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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 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
 are there on the parabola $y=x^{2}$, such that the tangent at these points intersects both coordinate axes at points with integer coordinates not exceeding 2020 in absolute value? | Answer: 44. Solution. The equation of the tangent to the parabola $y=x^{2}$ at the point $\left(x_{0}, y_{0}\right)$, where $y_{0}=x_{0}^{2}$, is $y-y_{0}=2 x_{0}\left(x-x_{0}\right)$. From this, we find the coordinates of the points of intersection of the tangent with the axes, namely $x_{1}=\frac{x_{0}}{2}, y_{1}=-x_... | 44 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain company, there are 100 shareholders, and any 66 of them own no less than $50 \%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? | 4. Let $M$ be the shareholder owning the largest percentage of shares - $\mathrm{x} \%$ of shares. Divide the other 99 shareholders into three groups A, B, C, each with 33 shareholders. Let them own a, b, c percentages of shares, respectively. Then $2(100-x)=2(a+b+c)=(a+b)+(b+c)+(c+a) \geq 50+50+50$, i.e., $x \leq 25$.... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. The age of a certain person in 1988 was equal to the sum of the digits of their birth year. How old was he
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9.2. The age of a certain person in 1988 was equal to the sum of the digits of their birth y... | Answer: 22.
Solution. The number of years a person has lived is equal to the sum of the digits of a four-digit number, each of which is no more than 9. Therefore, he is no more than 36 years old, and he was born in the 20th century. Let $x$ be the number of tens, $y$ be the number of units in his birth year.
Then, ac... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.2 A group of friends went for a morning run around a lake. During the run, one by one they realized they had miscalculated their strength, and switched from running to walking. One of the friends calculated that he had run one-eighth of the total distance that the entire group had run, and walked one-tenth of the to... | Solution 1: Let the person who ran cover $x$ part of the road, then $0<x<1$, and he walked $(1-x)$ part of the way. If there were $n$ people in total, then according to the condition, the total distance covered by the group (expressed in terms of parts) is on one side $n$, and on the other side $8 x + 10(1-x) = 10 - 2 ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4 For each integer from 10 to 2021, we found the product of the digits, and then added all the obtained results. What is the sum that was obtained? | Solution: Consider the product $(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9)$. If we expand the brackets, we get the products of pairs of digits that form all two-digit numbers. Similarly, in the product
$$
(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9) \cdot(0+1+2+\cdots+9)
$$
after expanding the brackets, we get all combinations... | 184275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3 There are 12 natural numbers. It is known that the sum of any three of them is not less than 100. Prove that the sum of all the numbers is not less than 406. | Solution. Arrange these numbers in non-decreasing order: $\mathrm{a}_{1} \leq \mathrm{a}_{2} \leq \mathrm{a}_{3} \leq \mathrm{a}_{4} \leq \ldots \leq \mathrm{a}_{12}$. By the condition $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3} \geq 100$, therefore $3 \mathrm{a}_{3} \geq 100, \mathrm{a}_{3} \geq \frac{100}{3}>33$. S... | 406 | Inequalities | proof | Yes | Yes | olympiads | false |
1. The midpoints of adjacent sides of a rectangle with a perimeter of 32 were connected by segments. The same operation was performed on the resulting quadrilateral: the midpoints of adjacent sides were connected by segments (see figure). How many times in total does one need to perform such an operation so that the pe... | Answer: 11.
Solution: After two operations, a quadrilateral is obtained, the sides of which are the midlines of triangles with bases parallel to the sides of the original rectangle. Therefore, this quadrilateral is a rectangle, and each of its sides is half the length of the corresponding side of the original rectangl... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A semicircle with diameter $A B$ and center at point $O$ is divided into three parts by points $C$ and $D$ such that point $C$ lies on the arc $A D$. Perpendiculars $D E$ and $D F$ are dropped from point $D$ to segments $O C$ and $A B$ respectively. It turns out that $D E$ is the angle bisector of triangle $A D C$, ... | Answer: $20^{\circ}$.
Solution. Triangle $A O D$ is isosceles ($O D=O A$, as radii), hence, $\angle O A D=\angle O D A$. Since $D O$ is the bisector of angle $A D F$, then $\angle O A D=$ $\angle O D F$. Calculation of angles in the right triangle $A F D$ shows that $\angle O A D=30^{\circ}$. Let $G$ be the point of i... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. Vlad and Dima decided to earn some money. Each of them decided to deposit 3000 rubles in the bank and withdraw all the money after a year.
Vlad chose the deposit "Confidence": the amount increases by $20\%$ over the year, but the bank charges a $10\%$ fee upon withdrawal.
Dima chose the deposit "Reliabil... | Answer: Dima will earn 120 rubles more.
Solution. Vlad's deposit amount will increase to $3000 \cdot 1.2$ rubles in a year, and after withdrawal, it will decrease to $3000 \cdot 1.2 \cdot 0.9=3240$ rubles.
