5&4\\\hline\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,740 |
1. On the board, 2020 quadratic equations are written:
$$
\begin{gathered}
2020 x^{2} + b x + 2021 = 0 \\
2019 x^{2} + b x + 2020 = 0 \\
2018 x^{2} + b x + 2019 = 0 \\
\ldots \\
x^{2} + b x + 2 = 0
\end{gathered}
$$
(each subsequent equation is obtained from the previous one by decreasing the leading coefficient and ... | Solution. According to the theorem converse to Vieta's theorem, the product of the roots of the first equation is $\frac{2021}{2020}$, the product of the roots of the second equation is $\frac{2020}{2019}$, the third is $-\frac{2019}{2018}, \ldots$, the two thousand and twentieth is $\frac{2}{1}$. Therefore, the produc... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,741 |
2. A team consisting of boys and girls from the Rostov region went to a CS:GO esports tournament. The average number of points scored by the girls turned out to be 22, by the boys - 47, and the average number of points for the entire team - 41. What is the percentage of girls in this team? | Solution. Let the number of girls be $x$, the number of boys be $y$, and the total points scored by them be $S_{1}$ and $S_{2}$ respectively. From the conditions, the following equations can be derived: $\frac{S_{1}}{x}=22, \frac{S_{2}}{y}=47$ and $\frac{S_{1}+S_{2}}{x+y}=41$. Then $S_{1}=22 x$ and $S_{2}=47 y$, from w... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,742 |
4. Each diagonal of the convex pentagon $A B C D E$ cuts off a triangle of unit area from it. Calculate the area of the pentagon $A B C D E$.
Lemma. Let in a convex quadrilateral $A B C D$ the diagonals $A C$ and $B D$ intersect at point $O$. If triangles $A B O$ and $C O D$ are equal in area, then $A B C D$ is a trap... | Solution. Since $S_{A B E}=S_{A B C}$, by the lemma proved above, we have $E C \| A B$. The other diagonals are also parallel to the corresponding sides. Let $P$ be the intersection point of $B D$ and $E C$. If $S_{B P C}=x$, then $S_{A B C D E}=$ $S_{A B E}+S_{E P B}+S_{E D C}+S_{B P C}=3+x\left(S_{E P B}=S_{A B E}=1\... | \frac{\sqrt{5}+5}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,744 |
5. In a computer game, one person can play as one of three factions: $T, Z$ or $P$. There is a network play mode in which 8 players are divided into two teams of 4 players each. How many different matches can there be, differing in the sets of factions? Matches are considered different if one match has a team that the ... | Solution. First, let's calculate the number of ways to form one team from the specified factions. Let's number the factions. The number of options is equal to the number of solutions to the equation $x_{1}+x_{2}+x_{3}=4, x_{i} \geqslant 0$, where $x_{i}$ is the number of players from faction $i$. The number of solution... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,745 |
1. There are candies in five bags. The first has 2, the second has 12, the third has 12, the fourth has 12, and the fifth has 12. Any number of candies can be moved from any bag to any other bag. What is the minimum number of moves required to ensure that all bags have an equal number of candies? | Answer: 4.
There are 50 candies in total and they should be 10 each. In four bags, there are 12 each, which is more than 10, and these bags participate in the redistributions to reduce to 10. Therefore, there are no fewer than 4 redistributions.
From the second to the fifth bag, two candies each go to the first.
Cri... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,747 |
3. There are seven red cubes, three blue ones, and nine green ones. Ten cubes were placed in a gift bag. In how many different ways could this have been done? | Answer: 31.
Let's put red cubes into the bag (8 ways from 0 to 7), now we place the blue cubes (4 ways from 0 to 3). Add the necessary number of green cubes (1 way). In total, $8 \times 4=32$.
One operation is impossible: 10 green. Therefore, there is one fewer way. | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,749 |
4. There is a five-pointed star. It has three equal angles at the vertices and the two remaining ones are also equal to each other. Is it true that among the five triangles at the vertices of the star, at least one is isosceles? | Answer: Correct.
Solution. We can always introduce notations such that the angles at vertices B and E, and C and D, are equal. Since the exterior angle of a triangle is equal to the sum of t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,750 |
5. Pete wants to color several cells of an $8 \times 8$ square so that for any vertex, there is a colored square to which it belongs. What is the minimum number of squares he must color? | Answer: 25
Let's mark 25 vertices of an $8 \times 8$ square (see the figure on the right). At each marked vertex, there must be a shaded square. Each square touches only one such vertex, so there must be at least 25. The example shown below indicates that the 25 shaded squares touch all the vertices of the grid.
 and obtained a set of the same 4 numbers (possibly in a different ord... | Answer: 3 numbers.
Solution. First, let's prove that at least one zero was written on the board. Indeed, suppose the numbers $a, b, c, d$ written on the board were non-zero, then their product $a b c d$ should equal the product of the new numbers $a b c d \sin \alpha \cos \alpha \operatorname{tg} \alpha \operatorname{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,753 |
11.3. The numbers $x$ and $y$ satisfy the inequalities $x^{5}>y^{4}$ and $y^{5}>x^{4}$. Prove that $x^{3}>y^{2}$. | Solution. We will prove that $x>1$ and $y>1$. The right-hand sides of the inequalities are non-negative, so both numbers $x$ and $y$ are positive. Multiplying the given inequalities and canceling by the positive number $(xy)^4$, we get: $xy>1$. This means that at least one of the (positive!) numbers $x$ and $y$ is grea... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,754 |
11.4. Quadrilateral $ABCD$ is inscribed in a circle. The perpendicular to side $BC$, drawn through its midpoint - point $M$, intersects side $AB$ at point $K$. The circle with diameter $KC$ intersects segment $CD$ at point $P (P \neq C)$. Prove that lines $MP$ and $AD$ are perpendicular. | Solution. Let $\omega$ be the circle constructed on $K C$ as its diameter, then point $M$ lies on $\omega$, since angle $C M K$ is a right angle. Therefore, $\angle C P M = \angle C K M = \alpha$ (they subtend the arc $C M$ of circle $\omega$). Let $T$ be the intersection point of lines $A D$ and $M P$. We will assume ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,755 |
11.5. In the castle, there are 25 identical square rooms, forming a $5 \times 5$ square. In these rooms, 25 people—liars and knights (liars always lie, knights always tell the truth)—settled, one person per room. Each of these 25 people said: “At least one of the neighboring rooms to mine is occupied by a liar.” What i... | Answer: 18 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 7 liars (thus showing that there are no more than 18 knights).
