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Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. The quadratic trinomial $P(x)$ is such that $P(P(x))=x^{4}-2 x^{3}+4 x^{2}-3 x+4$. What can $P(8)$ be? List all possible options. | Answer: 58.
Solution. Let $P(x)=a x^{2}+b x+c$. Then $P(P(x))=a\left(a x^{2}+b x+c\right)^{2}+b\left(a x^{2}+b x+c\right)+c=$
$a\left(a^{2} x^{4}+2 a b x^{3}+\left(b^{2}+2 a c\right) x^{2}+2 b c x+c^{2}\right)+b\left(a x^{2}+b x+c\right)+c$. Therefore,
\[
\begin{aligned}
& x^{4}-2 x^{3}+4 x^{2}-3 x+4=P(P(x))= \\
& =a... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.7. In a country, there are 110 cities. Between any two of them, there is either a road or there is not.
A motorist was in a certain city, from which there was exactly one road. After driving along this road, he arrived at a second city, from which there were already exactly two roads. Driving along one of t... | Answer: 107.
Solution. Let's number the cities in the order they are visited by the motorist: $1,2,3, \ldots, N$.
Suppose $N \geqslant 108$. From city 1, there is a road only to city 2, so from city 108, all roads lead to all 108 cities, except for 1 and 108. But then from city 2, there are at least three roads: to c... | 107 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
![](https://cdn.mathpix.com... | Answer: 20.
Solution. Lines $C_{1} Y$ and $Z X$ lie in parallel planes $B B_{1} C_{1} C$ and $A A_{1} D_{1} D$, so they do not intersect. Since these two lines also lie in the same plane $C_{1} X Z Y$, they are parallel. Similarly, lines $Y Z$ and $C_{1} X$ are parallel. Therefore, quadrilateral $C_{1} X Z Y$ is a par... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.2. On the plate, there were 15 doughnuts. Karlson took three times more doughnuts than Little Man, and Little Man's dog Bibbo took three times fewer than Little Man. How many doughnuts are left on the plate? Explain your answer. | Answer: 2 doughnuts are left.
From the condition of the problem, it follows that Little One took three times as many doughnuts as Bimbo, and Karlson took three times as many doughnuts as Little One. We can reason in different ways from here.
First method. If Bimbo took one doughnut, then Little One took three doughnu... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. Ladybugs gathered on a clearing. If a ladybug has 6 spots on its back, it always tells the truth, and if it has 4 spots, it always lies, and there were no other ladybugs on the clearing. The first ladybug said: "Each of us has the same number of spots on our backs." The second said: "Together, we have 30 spots on ... | Answer: 5 ladybugs.
If the first ladybug is telling the truth, then the second and third should also be telling the truth, as they should have the same number of spots as the first. However, the second and third ladybugs contradict each other, so at least one of them is lying, which means the first ladybug is also lyi... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) Cheburashka and Gena ate a cake. Cheburashka ate twice as slowly as Gena, but started eating a minute earlier. In the end, they each got an equal amount of cake. How long would it take Cheburashka to eat the cake alone? | Answer. In 4 minutes.
Solution.
First method. If Cheburashka eats twice as slowly as Gena, then to eat the same amount of cake as Gena, he needs twice as much time. This means that the time Cheburashka ate alone (1 minute) is half of the total time it took him to eat half the cake. Thus, he ate half the cake in 2 min... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In an equilateral triangle $ABC$ with side length $a$, points $M, N, P, Q$ are positioned as shown in the figure. It is known that $MA + AN = PC + CQ = a$. Find the measure of angle $NOQ$.
 | Answer: $60^{\circ}$
Solution. According to the problem, $A N=a-A M$, hence $A N=M C$. Similarly, $A P=Q C$. From these equalities and the equality $\angle A=\angle C=60^{\circ}$, it follows that $\triangle A N P=\triangle C M Q$. Therefore, $\angle A N P=\angle Q M C, \angle A P N=\angle M Q C$. By the theorem on the... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (7 points) On a chessboard, there were 21 kings. Each king was under attack by at least one of the others. After some kings were removed, no two of the remaining kings attack each other. What is the maximum number of kings that could remain?
a) Provide an example of the initial arrangement and mark the removed king... | Answer: b) 16.
Solution: Note that each king removed from the board could not have attacked more than 4 of the remaining ones (otherwise, some of the remaining ones would also attack each other). Therefore, the number of remaining kings cannot exceed the number of removed ones by more than 4 times, meaning it cannot b... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In the sum $(1-2-3+4)+(5-6-7+8)+\ldots+(2009-2010-2011+2012)+$ (2013-2014-2015), all numbers that end in 0 were crossed out. Find the value of the resulting expression. | 3. Answer: -3026. It is easy to see that in all parentheses except the last one, the sum is zero. Then the entire sum before cancellation was equal to -2016. At the same time, numbers that end in 0 and are divisible by 4 were included in it with a plus, while numbers that end in 0 and are not divisible by 4 were includ... | -3026 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. There are 10 red and 10 yellow apples with a total weight of 1 kg. It is known that the weights of any two apples of the same color differ by no more than 40 g. All apples were sequentially paired "red with yellow," with red apples chosen in non-decreasing order and yellow apples in non-increasing order (i.e., the h... | 6. Answer: 136 g. Solution Estimate. Note that the weights of the pairs also cannot differ by more than 40 g. Therefore, if the weight of the heaviest pair is more than 136 g, then the weight of each of the other nine pairs will be more than 96 g, but then the total weight of all apples will be more than 1 kg, which is... | 136 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.1. There are seven cards on the table. In one move, it is allowed to flip any five cards. What is the minimum number of moves required to flip all the cards? | 9.1. Answer. 3 moves.
