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10.1. (7 points)
Prove that $\sqrt{\frac{11 \ldots 1}{2 n \text { digits }}-\underbrace{22 \ldots .2}_{n \text { digits }}}=\underbrace{33 \ldots 3}_{n \text { digits }}$. | # Solution:
$$
\begin{aligned}
& \sqrt{\underbrace{11 \ldots 1}_{2 n \text { digits }}-\underbrace{22 \ldots 2}_{n \text { digits }}}=\sqrt{\frac{1}{9} \cdot \underbrace{99 \ldots 9}_{2 n \text { digits }}-\frac{2}{9} \cdot \underbrace{99 \ldots 9}_{n \text { digits }}}=\frac{1}{3} \sqrt{\left(10^{2 n}-1\right)-2\lef... | 42 | Number Theory | proof | Yes | Yes | olympiads | false |
5. Emperor Pea has a worm in one of his rejuvenating apples. He has 13 apples in total, and they are arranged in a circle in a special box for rejuvenating apples. To find the worm, Emperor Pea decided to use a balance scale. He knows that all the apples weigh the same, except for the one with the worm, which is heavie... | # Solution.
Let's number the apples clockwise. The neighboring apples will be numbered 1 and 2, 2 and 3, ..., 12 and 13, 13 and 1.
Weigh apples 1, 2, 3, 4, 5, 6 against 7, 8, 9, 10, 11, 12.
If they are equal, then the worm is in apple 13.
Suppose 1, 2, ..., 6 are heavier.
Then after the apples are put back and the... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. In 60 chandeliers (each chandelier has 4 lampshades), lampshades need to be replaced. Each electrician spends 5 minutes replacing one lampshade. A total of 48 electricians will be working. Two lampshades in a chandelier cannot be replaced simultaneously. What is the minimum time required to replace all the lampsha... | Answer: 25 minutes
Solution. Let's show how to proceed. First, 48 electricians replace one lampshade in 48 chandeliers, which takes 5 minutes, and 48 chandeliers have one lampshade replaced, while 12 have none replaced. Then, 12 electricians replace lampshades in the chandeliers that haven't been replaced yet, while t... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. Find the perimeter of a rectangle if the sum of the lengths of two of its sides is 10 dm, and the sum of the lengths of three of its sides is 14 dm. | Answer: 18 dm, 19 dm or 20 dm
Solution. If the sum of two adjacent sides is 10 dm, then the perimeter is 20 dm, which does not contradict the condition on the sum of three sides. If the sum of opposite sides is 10 dm, then each of these sides is 5 dm. In this case, the adjacent side to them is 4 dm or 4.5 dm. | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-1-1. Select the numbers that serve as counterexamples to the given statement: “If the sum of the digits of a natural number is divisible by 27, then the number itself is divisible by $27$.”
a) 81 ; b) 999 ; c) 9918 ; d) 18 . | Answer: only 9918.
Solution option 1. The statements about 81 and 18 are definitely not counterexamples, as for them, the premise "if the sum of the digits of a natural number is divisible by 27" is not satisfied.
The statement about 999 is not a counterexample, as it does not contradict our statement.
Finally, the ... | 9918 | Number Theory | MCQ | Yes | Yes | olympiads | false |
10-3-1. Non-negative integers $a, b, c, d$ are such that
$$
a b+b c+c d+d a=707
$$
What is the smallest value that the sum $a+b+c+d$ can take? | Answer: 108.
Solution variant 1. The given equality can be rewritten as
\[
(a+c)(b+d)=7 \cdot 101
\]
where the numbers 7 and 101 are prime. Therefore, either one of the expressions in parentheses is 1 and the other is 707, or one of the expressions in parentheses is 7 and the other is 101. In the first case, \(a+b+c... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10-4-1. Anya and Borya are playing rock-paper-scissors. In this game, each player chooses one of the figures: rock, scissors, or paper. Rock beats scissors, scissors beat paper, and paper beats rock. If the players choose the same figure, the game ends in a tie.
Anya and Borya played 25 rounds. Anya chose rock 12 time... | Answer: 16.