First, consider dividing the rooms into 6 groups (2 rooms, marked in gray, do not belong to any group). In each gro... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,756 |
8.2 Three numbers $x, y, z$ satisfy the relation $x^{2}+y^{2}=x y\left(z+\frac{1}{z}\right)$. Prove that at least one of the numbers $x$ or $y$ is equal to the product of the other two numbers. | Solution. According to the condition
$x^{2}-x y z+y^{2}-\frac{x y}{z}=0$
$x(x-y z)-y\left(\frac{x}{z}-y\right)=0$,
$x(x-y z)-\frac{y}{z}(x-y z)=0$,
$(x-y z)\left(x-\frac{y}{z}\right)=0$
Therefore, either $\mathrm{x}-\mathrm{yz}=0$, or $\mathrm{x}-\frac{\mathrm{y}}{\mathrm{z}}=0$, i.e. $\mathrm{y}-\mathrm{xz}=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,758 |
8.3 There are 11 different natural numbers, not greater than 20. Prove that among them, two numbers can be chosen such that one divides the other. | Solution. Consider the greatest odd divisor of each number. Among such divisors, there must be two that are the same (since there are only 10 odd numbers not greater than 20). Of the corresponding two numbers, the larger one is divisible by the smaller one. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,759 |
8.4 The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$ if it is known that $\mathrm{a}_{1}=1,... | Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,760 |
8.5 In triangle $\mathrm{ABC}$, the bisector $\mathrm{BL}$ is drawn, and a point $\mathrm{K}$ is chosen on its extension beyond point $\mathrm{L}$ such that $\mathrm{LK}=\mathrm{AB}$. It turns out that line $\mathrm{AK}$ is parallel to line $\mathrm{BC}$. Prove that $\mathrm{AB}>\mathrm{BC}$. | Solution. See fig.
Angle AKV is equal to KVC (since AK $\| \mathrm{BC}$, VK is a transversal). Therefore, triangle BAK is isosceles. Hence, AK $=\mathrm{AB}$. But then in triangle LKA, two sides are equal: $\mathrm{AK}=\mathrm{KL}(=\mathrm{AB})$. Triangle LBC is similar to triangle LKA, so it is also isosceles and $\m... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,761 |
11.1. Answer. $2022^{2022}>2023^{2021}$. | Solution. Consider the ratio $\frac{2023^{2021}}{2022^{2022}}=\frac{2023^{2022} \cdot 2023^{-1}}{2022^{2022}}=\left(\frac{2023}{2022}\right)^{2022} \cdot 2023^{-1}=\left(1+$ $\left.\frac{1}{2022}\right)^{2022} \cdot 2023^{-1}$.
Since $\left(1+\frac{1}{\alpha}\right)^{\alpha}2023^{2021}$. | 2022^{2022}>2023^{2021} | Inequalities | proof | Yes | Yes | olympiads | false | 14,762 |
10.1. Consider functions of the form $y=x^{2}+a x+b$, where $a+b=2021$. Prove that the graphs of all such functions have a common point. | Solution. $y$ (1) $=1+a+b=2022$. Therefore, each of the given graphs passes through the point with coordinates $(1 ; 2022)$. | (1;2022) | Algebra | proof | Yes | Yes | olympiads | false | 14,766 |
10.2. In the product of five natural numbers, each factor was decreased by 3. Could the product have increased exactly 15 times as a result? (N. Agakhanov, I. Bogdanov) Answer. Yes, it could. | Solution. As an example, the product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 48$ fits. After the specified operation, it becomes $(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot 45=720=15 \cdot 48$.
Remark. The given example is the only one. We will indicate how to come up with it. Suppose that four of the factors were equal to 1, ... | 1\cdot1\cdot1\cdot1\cdot48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,767 |
10.3. $\left(a_{n}\right)$ is an arithmetic progression with a common difference of 1. It is known that $\mathrm{S}_{2022}$ is the smallest among all $S_{n}$ (less than the sum of the first $n$ terms for any other value of n). What values can the first term of the progression take? | Answer: $a_{1}$ belongs to the interval ( $-2022 ; -2021$ ).
Solution. Since the difference of the progression is positive, the progression is increasing. Therefore, the described situation is possible if and only if the members of the progression from the first to the 2022nd are negative, and starting from the 2023rd... | -2022<a_{1}<-2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,768 |
10.4. Can the sum of 100 consecutive natural numbers end with the same digit as the sum of the next 98 numbers? | Answer: It cannot.
First solution. Note that the sum of 100 consecutive natural numbers is an even number, as it contains exactly 50 odd addends. The sum of 98 consecutive natural numbers is an odd number, as it contains exactly 49 odd addends. Therefore, these sums end in digits of different parity.