Obviously, one move is not enough. After two moves, there will be at least three cards that have been flipped twice, which means these cards will be in their original position.
Let's provide an example of flipping all the cards in three moves. Number the cards from 1 to 7 and flip cards numbered... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $k$ transfers. At the same time, the number of air routes from any city should not exceed four. What is the smallest $k$ for which this is p... | 9.4. Answer. $k=2$.
Note that at least two transfers will be required. Indeed, from an arbitrary city $A$ without a transfer, one can reach no more than 4 cities, and with exactly one transfer - no more than $4 \cdot 3=12$ cities (since one of the flights from each of these cities leads back to $A$). Therefore, if usi... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.8. A circle is circumscribed around triangle $A B C$. Let $A_{0}$ and $C_{0}$ be the midpoints of its arcs $B C$ and $A B$, not containing vertices $A$ and $C$. It turns out that the segment $A_{0} C_{0}$ is tangent to the circle inscribed in triangle $A B C$. Find the angle $B$. | 9.8. Let the incircle of triangle $ABC$ be denoted by $\omega$, its center by $I$, and the points of tangency of $\omega$ with $AC$ and $A_0 C_0$ by $K$ and $L$ respectively. Then, in triangles $IKC$ and $ILA_0$, angles $IKC$ and $ILA_0$ are right angles (as angles between a radius and a tangent), and $IK = IL$, since ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Triangle $A B C$ is equilateral. On side $A C$, point $M$ is marked, and on side $B C$, point $N$, such that $M C=B N=2 A M$. Segments $M B$ and $A N$ intersect at point $Q$. Find the angle $C Q B$.
# | # Answer: $90^{\circ}$.
## Solution:
Since $M C=B N$ and $A C=B C$, then $A M=A C-M C=B C-B N=N C$.
From the fact that $A M=N C, A B=A C$ and $\angle B A M=\angle C A N=60^{\circ}$, it follows that triangles $B A M$ and $A C N$ are congruent, so $\angle N A C=\angle M B A=\alpha$.
From triangle $A B M: \angle A M B... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the record of three two-digit numbers, there are no zeros, and in each of them, both digits are different. Their sum is 41. What could be their sum if the digits in them are swapped? | Answer: 113.
Solution. In all numbers, the tens digit is 1. Otherwise, the largest is at least 21, and the other two are at least 12. Their sum is no less than $12+12+21=44$, which is not equal to 41. The sum of the units digits is 11 (zeros are not allowed). Therefore, the sum with the digits rearranged is 113.
Ther... | 113 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the 5 brothers owns a plot of land. One day, they pooled their money, bought a neighbor's garden, and divided the new land equally among themselves. As a result, Andrey's plot increased by $10 \%$, Boris's plot - by $\frac{1}{15}$, Vladimir's plot - by $5 \%$, Grigory's plot - by $4 \%$, and Dmitry's plot - ... | Answer: By $5 \%$.
Solution. Let A, B, V, G, D be the areas of the plots of each brother, respectively. Then, according to the problem, $\frac{1}{10} \mathrm{~A}=\frac{1}{15} \mathrm{~B}=\frac{1}{20} \mathrm{~V}=\frac{1}{25} \mathrm{~G}=\frac{1}{30} \mathrm{~D}$ (*). Denoting $\mathrm{A}=x$, we find: $\mathrm{B}=1.5 x... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The participants of the Olympiad left 9 pens in the office. Among any four pens, at least two belong to the same owner. And among any five pens, no more than three belong to the same owner. How many students forgot their pens, and how many pens does each student have? | Answer. There are three students, each owning three pens.
Solution. No student owned more than three pens, as otherwise the condition "among any five pens, no more than three belonged to one owner" would not be met. There are a total of 9 pens, so there are no fewer than 3 students. On the other hand, among any four p... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter $\Gamma$, there is at least one colored cell? | Answer: 12.
Solution: Consider a $2 \times 3$ rectangle. It is obvious that at least 2 cells need to be colored in it. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown on t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$. | Answer: $45^{\circ}$.
Solution. Triangles $E T N$ and $A T B$ are similar, therefore, $T N: T B=T E: T A=E N: A B=1: 4$. Therefore, $C T=1 / 2 B T$. Since $H$ is the intersection point of the medians of triangle $A M N$, $E H=1 / 3 A E=E T$. Therefore, $H T=1 / 2 A T$.
Thus, right triangles $C T H$ and $B T A$ are si... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Find the sum:
$$
(-2021)+(-2020)+(-2019)+\ldots+2023+2024
$$ | Answer: 6069.