Dima's deposit amount will increase to $3000 \cdot 1.4$ rubles in a year, and after withdrawal, it will decrease... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. The Smeshariki Kros, Yozhik, Nyusha, and Barash ate a total of 86 candies, and each of them ate no fewer than 5 candies. It is known that:
- Nyusha ate more candies than each of the other Smeshariki;
- Kros and Yozhik together ate 53 candies.
How many candies did Nyusha eat | Answer: 28.
Solution. Krosh or Yozhik ate at least 27 candies (otherwise, they would have eaten no more than $26+26=52$ candies in total), so Nusha ate at least 28 candies. Considering that Barash ate at least 5 candies, we get that all of them together ate at least $53+28+5=86$ candies. Therefore, this is only possib... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A store sells four types of nuts: hazelnuts, almonds, cashews, and pistachios. Stepan wants to buy 1 kilogram of nuts of one type and another 1 kilogram of nuts of a different type. He calculated how much such a purchase would cost him depending on which two types of nuts he chooses. Five out of six possib... | Answer: 2290.
Solution. Let $a, b, c, d$ be the cost of 1 kilogram of hazelnuts, almonds, cashews, and pistachios, respectively. From the condition, it follows that the set $A=\{1900,2070,2110,2330,2500\}$ is contained in the set $B=\{a+b, b+c, c+d, d+a, a+c, b+d\}$.
Note that the 6 elements of set $B$ can be divided... | 2290 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.6. A magic square is a $3 \times 3$ table in which numbers are arranged so that the sums of all rows, columns, and the two main diagonals are the same. The figure shows a magic square in which all numbers except three have been erased. Find what the number in the upper left corner of the square is.
| $?$ | 3... | Answer: 14.
Solution. Let the unknown number be $x$, then the sums in all rows, columns, and on the main diagonals are $9+31+x=40+x$.
1) Considering the left column, we get that the number in the lower left corner of the square is 27.
2) Considering the main diagonal going up to the right, we get that the number in t... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. All 25 students in class 7A participated in a quiz consisting of three rounds. In each round, each participant scored a certain number of points. It is known that in each round, as well as in the total of all three rounds, all participants scored a different number of points.
Student Kolya from 7A was thi... | Answer: 10.
Solution. In the first round, 2 classmates overtook Kolya, in the second - 3, in the third - 4. Then, in the sum of all three rounds, he could be overtaken by no more than $2+3+4=9$ classmates, i.e., in the sum of the three rounds, he could not end up lower than 10th place.
Now let's provide an example of... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
84. a) Prove that there exists a pair of two-digit numbers such that if 20 is added to the first number and 15 is subtracted from the second, the resulting numbers will remain two-digit, and their product will be equal to the product of the original numbers? b) How many such pairs are there? | Answer: b) 16 pairs. Hint For part a), it is sufficient to provide a specific example (see problem 7.4). b) Let $a, b$ be the desired pair of numbers. Then $(a+20)(b-15)=a b$. From this, $20 b-15 a=20 \cdot 15 \Leftrightarrow 4 b-3 a=60$.
Since 60 and $3a$ are divisible by 3, $b$ must also be divisible by 3, i.e., $b=... | 16 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (po... | Answer: 20.
Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, throu... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.
.
. From this similarity and the cyclic nature of the pentagon
. It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
$ of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will tell them $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$, and separately inform them o... | Answer. For $k=2017$.
Solution. First, we prove that $k>2016$. Suppose the teacher used some $k \leqslant 2016$, thinking of the polynomial $P(x)$. Consider the polynomial $Q(x)=P(x)+(x-n_{1})(x-n_{2}) \ldots(x-n_{k})$. Note that the degree of the polynomial $Q(x)$ is also 2017, and its leading coefficient is also 1. ... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. What is the minimum number of unit-radius circles required to completely cover a triangle with sides $2 ; 3 ; 4$? | Answer: three circles. Solution. Let $A C=4, A B=2, B C=3$ and let $C_{1}, A_{1}$ and $B_{1}$ be the midpoints of sides $A B, B C$ and $A C$ respectively. Note that angle $B$ is obtuse, since $A C^{2}>A B^{2}+B C^{2}$. Therefore, points $B$ and $B_{1}$ lie inside the circle of radius 1 with center at point $O$ - the mi... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. 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.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | 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 |
3. After teacher Mary Ivanovna moved Vovochka from the first row to the second, Vanechka from the second row to the third, and Mashenka from the third row to the first, the average age of students sitting in the first row increased by one week, those sitting in the second row increased by two weeks, and those sitting i... | 3. Let there be x people in the third row. Since the average age is the sum of the ages divided by the number of people, after the rearrangement, the total age of the children in the first row increased by 12 weeks, in the second row by 24 weeks, and in the third row by -4x weeks. Since the total sum of the ages of all... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. From Zlatoust to Miass, "GAZ", "MAZ", and "KamAZ" set off simultaneously. "KamAZ", having reached Miass, immediately turned back and met "MAZ" 18 km from Miass, and "GAZ" - 25 km from Miass. "MAZ", having reached Miass, also immediately turned back and met "GAZ" 8 km from Miass. What is the distance from Zlatoust to... | Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" - $k$ km/h. For each pair of vehicles, we equate their travel time until they meet.
