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Task 4.2. Find the smallest number in which all digits are different, and the sum of all digits is 32. | Answer: 26789.
Solution. For a four-digit number with different digits, the maximum possible sum of digits is $9+8+7+6=30<32$, so the number we need must be at least five digits.
We will try to make the first digit as small as possible. It is clear that it is no less than 2. We will place 2 in the first position. The... | 26789 | Number Theory | 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 laid out a structure 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 diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | 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).
 \cdot 2=14$ rectangles.... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. Five consecutive natural numbers are written in a row. The sum of the three smallest of them is 60. What is the sum of the three largest? | Answer: 66.
Solution. The fifth number is 4 more than the first, and the fourth is 2 more than the second. Then the sum of the three largest numbers is $2+4=6$ more than the sum of the three smallest, and it is equal to $60+6=66$. | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.6. On an island, there live knights who always tell the truth, and liars who always lie. One day, 65 inhabitants of the island gathered for a meeting. Each of them, in turn, made the statement: "Among the previously made statements, there are exactly 20 fewer true statements than false ones." How many knights... | Answer: 23.
Solution. The first 20 people are liars, because before them, there were no more than 19 statements made, which means there cannot be exactly 20 fewer true statements than false ones.
Next, the 21st person tells the truth (since before him, there were exactly 20 false statements), so he is a knight. Then ... | 23 | Logic and Puzzles | 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 $5+3 \cdot 3-2=12$.
$ candies. Since the total number of candies for both boys did not change after the game, we get $7 x=3(50-x)$. Solvin... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. Numbers were placed in the cells of a square such that the sums of the numbers in each vertical, horizontal, and each diagonal of three cells are equal. Then some numbers were hidden. What is the sum of the numbers in the two shaded cells?
| 16 | | |
| :--- | :--- | :--- |
| | | 10 |
| 8 | | 12 | | Answer: 34.
Solution. Let $a$ be the number in the top right cell, and $b$ be the number in the central cell. The sum of the diagonal containing the number 8 and the sum of the numbers in the right column have a common term $a$, so $8+b=10+12$, from which $b=14$. The sums of the three numbers along both diagonals have... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total? | Answer: 38.
Solution. Let the pen cost $r$ rubles. Then the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bo... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. In each room of the hotel, no more than 3 people can be accommodated. The hotel manager knows that a group of 100 football fans, who support three different teams, will soon arrive. In one room, only men or only women can be accommodated; also, fans of different teams cannot be accommodated together. How m... | Answer: 37.
Solution. All fans are divided into six groups: men/women, supporting the first team; men/women, supporting the second team; men/women, supporting the third team. Only people from the same group can be housed in one room. We divide the number of people in each of the six groups by 3 with a remainder. If th... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. The numbers from 1 to 200 were arranged in a random order on a circle such that the distances between adjacent numbers on the circle are the same.
For any number, the following is true: if you consider 99 numbers standing clockwise from it and 99 numbers standing counterclockwise from it, there will be an... | Answer: 114.
Solution. Consider the number 2. Less than it is only the number 1. Since it is unique, it cannot be in any of the groups relative to the number 2. Therefore, 1 must be opposite 2.
Consider the number 4. The numbers less than it are three in number - an odd count. This means that one of them should not b... | 114 | Combinatorics | 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 8.5. The numbers $1, 2, 3, \ldots, 235$ were written on a board. Petya erased some of them. It turned out that among the remaining numbers, no number is divisible by the difference of any two others. What is the maximum number of numbers that could remain on the board? | Answer: 118.
Solution. 118 odd numbers could remain on the board: none of them is divisible by the difference of any two others, because this difference is even.
Suppose at least 119 numbers could remain. Consider 118 sets: 117 pairs $(1,2)$, $(3,4),(5,6), \ldots,(233,234)$ and one number 235. By the Pigeonhole Princ... | 118 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8. In how many ways can all natural numbers from 1 to 200 be painted in red and blue so that the sum of any two different numbers of the same color is never equal to a power of two? | Answer: 256.
## Solution.
Lemma. If the sum of two natural numbers is a power of two, then the powers of 2 in the prime factorizations of these two numbers are the same.
