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Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
^{3}-8^{3}=\left(\mathrm{a}^{2}-8\right)\left(\mathrm{a}^{4}+8 \mathrm{a}^{2}+64\right)$ is divisible by $n$. Also note that the number $\mathrm{y}=\mathrm{a}^{6}-25^{2}=\left(\mathrm{a}^{3}\right)^{2}-25^{2}=\left(\mathrm{a}^{3}... | 113 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.4. A magician and an assistant have a deck of cards; one side (the "back") of all cards is the same, while the other is painted in one of 2017 colors (the deck contains 1,000,000 cards of each color). The magician and the assistant are going to perform the following trick. The magician leaves the room, and the audie... | Answer. $n=2018$.
Solution. Let $k=2017$.
For $n=k+1$, the trick is easy to arrange. The magician and the assistant number the colors from 1 to $k$. The assistant, seeing the color of the last, $(k+1)$-th card (let its number be $a$), leaves the $a$-th card open. The magician, seeing which numbered card is open, can ... | 2018 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Petya runs down from the fourth floor to the first floor 2 seconds faster than his mother rides the elevator. Mother rides the elevator from the fourth floor to the first floor 2 seconds faster than Petya runs down from the fifth floor to the first floor. How many seconds does it take for Petya to run down from the ... | Answer: 12 seconds.
Solution. Between the first and fourth floors, there are 3 flights, and between the fifth and first floors, there are 4. According to the problem, Petya runs 4 flights 2 seconds longer than it takes his mother to ride the elevator, and 3 flights 2 seconds faster than his mother. Therefore, it takes... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the number line, points with integer coordinates are painted red and blue according to the following rules: a) points whose coordinate difference is 7 must be painted the same color; b) points with coordinates 20 and 14 should be painted red, and points with coordinates 71 and 143 - blue. In how many ways can all... | Answer. In eight ways.
Solution. From part a), it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers $0,1,2,3,4,5$, and 6. The point $0=14-2 \cdot 7$ must be colored the same as 14, i.e., red. Similarly, the point $1=71-10 ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given a rectangle $A B C D$. Point $M$ is the midpoint of side $A B$, point $K$ is the midpoint of side $B C$. Segments $A K$ and $C M$ intersect at point $E$. How many times smaller is the area of quadrilateral $M B K E$ compared to the area of quadrilateral $A E C D$? | Answer: 4 times.
Solution. Draw segments $MK$ and $AC$. Quadrilateral $MBKE$ consists of triangles $MBK$ and $MKE$, while quadrilateral $AEC D$ consists of triangles $AEC$ and $ACD$. We can reason in different ways.
1st method. Triangles $MBK$ and $ACD$ are right-angled, and the legs of the first are half the length ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. There are 40 pencils of four colors - 10 pencils of each color. They were distributed among 10 children so that each received 4 pencils. What is the smallest number of children that can always be selected to ensure that pencils of all colors are found among them, regardless of the distribution of pencils?
(I. Bo... | 10.1. Answer. 3 boys.
We will show that it is always possible to choose three boys such that they have pencils of all colors. Since there are 10 pencils of each color and each boy received 4 pencils, at least one boy must have received pencils of at least two different colors. It remains to add to him two boys who hav... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Plot the graph of the function $\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$. | Answer. The graph of the function will be the line $y = 4$.
## Solution.
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
, but in this case, the number must also be divisible by 2. Therefore, the digit 5 is not in this number. ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In triangle $A B C$, the median $B M$ is twice as short as side $A B$ and forms an angle of $40^{\circ}$ with it. Find the angle $A B C$. | Solution. Extend median $B M$ beyond point $M$ by the same length to get point $D$ (see figure). Since $A B = B D$, triangle $A B D$ is isosceles. Therefore, $\angle B A D = \angle B D A = (180^{\circ} - 40^{\circ}) : 2 = 70^{\circ}$.