Solution. By pairing numbers that differ in sign, we get that in each such pair the sum is 0, and without pairs, the numbers left are $0, 2022, 2023, 2024$. | 6069 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1.
In a certain three-digit number $N$, the last two digits were swapped and the resulting number was added to the original. The sum turned out to be a four-digit number starting with 173. Find the largest possible sum of the digits of the number $N$. | Answer: 20.
Solution: Let the original number be $\overline{a b c}$, and the last digit of the sum $\overline{a b c}+\overline{a c b}$ be $x$. Then, $100 a+10 b+c+100 a+10 c+b=200 a+11(c+b)=1730+x$. If $a1730+x$.
Therefore, $a=8$. Thus, $(800+10 b+c)+(800+10 c+b)=1730+x$, i.e., $11(b+c)=130+x$. From this, it follows ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
In the village, 7 people live. Some of them are liars who always lie, while the rest are knights (who always tell the truth). Each resident said whether each of the others is a knight or a liar. A total of 42 answers were received, and in 24 cases, a resident called a fellow villager a liar. What is the... | Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of "knight-liar," the phrase "He is a liar" will be said twice. Since this p... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1.
Given an equilateral triangle $A B C$ with an area of 210. Inside it, points for which vertex $A$ is neither the closest nor the farthest are painted red. What is the area of the painted part of the triangle? | Answer: 41.
Solution.
Consider a point $D$ for which vertex $B$ is the nearest, and vertex $C$ is the farthest. Let $M, N, K$ be the midpoints of sides $AB, BC$, and $AC$ respectively. $O$ i... | 41 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 2.
What is the smallest sum that nine consecutive natural numbers can have if this sum ends in 2050306? | Answer: 22050306.
Option 3.
What is the smallest sum that nine consecutive natural numbers can have if this sum ends in $1020156$?
Answer: 31020156. | 31020156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 6. Variant 1.
What is the largest root that the equation
$$
(x-a)(x-b)=(x-c)(x-d)
$$
can have if it is known that $a+d=b+c=2022$, and the numbers $a$ and $c$ are different? | Answer: 1011.
Solution. By expanding the brackets and combining like terms, we find $x=\frac{c d-a b}{c+d-a-b}$. Note that $c+d-a-b \neq 0$, because otherwise $c+d=a+b$, and considering the equality $a+d=b+c$, we would get $a=c$.
Considering that $d=2022-a, c=2022-b, x=\frac{2022^{2}-2022 a-2022 b}{2(2022-a-b)}=\frac... | 1011 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
In a $101 \times 101$ square, an $88 \times 88$ corner square is painted red. What is the maximum number of non-attacking queens that can be placed on the board without placing the figures on the red cells? A queen attacks along the horizontal, vertical, and diagonals of the square. Attacking through the... | Answer: 26.
Solution. Consider 13 horizontal and 13 vertical strips of size $1 \times 101$, not containing any red cells. Each of the queens stands on at least one of these strips, and no strip can contain more than one queen, so there are no more than 26 queens. To construct an example, consider two rectangles of siz... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1.
A line parallel to the leg $A C$ of the right triangle $A B C$ intersects the leg $B C$ at point $K$, and the hypotenuse $A B$ at point $N$. On the leg $A C$, a point $M$ is chosen such that $M K=M N$. Find the ratio $\frac{A M}{M C}$, if $\frac{B K}{B C}=14$. | Answer: 7.
Solution.
Drop the altitude $M H$ from point $M$ in the isosceles triangle $M N K$. Then $M H K C$ is a rectangle and $M C=K H=H N$. Let $M C=K H=H N=y$. Let $K B=x$, then $C K=C ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task No. 1.1
## Condition:
Five friends - Masha, Nastya, Irina, Olya, and Anya - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Irina: I was the first in line!
Olya: No one was after me.
Anya: Only one person was after me.
Masha: There wer... | Answer: 3
Exact match of the answer - 1 point
## Solution.
From the statements of Irina and Olya, it is clear that they were first and last, respectively. Since there was only one person after Anya, it was Olya. Nastya stood next to Irina, but she could not have stood in front of her, so Nastya was second. This mean... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 1.2
## Condition:
Five friends - Katya, Polina, Alyona, Lena, and Svetlana - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Polina: I stood next to Alyona.
Alyona: I was the first in line!
Lena: No one was after me.
Katya: There... | # Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.3
## Condition:
Five friends - Sasha, Yulia, Rita, Alina, and Natasha - meet in the park every day after buying ice cream from the little shop around the corner. One day, the girls had a conversation.
Sasha: There were five people in front of me.
Alina: There was no one after me.
Rita: I was the first i... | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 1.4
## Condition:
Five friends - Kristina, Nadya, Marina, Liza, and Galia - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Kristina: There were five people in front of me.
Marina: I was the first in line!
Liza: No one was behind ... | # Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 2.1
## Condition:
Karlson and Little together weigh 10 kg more than Freken Bok, and Little and Freken Bok weigh 30 kg more than Karlson. How much does Little weigh? Give your answer in kilograms. | Answer: 20
Exact match of the answer -1 point
Solution.