We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. While waiting for customers, the watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). The seller recorded on a piece of paper the weights of the weights he used. What is ... | Answer: 6.
Solution. With one or two weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, any of the given watermelons can be weighed. Indeed, $2=1+1, 4=3+1$, $6=5+1, 8=7+1, 11=10+1, 12=9+3, 13=10+3, 14=9+5, 15=10+5, 16=9+7, 17$ $=10+7, 18=9+9, 19=10+9, 20=10+10$. Thus, six different numbers could have been recorded.
... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Two types of tiles were laid on the wall in a checkerboard pattern. Several tiles fell off the wall. The remaining tiles are shown in the picture. How many striped tiles fell off? Be sure to explain your answer.
Method 2. Look at the number of fallen cells by rows: in the second row from the top, 2 tiles fell out, of which 1 ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Mowgli asked five monkeys to bring him some nuts. The monkeys collected an equal number of nuts and carried them to Mowgli. On the way, they quarreled, and each monkey threw one nut at each of the others. As a result, they brought Mowgli half as many nuts as they had collected. How many nuts did Mowgli receive? Be s... | Answer: 20 nuts.
## Solution.
Each monkey threw 4 nuts, so the monkeys threw a total of $5 \cdot 4=20$ nuts together.
If half of the nuts remained, it means that Mowgli brought as many nuts as were thrown, which is 20 nuts.
## Grading Criteria.
- Correct solution - 7 points.
- Very brief solution (such as "5 - $4=... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The distance from the home of Vintik and Shpuntik to school is 6 kilometers. Vintik and Shpuntik left for school at the same time, with Vintik spending half of the time riding a scooter at a speed of 10 km/h and then walking, while Shpuntik traveled half the distance by bicycle and then walked. They arrived at schoo... | Answer: 15 km/h.
## Solution:
First method.
Since Shtyubik rides twice as fast as he walks, he also covers twice the distance he walks (since he spends the same amount of time on both), which is 4 km. Since Shtyubik and Shpuntyk walk at the same speed, the last 2 km they walked together. Therefore, while Shtyubik wa... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In the notebook, all irreducible fractions with the numerator 15 are written down, but which are greater than $\frac{1}{16}$ and less than $\frac{1}{15}$. How many such fractions are written in the notebook? | Answer: 9 fractions.
## Solution:
We are looking for all suitable irreducible fractions of the form $\frac{15}{n}$. Since $\frac{1}{16}<\frac{15}{n}<\frac{1}{15}$, then $15 \cdot 15<n<15 \cdot 16$ or $225<n<240$ (with the fraction $\frac{15}{n}$ being irreducible, meaning $n$ is not divisible by 3 or 5). It is not di... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In each cell of a $2 \times 2$ table, a number was written, and all the numbers are different. It turned out that the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second colum... | # Answer: 0.
## Solution:
| $a$ | $b$ |
| :--- | :--- |
| $c$ | $d$ |
Let's denote the numbers in the table as shown on the left. According to the condition, $a+b=c+d$, $ac=bd$. Then $a=c+d-b$, substitute this into the product equality: $(c+d-b)c=bd$. Expand the brackets and move everything to the left: $c^2 + cd - ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $A B C$, the median $B M$ was drawn. It turned out that $A B=2 B M$ and $\angle M B A=40^{\circ}$. Find $\angle C B A$.
# | # Answer: $110^{\circ}$.
## Solution:
Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a p... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The organizers of a mathematics olympiad decided to photograph 60 participants. It is known that no more than 30 participants can fit in a single photograph, however, any two students must appear together in at least one photograph. What is the minimum number of photographs needed to achieve this? | # Answer: 6.
## Solution:
Example with 6 photos: divide 60 participants into 4 groups of 15 people (groups $A, B, B$, Г). Take 6 photos of all possible pairs of groups: $A+D, A+B, A+\Gamma, B+B, B+\Gamma, B+\Gamma$ - in each photo, there will be 30 people, and it is easy to see that in this way any two people will be... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.5. Five athletes came to training with their balls, and when leaving, each took someone else's. In how many ways is this possible. | Answer: 44.
Solution. First, let's assume that no two athletes have exchanged balls. Imagine them sitting around a round table. Then, depending on their different relative positions, they would choose the ball of the next (for example, clockwise) athlete. However, this is only possible if 5 are seated at one table, wh... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2 The sequence of numbers $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots$ is formed according to the rule: $\mathrm{x}_{1}=1, \mathrm{x}_{\mathrm{n}+1}=1+\frac{\mathrm{x}_{\mathrm{n}}^{2}}{\mathrm{n}}$ for $\mathrm{n}=1,2,3, \ldots$ Find $\mathrm{x}_{2019}$. | Answer: 2019
Reasoning. $x_{2}=1+\frac{1^{2}}{1}=2, x_{3}=1+\frac{2^{2}}{2}=3$. Hypothesis: $x_{n}=n$. We proceed by induction. If $x_{n}=n$, then $x_{n+1}=1+\frac{n^{2}}{n}=1+n$. The hypothesis is confirmed. | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 8.2. (7 points)
Twelve people are carrying 12 loaves of bread. Each man carries 2 loaves, each woman carries half a loaf, and each child carries a quarter of a loaf. How many men, women, and children were there? | Answer: 5 men, one woman, and 6 children.