Quadrilateral $A B C D$ is a parallelogram because its diagonals bisect each other. ... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. In the parliament of the island state of Promenade-and-Tornado, 2019 indigenous inhabitants were elected, who are divided into knights and liars: knights always tell the truth, liars always lie. At the first meeting, 2016 of them sat in the parliamentary seats arranged in the hall in a rectangle of $42 \times 48$,... | Solution. If two liars are adjacent in the hall, then the entire hall is filled with only liars, which corresponds to the maximum number of liars in the hall. For the minimum number of liars, each liar is adjacent only to knights. The rectangle $42 \times 48$ can be tiled with 224 squares of $3 \times 3$. The minimum n... | 227 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Dima was supposed to arrive at the station at 18:00. By this time, his father was supposed to pick him up in a car. However, Dima managed to catch an earlier train and arrived at the station at 17:05. He didn't wait for his father and started walking towards him. On the way, they met, Dima got into the car, and they... | Answer: 6 km/h
Solution. Dima arrived home 10 minutes earlier, during which time the car would have traveled the distance Dima walked twice. Therefore, on the way to the station, the father saved 5 minutes and met Dima at 17:55. This means Dima walked the distance from the station to the meeting point in 50 minutes, s... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. "Did you answer the previous questions honestly?"
40 gnomes answered "yes" to the first question, 50 to the second, 70 to the third, and 100 to the fourth. How many honest gnomes are there in the underground kingdom? | Answer: 40 honest gnomes.
## Solution.
On the 4th question, both an honest and a liar will answer "yes," so there are 100 gnomes in the underground kingdom.
An honest gnome will answer "yes" to one of the first three questions and "no" to two. A liar, on the other hand, will answer "yes" to two of the first three qu... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$, the median from vertex $A$ is perpendicular to the bisector of angle $B$, and the median from vertex $B$ is perpendicular to the bisector of angle $A$. It is known that side $AB=1$. Find the perimeter of triangle $ABC$. | Answer: 5.
Solution. Let $A M$ be the median drawn from vertex $A$. Then, in triangle $A B M$, the bisector of angle $B$ is perpendicular to side $A M$, i.e., the bisector is also an altitude. Therefore, this triangle is isosceles, $A B = B M = 1$. Hence, $B C = 2 B M = 2$. Similarly, from the second condition, we get... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. There are three vessels with volumes of 3 liters, 4 liters, and 5 liters, without any markings, a water tap, a sink, and 3 liters of syrup in the smallest vessel. Can you, using pourings, obtain 6 liters of a water-syrup mixture such that the amount of water is equal to the amount of syrup in each vessel? | Solution.
For example, as follows (see the table below, c - syrup, w - water, f - final mixture).
| |  | 4-liter container | 5-liter container |
| :---: | :---: | :---: | :---: |
| Pour the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. What is the maximum number of sides a polygon can have if each of its angles is either $172^{\circ}$ or $173^{\circ}$?
Let the number of angles with a degree measure of $172^{\circ}$ be $a$, and those with $173^{\circ}-b$. Then the sum of all angles of the polygon will be $172a + 173b$. On the other hand, the sum o... | Answer: 51.
Criteria: correct answer without explanation - 2 points. | 51 | 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 $4x$. 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. Drop a perpendicular from point $D$ in the isosceles triangle $ABD$, 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 inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles of $... | 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 |
Task No. 1.1
## Condition:
Kirill Konstantinovich's age is 48 years 48 months 48 weeks 48 days 48 hours. How many full years old is Kirill Konstantinovich? | Answer: 53
Exact match of the answer -1 point
## Solution.
48 months is exactly 4 years. 48 weeks is $48 \times 7=336$ days. Together with another 48 days, this totals 384 days, which is 1 year and another 18 or 19 days, depending on whether the year is a leap year. In any case, the remaining days plus another 48 ho... | 53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.3
Condition:
Anna Alexandrovna's age is 60 years 60 months 60 weeks 60 days 60 hours. How many full years old is Anna Alexandrovna | Answer: 66
Exact match of the answer -1 point
Solution by analogy with task №1.1.
# | 66 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.4
Condition:
Tatyana Timofeevna's age is 72 years 72 months 72 weeks 72 days 72 hours. How many full years old is Tatyana Timofeevna | Answer: 79
Exact match of the answer -1 point
Solution by analogy with task №1.1.
# | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 2.1
Condition:
Dmitry has socks in his wardrobe: 6 pairs of blue socks, 18 pairs of black socks, and 12 pairs of white socks. Dmitry bought some more pairs of black socks and found that now the black socks make up 3/5 of the total number of socks. How many pairs of black socks did Dmitry buy? | Answer: 9
Exact match of the answer -1 point
## Solution.
Dmitry had an equal number of pairs of black and all other pairs of socks, and after the purchase, it turned out that the black pairs make up three parts of all socks, while the other pairs make up two parts. This means that Dmitry bought half of the number o... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 2.3
## Condition:
Dmitry has socks in his wardrobe: 14 pairs of blue socks, 24 pairs of black socks, and 10 pairs of white socks. Dmitry bought some more pairs of black socks and found that now the black socks make up 3/5 of the total number of socks. How many pairs of black socks did Dmitry buy? | Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №2.1.
# | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.1
## Condition:
On a sheet of graph paper, there are three $5 \times 5$ squares, as shown in the figure. How many cells are covered by exactly two squares?
 | Answer: 15
Exact match of the answer -1 point
## Solution 1
Direct calculation. We will draw the squares and shade the cells that are covered exactly twice.
Solution 2.
The total area c... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.3
## Condition:
On a sheet of graph paper, there are three $5 \times 5$ squares, as shown in the figure. How many cells are covered by exactly two squares?
 | Answer: 13
Exact match of the answer -1 point
Solution by analogy with task №3.1.
# | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.2
## Condition:
A car number contains three letters and three digits, for example A123BE. The letters allowed for use are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Katya considers a number lucky if the second letter is a consonant, the first digit... | Answer: 288000
Exact match of the answer -1 point
Solution by analogy with task №5.1.
# | 288000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.3
## Condition:
A car number contains three letters and three digits, for example A123BE. The letters allowed for use are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Tanya considers a number lucky if the first letter is a consonant, the second lette... | Answer: 384000
Exact match of the answer -1 point
Solution by analogy with task №5.1.
# | 384000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.4
## Condition:
A car number contains three letters and three digits, for example A123BE. The letters allowed for use are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Kira considers a number lucky if the second letter is a vowel, the second digit is ... | Answer: 144000
Exact match of the answer -1 point
Solution by analogy with task №5.1.
# | 144000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.4
## Condition:
On the faces of a cube, 6 letters are drawn: A, B, V, G, D, E. The picture shows three images of the cube from different angles. Which letter is drawn on the face opposite the face with the letter $\mathrm{A}$?
![](https://cdn.mathpix.com/cropped/2024_05_06_4cfa7cdc7a8a51b9e752g-25.jpg?he... | Solution by analogy with problem №6.1.
## Condition:
Anya, Borya, and Vasya took the same test consisting of 6 questions, each of which can be answered with "yes" or "no." The answers are presented in the table:
| Question No. | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Anya | no |... | 3 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
# Task No. 8.1
## Condition:
For the "Handy Hands" club, Anton needs to cut several identical pieces of wire (the length of each piece is a whole number of centimeters). Initially, Anton took a piece of wire 10 meters long and was able to cut only 9 needed pieces from it. Then Anton took a piece 11 meters long, but i... | Answer: 111
Exact match of the answer -1 point
## Solution.
First, note that 9 pieces of 111 cm each make up 999 cm. Therefore, both the first and the second piece are enough for 9 parts, but the second piece is not enough for 10 parts: $10 \times 111=1110>1100$.
We will prove that if the length of the piece is not... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Currently, the mother is 24 years and 3 months old, and her daughter is 5 months old. After how many months will the number of years in the mother's age be equal to the number of months in the daughter's age? | Answer: 21.
Solution. Let $x$ be the required number of months. Then we get the equation: $24+(x+3) / 12=x+5$. From this, $x=21$. | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 2.
Currently, mom is 23 years and 8 months old, and her daughter is 9 months old. In how many months will the number of years in mom's age be equal to the number of months in her daughter's age? | Answer: 16.
Option 3.
Currently, the mother is 19 years and 4 months old, and her daughter is 1 month old. After how many months will the number of years in the mother's age be equal to the number of months in the daughter's age?
Answer: 20. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Variant 1.
In the game "Mathematical Running," nine teams participated (not necessarily equal in the number of participants). On average, there were 7 people in each team. After one team was disqualified, the average number of participants in the remaining teams decreased to 6. How many participants were in the dis... | Answer: 15.
Solution. The total number of participants before disqualification was $7 \cdot 9=63$. After the disqualification of participants, $6 \cdot 8=48$ remained. Therefore, the number of participants in the disqualified team was $63-48=15$. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Variant 1.
The height $A H$ and the bisector $C L$ of triangle $A B C$ intersect at point $O$. Find the angle $B A C$, if it is known that the difference between the angle $C O H$ and half the angle $A B C$ is $46^{\circ}$. | Answer: 92.