Second solution.... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,769 |
10.5. Circle $\omega$ is circumscribed around an acute-angled triangle $ABC$. A point $D$ is chosen on side $AB$, and a point $E$ is chosen on side $BC$ such that $AC \parallel DE$. Points $P$ and $Q$ on the smaller arc $AC$ of circle $\omega$ are such that $DP \parallel EQ$. The rays $QA$ and $PC$ intersect line $DE$ ... | Solution.
Since quadrilateral $AB C Q$ is inscribed and $A C \| D E$, we have $\angle B E X=\angle B C A=\angle B Q A=\angle B Q X$. Therefore, quadrilateral $X B E Q$ is inscribed, from which $\angle X B Q=\angle X E Q=\angle D E Q$. Similarly, quadrilateral $Y B D P$ is inscribed, and $\angle P B Y=\angle P D E$. By... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,770 |
7.1. On a long ribbon, the digits 201520152015 .... are written. Vasya cut out two pieces of the ribbon with scissors and formed a positive number from them, which is divisible by 45. Provide an example of such pieces and write down the number formed from them. | Answer: for example, you can cut out pieces "2" and "520" and form the number 2520, which is divisible by 45.
Other options are possible: from pieces 52 and 20, the number 5220 is formed; from pieces 1 and 2015, the number 12015; from pieces 201 and 15, the number 20115, and so on.
An example can be chosen based on t... | 2520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,771 |
7.3. For breakfast, Karlson ate $40 \%$ of the cake, and Little One ate 150 g. For lunch, Fräulein Bock ate $30 \%$ of the remainder and another 120 g, and Matilda licked the remaining 90 g of crumbs from the cake. What was the initial mass of the cake? | Answer: 750 g.
Solution. First method (solving "from the end").
1) $90+120=210$ (g) of the cake remained after Fröken Bok ate $30\%$ of the remainder.
Since Fröken Bok ate $30\%$ of the remainder, 210 g is $70\%$ of the remainder.
2) $210 \div 0.7 = 300$ (g) of the cake was left before Fröken Bok started her lunch.
... | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,773 |
7.5. On the board, 7 different odd numbers are written. Tanya calculated their arithmetic mean, and Dan ordered these numbers in ascending order and chose the number that ended up in the middle. If Tanya's number is subtracted from Dan's, the result is the number $\frac{3}{7}$. Did anyone make a mistake? | Answer: one of them must have made a mistake.
Solution. Let the numbers on the board be $a, b, c, d, e, f$ and $g$, with $a<b<c<d<e<f<g$. Then Tanya's number is $\frac{a+b+c+d+e+f+g}{7}$, and Danya's number is $d$. From the problem statement, it follows that $\frac{a+b+c+d+e+f+g}{7}-d=\frac{3}{7}$, which means $a+b+c+... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,775 |
2. Simplify the fraction: $\frac{x^{14}+x^{13}+\ldots+x+1}{x^{5}+x^{4}+x^{3}+x^{2}+x}$. | Solution. We will use the formula
$$
a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} \cdot b+a^{n-3} \cdot b+\ldots+a^{2} b^{n-3}+a b^{n-2}+b^{n-1}\right)
$$
to transform the given expression:
$$
\begin{aligned}
& \frac{x^{14}+x^{13}+\ldots+x+1}{x^{5}+x^{4}+x^{3}+x^{2}+x}=\frac{(x-1)\left(x^{14}+x^{13}+\ldots+x+1\right)}{(x-... | \frac{x^{10}+x^{5}+1}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,777 |
3. From a vessel containing 5 liters of a $6 \%$ acid solution, 1 liter was poured out, after which 2 liters of water were added. Find the concentration of the resulting solution. | Solution. After pouring out 1 liter, 4 liters of the solution remained in the container, with the amount of acid being $4 \cdot 0.06 = 0.24$ liters. After adding two liters of water, the volume of the solution became 6 liters, while the amount of acid remained unchanged. Thus, the concentration of the acid became $0.24... | 0.04 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,778 |
4. Points $M$ and $N$ are taken on the sides $BC$ and $CD$ of parallelogram $ABCD$, respectively. Diagonal $BD$ intersects sides $AM$ and $AN$ of triangle $AMN$ at points $E$ and $F$, respectively, dividing it into two parts. Prove that these parts have equal areas if and only if point $K$, defined by the conditions $E... | Solution. Connect point $K$ with points $A, B$ and $D$, denoting by $O, L, P$ the points of intersection of $K A$ and $B D, K B$ and $A M, K D$ and $A N$ respectively (see figure). Diagonal $B D$ has divided triangle $A M N$ into two parts. Let's compare them.
