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Problem 10.4. Given a rectangular grid of cells. It is known that:
- there are exactly 940 ways to cut out a $1 \times 2$ rectangle from it along the grid lines;
- there are exactly 894 ways to cut out a $1 \times 3$ rectangle from it along the grid lines.
How many ways are there to cut out a $1 \times 5$ rectangle f... | Answer: 802.
Solution. Let the table have $a$ rows and $b$ columns. Without loss of generality, $a \leqslant b$. It is clear that at least one of the numbers $a$ and $b$ is not less than 2.
- Suppose $a=1$. Then from the table $1 \times b$, a rectangle $1 \times 2$ can be cut in $b-1=940$ ways, and a rectangle $1 \ti... | 802 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. Several sweet-tooths participated in a candy eating contest. Each participant ate an integer number of candies, and any two participants ate a different number of candies. Summarizing the contest, the jury ranked all the people in descending order of the number of candies eaten (for example, the winner at... | Answer: 21.
Solution. Let there be $n$ sweet-tooth participants in total, and they ate $S$ candies in total. From the condition, it follows that the winner ate $\frac{S}{15}$ candies, the person in third place $-\frac{S}{21}$ candies, and the person in last place $-\frac{S}{22}$ candies.
All participants, except the ... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. Given trapezoid $ABCD$. Segments $AB$ and $CD$ - its lateral sides - are equal to 24 and 10, respectively. Points $X$ and $Y$ are marked on side $AB$ such that $AX=6, XY=8, YB=10$. It is known that the distances from points $X$ and $Y$ to line $CD$ are 23 and 27, respectively.
 (1 point) 100.
(b) (3 points) 260.
Fig. 7: to the solution of problem 10.6
Solution. We will prove that the distances from points $A$ and $B$ to the line $C D$ are 20 and 32,... | 260 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. On an island, there are 23 knights and 200 liars; all residents have different names. A tourist who knows this asked each of the 223 residents to write down 200 names of liars on a piece of paper. Each knight correctly wrote down 200 names of liars, while each liar wrote down an arbitrary list of 200 name... | Answer: 16.
Solution. First, let's outline a strategy for the liars that allows identifying no more than 16 of them. Let's call the knights $R_{1}, R_{2}, \ldots, R_{23}$, and the liars $-L_{1}, L_{2}, \ldots, L_{200}$. All knights will list $L_{1}, L_{2}, \ldots, L_{200}$ in their lists. For each integer $0 \leqslant... | 16 | Logic and Puzzles | 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 vote was 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 voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | 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), the right triangles $X A B$ and $Y D A$ are similar by the acut... | 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 as $x$ (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 $D Z$.
. Find the sum of all angles with vertex O, whose sides are rays OA, OC, OD, OE, OF, OB, if $\angle \mathrm{COF}=97^{\circ}, \angle \mathrm{DOE}=35^{\c... | Answer: $1226^{\circ}$.
Instructions. There are four pairs of adjacent angles and a straight angle, which sum up to $180^{\circ}$ or are equal to $180^{\circ}$. There are two pairs of angles whose sum equals $\angle \mathrm{COF}$, and the angle itself. And there is still $\angle \mathrm{DOE}$. Therefore, the sum of th... | 1226 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be t... | 22 | Logic and Puzzles | 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 |
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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\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 $B C=9$, $C D=30$.
. 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 |
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. 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 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 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 № 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. 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 |
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 |
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 |
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 |
9.2. 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 |
11.3. 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 |
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 |
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. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
We will call suc... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Winnie-the-Pooh stocked up on chocolate bars for the winter: $60\%$ of the total number were "Snickers", $30\%$ were "Mars", and $10\%$ were "Bounty". In the spring, it turned out that the number of "Bounty" bars eaten by Winnie-the-Pooh was $120\%$ of the number of "Mars" bars eaten and $30\%$ of the number of "Sni... | # Solution.
Let there be $3 k$ "Bounty" chocolate bars in total. Then there were $9 k$ "Mars" bars and $18 k$ "Snickers" bars. Since $k$ "Bounty" bars were eaten, $\frac{k}{1.2}=\frac{5 k}{6}$ "Mars" bars were eaten. Therefore, $k$ is divisible by 6. "Snickers" bars eaten were $\frac{k}{0.3}=\frac{10 k}{3}$, and the r... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Anya, Borya, Vika, and Gena are on duty at school for 20 days. It is known that exactly three of them are on duty each day. Anya was on duty 15 times, Borya - 14 times, Vika - 18 times. How many times was Gena on duty | Answer: 13.
Solution. Since 3 people are on duty in school every day, a total of $3 \cdot 20=60$ people are needed for the duty. Therefore, Gena was on duty $60-15$ (Anya's duties) -14 (Borya's duties) -18 (Vika's duties) $=13$ times.
| Anya | $\checkmark$ | $\checkmark$ | $\checkmark$ | | $\checkmark$ | $\checkmark... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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