Solution. Let the halves of the angles $A, B, C$ of triangle $ABC$ be denoted by $x, y$, and $z$ respectively. Then, $\angle COH = 90^{\circ} - z$ and $46^{\circ} = 90^{\circ} - z - y$. Since $x + y + z = 90^{\circ}$, we have $x = 44$ and $\angle BAC = 92^{\circ}$. | 92 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 4. Option 1.
Find the number of four-digit numbers where the digit in the units place is exactly 1 more than the digit in the tens place. The number cannot start with zero. | Answer: 810.
Solution. The leading digit of the number can be chosen in 9 ways (any digit except zero). The digit in the hundreds place can be chosen in 10 ways (any digit will do). The digit in the tens place can be any digit from 0 to 8, and the digit in the units place is uniquely determined by the chosen digit in ... | 810 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
An artistic film with a duration of 192 minutes consists of four parts. It is known that the duration of any two parts differs by at least 6 minutes. What is the maximum duration that the shortest part can have? Express your answer in minutes. | Answer: 39.
Solution. Let the shortest part be $x$ minutes, then the second (in terms of duration) is no less than $x+6$, the third is no less than $x+12$, and the fourth is no less than $x+18$. Therefore, the entire film lasts no less than $4 x+36$ minutes. Solving the inequality $192 \geq 4 x+36$, we get $x \leq 39$... | 39 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
In a convex $n$-gon, a diagonal is highlighted. The highlighted diagonal is intersected by exactly 14 other diagonals of this $n$-gon. Find the sum of all possible values of $n$. A vertex of the $n$-gon is not considered an intersection. | Answer: 28.
Solution. Let there be $x$ sides on one side of the diagonal, and $-n-x$ on the other. Then there are $-x-1$ vertices on one side, and $-n-x-1$ on the other. They can be the endpoints of the required diagonals. We get $14=(x-1)(n-x-1)$.
The following cases are possible:
$x-1=14, n-x-1=1$, then $n=17$
$x... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1.
On the Island of Misfortune, there live truth-tellers, who always tell the truth, and liars, who always lie. One day, 2023 natives, among whom $N$ are liars, stood in a circle, and each said: "Both of my neighbors are liars." How many different values can $N$ take? | Answer: 337.
Solution: Both neighbors of a knight must be liars, and the neighbors of a liar are either two knights or a knight and a liar. Therefore, three liars cannot stand in a row (since in this case, the middle liar would tell the truth). We can divide the entire circle into groups of liars/knights standing in a... | 337 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.6. On the board, the expression $\cos x$ is written. It is allowed to add or multiply several expressions written on the board (an expression can be used multiple times) and write the new obtained expression on the board. Is it possible to get an expression that takes the value 0 when $x=\pi$ after several actions?
... | Answer. Yes, it is possible.
Solution. The first action is to append $\cos ^{2} x$, the second is to append $\cos ^{2} x+\cos x$. Since $\cos \pi=-1$, the value of the last expression at $x=\pi$ is 0.
Comment. An answer without presenting the required expression - 0 points.
Only presenting any correct required expre... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Three lines intersect to form 12 angles, and $n$ of them turn out to be equal. What is the maximum possible value of $n$? | Answer: 6.
Solution: Three lines limit a certain triangle. If this triangle is equilateral, then out of twelve angles, six are $60^{\circ}$, and the other six are $120^{\circ}$.
Can any external angle of the triangle be equal to its internal angle? It is equal to the sum of the non-adjacent internal angles, so it is ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the coefficients of the polynomial obtained after expanding the brackets and combining like terms in the expression
$\left(2 x^{2021}-x^{2020}+x^{2019}\right)^{11}-29$. | Solution. First, note that the sum of the coefficients of any polynomial in canonical form $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}$ is $P(1)$. Next, note that after raising the bracket to the 11th power, the polynomial $Q(x)=\left(2 x^{2021}-x^{2020}+x^{2019}\right)^{11}-29$ will take a canonical form (s... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1
Masha wrote the number 547654765476 on a piece of paper. She erased several digits so that the resulting number is the largest possible multiple of 9. What is this number? | Answer: 5476547646.
Solution: The sum of the digits of the original number is $3 \cdot(7+6+5+4)=66$. From the divisibility rule by 9, it follows that the sum of the erased digits must give a remainder of 3 when divided by 9. It is impossible to select digits with a sum of 3. Digits with a sum of $3+9=12$ can be chosen... | 5476547646 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 4. Variant 1
A number with the sum of its digits equal to 2021 was divided by 7, resulting in a number that is written only with the digit 7. How many digits 7 can be in it? If there are multiple answers, indicate their sum. | Answer: 503.