Solution version 1. Note that, since there were no draws, when Anya chose rock, Borya must have chosen scissors or paper. Anya chose rock 12 times. Borya chose scissors or paper a total of $9+3=12$ times. Therefore, all these cases must have occurred in the rounds where Anya chose rock.
This means that in... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-6-1. In the figure, there are two circles with centers $A$ and $B$. Additionally, points $A, B$, and $C$ lie on the same line, and points $D, B$, and $E$ also lie on the same line. Find the degree measure of the angle marked with a “?”.
. Angles $ABE$ and $DBC$ are equal as vertical angles. Note that triangle $DBC$ is isosceles wi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-7-1. At a large round table, 90 people are feasting, facing the center of the table: 40 barons, 30 counts, and 20 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?
For example, for a count to stand... | Answer: 86.
Solution Variant 1. We ask half of the revelers, standing every other person, to step forward. Then all the people will be divided into two circles: an inner and an outer one. Note that a person in the inner circle has both neighbors in the outer circle, and they are also neighbors there. Similarly for a p... | 86 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-7-3. At a large round table, 55 people are feasting, facing the center of the table: 25 barons, 20 counts, and 10 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?
For example, for a count to stand... | Answer: 52.
Solution variant 3. We will place people in a new circle every other person: $1,3,5,7$, $\ldots, 55,2,4,6, \ldots, 54$, back to 1. It is sufficient to find the maximum number of neighbors with the same title in the new circle.
It is clear that there are neighbors of different titles in the new circle. It ... | 52 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-8-1. There is a magical grid sheet of size $2000 \times 70$, initially all cells are gray. The painter stands on a certain cell and paints it red. Every second, the painter takes two steps: one cell to the left and one cell down, and paints the cell red where he ends up after the two steps. If the painter is in the ... | Answer: 14000.
Solution version 1. We need to understand how many cells will be painted by the time the painter returns to the initial cell. Note that after every 2000 moves, the painter returns to the starting column, and after every 140 moves, he returns to the starting row. Therefore, he will return to the initial ... | 14000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.3. Inside parallelogram $A B C D$, a point $E$ is chosen such that $A E=D E$ and $\angle A B E=90^{\circ}$. Point $M$ is the midpoint of segment $B C$. Find the angle $D M E$. | Answer: $90^{\circ}$.
First solution. Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, its median $E N$ is also an altitude, that is, $E N \perp A D$. Therefore, $N E \perp B C$ (see Fig. 2).
Since $A D \| B C$ and $B M = M C = A N = N D = A D / 2$, quadrilaterals $A B M N$ and $B M D N$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. A confectionery factory produces $N$ types of candies. For New Year, the factory gave each of 1000 school students a gift containing candies of several types (the compositions of the gifts could be different). Each student noticed that for any 11 types of candies, they received a candy of at least one of these typ... | Answer. $N=5501$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{N}$ be the sets of students who did not receive candies of the 1st, 2nd, ..., $N$-th types, respectively. According to the problem, all these sets are distinct; moreover, each student is contained in no more than ten of them. Therefore, the total number of elem... | 5501 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. The numbers $1,2, \ldots, 2016$ are written on a board. It is allowed to erase any two numbers and replace them with their arithmetic mean. How should one proceed to ensure that the number 1000 remains on the board? | 6. Solution. We will first act according to the following scheme: $1,2,3, \ldots, n-3, n-2, n-1, n \rightarrow 1,2,3, \ldots, n-3, n-1, n-1 \rightarrow 1,2,3, \ldots, n-3, n-1$. Acting in this way, we will arrive at the set $1,2,3, \ldots, 1000,1002$. Then we will act according to the same scheme from the other end:
\... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Karlson counts 200 buns baked by Fräulein Bock: «one, two, three, ..., one hundred and nine, one hundred and ten, ..., one hundred and ninety-eight, one hundred and ninety-nine, two hundred». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer: 443 words.