Let's denote Carlson's mass by K, Freken Bok's mass by F, and Little One's mass by M. Then from the condition, it follows that $K+M=F+10$ and $M+F=K+30$. Adding these equations term by term, we get $K+2M+F=F+K+40$. From this, $M=20$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 3.1
## Condition:
Polina makes jewelry on order for a jewelry store. Each piece of jewelry consists of a chain, a stone, and a pendant. The chain can be silver, gold, or iron. Polina has stones - a diamond, an emerald, and a quartz - and pendants in the shape of a star, a sun, and a moon. Polina is only sa... | # Answer: 24
Exact match of the answer - 1 point
## Solution.
Notice that the iron ornament with the sun is definitely in the middle, as it must be between the gold and silver ornaments. For the first position, we can choose a silver or gold chain. After choosing a chain for the first position, the chain for the thi... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.2
## Condition:
Artyom makes watches for a jewelry store on order. Each watch consists of a bracelet and a dial. The bracelet can be leather, metal, or nylon. Artyom has round, square, and oval dials. Watches can be mechanical, quartz, or electronic.
Artyom is only satisfied when the watches are arranged ... | Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №3.1
# | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.3
## Condition:
Alina makes phone cases on order for a tech store. Each case has a pattern and a charm.
The case can be silicone, leather, or plastic. Alina has charms in the shapes of a bear, a dinosaur, a raccoon, and a fairy, and she can draw the moon, the sun, and clouds on the case.
Alina is only sa... | Answer: 36
Exact match of the answer - 1 point
Solution by analogy with task №3.1
# | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.4
## Condition:
Anton makes watches for a jewelry store on order. Each watch consists of a bracelet, a precious stone, and a clasp.
The bracelet can be silver, gold, or steel. Anton has precious stones: zircon, emerald, quartz, diamond, and agate, and clasps: classic, butterfly, and buckle. Anton is only ... | Answer: 48
Exact match of the answer -1 point
Solution by analogy with task №3.1
# | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.1
## Condition:
Yasha and Grisha are playing a game: first, they take turns naming a number from 1 to 105 (Grisha names the number first, the numbers must be different). Then each counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The one with... | # Answer: 104
## Exact match of the answer -1 point
## Solution.
We will show that by naming the number 104, Grisha will win. Consider a rectangle with sides a and b. Its perimeter \( P = 2(a + b) \Rightarrow (a + b) = P / 2 \), then the length of the smaller side can take integer values from 1 to the integer part o... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.2
## Condition:
Sasha and Misha are playing a game: first, they take turns naming a number from 1 to 213 (Misha names the first number, the numbers must be different). Then each counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The winner is ... | Answer: 212
Exact match of the answer -1 point
Solution by analogy with task №6.1
## Condition:
Oksana and Seryozha are playing a game: they take turns naming a number from 1 to 165 (Oksana goes first, the numbers must be different). Then each counts the number of different rectangles with integer sides whose perim... | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.4
## Condition:
Dima and Vlad are playing a game: first, they take turns naming a number from 1 to 97 (Dima names the first number, the numbers must be different). Then each of them counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The winner... | Answer: 96
Exact match of the answer -1 point
Solution by analogy with task №6.1
# | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Assignment No. 7.1
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 300 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Sna... | # Answer: 80
## Exact match of the answer -1 point
## Solution.
First, Hermione drinks $300 / 2=150$ ml of the wisdom potion, and then she pours 150 ml into the second jug, after which the second jug contains 150 ml of the wisdom potion and 300 ml of the beauty potion. When she drinks half of the contents of this ju... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.2
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 600 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape was... | Answer: 40
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 7.3
## Condition:
Potions teacher Severus Snape has prepared three potions, each with a volume of 400 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape... | Answer: 60
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.4
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 480 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape was... | Answer: 50
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Given a sequence $x_{n}$ such that $x_{1}=1, x_{2}=2, x_{n+2}=\left|x_{n+1}\right|-x_{n}$. Find $x_{2015}$. | 5. Answer. ${ }^{x_{2015}}=0$.
We will show that the given sequence is periodic with a period of 9. Let's find
$x_{1}=1, x_{2}=2, x_{3}=1, x_{4}=-1, x_{5}=0, x_{6}=1, x_{7}=1, x_{8}=0, x_{9}=-1, x_{10}=1, x_{11}=2, \ldots$.
Since the sequence is completely determined by any two consecutive terms, we have obtained th... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Several different real numbers are written on the board. It is known that the sum of any three of them is rational, while the sum of any two of them is irrational. What is the largest number of numbers that can be written on the board? Justify your answer. | Solution: The set of three numbers $\sqrt{2}, \sqrt{3}, -\sqrt{2}-\sqrt{3}$, as is easily verified, satisfies the condition of the problem. We will show that no more than three numbers can be written on the board.
Assume the contrary, and let $a_{i} (i=\overline{1,4})$ be some four numbers from this set. Then the numb... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.6. At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote in "VK" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each of the brunettes ... | Solution: According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times but... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false |
Problem 7.1. Denis thought of four different natural numbers. He claims that
- the product of the smallest and the largest numbers is 32;
- the product of the two remaining numbers is 14.
What is the sum of all four numbers? | Answer: 42.