Proof of the lemma. Let the sum of the numbers $a=2^{\alpha} \cdot a_{1}$ and $b=2^{\beta} \cdot b_{1}$, where $a_{1}$ and $b_{1}$ are odd, be $2^{... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. Denis has cards with numbers from 1 to 50. How many ways are there to choose two cards such that the difference between the numbers on the cards is 11, and the product is divisible by 5?
The order of the selected cards does not matter: for example, the way to choose cards with numbers 5 and 16, and the wa... | Answer: 15.
Solution. For the product to be divisible by 5, it is necessary and sufficient that one of the factors is divisible by 5. Let $n$ be the chosen number divisible by 5, then its pair should be $n-11$ or $n+11$, and both chosen numbers must be natural numbers. Clearly, for $n=5,10,40,45,50$ there is only one ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Traders Andrey and Boris bought 60 bags of potatoes each from the same farmer. All the bags cost the same.
Andrey sold all his bags, increasing their price by $100 \%$. Boris first increased the price by $60 \%$, and after selling 15 bags, increased the price by another $40 \%$ and sold the remaining 45 b... | Answer: 250.
Solution. Let the bag cost the farmer $x$ rubles. Andrei and Boris spent the same amount on buying 60 bags.
From the condition, it follows that Andrei sold 60 bags for $2 x$ rubles each, i.e., he received $60 \cdot 2 x$ rubles. Boris, on the other hand, sold 15 bags at a price of $1.6 x$ rubles and then ... | 250 | Algebra | 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 $ABC$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $AC$, point $K$ is marked, and on side $BC$, points $L$ and $M$ are marked such that $KL=KM$ (point $L$ lies on segment $BM$).
Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$.
... | 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 9.8. Integers $n$ and $m$ satisfy the inequalities $3 n - m \leq 26$, $3 m - 2 n < 46$. What can $2 n + m$ be? List all possible options. | Answer: 36.
Solution. Since the numbers $m$ and $n$ are integers, the values of the expressions in the condition $3 n-m$, $n+m$, $3 m-2 n$ are also integers. Then
$$
\left\{\begin{array}{l}
3 n-m \leqslant 4 \\
n+m \geqslant 27 \\
3 m-2 n \leqslant 45
\end{array}\right.
$$
By adding three times the first inequality ... | 36 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Task 10.1. Find the largest 12-digit number $N$ that satisfies the following two conditions:
- In the decimal representation of the number $N$, there are six digits «4» and six digits «7»;
- In the decimal representation of the number $N$, no four consecutive digits form the number «7444». | Answer: 777744744744.
Solution. It is clear that the number 777744744744 meets the condition of the problem.
Suppose there exists a larger number, then it must start with four sevens. In this number, we select the largest number of consecutive fours, let there be $k$ of them. If $k \geqslant 3$, then by selecting amo... | 777744744744 | Combinatorics | 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?
 contain exactly 50 perfect squares? | Answer: 4486.
Solution. Let's choose the largest natural $n$ such that $n^{2} \leqslant a$. Then
$$
n^{2} \leqslant a\frac{49}{\sqrt{3}-1}=\frac{49(\sqrt{3}+1)}{2}>\frac{49(1.7+1)}{2}>\frac{132}{2}=66
$$
From the fact that $n+1>66$, it follows that $n \geqslant 66$ and $a \geqslant 66^{2}$.
If on the corresponding ... | 4486 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 11.1. The product of nine consecutive natural numbers is divisible by 1111. What is the smallest possible value that the arithmetic mean of these nine numbers can take? | Answer: 97.
Solution. Let these nine numbers be $-n, n+1, \ldots, n+8$ for some natural number $n$. It is clear that their arithmetic mean is $n+4$.
For the product to be divisible by $1111=11 \cdot 101$, it is necessary and sufficient that at least one of the factors is divisible by 11, and at least one of the facto... | 97 | Number Theory | 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 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 11.6. The quadratic trinomial $P(x)$ is such that $P(P(x))=x^{4}-2 x^{3}+4 x^{2}-3 x+4$. What can $P(8)$ be? List all possible options. | Answer: 58.