Solution. By multiplying $777 \ldots 77$ and 7 in a column, we get 54...439. Let $x$ be the number of sevens in the original number, then the number of fours in the resulting product is $x-2$. Its sum of digits is $4(x-2)+5+3+9=4x+9$, but on the other hand, it is equal to 2021, so $x=\frac{2021-9}{4}=$ $\... | 503 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1
Unit cubes were used to build a large cube. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 4 neighbors is 132. Find the number of cubes that have exactly 5 neighbors. | Answer: 726.
Solution: Cubes that have exactly 4 neighbors touch exactly one edge of the large cube. There are 12 edges in total, so 11 such cubes touch each edge. Therefore, the number of cubes that are strictly inside each face is $11 \cdot 11 = 121$. Since there are 6 faces, the answer is: $121 \cdot 6 = 726$.
## ... | 726 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1
In the garden, there are 46 trees - apple trees and pears. It turned out that among any 28 trees, there is at least one apple tree, and among any 20 trees, there is at least one pear. How many pears are there in the garden? | Answer: 27.
Solution: Since among 28 trees there is at least one apple tree, the number of pears is no more than 27. Since among any 20 trees there is at least one pear, the number of apple trees is no more than 19. There are 46 trees in total, so there are 19 apple trees and 27 pears.
## Variant 2
In the garden, th... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
In trapezoid $A B C D(A D \| B C)$, the bisectors of angles $D A B$ and $A B C$ intersect on side $C D$. Find $A B$, if $A D=5, B C=2$. | Answer: 7.
Solution.
Mark point $L$ on side $AB$ such that $LB = BC$. Let $K$ be the intersection point of the angle bisectors of $\angle DAB$ and $\angle ABC$. Then triangles $LBK$ and $BC... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.2. Masha eats a bowl of porridge in 12 minutes, and Bear eats it twice as fast. How long will it take them to eat six bowls of porridge? | Answer: 24 min.
Solution: In 12 minutes, Masha eats one bowl of porridge, and Bear eats 2 bowls. In total, 3 bowls of porridge are eaten in 12 minutes. Six bowls of porridge will be eaten in twice the time, i.e., $12 * 2=24$ min.
Comment: A correct answer without justification - 3 points. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.5. The hedgehogs collected 65 mushrooms and divided them so that each hedgehog got at least one mushroom, but no two hedgehogs had the same number of mushrooms. What is the maximum number of hedgehogs that could be | Answer: 10
Solution: If there were 11 hedgehogs, then together they would have collected no less than $1+2+3+\ldots+10+11=66$ mushrooms, which exceeds the total number of mushrooms collected. Therefore, there were fewer than 11 hedgehogs. We will show that there could have been 10 hedgehogs. Suppose the first found 1 ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. Given an odd number $n>10$. Find the number of ways to arrange the natural numbers $1,2,3, \ldots, n$ in a circle in some order so that each number is a divisor of the sum of the two adjacent numbers. (Ways that differ by rotation or reflection are considered the same.) (D. Khramov) | Answer. Two ways
Solution. Consider an arbitrary arrangement of numbers from 1 to $n$ that satisfies the conditions. Suppose that two even numbers $x$ and $y$ are adjacent, and the next number is $z$. Since $x+z$ is divisible by $y$, the number $z$ is also even. Continuing this movement around the circle, we get that ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. What values can the sum $f_{2}\left(x_{1}\r... | Answer: Only 0.
Solution: Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then
$f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$, since $f_{1}\left(x_{1}\right)=0$. Similarly, we obtain the equalities $f_{3}\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, $2, \ldots, 11$ kg. He also has a bag that will tear if more than 11 kg is placed in it. Archimedes has learned the weights of all the ingots and wants to prove to Hiero that the first ... | Answer. In 2 loads.
Solution. We will show that Archimedes can use the bag only twice. Let him first put in the bag ingots weighing 1, 2, 3, and 5 kg, and then ingots weighing 1, 4, and 6 kg. In both cases, the bag will not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if A... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. What values can the sum $f_{2}\left(x_{1}\... | Answer: Only 0.