Solution. One word will be required to pronounce 29 numbers: $1,2,3,4,5,6,7$, $8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required for their pronunciation is $2 \c... | 443 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. Lёsha colors cells inside a $6 \times 6$ square drawn on graph paper. Then he marks the nodes (intersections of the grid lines) to which the same number of colored and uncolored squares are adjacent. What is the maximum number of nodes that can be marked? | Answer: 45.
## Solution.
Estimation. Each grid node belongs to one, two, or four squares.
The corner vertices of the original square are adjacent to only one small square, so Lёsha will not be able to mark them. Therefore, the maximum number of marked nodes does not exceed $7 \cdot 7-4=45$.
Example.
, the third is in the 49th position, and so on. The ninth nine (from the number 89) is in the $9+2 \cdot 80=$ 169th position (numbers from 1 to 9 occ... | 174 | Number Theory | proof | Yes | Yes | olympiads | false |
3.1. From a square grid with a side of 40, a rectangle $36 \times 37$ was cut out, adjacent to one of the corners of the square. Grisha wants to color a five-cell cross in the remaining piece. In how many ways can he do this?
. Due to the cut-out, $36 \times 37-1$ ways are lost (the center of the cross cannot be in the cut-out rec... | 113 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.1. Every evening, starting from September 1st, little Anton ate one pastry. After eating another pastry, he noticed that during this entire time, he had eaten 10 delicious pastries (the rest seemed tasteless to him). But among any seven consecutive pastries he ate, no fewer than three turned out to be delicious. What... | Answer: 26
Solution. We will show that among 27 pastries, there will be no fewer than 11 delicious pastries. Let's number the pastries from 1 to 27. Note that among the first seven pastries, at least three are delicious, among the next seven as well, and among the pastries numbered from 15 to 21, there are at least th... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. Find the smallest natural number $n$ such that the natural number $n^{2}+14 n+13$ is divisible by 68. | Answer: 21
Solution. Note that $n^{2}+14 n+13=(n+1)(n+13)$, and $68=4 \cdot 17$. Therefore, one of the numbers $n+1$ and $n+13$ must be divisible by 17. Moreover, $n$ must be odd; otherwise, $n+1$ and $n+13$ would both be odd. The smallest such natural number $n$ is $2 \cdot 17-13=21$. Note that it works because $21^{... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. On September 2nd, Robin Bobin ate 12 chickens, and starting from September 3rd, he ate every day as many as he had already eaten on average in September. On September 15th, Robin Bobin ate 32 chickens. How many chickens did he eat on September 1st? | Answer: 52
Solution. Let at the beginning of the $k$-th day he on average ate $x$ chickens. This means that by this point, he had eaten $(k-1) x$ chickens. Then on the $k$-th day, he also ate $x$ chickens, which means that over $k$ days, he ate $k x$ chickens. This means that after the $k$-th day, the average number o... | 52 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. In quadrilateral $A B C D$, $A B=B C=C D$. Let $E$ be the intersection point of $A B$ and $C D$ (with $B$ between $A$ and $E$). It turns out that $A C=C E$ and $\angle D B C=15^{\circ}$. Find $\angle A E D$. | Answer: $50^{\circ}$
Solution. We will solve the general problem. Let $\angle D B C=\alpha$, and $\angle B E C=\beta$. From the isosceles property of $\triangle D C B$, we get $\angle B D C=\alpha$. From the isosceles property of $\triangle A C E$, we get $\angle B E C=\angle B A C$, and from the isosceles property of... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. There are 15 rectangular sheets of paper. In each move, one of the sheets is chosen and divided by a straight cut, not passing through the vertices, into two sheets. After 60 moves, it turned out that all the sheets are either triangles or hexagons. How many hexagons? | Answer: 25
Solution. Let after 60 cuts, $x$ hexagons and $75-x$ triangles are formed (there were 15 sheets, and with each cut, the number of sheets increases by 1). Now let's count the number of sides. Initially, there were $15 \cdot 4=60$ sides, and with each cut, the number of sides increases by 4 (the cut divides t... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (po... | Answer: 20.
Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, throu... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.
.