Solution. There are two ways to represent the number 14 as the product of two numbers: $1 \cdot 14$ and $2 \cdot 7$; therefore, the second and third largest numbers are 1 and 14 or 2 and 7. The first case does not work for us, as 1 cannot be the second largest number. So, the second largest number is 2, an... | 42 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. On Monday, 5 people in the class received fives in math, on Tuesday, 8 people received fives, on Wednesday - 6 people, on Thursday - 4 people, on Friday - 9 people. No student received fives on two consecutive days. What is the minimum number of students that could have been in the class | Answer: 14.
Solution. Let's consider pairs of consecutive days.
- On Monday and Tuesday, 5s were received by $5+8=13$ people.
- On Tuesday and Wednesday, 5s were received by $8+6=14$ people.
- On Wednesday and Thursday, 5s were received by $6+4=10$ people.
- On Thursday and Friday, 5s were received by $4+9=13$ people... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. At a tribal council meeting, 60 people spoke in turn. Each of them said only one phrase. The first three speakers said the same thing: "I always tell the truth!" The next 57 speakers also said the same phrase: "Among the previous three speakers, exactly two told the truth." What is the maximum number of sp... | Answer: 45.
Solution. Note that among any four consecutive speakers, at least one lied (if the first three told the truth, then the fourth definitely lied). By dividing 60 people into 15 groups of four consecutive speakers, we get that at least 15 people lied, meaning no more than 45 people told the truth.
To constru... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.6. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively. It is known that $A B=B D, \angle A B D=46^{\circ}, \angle D E C=90^{\circ}$. Find $\angle B D E$, if it is known that $2 D E=A D$. | Answer: $67^{\circ}$.
Solution. Draw the height $B M$ in triangle $A B D$. Since this triangle is isosceles, $M$ bisects $A D$ (Fig. 7.6). This means that $M D = A D / 2 = D E$.
Then the right triangles $B D M$ and $B D E$ are congruent by the leg ( $D M = D E$ ) and the hypotenuse ( $B D$ - common). From this, it is... | 67 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In the classroom, there are several single desks (no more than one person can sit at each desk; there are no other desks in the classroom). During the break, a quarter of the students went out into the corridor, and the number of people left in the classroom was equal to 4/7 of the total number of desks. H... | Answer: 21.
Solution. Note that the number of remaining students is $\frac{3}{4}$ of the initial number, so it must be divisible by 3. Let's denote it by $3x$. Let $y$ be the number of desks. Then
$$
3 x=\frac{4}{7} y
$$
from which $21 x=4 y$. Since 4 and 21 are coprime, $y$ must be divisible by 21. From the problem... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.8. Tanya and Vера are playing a game. Tanya has cards with numbers from 1 to 30. She arranges them in some order in a circle. For each pair of adjacent numbers, Vера calculates their difference, subtracting the smaller number from the larger one, and writes down the resulting 30 numbers in her notebook. After... | Answer: 14.
Solution. We will prove that Tanya cannot get 15 candies. Let's look at the card with the number 15. The number 15 differs from all the remaining numbers, except for the number 30, by no more than 14. Thus, at least one of the differences involving the number 15 will be no more than 14.
Let's provide an e... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Plot the graph of the function $y=(4 \sin 4 x-2 \cos 2 x+3)^{0.5}+(4 \cos 4 x+2 \cos 2 x+3)^{0.5}$. | Solution:
$\mathbf{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x+1}$
$\mathrm{y}=2 \sin ^{2} x+1+2 \cos ^{2} x+1, \m... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest natural number $A$, which is divisible by $p$, ends in $\boldsymbol{p}$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and is the cube of a natural number. | Solution. Let $2 \mathrm{p}+1=\mathrm{n}^{3}$. Then $(\mathrm{n}-1)\left(\mathrm{n}^{2}+n+1\right)=2 \mathrm{p}$. The number $2 \mathrm{p}$ can only have the following positive divisors: 1, 2, p, 2p. The number n is clearly odd, so n-1 is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=$ ... | 11713 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The numbers from 1 to 25 (each exactly once) were placed in the cells of a $5 \times 5$ square such that the sum of the numbers in each row and each column is the same. A boy erased all the numbers except for four (see the picture). Find the number marked with a circle $\bigcirc$.
=$ $65-51=14$. Answer: 14. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In grandmother's garden, apples have ripened: Antonovka, Grushovka, and White Naliv. If there were three times as many Antonovka apples, the total number of apples would increase by $70 \%$. If there were three times as many Grushovka apples, it would increase by $50 \%$. By what percentage would the total number of... | Solution. First method. If the quantity of each type of apple were three times as much, the total number of apples would increase by $200 \%$. Of this increase, $70 \%$ is due to Antonovka, and $50 \%$ is due to Grushovka. Therefore, the increase due to White Naliv would be $200 \%-70 \%-50 \%=80 \%$.