Replace triangle $A F E$ with a figure of equal area locat... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,779 |
5. Petya and Vasya made 5 shots each in the shooting range. With their first three shots, they scored the same number of points, and with the last three shots, Petya scored three times more points than Vasya. The targets have bullet holes worth $10,9,9,8,8,5,4,4,3,2$ points. Where did each of them hit with their third ... | Solution. With the last three shots, Vasya could not score more than 9 points (otherwise, Petya would have scored no less than 30 with his last three shots). Vasya also could not score less than 9 points, as the smallest sum for three shots is $2+3+4=9$. Therefore, Vasya scored 2, 3, and 4 points, while Petya scored 10... | Petyascored10,Vasyascored2points | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,780 |
1. There are $N$ coins on the table. Sasha, Misha, and Vitalik take turns picking up coins from the table. Sasha can only take 1 coin per turn. Misha can take exactly 1, 3, 5, 7, or 9 coins per turn. Vitalik can take exactly 2, 4, 6, 8, or 10 coins per turn (if only 1 coin is left on the table, he skips his turn). The ... | Solution. a) Sasha chooses the following order of moves - first Sasha, then Misha, and third Vitalik. After Sasha's move, there are 2022 coins left on the table. Misha, before his move, has an even number of coins, and therefore cannot take the last one. After his move, there is an odd number of coins left on the table... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,781 |
2. In a correspondence mathematics olympiad, out of 500 participants, exactly 30 did not like the problem conditions, exactly 40 did not like the organization of the event, and finally, exactly 50 did not like the method of determining the winners of the olympiad. We will call an olympiad participant "significantly dis... | Solution. In total, there were $120=30+40+50$ "dissatisfactions" expressed, so there cannot be more than 60 "significantly dissatisfied" individuals, because otherwise they would have more than 120 "dissatisfactions" in total, which contradicts the condition. It remains to show that there could have been 60 of them. Fo... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,782 |
3. The sides of the quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ have the following lengths: $A B=9, B C=2$, $C D=14, D A=5$. Find the length of the diagonal $\boldsymbol{A} \boldsymbol{C}$, if it is known that it is an integer. | Solution. Apply the triangle inequality to $\triangle ABC$ and to $\triangle ACD$: for the first, we will have that $AB + BC > AC$, that is, $AC < AB + BC = 11$; for the second, we will have that $AC + CD > AD$, that is, $AC > CD - DA = 9$. Therefore, $9 < AC < 11$, from which $AC = 10$.
Answer: $AC = 10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,783 |
4. There are three pairwise distinct natural numbers $a, b, c$. Prove that the numbers $2023 + a - b, 2023 + b - c$, and $2023 + c - a$ cannot be three consecutive natural numbers. | Solution. If all three numbers are added and divided by 3, the result is their arithmetic mean, which on one side is
$$
(2023+a-b+2023+b-c+2023+c-a) / 3=2023
$$
and on the other side, it equals the middle of these three consecutive numbers.
Thus, one of the numbers equals 2023. For example, this is $2023+a-b=2023$, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,784 |
5. Grandfather, father, and grandson ran the distance from home to the stadium and back. The grandson ran there and back at the same speed. Grandfather ran there twice as fast as the grandson, and back three times slower. Father ran there twice as slow as the grandson, and back three times faster. In what order will th... | Solution. Let the grandson's speed be denoted by $x$, and the distance by $S$. Then the times for the grandson, father, and grandfather to complete their runs are respectively:
$$
t_{1}=\frac{S}{x}+\frac{S}{x}, t_{2}=\frac{S}{\frac{1}{2} x}+\frac{S}{3 x}, t_{3}=\frac{S}{2 x}+\frac{S}{\frac{1}{3} x}
$$
Thus, we need t... | The\grson\came\in\first,\the\father\,\\the\grfather\last | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,785 |
1. The trainer Bronislav Rasplyany was asked how many lions he has in the circus, to which he replied that he has 5 times fewer lions than non-lions. When asked how many tigers he has, he replied that he has 5 times more tigers than non-tigers. Could there be leopards in the circus of Bronislav Rasplyany? Explain your ... | Answer: they cannot.
## Solution:
Lions make up $1 / 6$ of the total number of animals (if there are $X$ lions, then there are $5X$ non-lions, making a total of $6X$ animals in the circus, and thus $X$ out of $6X$ is $1 / 6$). Similarly, we can say that tigers make up $5 / 6$ of the total number of animals. Therefore... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,786 |
2. The bisector of angle $B A D$ of a rectangular trapezoid $A B C D$ (with bases $A D$ and $B C, \angle B A D=90^{\circ}$) intersects the lateral side $C D$ at point $E$. Find the ratio $C E: E D$, if $A D+B C=A B$.
## Answer: $1: 1$. | # Solution:
Let's extend the bisector $A E$ until it intersects with line $B C$ (denote the intersection point as $F$). Since $\angle A B C$ is a right angle and $\angle F A B=45^{\circ}$, then $\angle A F B=45^{\circ}$, so $F A B$ is an isosceles right triangle, and $A B=B F$.
Therefore, $C F=B F-B C=A B-B C=A D$.
... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,787 |
3. Mario brought a round pizza with an area of $4 \mu^{2}$ and cut it into 4 pieces with two perpendicular straight cuts. Each cut passed 50 cm from the center of the pizza. His brother Luigi took the largest and the smallest piece, while the other two pieces went to Mario. Find the total area of the pieces that Mario ... | Answer: $1.5 m^{2}$.
## Solution:
Let's make two more cuts, symmetrically (relative to the center) to the ones already made. We will denote the areas of some regions as shown in the figure on the right (the shaded area is what Mario received). Then we notice that Mario received pieces with areas $x+y$ and $x+y$ (a to... | 1.5^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,788 |
4. Unlucky Emelya was given several metal balls, from which he broke 3 of the largest ones (their mass was $35 \%$ of the total mass of all the balls), then lost 3 of the smallest ones, and brought the remaining balls (their mass was $8 / 13$ of the unbroken ones) home. How many balls were given to Emelya | Answer: 10 balls.
## Solution:
Let the total mass of all the balls be M. Of these, Emelya broke balls with a total mass of $\frac{35}{100} M = \frac{7}{20} M$, did not break balls with a total mass of $\frac{13}{20} M$, brought home balls with a total mass of $\frac{8}{13} \cdot \frac{13}{20} M = \frac{8}{20} M$, and... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,789 |
5. On some cells of an $8 \times 8$ chessboard, one candy was placed. It turned out that in each row and each column, there is an even number of candies. Prove that then there is also an even number of candies on all the white cells.