. From this similarity and the cyclic nature of the pentagon
. It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
# | # Answer: 110
## Solution
We will call the equilateral triangles that make up the original hexagon unit triangles. Consider each of the highlighted triangles inside the hexagon separately. The smallest one coincides with the unit triangle and has an area of 10. The medium triangle is inscribed in a hexagon made up of... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. In the chat of students from one of the schools, a poll was being held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the poll.
Then, 80 more people participated in the poll, voting only for October 22. After that, th... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 11.8. 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 $D Z$.
. In which year will this property reoccur for the first time?
(I. V. Raskina) | Answer: in 2022.
Solution. In 2010, 2011, ..., 2019, and in 2021, the year number contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be decreased by rearranging the digits.
## Comment.
3 points. For th... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. What angle do the clock hands form at 12:20? | Answer: $110^{\circ}$.
Solution. At 12:00, the clock hands coincide. After this, in 20 minutes, the minute hand travels $1 / 3$ of the circumference, i.e., it describes an angle of $120^{\circ}$. The hour hand moves 12 times slower than the minute hand (since it describes one circle in 12 hours). Therefore, in 20 minu... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.5. A square plot of 14 by 14 cells needs to be paved with rectangular tiles of size $1 \times 4$. The tiles can only be laid along the grid (not diagonally), and the tiles cannot be broken. What is the maximum number of tiles required? Will there be any uncovered area left? | # Solution:
Evaluation. The total number of cells on the plot is $14 \times 14=196$. Dividing by the number of cells in one tile, 196:4 = 49. Therefore, the number of tiles that can be cut from the $14 \times 14$ plot is no more than 49.
We will color the cells of the plot in 4 colors, as shown in the diagram. Clearl... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Six numbers are written in a row on the board. It is known that each number, starting from the third, is equal to the product of the two preceding numbers, and the fifth number is equal to 108. Find the product of all six numbers in this row. | 4. Answer: 136048896.
Solution. Let the first number be $x$, and the second number be $y$. Then the third number is $xy$.
The fourth number is $xy^2$. The fifth number is $x^2 y^3 = 108$. The sixth number is $x^3 y^3$.
The product of all six numbers is $x^8 y^{12} = (x^2 y^3)^4 = 108^4 = 136048896$.
## Grading Crit... | 136048896 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In Grandfather Frost's bag, there are chocolate and jelly candies, a total of 2023 pieces. Chocolate candies make up 75% of the jelly candies. How many chocolate candies are in Grandfather Frost's bag? | Answer: 867. Solution. Chocolate candies make up 3/4 of the quantity of jelly candies. Then the total number of candies is 7/4 of jelly candies. Therefore, the number of jelly candies is 2023×4:7=1156, and the number of chocolate candies is 2023-1156=867.
Grading criteria. A correct and justified solution - 7 points. ... | 867 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right ... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
.
What is the length of the path along the arrows if the length of segment ... | Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$. | 219 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.8. A numismatist has 100 coins that look identical. He knows that among them, 30 are genuine and 70 are counterfeit. Moreover, he knows that the weights of all genuine coins are the same, while the weights of all counterfeit coins are different, and any counterfeit coin is heavier than a genuine one; however, the ex... | Answer: 70.
Solution: 1. We will show that the numismatist can find the genuine coin in 70 weighings. Stack all 100 coins in a pile. With each weighing, he will select two coins from the pile and compare them. If their weights are equal, then both coins are genuine, and the required coin is found. If not, then the hea... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. Twelve different natural numbers are written in a circle, one of which is 1. Any two adjacent numbers differ by either 10 or 7. What is the greatest value that the largest written number can take? | Answer: 58.
Solution. For convenience, let's number all the numbers in a circle clockwise, starting from the number 1: $a_{1}=1, a_{2}, \ldots, a_{12}$. Notice that for all $1 \leqslant i \leqslant 6$ we have
$$
\begin{gathered}
a_{i+1}-a_{1}=\left(a_{i+1}-a_{i}\right)+\left(a_{i}-a_{i-1}\right)+\ldots+\left(a_{2}-a_... | 58 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. Let $\alpha$ and $\beta$ be the real roots of the equation $x^{2}-x-2021=0$, with $\alpha>\beta$. Denote
$$
A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7
$$
Find the greatest integer not exceeding $A$. | Answer: -6055.