Second method. S... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Three mryak are more expensive than five bryak by 10 rubles. And six mryak are more expensive than eight bryak by 31 rubles. By how many rubles are seven mryak more expensive than nine bryak | 3. Answer: by 38 rubles. Solution. By adding 3 bryaks and myraks, we increase the price difference between myraks and bryaks by 21 rubles. This means that one myrak is 7 rubles more expensive than one bryak. Then seven myraks are more expensive than nine bryaks by $31+7=38$ rubles. Criteria: correct solution - 7 points... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Two very small fleas are jumping on a large sheet of paper. The first jump of the fleas is along a straight line towards each other (their jumps may have different lengths). The first flea first jumps to the right, then up, then to the left, then down, then to the right again, and so on. Each jump is 1 cm longer tha... | 4. Answer: 2 meters. Solution: Let's observe the first four jumps of the first flea: 2 cm more to the left than to the right, and 2 cm more down than up. That is, after four jumps, the first flea moves 2 cm to the left and 2 cm down. Therefore, after 100 jumps, it will move 50 cm to the left and 50 cm down. Similarly, ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13. Find the 2013-th term of the sequence. | Answer: 130
Solution. Let's calculate the first few terms of the sequence. We get: $a_{1}=934 ; a_{2}=16 \times 13=208$; $a_{3}=10 \times 13=130 ; a_{4}=4 \times 13=52 ; a_{5}=7 \times 13=91 ; a_{6}=10 \times 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the seq... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.3. Point $F$ is the midpoint of side $B C$ of square $A B C D$. A perpendicular $A E$ is drawn to segment $D F$. Find the angle $C E F$.
---
Translation provided as requested, maintaining the original format and line breaks. | Answer: $45^{\circ}$.
Solution. Let the line $A E$ intersect the side $C D$ of the square at point $M$ (see Fig. 10.3). Then triangles $A D M$ and $D C F$ are equal (by the leg and acute angle). Therefore, point $M$ is the midpoint of side $C D$. Then triangle $C F M$ is a right and isosceles triangle, so $\angle C M ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.4. Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0 ; 1]$. | Answer: 2.
Solution. The first method. The value 2 is achieved, for example, if $a=c=1, b=d=0$. We will prove that with the given values of the variables $a+b+c+d-ab-bc-cd-da \leqslant 2$.
Notice that $a+b+c+d-ab-bc-cd-da=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y$, then we need to prove that $x+y-xy \leqslant 2$, if... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.5. On side $AB$ of triangle $ABC$, point $K$ is marked, and on side $AC$, point $M$ is marked. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find angle $BAC$. | Answer: $60^{\circ}$.
Solution. To both sides of the equality $S_{A K P M}=S_{B P C}$, add the area of triangle $B P K$ (see Fig. 10.5). We get that $S_{A B M}=S_{B C K}$. Therefore, $\frac{1}{2} B C \cdot B K \sin \angle B = \frac{1}{2} A B \cdot A M \sin \angle A$. Then $\frac{B K}{A M} = \frac{A B \sin \angle A}{B ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. There was a whole number of cheese heads on the kitchen. At night, rats came and ate 10 heads, and everyone ate equally. Several rats got stomachaches from overeating. The remaining seven rats the next night finished off the remaining cheese, but each rat could eat only half as much cheese as the night before. How m... | Solution. Let there be $k$ rats in total $(k>7)$, then each rat ate $\frac{10}{k}$ pieces of cheese on the first night. On the second night, each rat ate half as much, that is, $\frac{5}{k}$ pieces. Then seven rats ate $\frac{35}{k}$ pieces. This is an integer. The only divisor of the number 35 that exceeds 7 is the nu... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In an equilateral triangle, 3 points were marked on each side so that they divide the side into 4 equal segments. These points and the vertices of the triangle were painted. How many isosceles triangles exist with vertices at the painted points? | # Answer. 20
Solution. Let the vertices of the triangle be denoted as A, B, C. Suppose points M, N (M closer to A) are taken on side AB, points P, Q (P closer to A) on side AC, and points K, L (K closer to B) on side BC. Then the following cases are possible for isosceles triangles with the vertex at point A: ABC, AKL... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1 From a three-digit number $A$, which does not contain zeros in its notation, a two-digit number $B$ was obtained by writing the sum of the first two digits instead of them (for example, the number 243 turns into 63). Find $A$ if it is known that $A=3 B$. | Answer: 135.
Solution: The last digit of the numbers $B$ and $A=3B$ is the same, so this digit is 5. Moreover, $A$ is divisible by 3, which means $B$ is also divisible by 3 (the sums of the digits are the same). Therefore, $B$ is one of the numbers $15, 45, 75$. By checking, we find that the number $B=45$ satisfies th... | 135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 4 blue points. And on any segment with endpoints at blue points, containing 3 blue points inside, there are at least 2 red point... | Answer: 4.