Solution: Let $x$ be the number of men, $y$ be the number of women, and $z$ be the number of children; $x, y, z$ are natural numbers.
Then $x+y+z=12$ and $2 x+\frac{y}{2}+\frac{z}{4}=12$. From the last equation, it follows that $8 x+2 y+z=48$.
Transform the last equation: $7... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Solve the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
Answer: 2 | Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Acting similarly, we get that $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this e... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The median $\mathrm{AA}_{0}$ of triangle $\mathrm{ABC}$ is laid off from point $\mathrm{A}_{0}$ perpendicular to side $\mathrm{BC}$ to the outside of the triangle. Denote the second endpoint of the constructed segment as $\mathrm{A}_{1}$. Similarly, points $\mathrm{B}_{1}$ and $\mathrm{C}_{1}$ are constructe... | Answer._ $\triangle \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$ is equilateral, all angles are $60^{\circ}$.
## Solution.
Since $\triangle \mathrm{ABC}$ is isosceles, $\mathrm{BB}_{0}$ is the perpendicular bisector of the base AC. Therefore, $\mathrm{B}_{1}$ lies on this perpendicular and $\mathrm{CB}_{0} \perp \m... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. The math teacher agreed with the eleven students who came to the elective that he would leave the classroom, and the students would agree among themselves who would be a liar (always lie) and who would be a knight (always tell the truth). When the teacher returned to the class, he asked each student to say about ... | Answer: 7.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of "knight-liar," the phrase "He is a liar" will be said twice. Since this p... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. A circle is inscribed with 103 numbers. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 103 written numbers? | Answer: 42.
Solution. We will show that there will be 3 consecutive numbers, among which there are at least 2 positive ones. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them, there are at least 6 positive numbers. But th... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2. In the bottom left corner of a $7 \times 7$ chessboard, there is a king. In one move, he can move one square to the right, or one square up, or one square diagonally to the right and up. In how many different ways can the king travel to the top right corner of the board, if he is forbidden to visit the central ce... | Answer: 5020.
Solution. We will construct a $7 \times 7$ table, in each cell of which we will write the number equal to the number of allowed paths by which the king can reach this cell from the bottom left corner. First, we fill the left column and the bottom row with ones and write 0 in the central cell (as per the ... | 5020 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. At competitions, an athlete's performance is evaluated by 7 judges, each of whom gives a score (an integer from 0 to 10). To obtain the final score, the best and worst scores from the judges are discarded, and the arithmetic mean is calculated. If the average score were calculated based on all seven scores, the a... | # Answer. 5.
Solution. Suppose there are no fewer than six dancers. Let $A, a, S_{A}$ be the best score, the worst score, and the sum of all non-discarded scores of the winner, respectively, and $B, b, S_{B}$ be the same for the last athlete. Instead of averages, we can arrange the dancers by the sum of all scores or ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. In a regular pentagon $A B C D E$, a point $F$ is marked on side $A B$, and a point $G$ is marked on side $B C$ such that $F G=G D$. Find the angle $C D G$, if the angle $F D E$ is $60^{\circ}$. | Answer: $6^{\circ}$.
Solution. The angles of a regular pentagon are each $108^{\circ}, \angle E D A = 36^{\circ}, \angle F D E = 60^{\circ}$. Therefore, $\angle A D F = 24^{\circ}$.
Additio... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the maximum value of the expression $x^{2}+y^{2}+z^{2}$, if $x, y, z$ are integers satisfying the system
$$
\left\{\begin{array}{l}
x y+x+y=20 \\
y z+z+y=6 \\
x z+x+z=2
\end{array}\right.
$$ | Solution. Adding 1 to both sides of each equation in the system, we get:
$$
\left\{\begin{array}{l}
x y + x + y + 1 = 21 \\
y z + z + y + 1 = 7 \\
x z + x + z + 1 = 3
\end{array}\right.