Solution. By Vieta's theorem, we have $\alpha+\beta=1$ and $\alpha \beta=-2021$. Also, $\beta^{2}-\beta-2021=0$, since $\beta$ is a root of the equation. Therefore,
$A=\alpha^{2}-2 \beta^{2}+2 \alpha \beta+3 \beta+7=(\alpha+\beta)^{2}-3\left(\beta^{2}-\beta-2021\right)-6063+7=1^{2}-3 \cdot 0-6063+7=-60... | -6055 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. Let $k_{1}$ be the smallest natural number that is a root of the equation
$$
\sin k^{\circ}=\sin 334 k^{\circ}
$$
(a) (2 points) Find $k_{1}$.
(b) (2 points) Find the smallest root of this equation that is a natural number greater than $k_{1}$. | # Answer:
(a) (2 points) 36.
(b) (2 points) 40.
Solution.
$$
0=\sin 334 k^{\circ}-\sin k^{\circ}=2 \sin \frac{333 k^{\circ}}{2} \cos \frac{335 k^{\circ}}{2}
$$
- Let $\sin \frac{333 k^{\circ}}{2}=0$. This is equivalent to $\frac{333 k}{2}=180 \mathrm{~m}$ for some integer $m$, that is, $333 k=360 m$ and $37 k=40 m... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. In a sports school, 55 people are training, each of whom is either a tennis player or a chess player. It is known that there are no four chess players who would have the same number of friends among the tennis players. What is the maximum number of chess players that can train in this school? | Answer: 42.
Solution. Let there be $a$ tennis players and $55-a$ chess players in the school. Each of the chess players has a number of tennis player friends that is no less than 0 and no more than $a$, meaning it can take on $a+1$ values. If there were more than $3(a+1)$ chess players, by the pigeonhole principle, th... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Points $A, B, C, D, E, F$ are arranged clockwise on a circle, as shown in the figure. Chords $A D$ and $C E$ intersect at point $X$ at a right angle, and chords $A D$ and $B F$ intersect at point $Y$.
It is known that $C X=12, X E=27, X Y=15, B Y=9, Y F=11$.
 (2 points) 36.
(b) (2 points) 19.5.
Solution. Let $A Y=a, X D=b$ (Fig. 8). We use the fact that the products of the segments of intersecting chords through a given point inside a circle are equal. This means that $324=12 \cdot 27=a(15+b)$ and $99=9 \cdot 11=b(a+15)$. Subtracting the second equation fro... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Given the set of numbers $\{-1,-2,-3, \ldots,-26\}$. On the board, all possible subsets of this set were written, each containing at least 2 numbers. For each written subset, the product of all numbers belonging to that subset was calculated. What is the sum of all these products? | Answer: 350.
Solution. Consider the expression $(1-1)(1-2)(1-3) \ldots(1-26)$, the value of which is 0. Expand this expression without combining like terms, resulting in the sum $1^{26}-1^{25}(1+2+3+\ldots+26)+S$. Notice that the sum $S$ coincides with the sum from the problem's condition. Therefore, it equals $0-1+(1... | 350 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.7. All vertices of a regular tetrahedron $ABCD$ are on the same side of the plane $\alpha$. It turns out that the projections of the vertices of the tetrahedron onto the plane $\alpha$ form the vertices of a certain square. Find the value of $AB^2$ if it is known that the distances from points $A$ and $B$ to... | Answer: 32.
Fig. 9: to the solution of problem 11.7
Solution. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the projections of points $A, B, C, D$ onto the plane $\alpha$. Let $M$ be the midpoint of s... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. In each cell of a strip $1 \times N$ there is either a plus or a minus. Vanya can perform the following operation: choose any three cells (not necessarily consecutive), one of which is exactly in the middle between the other two cells, and change the three signs in these cells to their opposites. A number... | Answer: 1396.
Solution. We will prove that all the numbers under consideration are positive, except for 4 and 5. Then the answer to the problem will be $1398-2=1396$.