Solution: Note that on a segment with endpoints at red points, not containing other red points, there cannot be 5 blue points. Indeed, in this case, between the outermost blue points, there would be 3 blue points, which means there would be at least 2 more red points. Therefore, on such a segment, there are... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the castle, there are 9 identical square rooms, forming a $3 \times 3$ square. Nine people, consisting of liars and knights (liars always lie, knights always tell the truth), each occupied one of these rooms. Each of these 9 people said: "At least one of the neighboring rooms to mine is occupied by a liar." Wha... | Answer: 6 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 3 liars (thus showing that there are no more than 6 knights). Suppose there are no more than 2 liars, then there will be a "vertical row" of rooms where only knights live. But then... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.1. The lines containing the bisectors of the exterior angles of a triangle with angles of 42 and 59 degrees intersected pairwise and formed a new triangle. Find the degree measure of its largest angle. | Answer: 69
Solution. Let the vertices of the given triangle be $A, B, C$, and its angles be $\alpha, \beta$, and $\gamma$, respectively. Additionally, let $A_{1}, B_{1}$, and $C_{1}$ be the points of intersection of the external angle bisectors of angles $B$ and $C$, $A$ and $C$, and $A$ and $B$, respectively. Then, $... | 69 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.1. For what largest $k$ can we assert that in any coloring of $k$ cells in black in a white $7 \times 7$ square, there will necessarily remain a completely white $3 \times 3$ square with sides along the grid lines? | Answer: 3
Solution. Let's highlight four $3 \times 3$ squares that are adjacent to the corners of the $7 \times 7$ square. These squares do not overlap, so if no more than three cells are sha... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. A pedestrian left point $A$ for point $B$. At the same time, a cyclist left point $B$ for point $A$. After one hour, the pedestrian was three times farther from the cyclist than from point $A$. Another 30 minutes later, they met, after which both continued their journey. How many hours did it take the pedestrian t... | Answer: 9
Solution. Let the distance from $A$ to $B$ be 1 km, the pedestrian moves at a speed of $x$ km/h, and the cyclist at $y$ km/h. Then in one hour, the pedestrian has walked $x$ km, the cyclist has traveled $y$ km, and the distance between them is $1-x-y$, which should be three times $x$. Therefore, $3 x=1-x-y, ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. Petya wrote a natural number $A$ on the board. If it is multiplied by 8, the result is the square of a natural number. How many three-digit numbers $B$ exist such that $A \cdot B$ is also a square of a natural number? | # Answer: 15
Solution. If $8 A$ is the square of a natural number, then any prime number greater than 2 must appear in $A$ with an even exponent, and the number 2 must appear with an odd exponent. Therefore, in $B$, any prime number greater than 2 must also appear with an even exponent, and the number 2 must appear wi... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. How many six-digit numbers exist that consist only of the digits 1 and 2, given that each of them appears? | Answer: 62
Solution. There are a total of $64=2^{6}$ six-digit numbers consisting of the digits 1 and 2, since there are 6 positions and for each position, there are two options to place a digit. However, two numbers - the one consisting only of ones and the one consisting only of twos - do not satisfy the condition, ... | 62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. Coins come in denominations of 50 kopecks, 1 ruble, 2 rubles, 5 rubles, 10 rubles. $V$ has several coins in his wallet. It is known that no matter which 20 coins are taken out of the wallet, there will be at least one 1-ruble coin, at least one 2-ruble coin, and at least one 5-ruble coin. What is the maximum numbe... | Answer: 28
Solution. Example: 9 coins of 1 ruble, 9 coins of 2 rubles, 9 coins of 5 rubles, and 1 coin of 10 rubles. Note that the wallet contains a total of $9+9+1=19$ coins that are not 1 ruble, so among any 20 coins, there will definitely be a 1-ruble coin. Similarly, this can be verified for all other denomination... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. The real number a is such that $2a - 1/a = 3$. What is the value of $16a^4 + 1/a^4$? | Answer: 161
Solution. Squaring the equality from the condition, we get $4 a^{2}-4+1 / a^{2}=9$, that is, $4 a^{2}+1 / a^{2}=13$. Squaring again, we obtain $16 a^{4}+8+1 / a^{4}=169$. Therefore, $16 a^{4}+1 / a^{4}=161$. | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8-1. Kolya made a figure from four identical blocks as shown in the picture. What is the surface area of this figure? Express your answer in square centimeters.
 | Answer: 64.
Solution. The surface area of one block is 18 cm². Out of this area, 2 cm² is "lost" at the joints with other blocks, leaving a total area of $18-2=16$ cm². Since there are 4 blocks, the answer is $4 \cdot 16=64$ cm². | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-2. On a distant planet in the mangrove forests, there lives a population of frogs. The number of frogs born each year is one more than the difference between the number of frogs born in the previous two years (the larger number minus the smaller number).
For example, if 5 frogs were born last year and 2 in the year ... | Answer: 2033.
Solution: This is a fairly simple problem where the only thing you need to do is correctly understand the condition and apply it the necessary number of times. The ability to correctly understand the condition is an important skill in solving problems!
By applying the algorithm described in the conditio... | 2033 | Other | math-word-problem | Yes | Yes | olympiads | false |
8-3. Olya bought three gifts and packed them in three rectangular boxes: blue, red, and green. She tried to place these gifts in different ways: one on the table, and two on top of each other on the floor. Some distances are given in the diagram. Find the height of the table $h$. Express your answer in centimeters.
![... | Answer: 91.