Solution. Let $P(x)=a x^{2}+b x+c$. Then $P(P(x))=a\left(a x^{2}+b x+c\right)^{2}+b\left(a x^{2}+b x+c\right)+c=$
$a\left(a^{2} x^{4}+2 a b x^{3}+\left(b^{2}+2 a c\right) x^{2}+2 b c x+c^{2}\right)+b\left(a x^{2}+b x+c\right)+c$. Therefore,
\[
\begin{aligned}
& x^{4}-2 x^{3}+4 x^{2}-3 x+4=P(P(x))= \\
& =a... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.7. In a country, there are 110 cities. Between any two of them, there is either a road or there is not.
A motorist was in a certain city, from which there was exactly one road. After driving along this road, he arrived at a second city, from which there were already exactly two roads. Driving along one of t... | Answer: 107.
Solution. Let's number the cities in the order they are visited by the motorist: $1,2,3, \ldots, N$.
Suppose $N \geqslant 108$. From city 1, there is a road only to city 2, so from city 108, all roads lead to all 108 cities, except for 1 and 108. But then from city 2, there are at least three roads: to c... | 107 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
 | Answer: $60^{\circ}$
Solution. According to the problem, $A N=a-A M$, hence $A N=M C$. Similarly, $A P=Q C$. From these equalities and the equality $\angle A=\angle C=60^{\circ}$, it follows that $\triangle A N P=\triangle C M Q$. Therefore, $\angle A N P=\angle Q M C, \angle A P N=\angle M Q C$. By the theorem on the... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (7 points) On a chessboard, there were 21 kings. Each king was under attack by at least one of the others. After some kings were removed, no two of the remaining kings attack each other. What is the maximum number of kings that could remain?
a) Provide an example of the initial arrangement and mark the removed king... | Answer: b) 16.
Solution: Note that each king removed from the board could not have attacked more than 4 of the remaining ones (otherwise, some of the remaining ones would also attack each other). Therefore, the number of remaining kings cannot exceed the number of removed ones by more than 4 times, meaning it cannot b... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In the sum $(1-2-3+4)+(5-6-7+8)+\ldots+(2009-2010-2011+2012)+$ (2013-2014-2015), all numbers that end in 0 were crossed out. Find the value of the resulting expression. | 3. Answer: -3026. It is easy to see that in all parentheses except the last one, the sum is zero. Then the entire sum before cancellation was equal to -2016. At the same time, numbers that end in 0 and are divisible by 4 were included in it with a plus, while numbers that end in 0 and are not divisible by 4 were includ... | -3026 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. There are 10 red and 10 yellow apples with a total weight of 1 kg. It is known that the weights of any two apples of the same color differ by no more than 40 g. All apples were sequentially paired "red with yellow," with red apples chosen in non-decreasing order and yellow apples in non-increasing order (i.e., the h... | 6. Answer: 136 g. Solution Estimate. Note that the weights of the pairs also cannot differ by more than 40 g. Therefore, if the weight of the heaviest pair is more than 136 g, then the weight of each of the other nine pairs will be more than 96 g, but then the total weight of all apples will be more than 1 kg, which is... | 136 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.8. A circle is circumscribed around triangle $A B C$. Let $A_{0}$ and $C_{0}$ be the midpoints of its arcs $B C$ and $A B$, not containing vertices $A$ and $C$. It turns out that the segment $A_{0} C_{0}$ is tangent to the circle inscribed in triangle $A B C$. Find the angle $B$. | 9.8. Let the incircle of triangle $ABC$ be denoted by $\omega$, its center by $I$, and the points of tangency of $\omega$ with $AC$ and $A_0 C_0$ by $K$ and $L$ respectively. Then, in triangles $IKC$ and $ILA_0$, angles $IKC$ and $ILA_0$ are right angles (as angles between a radius and a tangent), and $IK = IL$, since ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Triangle $A B C$ is equilateral. On side $A C$, point $M$ is marked, and on side $B C$, point $N$, such that $M C=B N=2 A M$. Segments $M B$ and $A N$ intersect at point $Q$. Find the angle $C Q B$.
# | # Answer: $90^{\circ}$.
## Solution:
Since $M C=B N$ and $A C=B C$, then $A M=A C-M C=B C-B N=N C$.
From the fact that $A M=N C, A B=A C$ and $\angle B A M=\angle C A N=60^{\circ}$, it follows that triangles $B A M$ and $A C N$ are congruent, so $\angle N A C=\angle M B A=\alpha$.