# | # Solution:
Let the lower left cell of the board be black (if it's white, it doesn't change anything - we will just prove that the number of candies on all black cells is even). We will divide all the cells of the chessboard into 4 groups, as shown in the figure (white cells get numbers 2 and 3, black cells - 1 and 4)... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,790 |
1. On the road between cities A and B, there are some poles in certain places, and two numbers are written on each pole: how many kilometers from the pole to city A, and how many kilometers from the pole to city B. A tourist walking from one city to the other saw a pole where one of the numbers was three times the othe... | Answer: 7.
Solution: Without loss of generality, we can assume that the tourist saw the first post when he was closer to city A. Let the distance from the first post to A be $x$ kilometers, then the distance from it to B is $3x$. If the numbers on the second post are $y$ and $3y$, then $x + 3x = y + 3y$, since the sum... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,791 |
2. For 7 real numbers, it is known that the sum of any 3 of them is less than the sum of the remaining 4. Prove that all these numbers are positive. | First solution. Let's write down the numbers from the condition in non-decreasing order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{7}$. According to the condition, $a_{5}+a_{6}+a_{7}\left(a_{7}-a_{4}\right)+\left(a_{6}-a_{3}\right)+\left(a_{5}-a_{2}\right)$. Since each of the parentheses is non-negative, $a_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 14,792 |
3. Find the smallest real number $x$ such that $45 \%$ of $x$ and $24 \%$ of $x$ are natural numbers. | Answer: $\frac{100}{3}$.
First solution. Since $\frac{45}{100} x$ is a natural number, $x$ is a positive rational number. Let $x=\frac{m}{n}$, where the fraction $\frac{m}{n}$ is irreducible. The numbers $\frac{45}{100} \cdot \frac{m}{n}=\frac{9 m}{20 n}$ and $\frac{24}{100} \cdot \frac{m}{n}=\frac{6 m}{25 n}$ are int... | \frac{100}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,793 |
4. All cells of an infinite grid plane are squares with a side length of 1. Each node of this grid is painted in one of two colors, and there is a node of each color. Prove that there will be two nodes of different colors, the distance between which is 5. | Solution. Assume the opposite, then any two nodes at a distance of 5 from each other are of the same color. Let the nodes be colored in blue and red. Introduce a coordinate system, choosing two arbitrary perpendicular lines of the grid as axes. The distance between the pairs of points $(0,0)$ and $(3,4)$, $(3,4)$ and $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,794 |
5. In trapezoid $ABCD$, points $N$ and $F$ are chosen on the lateral sides $AB$ and $CD$ respectively. Prove that if $\angle BAF = \angle CDN$, then $\angle AFB = \angle DNC$. | The first solution. Since angles $N A F$ and $F D N$ are equal, subtend $N F$, and lie on the same side of $N F$, quadrilateral $A N F D$ is cyclic. Therefore, $\angle N F C = 180^{\circ} - \angle N F D = \angle B A D = 180^{\circ} - \angle A B C$, where the last equality follows from the parallelism of lines $A D$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,795 |
9.1. A four-digit number $n$ was doubled and 1000 was added to the result. As a result, a four-digit number was obtained, written with the same digits as $n$, but in reverse order. Find all possible values of $n$. | Answer: 2996.
Solution. Let $a, b, c, d$ be the number of thousands, hundreds, tens, and units in the number $n$ respectively. Write the condition in the form:
| $a$ | $b$ | $c$ | $d$ |
| ---: | ---: | ---: | ---: |
| $a$ | $b$ | $c$ | $d$ |
| $+\quad 1$ | 0 | 0 | 0 |
| $d$ | $c$ | $b$ | $a$ |
Since $a$ is the last ... | 2996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,796 |
9.2. In the basket, there are oranges and bananas. If you add as many oranges as there are currently bananas (in pieces), then the percentage of oranges
will be twice as much as it would be if you added as many bananas as there are currently oranges. What is the current percentage of oranges in the basket? | Answer: 50.
Solution. Let $a$ be the number of oranges, and $b$ be the number of bananas in the basket. If we add as many oranges as there are currently bananas, then there will be $a+b$ oranges out of $a+2b$ fruits, and the proportion of oranges will be $(a+b)/(a+2b)$. If we add as many bananas as there are currently... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,797 |
9.3. Find the value of the expression $x-\sqrt{2022 x}+2023$,
if $x-\sqrt{\frac{2022}{x}}=2023$. | Answer: 2024.
Solution. Transform the condition $x-\sqrt{\frac{2022}{x}}=2023$ given that $x>0$. We get: $x^{2}-\sqrt{2022 x}=2023 x ; x^{2}-2022 x-x-\sqrt{2022 x}=0$;
$$
(x-\sqrt{2022 x})(x+\sqrt{2022 x})-(x+\sqrt{2022 x});(x-\sqrt{2022 x}-1)(x+
$$
$\sqrt{2022 x})=0$.
The second factor is positive for all $x>0$, w... | 2024 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,798 |
9.4. Two circles intersect at points $A$ and $B$. On the arc $A B$ of one circle, which lies inside the second circle, a point $C$ is taken. The points of intersection of $A C$ and $B C$ with the second circle, different from $A$ and $B$, are denoted by $E$ and $D$ respectively. Prove that the lines $D E$ and $O C$ are... | Solution. Draw the tangent line to the first circle at point $C$, denote it as $l$. The angle between $l$ and $A E$
is equal to the angle $A B C$ (as the angle between the tangent and the ch... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,799 |
9.5. At competitions, an athlete's performance is evaluated by 7 judges, each of whom gives a score (an integer from 0 to 10). To obtain the final score, the best and worst scores from the judges are discarded, and the arithmetic mean is calculated. If the average score were calculated based on all seven scores, the at... | Answer: 5.