For each $N$, we will number the cells of the strip $1 \times N$ from left to right with numbers from 1 to $N$.
- Let $N=3$. By applying the operation... | 1396 | Combinatorics | 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 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 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?
This results in 8 rectan... | 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 |
8.5. In the parliament of the island state of Promenade-and-Tornado, only the indigenous inhabitants of the island can be elected, who are divided into knights and liars: knights always tell the truth, liars always lie. A secret ballot on 8.09.19 re-elected 2019 deputies. At the first meeting, all deputies were present... | Solution. A rectangle $42 \times 48$ can be tiled with $3 \times 3$ squares (224 are required). In each such square, there must be at least one knight (otherwise - if the square contains only liars, the liar in the central cell would have told the truth, which is impossible). Thus, the minimum number of knights in the ... | 225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. It is known that $\frac{1}{\cos (2022 x)}+\operatorname{tg}(2022 x)=\frac{1}{2022}$.
Find $\frac{1}{\cos (2022 x)}-\operatorname{tg}(2022 x)$. | Answer. 2022.
## Solution.
Consider the product $\left(\frac{1}{\cos \alpha}+\operatorname{tg} \alpha\right)\left(\frac{1}{\cos \alpha}-\operatorname{tg} \alpha\right)=\frac{1}{\cos ^{2} \alpha}-\operatorname{tg}^{2} \alpha=\frac{1-\sin ^{2} \alpha}{\cos ^{2} \alpha}=1$. Therefore, the desired expression is 2022.
Co... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. One hundred and one non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of negative numbers that can be among these 101 written numbers? | Answer: 67.
Solution: Consider any 3 consecutive numbers. All of them cannot be negative. Among them, there is a positive one. Fix this positive number and its neighbor (this number can be negative), and divide the remaining 99 numbers into 33 groups of 3 consecutive numbers. In each such group, there will be no more ... | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Martians love to dance dances where they have to hold hands. In the dance "Pyramid," no more than 7 Martians can participate, each of whom has no more than three hands. What is the maximum number of hands that the dancers can have if any hand of one Martian holds exactly one hand of another Martian? | Solution. In each handshake, two hands are involved, so the total number of hands shaken will be even. The maximum variant (7 Martians with three hands) gives 21 hands. Therefore, the greatest number of hands will be achieved if 6 Martians have three hands each and one has two hands. The dancers have 20 hands.
 for a game. Each had to say one of the following phrases: "There is a liar below me" or "There is a k... | Answer: 17
Solution. The formation cannot consist of only liars (in such a case, the 3-7th person would be telling the truth). Consider the tallest knight, let's call him P. He could only say the first phrase, so the 1st and 2nd places must be occupied by liars, and P stands from the 3rd to the 7th place. Moreover, th... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3 On the Island of Truth and Lies, there are knights who always tell the truth, and liars who always lie. One day, 17 residents of the island lined up by height (from tallest to shortest, the tallest being the first) for a game. Each had to say one of the phrases: "There is a liar below me" or "There is a knight abov... | Answer: 14
Solution. The formation cannot consist of only liars (in such a case, 3-6 would be telling the truth). Consider the tallest knight, let's call him P. He could only say the first phrase, so the 1st and 2nd positions must be occupied by liars, and P stands from the 3rd to the 6th position. Moreover, there is ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. Each student in the 7th grade attends a math club or an informatics club, and two students attend both the math club and the informatics club. Two-thirds of all students attend the math club, and $40 \%$ attend the informatics club. How many students are there in this class? | Answer: 30.