From triangle $A B M: \angle A M B... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the record of three two-digit numbers, there are no zeros, and in each of them, both digits are different. Their sum is 41. What could be their sum if the digits in them are swapped? | Answer: 113.
Solution. In all numbers, the tens digit is 1. Otherwise, the largest is at least 21, and the other two are at least 12. Their sum is no less than $12+12+21=44$, which is not equal to 41. The sum of the units digits is 11 (zeros are not allowed). Therefore, the sum with the digits rearranged is 113.
Ther... | 113 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$. | Answer: $45^{\circ}$.
Solution. Triangles $E T N$ and $A T B$ are similar, therefore, $T N: T B=T E: T A=E N: A B=1: 4$. Therefore, $C T=1 / 2 B T$. Since $H$ is the intersection point of the medians of triangle $A M N$, $E H=1 / 3 A E=E T$. Therefore, $H T=1 / 2 A T$.
Thus, right triangles $C T H$ and $B T A$ are si... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Find the sum:
$$
(-2021)+(-2020)+(-2019)+\ldots+2023+2024
$$ | Answer: 6069.
Solution. By pairing numbers that differ in sign, we get that in each such pair the sum is 0, and without pairs, the numbers left are $0, 2022, 2023, 2024$. | 6069 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1.
In a certain three-digit number $N$, the last two digits were swapped and the resulting number was added to the original. The sum turned out to be a four-digit number starting with 173. Find the largest possible sum of the digits of the number $N$. | Answer: 20.
Solution: Let the original number be $\overline{a b c}$, and the last digit of the sum $\overline{a b c}+\overline{a c b}$ be $x$. Then, $100 a+10 b+c+100 a+10 c+b=200 a+11(c+b)=1730+x$. If $a1730+x$.
Therefore, $a=8$. Thus, $(800+10 b+c)+(800+10 c+b)=1730+x$, i.e., $11(b+c)=130+x$. From this, it follows ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1.
Given an equilateral triangle $A B C$ with an area of 210. Inside it, points for which vertex $A$ is neither the closest nor the farthest are painted red. What is the area of the painted part of the triangle? | Answer: 41.
Solution.
Consider a point $D$ for which vertex $B$ is the nearest, and vertex $C$ is the farthest. Let $M, N, K$ be the midpoints of sides $AB, BC$, and $AC$ respectively. $O$ i... | 41 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 2.
What is the smallest sum that nine consecutive natural numbers can have if this sum ends in 2050306? | Answer: 22050306.
Option 3.
What is the smallest sum that nine consecutive natural numbers can have if this sum ends in $1020156$?
Answer: 31020156. | 31020156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 6. Variant 1.
What is the largest root that the equation
$$
(x-a)(x-b)=(x-c)(x-d)
$$
can have if it is known that $a+d=b+c=2022$, and the numbers $a$ and $c$ are different? | Answer: 1011.
Solution. By expanding the brackets and combining like terms, we find $x=\frac{c d-a b}{c+d-a-b}$. Note that $c+d-a-b \neq 0$, because otherwise $c+d=a+b$, and considering the equality $a+d=b+c$, we would get $a=c$.
Considering that $d=2022-a, c=2022-b, x=\frac{2022^{2}-2022 a-2022 b}{2(2022-a-b)}=\frac... | 1011 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
In a $101 \times 101$ square, an $88 \times 88$ corner square is painted red. What is the maximum number of non-attacking queens that can be placed on the board without placing the figures on the red cells? A queen attacks along the horizontal, vertical, and diagonals of the square. Attacking through the... | Answer: 26.
Solution. Consider 13 horizontal and 13 vertical strips of size $1 \times 101$, not containing any red cells. Each of the queens stands on at least one of these strips, and no strip can contain more than one queen, so there are no more than 26 queens. To construct an example, consider two rectangles of siz... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 2.1
## Condition:
Karlson and Little together weigh 10 kg more than Freken Bok, and Little and Freken Bok weigh 30 kg more than Karlson. How much does Little weigh? Give your answer in kilograms. | Answer: 20
Exact match of the answer -1 point
Solution.