Solution. Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then
$$
f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1},
$$
since $f_{1}\left(x_{1}\right)=0$. Similarly, we obtain the equalities $f... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. A $100 \times 100$ grid is given, with cells colored black and white. In each column, there are an equal number of black cells, while in each row, the number of black cells is different. What is the maximum possible number of pairs of adjacent cells of different colors? (I. Bogdanov) | Answer. $6 \cdot 50^{2}-5 \cdot 50+1=14751$ pairs.
Solution. Let the side length of the table be $2 n = 100$ (so $n=50$) and number the rows from top to bottom and the columns from left to right with numbers from 1 to $2 n$.
In each row, there can be from 0 to $2 n$ black cells. Since the number of black cells in all... | 14751 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. What different values can the digit U take in the puzzle U$\cdot$LAN + U$\cdot$DE = 2020? Justify your answer. (Identical digits are replaced by the same letters, different digits by different letters.) | Answer: two. $\mathrm{V}=2, \mathrm{y}=5$.
Solution: Factor out the common factor: У$\cdot$(ЛАН + ДЭ) $=2020$. Note that У, Л, and Э are not equal to 0.
Factorize the right-hand side: $2020=1 \cdot 2 \cdot 2 \cdot 5 \cdot 101$. Since У is a digit, consider all possible values: $\mathrm{V}=1,2,4,5$.
1) $У=1$. Then ЛА... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. A student-entrepreneur bought several packs of masks from a pharmacy and sold them to classmates at a higher price, making a profit of 1000 rubles. With all the money earned, he again bought masks from the pharmacy (at the same price as the first time) and sold them to classmates (at the same price as the first ti... | Answer: 2000 rubles.
Solution 1: Let the package of masks in the pharmacy cost x rubles, and the student sold the masks for y rubles, and bought a packages of masks the first time. Then, according to the condition, $a(y-x)=1000$.
The revenue amounted to ay rubles, so the second time the student was able to buy $\frac... | 2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. In triangle $M P Q$, a line parallel to side $M Q$ intersects side $M P$, median $M M_{1}$, and side $P Q$ at points $D, E$, and $F$ respectively. It is known that $D E=5$, and $E F=7$. What is the length of $M Q$? | Answer: 17.
Solution: Draw a line through point $E$ parallel to $P Q$ ( $K$ and $L$ - the points of intersection of this line with sides $M P$ and $M Q$ respectively). Since $M M_{1}$ is a median, then $L E = E K$, and since $D F \parallel M Q$, it follows that $D E$ is the midline of triangle $M K L$. Therefore, $M L... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest 10-digit number, the sum of whose digits is greater than that of any smaller number. | Answer: 1999999999.
Solution. Among 9-digit numbers, the largest sum of digits is for the number 999999 999, which is 81. Since the sought 10-digit number is greater than 999999 999, we need to find the smallest number with a sum of digits no less than 82. If the first digit of this number is 1, then the sum of the re... | 1999999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 9 times fewer points than all the others combined. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament? | # Answer: 15 chess players.
Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. They drew the other half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Side $A B$ of triangle $A B C$ is greater than side $B C$, and angle $B$ is $40^{\circ}$. A point $P$ is taken on side $A B$ such that $B P = B C$. The bisector $B M$ intersects the circumcircle of triangle $A B C$ at point $T$. Find the angle $M P T$. | Answer: $20^{\circ}$.
Solution. (Fig. 3.) In the quadrilateral $A P M T$, the angle at vertex $A$ is measured by half the arc $T C B$. Triangles $P M B$ and $C M B$ are equal by two sides and the angle between them, so $\angle P M B = \angle C M B = \angle A M T$. The angle $\angle A M T$ is measured by half the sum o... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. Does there exist a natural number $n$, greater than 1, such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number? | Answer: Yes, it exists.
Solution. For example, $n=2^{8}=256$.
Indeed, $\sqrt{n \sqrt{n \sqrt{n}}}=\sqrt{n \sqrt{n \cdot n^{\frac{1}{2}}}}=\sqrt{n \sqrt{n^{\frac{3}{2}}}}=\sqrt{n \cdot n^{\frac{3}{4}}}=\sqrt{n^{\frac{7}{4}}}=n^{\frac{7}{8}}$. Then, for $n=2^{8}$, the value of this expression is $\left(2^{8}\right)^{\f... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.6. Each cell of a $7 \times 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the numbe... | Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones.
In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones.