Solution: Note that $40\%$ is $\frac{2}{5}$ of all students. Adding $\frac{2}{3}$ and $\frac{2}{5}$ gives $\frac{16}{15}$. The two students who are involved in both the math and informatics clubs were counted twice. Therefore, these two students represent $\frac{1}{15}$ of the entire class, meaning there a... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Vovochka is playing a computer game. If he scores less than 1000 points, the computer will add $20 \%$ of his score. If he scores from 1000 to 2000 points, the computer will add $20 \%$ of the first thousand points and $30 \%$ of the remaining points. If Petya scores more than 2000 points, the computer will add $20 ... | # Solution
It is clear that Petya scored more than 1000 points (otherwise his result would not exceed 1200) and less than 2000 (otherwise the result would not be less than 2500). Let's discard the 1200 points (the first thousand plus the prize for it). The remaining 1170 points constitute $130 \%$ of the points Petya ... | 470 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. Students from 9A, 9B, and 9C gathered for a line-up. Maria Ivanovna decided to count the number of attendees from each class. She found that there were 27 students from 9A, 29 students from 9B, and 30 students from 9C. Ilya Grigoryevich decided to count the total number of attendees from all three classes,... | Answer: 29.
Solution. Let $a, b, c$ be the actual number of students from 9 "A", 9 "B", 9 "C" respectively. From Maria Ivanovna's counts, it follows that $a \leqslant 29, b \leqslant 31, c \leqslant 32$, and from Ilya Grigoryevich's counts, it follows that $a+b+c \geqslant 92$. We get that
$$
92 \leqslant a+b+c \leqs... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. An empty $3 \times 51$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 153$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 152$, the num... | Answer: (a) 152. (b) 76.
Solution. Let's color the table in black and white so that the cell where the number 1 should be is black. The third rule in the condition means that consecutive numbers in the table are placed in a "snake" pattern, and the colors of their cells alternate, that is, all odd numbers, including 1... | 152 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.5. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees doe... | Answer: (a) $107^{\circ}$. (b) $47^{\circ}$.
Fig. 5: to the solution of problem 9.5
Solution. Given $A B=B X$ and $\angle B=34^{\circ}$, therefore $\angle A X B=\angle X A B=\frac{1}{2}\lef... | 107 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. The graph of the line $y=k x+l$ intersects the $O x$ axis at point $B$, the $O y$ axis at point $C$, and the graph of the function $y=1 / x$ at points $A$ and $D$, as shown in the figure. It turns out that $A B=B C=C D$. Find $k$, given that $O C=3$.
Fig. 6: to the solution of problem 9.6
Solution. Draw a line through point $D$ parallel to the $O y$ axis; let it intersect the $O x$ axis at point $D_{1}$ (Fig. 6). It is easy ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In front of a sweet-tooth lie five boxes of candies: the first box contains 11 candies, the second - 22 candies, the third - 33 candies, the fourth - 44 candies, the fifth - 55 candies. In one move, the sweet-tooth can take four candies from one box and distribute them, one candy to each of the remaining f... | Answer: 159.
Solution. Consider the number of candies in the boxes. Initially, these are 5 natural numbers that give different remainders when divided by 5. We will prove that as long as the sweet-tooth does not take candies from any box, these 5 numbers will always give different remainders when divided by 5. For thi... | 159 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.2.2. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=11$, $NC=39$, $AM=12$, $MB=3$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 705. (b) 298.5. | 705 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.2.3. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=3$, $NC=39$, $AM=10$, $MB=5$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 585. (b) 202.5. | 585 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.2.4. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=9$, $NC=39$, $AM=10$, $MB=5$.
(a) (1 point) Find the area of rectangle $ABCD$.
(b) (3 points) Find the area of triangle $MNC$.
 675. (b) 247.5. | 675 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.1. An empty $3 \times 51$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 153$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 152$, the n... | Answer: (a) 152. (b) 76. | 152 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.2. An empty $3 \times 53$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1,2,3, \ldots, 159$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 158$, the num... | Answer: (a) 158. (b) 79. | 158 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.3. An empty $3 \times 55$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1,2,3, \ldots, 165$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 164$, the num... | Answer: (a) 164 . (b) 82. | 82 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.4.4. An empty $3 \times 57$ table is drawn on the board. Masha wants to fill its cells with numbers, following these rules:
- each of the numbers $1, 2, 3, \ldots, 171$ must be present in the table;
- the number 1 must be in the bottom-left cell of the table;
- for any natural number $a \leqslant 170$, the n... | Answer: (a) 170 . (b) 85. | 170 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.1. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 34^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $107^{\circ}$. (b) $47^{\circ}$. | 107 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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