$$
Factoring the left-hand sides, we obtain:
$$
\left\{\begin{array}{l}
(x+1)(y+1) = 21 \\
(y+1)(z+1) = 7 \\
(z+1)(x+1) = 3
\end{ar... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. During a physical education class, all students in the class lined up in a row. It turned out that boys and girls alternated in the row. It is known that exactly $52 \%$ of the students are girls. Find the number of boys in the class. Justify your answer. | Solution: Since there are more than half girls, the first and last student in the row are girls. Let's remove the last girl from the row - there will be an equal number of boys and girls left. In this row, all the boys in the class make up $48 \%$. There are as many girls, another $48 \%$. The remaining $100-2 \cdot 48... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. While Cheburashka eats two portions of ice cream, Winnie-the-Pooh manages to eat five such portions, and while Winnie-the-Pooh eats three portions, Karlson eats seven. Working together, Cheburashka and Karlson ate 82 portions. How many portions did Winnie-the-Pooh eat during this time? Justify your answer. | Solution: Suppose Winnie-the-Pooh ate exactly 15 portions of ice cream. Since $15=3 \cdot 5$, Cheburashka ate $3 \cdot 2=6$ portions, and Karlson ate $-5 \cdot 7=35$ portions. Together, Karlson and Cheburashka ate $6+35=41$ portions, which is half of the 82 portions given in the problem. Therefore, the feast actually l... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the equations $x^{2}+a x+b=0$ and $x^{3}+b x+a=0$ have a common root and $a>b>0$. Find it. | Answer: -1.
Solution. Multiply the first equation by $x$ and subtract the second from it. The common root of the original equations will also be a root of the resulting equation
$$
\left(x^{3}+a x^{2}+b x\right)-\left(x^{3}+b x+a\right)=0 \quad \Longleftrightarrow \quad a\left(x^{2}-1\right)=0
$$
The last equation h... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$. | Answer: $A C=13$.
Solution. (Fig. 3). First, we prove that $M I=M A$ (trident lemma).
Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is equa... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. There are 55 boxes standing in a row, numbered in order from 1 to 55. Each box contains no more than 10 balls, and the number of balls in any two adjacent boxes differs by exactly 1. It is known that the boxes numbered $1,4,7,10, \ldots, 55$ contain a total of 181 balls. What is the minimum total number of balls t... | Solution. Since the number of balls in two adjacent boxes differs by 1, the parity of this number always differs. Therefore, throughout the entire row of boxes, the parity alternates - the number of balls in boxes with different parity numbers also has different parity. We will divide the boxes into pairs: 1 with 4, 7 ... | 487 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. Pasha is playing a computer game. The game takes place on an infinite grid. Each cell contains either a treasure or a natural number. The number indicates the distance to the nearest treasure in cells (if it takes $A$ steps vertically and $B$ steps horizontally to reach the treasure, the cell contains the number $... | Solution. a) Note that if Pasha finds a certain number $K$ in a cell, then at least one treasure must be located in the cells along the perimeter of a square rotated by $45^{\circ}$ relative to the grid lines, where the side of the square is exactly $K+1$ cells. At the same time, there cannot be any treasures inside su... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the smallest six-digit number that is a multiple of 11, where the sum of the first and fourth digits is equal to the sum of the second and fifth digits and is equal to the sum of the third and sixth digits. | Solution. Let the desired number have the form $\overline{1000 x y}$, where $x, y$ are some digits. Then, by the condition, the sum of the first and fourth digits is 1, from which $x=y=1$. But the number 100011 is not divisible by 11. Therefore, we will look for a number of the form $\overline{1001 x y}$. Then $x=y=2$,... | 100122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. Evgeny is tiling the floor of his living room, which measures 12 by 16 meters. He plans to place 1 m $\times$ 1 m square tiles along the perimeter of the room, and the rest of the floor will be covered with 2 m $\times$ 2 m square tiles. How many tiles will he need in total? | Solution. Let's step back 1 meter from the living room's boundaries. We will be left with a rectangle of size $10 \times 14$, which needs to be covered with $2 \times 2$ tiles. The area of the rectangle is 140, and the area of one tile is 4, so we will need $140: 4=35$ large tiles. The area of the boundary strip is $12... | 87 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.4. Daniil and Pasha were driving in a car along a straight road at a constant speed. While Daniil was driving, Pasha had nothing to do and was looking at the kilometer markers. Pasha noticed that exactly at noon they passed a marker with the number $X Y$ (where $X, Y$ are some digits), at $12:42$ - a marker with the ... | Solution. Method 1. In 42 minutes from 12:00 to 12:42, the car traveled no more than 100 km, and in the remaining 18 minutes, it traveled even less (in $42 / 18=7 / 3$) times. Therefore, by 13:00, the car had not reached the sign with the number 200. Then $X=1$. In this case, regardless of the number $Y$, the car trave... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.4. Ivan on a tractor and Petr in a "Mercedes" left point A for point B. Petr arrived at point B, waited for 10 minutes, and called Ivan to find out that Ivan had only covered a third of the distance and was currently passing by a cafe. Petr drove to meet him. Not noticing Ivan, he reached the cafe and spent half an h... | Solution. From the fact that Ivan had traveled a third of the distance by the time Peter had traveled the whole distance and waited for 10 minutes, we can conclude that by the time Ivan reaches point B, Peter could have traveled the route three times and waited for half an hour. Instead, he traveled the route $1+2 / 3+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 11.1. Solve the equation
$$