Let's denote Carlson's mass by K, Freken Bok's mass by F, and Little One's mass by M. Then from the condition, it follows that $K+M=F+10$ and $M+F=K+30$. Adding these equations term by term, we get $K+2M+F=F+K+40$. From this, $M=20$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 3.1
## Condition:
Polina makes jewelry on order for a jewelry store. Each piece of jewelry consists of a chain, a stone, and a pendant. The chain can be silver, gold, or iron. Polina has stones - a diamond, an emerald, and a quartz - and pendants in the shape of a star, a sun, and a moon. Polina is only sa... | # Answer: 24
Exact match of the answer - 1 point
## Solution.
Notice that the iron ornament with the sun is definitely in the middle, as it must be between the gold and silver ornaments. For the first position, we can choose a silver or gold chain. After choosing a chain for the first position, the chain for the thi... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.3
## Condition:
Alina makes phone cases on order for a tech store. Each case has a pattern and a charm.
The case can be silicone, leather, or plastic. Alina has charms in the shapes of a bear, a dinosaur, a raccoon, and a fairy, and she can draw the moon, the sun, and clouds on the case.
Alina is only sa... | Answer: 36
Exact match of the answer - 1 point
Solution by analogy with task №3.1
# | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.4
## Condition:
Anton makes watches for a jewelry store on order. Each watch consists of a bracelet, a precious stone, and a clasp.
The bracelet can be silver, gold, or steel. Anton has precious stones: zircon, emerald, quartz, diamond, and agate, and clasps: classic, butterfly, and buckle. Anton is only ... | Answer: 48
Exact match of the answer -1 point
Solution by analogy with task №3.1
# | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.1
## Condition:
Yasha and Grisha are playing a game: first, they take turns naming a number from 1 to 105 (Grisha names the number first, the numbers must be different). Then each counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The one with... | # Answer: 104
## Exact match of the answer -1 point
## Solution.
We will show that by naming the number 104, Grisha will win. Consider a rectangle with sides a and b. Its perimeter \( P = 2(a + b) \Rightarrow (a + b) = P / 2 \), then the length of the smaller side can take integer values from 1 to the integer part o... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.2
## Condition:
Sasha and Misha are playing a game: first, they take turns naming a number from 1 to 213 (Misha names the first number, the numbers must be different). Then each counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The winner is ... | Answer: 212
Exact match of the answer -1 point
Solution by analogy with task №6.1
## Condition:
Oksana and Seryozha are playing a game: they take turns naming a number from 1 to 165 (Oksana goes first, the numbers must be different). Then each counts the number of different rectangles with integer sides whose perim... | 164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.4
## Condition:
Dima and Vlad are playing a game: first, they take turns naming a number from 1 to 97 (Dima names the first number, the numbers must be different). Then each of them counts the number of different rectangles with integer sides, the perimeter of which is equal to the named number. The winner... | Answer: 96
Exact match of the answer -1 point
Solution by analogy with task №6.1
# | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Assignment No. 7.1
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 300 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Sna... | # Answer: 80
## Exact match of the answer -1 point
## Solution.
First, Hermione drinks $300 / 2=150$ ml of the wisdom potion, and then she pours 150 ml into the second jug, after which the second jug contains 150 ml of the wisdom potion and 300 ml of the beauty potion. When she drinks half of the contents of this ju... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.2
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 600 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape was... | Answer: 40
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 7.3
## Condition:
Potions teacher Severus Snape has prepared three potions, each with a volume of 400 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape... | Answer: 60
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.4
## Condition:
Professor Severus Snape has prepared three potions, each with a volume of 480 ml. The first potion makes the person who drinks it smart, the second makes them beautiful, and the third makes them strong. To ensure the potion works, one needs to drink at least 30 ml of it. Severus Snape was... | Answer: 50
Exact match of the answer -1 point
Solution by analogy with task №7.1
# | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.6. At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote in "VK" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each of the brunettes ... | Solution: According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times but... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false |
Problem 7.1. Denis thought of four different natural numbers. He claims that
- the product of the smallest and the largest numbers is 32;
- the product of the two remaining numbers is 14.
What is the sum of all four numbers? | Answer: 42.