Solution: Let the height of the blue rectangle be $b$, the height of the red rectangle be $r$, and the height of the green rectangle be $g$. Then, according to the condition, $h+b-g=111, h+r-b=80, h+g-r=82$. Adding all these equations, we get $3h=273$, from which $h=91$. | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-4. There are 50 parking spaces on a car park, numbered from 1 to 50. Currently, all parking spaces are empty. Two cars, a black one and a pink one, have arrived at the parking lot. How many ways are there to park these cars such that there is at least one empty parking space between them?
If the black and pink cars ... | Answer: 2352.
Solution I. Carefully consider the cases of the placement of the black car. If it parks in spot number 1 or 50, the pink car can park in any of the 48 spots (numbered from 3 to 50 or from 1 to 48, respectively). If the black car parks in a spot numbered from 2 to 49, the pink car has only 47 options (all... | 2352 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-6. In a kindergarten, 150 children are standing in a circle. Each child is looking at the teacher standing in the center of the circle. Some children are wearing blue jackets, and the rest are wearing red ones. There are 12 children in blue jackets whose left neighbor is in a red jacket. How many children have a left... | Answer: 126.
Solution I. Let each of the children say whether the color of their jacket is the same as that of the child to their left. We need to find out how many children will say "The same."
Let's find out how many children said "Different." Note that if we go around the circle, the jacket colors of such children... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-8. The numbers $a, b$, and $c$ (not necessarily integers) are such that
$$
a+b+c=0 \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=100
$$
What is $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ ? | Answer: -103.
Solution. We have
$$
100=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{-b-c}{b}+\frac{-a-c}{c}+\frac{-a-b}{a}=-3-\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)
$$
from which $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}=-103$.
## Variants of tasks | -103 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. We will call a natural number interesting if it is the product of exactly two (distinct or equal) prime numbers. What is the greatest number of consecutive numbers, all of which are interesting | Solution. One of four consecutive numbers is divisible by 4. However, among the numbers divisible by 4, only the number 4 itself is interesting. But the numbers 3 and 5 are not interesting, so four consecutive interesting numbers do not exist. An example of three consecutive interesting numbers: $33,34,35$. Answer: thr... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangular grid of size 1 x 60. In how many ways can it be cut into grid rectangles of size 1 x 3 and 1 x 4? | 3. Answer: 45665.
Solution. Let $x$ be the number of rectangles of size 1 x $3, y$ be the number of rectangles of size $1 \times 4$. Then $3 x+4 y=60$, where $x, y$ are non-negative integers. The value $\mathrm{x}=\frac{60-4 \mathrm{y}}{3}=20-\frac{4 \mathrm{y}}{3}$ is an integer if $\mathrm{y}=3 \mathrm{t}(\mathrm{t}... | 45665 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the board, there are 2010 non-zero numbers $\mathrm{a}_{1}$, $\mathrm{a}_{2}, \ldots \mathrm{a}_{2010}$ and the products of all pairs of adjacent numbers: $\mathrm{a}_{1} \cdot \mathrm{a}_{2}, \mathrm{a}_{2} \cdot \mathrm{a}_{3}, \ldots \mathrm{a}_{2009} \cdot \mathrm{a}_{2010}$. What is the maximum number of neg... | 4. Answer: 3014.
Solution. Consider the triples of numbers ( $\left.\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{1} \cdot \mathrm{a}_{2}\right)$ ), ( $\left.\mathrm{a}_{3}, \mathrm{a}_{4}, \mathrm{a}_{3} \cdot \mathrm{a}_{4}\right), \ldots,\left(\mathrm{a}_{2009}, \mathrm{a}_{2010}, \mathrm{a}_{2009} \cdot \mathrm{a}_{... | 3014 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the required trio of people. How man... | Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. The part of the graph of a linear function located in the second coordinate quadrant, together with the coordinate axes, forms a triangle. By what factor will its area change if the slope of the function is doubled and the y-intercept is halved? | Answer: It will decrease by eight times.
Solution. Let the original linear function be defined by the equation $y=k x+b$. From the condition of the problem, it follows that $k>0$ and $b>0$ (see Fig. 8.2). The points of intersection of its graph with the axes are: $A(0 ; b)$ and $B\left(-\frac{b}{k} ; 0\right)$. Since ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. The height $CH$, dropped from the vertex of the right angle of triangle $ABC$, bisects the bisector $BL$ of this triangle. Find the angle $BAC$.
---
The provided text has been translated from Russian to English, maintaining the original formatting and structure. | Answer: $30^{\circ}$.
Solution. Let $C H$ and $B L$ intersect at point $K$ (see Fig. 8.3). Then $C K$ is the median of the right triangle $B C L$, drawn to the hypotenuse, so $C K = 0.5 B L = B K$. Therefore, $\angle K C B = \angle K B C = \angle K B H$. Since the sum of these three angles is $90^{\circ}$ (from triang... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. On the Island of Liars and Knights, a circular arrangement is called correct if each person standing in the circle can say that among their two neighbors, there is a representative of their tribe. Once, 2019 natives formed a correct circular arrangement. A liar approached them and said: "Now we can also form a cor... | Answer: 1346.
Solution. We will prove that a correct arrangement in a circle is possible if and only if the number of knights is at least twice the number of liars.
Indeed, from the problem's condition, it follows that in such an arrangement, each liar has two knights as neighbors, and among the neighbors of any knig... | 1346 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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