x^{4}+2 x \sqrt{x-1}+3 x^{2}-8 x+4=0
$$ | Solution. Let's check that $x=1$ is a root. Indeed,
$$
1^{4}+2 \sqrt{1-1}+3 \cdot 1^{2}-8+4=1+3-8+4=0
$$
We will show that there are no other roots. Since the left side of the equation is defined only for $x \geqslant 1$, it is sufficient to consider the case $x>1$. Using the valid inequalities in this case $x^{4}>x^... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. In how many ways can all natural numbers from 1 to $2 n$ be arranged in a circle so that each number is a divisor of the sum of its two neighboring numbers? (Ways that differ only by rotation or symmetry are considered the same) | Solution. Note that if the numbers in the circle do not alternate in parity, then some two even numbers are adjacent. An even number is a divisor of the sum of the even number adjacent to it and the second adjacent number, which means the latter is also even. Continuing this reasoning around the circle, we get that all... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
36. A pair of natural numbers $a>b$ is called good if the least common multiple (LCM) of these numbers is divisible by their difference. Among all natural divisors of the number $n$, exactly one good pair was found. What can $n$ be? | 36. Answer: $n=2$. Two odd numbers cannot form a good pair. Therefore, the number $n$ has an even divisor, and thus it is even. But then it has two good pairs of divisors: $(2,1)$ and $(n, n / 2)$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
40. At the vertices of a convex 2020-gon, numbers are placed such that among any three consecutive vertices, there is both a vertex with the number 7 and a vertex with the number 6. On each segment connecting two vertices, the product of the numbers at these two vertices is written. Andrey calculated the sum of the num... | 40. Answer: 1010. This answer is achieved if the numbers 7 and 6 alternate.
We can assume that all numbers are equal to 6 or 7. Indeed, if a number is different from 6 and 7, then among its neighbors, one is 6 and one is 7, and the same is true for the neighbors two steps away. Then, replacing this number with 7 does ... | 1010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On a circle, 2021 points are marked. Kostya marks a point, then marks the next point to the right, then he marks the point to the right of the last marked point, skipping one, then the point to the right of the last marked point, skipping two, and so on. On which move will a point be marked for the second time? | Answer: on the 67th move. The question is about the smallest natural $b$ for which $(a+1)+(a+2)+\cdots+(b-1)+b$ is divisible by $2021=43 \cdot 47$. This sum is equal to $b(b+1) / 2-a(a+1) / 2=$ $(b-a)(b+a+1) / 2$ and it is not hard to see that the smallest $b=(b-a) / 2+(b+a) / 2$ is achieved with the factorization $(66... | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find all integers $x, y$ for which $x+y, 2x+3y$ and $3x+y$ are perfect squares. | Answer: $x=y=0$. Let $x+y=a^{2}, 2 x+3 y=b^{2}$ and $3 x+y=c^{2}$. Note that $2 b^{2}+c^{2}=7 a^{2}$. We will prove that this equation has a unique solution: $a=b=c=0$. Indeed, let $(a, b, c)$ be a solution with the minimal sum $a^{2}+b^{2}+c^{2}$, in which not all numbers $a, b, c$ are equal to 0. Then $2 b^{2}+c^{2}:... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On the table, there are 100 weights of different masses. A weight is called successful if its mass is equal to the sum of the masses of some two other weights on the table. What is the smallest number of successful weights that would allow us to assert with certainty that the masses of some two weights differ by at ... | Answer: 87. Let there be 87 lucky numbers. If one of the two addends of a lucky number $a=b+c$ is itself lucky, for example, $b=d+e$, then $a=c+d+e \geqslant 3 \min (c, d, e)$, which means there will be two numbers differing by at least a factor of three. If, however, all the addends of the eighty-seven lucky numbers a... | 87 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
25. The sequence $\left(x_{n}\right)$ is defined by the conditions: $x_{1}=1$, and for each natural number $n$, the number $x_{n+1}$ is the largest number that can be obtained by rearranging the digits of the number $x_{n}+1$. Find the smallest $n$ for which the decimal representation of the number $x_{n}$ has exactly ... | 25. Answer: the smallest such number $n$ is $18298225=9(1+2+3+\ldots+2016)+1$.
To understand how this sequence is constructed, let's write out the first few terms. Initially, the sequence contains single-digit numbers:
$$
1, \quad 2, \quad 3, \quad \ldots, \quad 9 .
$$
Next come the two-digit numbers:
. The cards are lying on the table face down. A set of $n$ cards is called good if each number appears exactly once. Baron Munchausen claims that he can point out 80 sets of $n$ cards, at least one of which is guarantee... | Answer: $n=7$. We will present an algorithm for how to indicate $2^{n-1}$ sets on $2n$ cards, one of which is suitable. (In our case, this is $2^{6}=64<80$ sets.) Imagine that identical cards are connected by (invisible to us for now) red edges. We will arbitrarily pair the cards with blue edges. The red-blue graph is ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
31. Let's call a number complex if it has at least two different prime divisors. Find the largest natural number that cannot be represented as the sum of two complex numbers. | 31. The numbers $6, 12, 15$, and 21 are composite and give all possible remainders when divided by 4. Therefore, from any number $n>23$, one of these numbers can be subtracted so that the result is a number of the form $4k+2=2(2k+1)$. Clearly, this difference will be a composite number. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
34. In the gym, 200 schoolchildren gathered. Every pair of acquaintances shook hands. It turned out that any two strangers made at least 200 handshakes in total. Prove that there were at least 10000 handshakes in total. | 34. Let's write the solution in the language of graphs. If the degree of a vertex is not less than 100, we call it rich; if it is less than 100, we call it poor. From the condition, it follows that all poor vertices are pairwise adjacent to each other.