Solution: Suppose there are no fewer than six dancers. Let $A, a, S_{A}$ be the best score, the worst score, and the sum of all non-discarded scores of the winner, respectively, and $B, b, S_{B}$ be the same for the last athlete. Instead of averages, the dancers can be ranked by the sum of all scores or the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,800 |
7.1 Nine numbers are written in one line. The sum of any two adjacent ones is negative, while the sum of all nine numbers is positive. Can the sign of the middle (i.e., the fifth in order) of these numbers be uniquely determined from this information? What about its neighbors to the left and right? Justify your answers... | Solution. The sum of any four consecutive numbers is negative, as it represents the sum of two pairs of adjacent numbers. Then the sum of all 8 numbers, except the central one, is negative. Since the sum of all 9 numbers is positive, the middle number must be positive. Therefore, its neighbors to the left and right are... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,802 |
7.2 Draw on the plane a figure that cannot cover a semicircle of radius 1, but two copies of which can cover a circle of radius 1. | Solution. For example, a ring. The outer circle has a radius of 1, and the inner circle has a sufficiently small radius (so that the thickness of the ring allows it to contain this inner circle), for example, a radius of 0.1. Then such a figure does not contain a segment of length 2, and therefore will not cover a semi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,803 |
7.3 In a pond with carp, 30 pikes were released, which began to gradually eat each other. A pike is considered full if it has eaten three other pikes (hungry or full), and each full pike ate exactly one carp for dessert (hungry pikes did not eat carp). What is the maximum number of carp that could have been eaten? Just... | Solution. The maximum number of pikes eaten is 29, so no more than $29: 3=9.6666 \ldots$ pikes can be satiated, which means no more than 9 pikes (and, accordingly, no more than 9 roaches can be eaten). 9 roaches can be eaten as follows (the method is not unique). Choose 9 pikes and denote them as $A_{1}, A_{2}, \ldots,... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,804 |
7.4 On the table, there are 2020 boxes, some of which contain candies, while the others are empty. On the first box, it is written: "All boxes are empty." On the second: "At least 2019 boxes are empty." On the third: "At least 2018 boxes are empty," and so on, up to the 2020th, which says: "At least one box is empty." ... | Solution. Note that if the inscription "At least $A$ boxes are empty" is true for some $A$, then all subsequent inscriptions are also true. This means that the boxes with candies are all boxes starting from some number $N$. Then the number of boxes without candies is exactly $N-1$. The inscription on the $N$-th box rea... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,805 |
7.5 A businessman was driving to a business meeting. He calculated that if he traveled at a speed of 90 km/h, he would arrive an hour earlier, and if he traveled at 60 km/h, he would be an hour late. What is the minimum speed he should travel to arrive on time? Justify your answer. | Solution. Method 1. Suppose that three businessmen, each in their own car, were heading to a meeting. The first (Speedster) at a speed of 90 km/h, the second (Turtle) at a speed of 60 km/h, and the third - Punctual, whose speed we need to find. Consider the moment when Speedster arrives at the meeting place. The meetin... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,806 |
7.6 On a horizontal line, points $A$ and $B$ are marked, the distance between which is 4. Above the line, two semicircles with a radius of 2 are constructed, centered at points
A and B. Additionally, one circle, also with a radius of 2, is constructed, for which the point of intersection of these semicircles is the lo... | Solution. Method 1. Let the midpoint of segment $A B$ be $E$, the center of the circle be $O$, and the second points of intersection of the circle and the semicircles be $C$ and $D$ (see figure). Since the radii of the semicircles and the radius of the circle are equal, segments $A D, E O$, and $B C$ will be equal to e... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,807 |
10.1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2016$ and $a \neq c$ (the numbers themselves are not given). | Solution. Expanding the brackets $\mathrm{x}^{2}-(\mathrm{a}+\mathrm{b}) \mathrm{x}+\mathrm{ab}=\mathrm{x}^{2}-(\mathrm{c}+\mathrm{d}) \mathrm{x}+\mathrm{cd}$, i.e., $\quad(c+d-a-b) x=c d-a b$. By the condition $c - a = d - b$ - let's denote this by $r$. Then $a=c-r, d=b+r$ and $c d-a b=c(b+r)-(c-r) b=(b+c) r=2016 r$. ... | 1008 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,808 |
10.2. Vasya writes down a sequence of numbers: $a_{1}=1, a_{2}=1, \ldots, a_{n+1}=1+\frac{a_{n-1}}{a_{n}}$ for $n \geq 2$. He noticed that $a_{5}>2$, $a_{6}<2$, $a_{7}>2$, $a_{8}<2$ and made the following hypothesis: all terms of the sequence with odd indices, starting from index 5, are greater than 2, and all terms wi... | Solution. We proceed by induction. The base case is already stated. Suppose that for a natural number $k \geq 3$ we have $a_{2 k-1}>2, a_{2 k}<2$. Then,
$a_{2 k+1}=1+\frac{a_{2 k-1}}{a_{2 k}}>1+1=2$, since $a_{2 k-1}>2>a_{2 k}$,
$a_{2 k+2}=1+\frac{a_{2 k}}{a_{2 k+1}}<1+1=2$ since $a_{2 k}<2<a_{2 k+1}$.
Thus, the ind... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,809 |
10.4. Let AL be the bisector of an acute-angled triangle $\mathrm{ABC}$, and let $\omega$ be the circumcircle of this triangle. Denote by P the intersection point of the extension of the altitude BH of triangle $\mathrm{ABC}$ with the circle $\omega$. Prove that if $\angle \mathrm{BLA}=\angle \mathrm{BAC}$, then $\math... | Solution. See fig.