Th... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.3. On the $O x$ axis, points $0,1,2, \ldots, 100$ were marked and graphs of 200 different quadratic functions were drawn, each of which passes through two of the marked points and touches the line $y=-1$. For each pair of graphs, Oleg wrote on the board a number equal to the number of common points of these graphs. ... | Answer: Could not.
Solution. Each of our 200 polynomials corresponds to two integer points $a$ and $b$ on the $O x$ axis. Without loss of generality, we will assume that $a < b$. The width of the polynomial is $w=b-a$, and the axis is $c=\frac{a+b}{2}$. If the width is $w>0$ and the axis is $c$, then it is written as ... | 39698 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. In Flower City, there live $99^{2}$ dwarfs. Some of the dwarfs are knights (always tell the truth), while the rest are liars (always lie). The houses in the city are located in the cells of a $99 \times 99$ square (a total of $99^{2}$ houses, arranged in 99 vertical and 99 horizontal streets). Each house is inhab... | Answer: 75.
Solution: Example. Let's show that if $k=74$, we cannot guarantee finding the house of Znayka. Place Znayka and the liar Neznayka in houses with numbers $(50 ; 49)$ and $(49 ; 50)$, respectively. We will show that it might be such that from the answers of the residents, we cannot uniquely determine which o... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Alyosha and Vitya set off from point $N$ to point $M$, the distance between which is 20 km. Unfortunately, they have only one bicycle between them. From $N$, Alyosha sets off on the bicycle, while Vitya starts walking. Alyosha can leave the bicycle at any point along the road and continue on foot. When Vitya reaches... | Answer. 12 km from point $N$.
Solution. Let $x$ (km) be the distance from $N$ to the point where Alyosha leaves the bicycle. Then Alyosha will spend $\frac{x}{15}+\frac{20-x}{4}$ hours on the entire journey, and Vitya will spend $\frac{x}{5}+\frac{20-x}{20}$ hours. By setting up and solving the equation, we find $x=12... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In an equilateral triangle $\mathrm{ABC}$, the height $\mathrm{BH}$ is drawn. On the line $\mathrm{BH}$, a point $\mathrm{D}$ is marked such that $\mathrm{BD}=\mathrm{AB}$. Find $\angle \mathrm{CAD}$. | Answer: $15^{\circ}$ or $75^{\circ}$.
Solution. Obviously, $\triangle \mathrm{ADH}=\Delta \mathrm{CDH}$ (right-angled, by two legs), hence $\mathrm{AD}=\mathrm{CD}$. Therefore, $\triangle \mathrm{BDA}=\triangle \mathrm{BDC}$ (by three sides) $\Rightarrow \angle \mathrm{DAB}=\angle \mathrm{DCB}, \angle \mathrm{BDA}=\an... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Indicate the largest possible number, in decimal notation, in which all digits are different, and the sum of its digits is 37.
# | # Answer: 976543210.
Sketch of the solution. The sum of all ten digits is 45, so to achieve the maximum, we exclude only one digit. It will be 8. Among the nine-digit numbers, the larger one has the higher significant digits. Hence the answer. | 976543210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. There is a contour of a square with a side of 20 cm, it was cut into two groups of equal segments consisting of three and four segments. What is the length of these segments? Find all possible answers. | Answer: 20 cm and 5 cm.
Sketch of the solution.
If each side of the square is cut into at least two segments, there will be 8 segments, and their $4+3=7$. This means one of the segments is equal to the side of the square, and there are three such segments, each 20 cm long. Four segments are obtained by cutting one si... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. A rectangle $3 \times 100$ consists of 300 squares $1 \times 1$. What is the maximum number of diagonals that can be drawn in the squares so that no two diagonals share a common endpoint? (In one square, two diagonals can be drawn without sharing endpoints. Common internal points are allowed.) | Answer: 200.
Example. Let's number the rows and columns containing the squares. In each square with both odd numbers, we will draw two diagonals.
Another example. In all cells of the first and third rows, we will draw parallel diagonals.
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-8. In a chess club, 90 children attend. During the session, they divided into 30 groups of 3 people, and in each group, everyone played one game with each other. No other games were played. In total, there were 30 games of "boy+boy" and 14 games of "girl+girl". How many "mixed" groups were there, that is, groups in w... | Answer: 23.
Solution: There were a total of 90 games, so the number of games "boy+girl" was $90-30-14=46$. In each mixed group, two "boy+girl" games are played, while in non-mixed groups, there are no such games. In total, there were exactly $46 / 2=23$ mixed groups.
## Grade 5 | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 |
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