We will prove that each poor vertex can be paired with a non-adjac... | 10000 | Combinatorics | proof | Yes | Yes | olympiads | false |
6. Find the minimum value of the expression $\left[\frac{7(a+b)}{c}\right]+\left[\frac{7(a+c)}{b}\right]+\left[\frac{7(b+c)}{a}\right]$, where $a, b$ and $c$ are arbitrary natural numbers. | Answer: 40. Evaluation: the sum of the fractions under the integer parts is no less than $14 \cdot 3=42$ (we add three inequalities of the form $7(a / b+b / a) \geqslant 14$, and each integer part is greater than the fraction reduced by 1. Therefore, the desired value is greater than $42-3=39$, i.e., not less than 40. ... | 40 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
24. A polynomial of degree 10 has three distinct roots. What is the maximum number of zero coefficients it can have?
(A. Khryabrov) | 24. Answer: Answer: 9 zero coefficients. For example, the polynomial $x^{10}-x^{8}$ has roots $0,1,-1$. If a polynomial has only one non-zero coefficient, it is of the form $a x^{10}$, and therefore has exactly one root. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the bank of the river stand 10 sheikhs, each with a harem of 100 wives. Also at the bank stands an $n$-person yacht. By law, a woman should not be on the same bank, on the yacht, or even at a transfer point with a man if her husband is not present. What is the smallest $n$ for which all the sheikhs and their wive... | Answer: 10. Example: first, all 1000 wives move quietly, then one returns and 10 sheikhs leave. Finally, another returns and picks up the first. Evaluation: let the number of places not exceed 9. Consider the moment when the first sheikh appears on the other shore - or several at once, but not all, for they could not h... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Given a prime number $p$. All natural numbers from 1 to $p$ are written in a row in ascending order. Find all $p$ for which this row can be divided into several blocks of consecutive numbers so that the sums of the numbers in all blocks are equal. | Answer: $p=3$. Let $k-$ be the number of blocks, $S-$ be the sum in each block. Since $p(p+1) / 2=k S$, either $k$ or $S$ is divisible by $p$. Clearly, $k<p$, so $S$ is a multiple of $p$. Let the leftmost group consist of numbers from 1 to $m$. Then $m(m+1) / 2$ is a multiple of $p$, from which $m \geqslant p-1$, i.e.,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The cells of a $100 \times 100$ table are painted white. In one move, it is allowed to select any 99 cells from one row or one column and repaint each of them to the opposite color - from white to black, and from black to white. What is the minimum number of moves required to obtain a table with a checkerboard patte... | Answer: in 100 moves. Evaluation: to repaint each of the cells on the black diagonal, a move is required. Example: repaint all rows and columns with odd numbers. In this case, in the $k$-th row and $u$-th column, we will not repaint their common cell. It is easy to see that as a result, we will get a chessboard colorin... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A $10 \times 10$ table is filled with numbers from 1 to 100: in the first row, the numbers from 1 to 10 are written in ascending order from left to right; in the second row, the numbers from 11 to 20 are written in the same way, and so on; in the last row, the numbers from 91 to 100 are written from left to right. C... | 1. Answer: Yes. If $x$ is in the center of the fragment, then the sum of the numbers in it is $7x$ and then with 646566 $x=65$ the sum will be exactly 455 (see figure). | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Given $n$ distinct natural numbers, any two of which can be obtained from each other by permuting the digits (zero cannot be placed in the first position). For what largest $n$ can all these numbers be divisible by the smallest of them? | Answer: 9. It is clear that there cannot be more than nine numbers.
We will use a known property of the period of a purely periodic rational fraction $\alpha=$ $a / b<1$ with coprime $(a, b)$: the length of the period is the smallest natural $t$ for which $\left(10^{t}-1\right) \vdots b$, and the period $T$ itself is ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
34. A company gathered for a meeting. Let's call a person sociable if in this company they have at least 20 acquaintances, and at least two of them are acquainted with each other. Let's call a person shy if in this company they have at least 20 strangers, and at least two of them are strangers to each other. It turned ... | 34. Answer: 40.