We will prove that $\angle \mathrm{PBC}=\angle \mathrm{PCB}$. Let $\angle \mathrm{BAC}=2 \alpha$. Then $\angle \mathrm{ABH}=\frac{\pi}{2}-2 \alpha$ (from the right triangl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,811 |
10.5. 10.5 A - a four-digit number composed of non-zero digits, B the number written with the same digits in reverse order. It is known that the sum A+B is divisible by 109. What can the sum of the digits of A be? | Solution. Let $\mathrm{A}=\overline{a b c d}(a, b, c, d$ - digits), i.e., $\mathrm{A}=1000 a+100 b+10 c+d$. Then $\mathrm{B}=1000 d+100 c+10 b+a$ and $\mathrm{A}+\mathrm{B}=1001 a+110 b+110 c+1001 d$.
In the last sum, we isolate the largest possible part that is definitely divisible by 109:
$\mathrm{A}+\mathrm{B}=(98... | 14,23,28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,812 |
1. Variant 1.
At the entrance to the amusement park, they sell children's and adult tickets. One children's ticket costs 600 rubles. Alexander bought 2 children's and 3 adult tickets, while Anna bought 3 children's and 2 adult tickets. It is known that Alexander paid 200 rubles more than Anna. How much did Alexander p... | Answer: 3600
## Solution.
Let $A$ be the cost of a children's ticket, and $B$ be the cost of an adult ticket. We calculate the difference $3B + 2A - 2B - 3A = B - A = 200$ rubles. This means the difference between the cost of an adult ticket and a children's ticket is 200 rubles. Then Alexander paid for the tickets $... | 3600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,815 |
2. Variant 1.
Given the set $A=\{1,2,3, \ldots, 1002\}$. Petya and Vasya are playing a game. Petya names a number $n$, and Vasya selects a subset from $A$ consisting of $n$ elements. Vasya wins if there are no two coprime numbers in the chosen subset; otherwise, Petya wins. What is the smallest $n$ that Petya should n... | # Solution.
1) Estimation. Note that the set $A$ contains 501 even numbers. If Petya names $n \leq 502$, then Vasya can choose $n$ even numbers from this set and win.
2) Now we will show that for $n=502$, Vasya will lose. For this, it is sufficient to show that in the chosen subset, there will necessarily be two conse... | 502 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,816 |
3. Variant 1.
On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, 2022 natives gathered around a round table, and each of them made the following statement:
"I am sitting next to a knight and a liar!"
It is known that three knights made a mistake (i.e., accid... | Answer: 1349
## Solution.
Let's divide all the people sitting at the table into blocks consisting of natives of the same type sitting in a row. Then, blocks of liars can only consist of 1 person; otherwise, a liar sitting at the edge of this block would tell the truth. If a block of knights consists of two people, th... | 1349 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,817 |
5. Variant 1.
On the coordinate plane, the graphs of three reduced quadratic trinomials are drawn, intersecting the y-axis at points $-15, -6, -27$ respectively. For each trinomial, the coefficient of $x$ is a natural number, and the larger root is a prime number. Find the sum of all roots of these trinomials. | Answer: -9.
## Solution.
The parabola intersects the $O y$ axis at a point whose ordinate is equal to the free term. Therefore, our trinomials have the form $x^{2}+a x-15, x^{2}+b x-6, x^{2}+c x-27$. Let's denote their larger roots by $p, q, r$ respectively. By Vieta's theorem, the second root of the first trinomial ... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,819 |
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
 | Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides som... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,820 |
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cine... | Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats ... | 1035 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,821 |
8. Variant 1.
On side $A B$ of parallelogram $A B C D$, a point $F$ is chosen, and on the extension of side $B C$ beyond vertex $B$, a point $H$ is chosen such that $A B / B F = B C / B H = 5$. Point $G$ is chosen so that $B F G H$ is a parallelogram. $G D$ intersects $A C$ at point $X$. Find $A X$, if $A C = 100$. | Answer: 40.
Solution.
In parallelogram $ABCD$, draw diagonal $BD$, and in parallelogram $BFGH$, draw diagonal $GB$. Let $BD$ intersect $AC$ at point $O$. We will prove that $AC \| GB$. Tri... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,822 |
1. (7 points) Represent the number 2017 as the sum of five natural numbers such that all digits used in these five numbers are distinct. | Solution. One possible example: $2017=1976+30+4+2+5$. There are other representations as well.
Criteria. At least one correct representation is provided, even without any explanations: 7 points.
The required representation is not obtained: 0 points. | 2017=1976+30+4+2+5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,823 |
2. (7 points) In rectangle $A B C D$, side $A B$ is equal to 6, and side $B C$ is equal to 11. Bisectors of the angles from vertices $B$ and $C$ intersect side $A D$ at points $X$ and $Y$ respectively. Find the length of segment $X Y$.
 The knightly tournament lasts exactly 7 days. By the end of the fourth day, Sir Lancelot had not yet faced one quarter of the total number of participants. And by this time, Sir Tristan had fought exactly one seventh of the knights that Sir Lancelot had faced. What is the minimum number of knights that co... | Answer: 20.
Solution. Let Lancelot not have fought with $x$ knights. Then the total number of knights is $4 x$, and Lancelot fought with $3 x-1$ knights (the total number minus $x$ and Lancelot himself). Then Tristan fought with $\frac{3 x-1}{7}$ knights. To find the smallest possible number of knights, we need to fin... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,825 |
4. (7 points) Vova placed several (possibly 0) chess pieces on an $8 \times 8$ board. Lena noticed that in each $2 \times 2$ square, the same number of pieces is placed. And Vlad noticed that in each $3 \times 1$ (or $1 \times 3$) rectangle, the same number of pieces is placed. How many pieces were placed on the board?... | Answer: 0 or 64.