Evaluation: Suppose that a certain person (let's call him Kostya) is acquainted with at least 20 people. If some of Kostya's acquaintances know each other, then Kostya is sociable. Let's consider the case where all of Kostya's acquaintances do not know each other. In this case, if Kostya is acquainted ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
50. In the city, 2019 metro stations have been built. Some pairs of stations are connected by tunnels, and from any station, you can reach any other station via the tunnels. The mayor ordered the organization of several metro lines: each line must include several different stations, sequentially connected by tunnels (t... | 50. Answer: $k=1008$. Let's provide an example of a connected graph with 2019 vertices that cannot be covered by 1008 simple paths: one vertex is connected to 2018 vertices of degree 1. Any simple path contains no more than two pendant vertices, so at least 1009 paths are required for coverage.
We will prove that 1009... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
52. Olya wrote fractions of the form $1 / n$ on cards, where $n-$ are all possible divisors of the number $6^{100}$ (including one and the number itself). She arranged these cards in some order. After that, she wrote the number on the first card on the board, then the sum of the numbers on the first and second cards, t... | 52. Answer: two denominators. Let's represent all fractions in the form $a_{n} / 6^{100}$, then $a_{1}, a_{2}, \ldots$ are again all divisors of the number $6^{100}$, each appearing once. Let the partial sums be denoted by $S_{n} / 6^{100}$. Then the denominator of the irreducible representation of the partial sum depe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. A grasshopper starts moving in the top-left cell of a $10 \times 10$ square. It can jump one cell down or to the right. Additionally, the grasshopper can fly from the bottom cell of any column to the top cell of the same column, and from the rightmost cell of any row to the leftmost cell of the same row. Prove that... | 13. Consider the diagonal running from the bottom-left corner to the top-right corner. We will paint all 10 cells on this diagonal red. Note that from any red cell, without making any jumps, you can only move to cells that are to the right of it, below it, or to the right and below it. Therefore, it is impossible to mo... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false |
66. An isosceles triangle \(ABC\) with a perimeter of 12 is inscribed in a circle \(\omega\). Points \(P\) and \(Q\) are the midpoints of the arcs \(ABC\) and \(ACB\) respectively. The tangent to the circle \(\omega\) at point \(A\) intersects the ray \(PQ\) at point \(R\). It turns out that the midpoint of segment \(A... | 66. Answer: 4. Let $I_{A}, I_{B}, I_{C}$ be the centers of the excircles of triangle $ABC$, touching sides $BC, CA$, and $AB$ respectively. Then the lines $A I_{A}, B I_{B}, C I_{C}$ will be the angle bisectors of triangle $ABC$, and the lines $I_{B} I_{C}, I_{C} I_{A}, I_{A} I_{B}$ will be its external angle bisectors... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
49. What is the maximum number of solutions that the equation $\max \left\{a_{1} x+b_{1}, \ldots, a_{10} x+b_{10}\right\}=0$ can have, if $a_{1}, \ldots, a_{10}, b_{1}, \ldots, b_{10}$ are real numbers, and all $a_{i}$ are not equal to 0? | 49. Answer: 2 solutions. For example, 5 functions $-x-1$ and 5 functions $x-1$.
Suppose this equation has three roots: $u<v<w$. At point $v$, one of the linear functions $a_{i} x+b_{i}$ is zero. On the other hand, its values at points $u$ and $w$ do not exceed zero. However, such a linear function can only be a consta... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
52. In the table, there are 25 columns and 300 rows, and Kostya painted all its cells in three colors. Then Lesha, looking at the table, names one of the three colors for each row and marks all the cells of this color in that row. (If there are no cells of the specified color in the row, he marks nothing in it.) After ... | 52. Answer: Two columns. To leave at least two columns, Lesha must for each row name a color that does not appear in the first two cells of that row. With this strategy, the first two columns will not be crossed out.
Note now that $300=C_{25^{2}}$. Using this observation, we can associate each row with its own pair of... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
55. In a $10 \times 10$ grid (the sides of the cells have a unit length), $n$ cells were chosen, and in each of them, one of the diagonals was drawn and an arrow was placed on this diagonal in one of two directions. It turned out that for any two arrows, either the end of one coincides with the beginning of the other, ... | 55. Answer: when $n=48$.
For each arrow, consider the three-cell corner obtained by removing from a $2 \times 2$ square, centered at the end of the arrow, a $1 \times 1$ square, the diagonal of which is this arrow. Note that such corners do not intersect and are contained within a $12 \times 12$ square. Therefore, the... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
46. In a convex quadrilateral $A B C D$, point $M$ is the midpoint of side $A D$, $C M \| A B$, $A D=B D$ and $3 \angle B A C=\angle A C D$. Find the angle $A C B$.
(S. Berlov) | 46. Answer: $\angle A C B=90^{\circ}$.
Notice that $\angle B A C=\angle A C M$, hence $\angle D C M=\angle A C D-\angle A C M=2 \angle A C M$. Let $N$ be the midpoint of segment $A B$. Then $D N$ is the median, bisector, and altitude of the isosceles triangle $A B D$. Next, point $C$ lies on the midline of this triang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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