Solution. Suppose that in each $2 \times 2$ square there are $m$ figures, and in each $1 \times 3$ rectangle there are $n$ figures. Let's select a $2 \times 6$ rectangle from the board. On one hand, this rectangle can be divided into three $2 \times 2$ squares, and thus it contains $3m$ figures. On th... | 0or64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,826 |
5. (7 points) Three schoolgirls entered a store. Anya bought 2 pens, 7 pencils, and 1 notebook, Varya - 5 pens, 6 pencils, and 5 notebooks, Sasha - 8 pens, 4 pencils, and 9 notebooks. They all paid equally, but one of them used a discount when paying. Who? (Explain your answer). | Answer: Varya.
Solution. Let's denote the cost of a pen, a pencil, and a notebook by the symbols $\mathrm{P}, \mathrm{K}$, and L, respectively. We also denote the total cost of purchases (excluding the discount) by A, B, and C for Anya, Varya, and Sasha, respectively. According to the problem,
$$
A=2 \mathrm{P}+7 \ma... | Varya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,827 |
6. (7 points) In triangle $A B C$, the median $A M$ is drawn. Find the angle $A M C$, if angles $B A C$ and $B C A$ are equal to $45^{\circ}$ and $30^{\circ}$ respectively.
Answer: $135^{\circ}$. | Solution. Let $B H$ be the height of triangle $A B C$. According to the problem, angle $B A C$ is $45^{\circ}$, so $B H = A H$. In triangle $C B H$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B C = 2 B H$. The median $H M$ of the right triangle $B H C$ is equal to half the hypotenuse $B C$.
 | Solution. Let's denote the numbers in the clockwise order as $a_{1}, a_{2}, \ldots, a_{2 n}$; let $S>0$ be the sum of all numbers, and set $S_{i}=a_{i}+a_{i+1}+\ldots+a_{i+n-1}$ (all indices are considered modulo $2 n$, so $a_{2 n+i}=a_{i}$ and $S_{2 n+i}=S_{i}$). Then we need to prove that for some $i$ both sums $S_{i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,829 |
9.6. Petya and Vasya came up with ten quadratic trinomials. Then Vasya sequentially named consecutive natural numbers (starting from some number), and Petya substituted each named number into one of the trinomials of his choice and wrote down the obtained values on the board from left to right. It turned out that the n... | Answer: 20 numbers.
Solution. We will show that Petya could substitute no more than two numbers into each quadratic polynomial $P(x)$. Indeed, let the $n$-th term of the resulting arithmetic progression be $a n+b$, and the $n$-th of Vasya's consecutive numbers be $k+n$. Then Petya could substitute this number into $P(... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,830 |
9.7. On the sides of an acute-angled triangle $A B C$, squares $C A K L$ and $C B M N$ are constructed outside it. The line $C N$ intersects the segment $A K$ at point $X$, and the line $C L$ intersects the segment $B M$ at point $Y$. The point $P$, lying inside triangle $A B C$, is the intersection of the circumcircle... | Solution. Let $Q$ be the intersection point of lines $KL$ and $MN$ (see Fig. 1). Since $\angle QLC = \angle NMY = 90^\circ$, quadrilateral $QLYM$ is cyclic. Similarly, quadrilateral $QNXL$ is cyclic. Thus, $Q$ is the second intersection point of the circumcircles $\omega_1$ and $\omega_2$ of triangles $KXN$ and $LYM$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,831 |
9.8. From a $55 \times 55$ square grid, 400 three-cell corners $\square$ (oriented in any direction) and 500 cells were cut out along the grid lines. Prove that some two of the cut-out figures share a boundary segment.
(S. Berlov) | Solution. First solution. Add a border to each figure as shown in Fig. 2. Suppose the cut-out figures do not share any sides. Then the figures with added borders do not overlap. Indeed, the border of figure $F$ consists precisely of those points whose distance to $F$ is no greater than the distance to any cell not shar... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,832 |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,835 |
Problem 4.4. Four girls: Katya, Olya, Liza, and Rita, stood in a circle in some order. They were wearing dresses of different colors: pink, green, yellow, and blue. It is known that:
- Katya was not wearing a pink or blue dress;
- the girl in the green dress is standing between Liza and the girl in the yellow dress;
-... | Answer: Katya is wearing a green dress, Olya - a blue one, Liza - a pink one, Rita - a yellow one. Solution. From the first and fourth statements, it follows that Katya, Olya, and Rita are not wearing pink dresses. Therefore, Liza is in the pink dress.
From the third statement, it follows that Rita is not wearing a gr... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,836 |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,837 |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
![](https://cdn.mathpix.com/cropped/2024_05_06_4984b677816e04f7f1c9g-11.jpg?height=309&width=313&top_left_y=92&top_left... | Answer: 27.
Solution. The area of the entire square is $6 \cdot 6=36$ sq. cm.
Divide the triangle located in the middle of the square into two smaller triangles, as shown in the picture on the left. Then the dark gray triangles can be combined into a rectangle $1 \times 3$, and the light gray triangles into a rectang... | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,838 |
Problem 5.6. On the board, there is one three-digit number and two two-digit numbers. The sum of the numbers that have a seven in their notation is 208. The sum of the numbers that have a three in their notation is 76. Find the sum of all three numbers. | Answer: 247.
Solution. Let the three-digit number be $A$, the two-digit numbers be $-B$ and $C$. Among the numbers whose sum is 76, there cannot be a three-digit number. The sum cannot consist of a single number either, because otherwise that number would be 76, but it does not contain a three. Therefore, 76 is the su... | 247 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,839 |
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