Solution. There are two ways to represent the number 14 as the product of two numbers: $1 \cdot 14$ and $2 \cdot 7$; therefore, the second and third largest numbers are 1 and 14 or 2 and 7. The first case does not work for us, as 1 cannot be the second largest number. So, the second largest number is 2, an... | 42 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. On Monday, 5 people in the class received fives in math, on Tuesday, 8 people received fives, on Wednesday - 6 people, on Thursday - 4 people, on Friday - 9 people. No student received fives on two consecutive days. What is the minimum number of students that could have been in the class | Answer: 14.
Solution. Let's consider pairs of consecutive days.
- On Monday and Tuesday, 5s were received by $5+8=13$ people.
- On Tuesday and Wednesday, 5s were received by $8+6=14$ people.
- On Wednesday and Thursday, 5s were received by $6+4=10$ people.
- On Thursday and Friday, 5s were received by $4+9=13$ people... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. At a tribal council meeting, 60 people spoke in turn. Each of them said only one phrase. The first three speakers said the same thing: "I always tell the truth!" The next 57 speakers also said the same phrase: "Among the previous three speakers, exactly two told the truth." What is the maximum number of sp... | Answer: 45.
Solution. Note that among any four consecutive speakers, at least one lied (if the first three told the truth, then the fourth definitely lied). By dividing 60 people into 15 groups of four consecutive speakers, we get that at least 15 people lied, meaning no more than 45 people told the truth.
To constru... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.6. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively. It is known that $A B=B D, \angle A B D=46^{\circ}, \angle D E C=90^{\circ}$. Find $\angle B D E$, if it is known that $2 D E=A D$. | Answer: $67^{\circ}$.
Solution. Draw the height $B M$ in triangle $A B D$. Since this triangle is isosceles, $M$ bisects $A D$ (Fig. 7.6). This means that $M D = A D / 2 = D E$.
Then the right triangles $B D M$ and $B D E$ are congruent by the leg ( $D M = D E$ ) and the hypotenuse ( $B D$ - common). From this, it is... | 67 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In the classroom, there are several single desks (no more than one person can sit at each desk; there are no other desks in the classroom). During the break, a quarter of the students went out into the corridor, and the number of people left in the classroom was equal to 4/7 of the total number of desks. H... | Answer: 21.
Solution. Note that the number of remaining students is $\frac{3}{4}$ of the initial number, so it must be divisible by 3. Let's denote it by $3x$. Let $y$ be the number of desks. Then
$$
3 x=\frac{4}{7} y
$$
from which $21 x=4 y$. Since 4 and 21 are coprime, $y$ must be divisible by 21. From the problem... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.8. Tanya and Vера are playing a game. Tanya has cards with numbers from 1 to 30. She arranges them in some order in a circle. For each pair of adjacent numbers, Vера calculates their difference, subtracting the smaller number from the larger one, and writes down the resulting 30 numbers in her notebook. After... | Answer: 14.
Solution. We will prove that Tanya cannot get 15 candies. Let's look at the card with the number 15. The number 15 differs from all the remaining numbers, except for the number 30, by no more than 14. Thus, at least one of the differences involving the number 15 will be no more than 14.
Let's provide an e... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest natural number $A$, which is divisible by $p$, ends in $\boldsymbol{p}$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and is the cube of a natural number. | Solution. Let $2 \mathrm{p}+1=\mathrm{n}^{3}$. Then $(\mathrm{n}-1)\left(\mathrm{n}^{2}+n+1\right)=2 \mathrm{p}$. The number $2 \mathrm{p}$ can only have the following positive divisors: 1, 2, p, 2p. The number n is clearly odd, so n-1 is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=$ ... | 11713 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The numbers from 1 to 25 (each exactly once) were placed in the cells of a $5 \times 5$ square such that the sum of the numbers in each row and each column is the same. A boy erased all the numbers except for four (see the picture). Find the number marked with a circle $\bigcirc$.
=$ $65-51=14$. Answer: 14. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In grandmother's garden, apples have ripened: Antonovka, Grushovka, and White Naliv. If there were three times as many Antonovka apples, the total number of apples would increase by $70 \%$. If there were three times as many Grushovka apples, it would increase by $50 \%$. By what percentage would the total number of... | Solution. First method. If the quantity of each type of apple were three times as much, the total number of apples would increase by $200 \%$. Of this increase, $70 \%$ is due to Antonovka, and $50 \%$ is due to Grushovka. Therefore, the increase due to White Naliv would be $200 \%-70 \%-50 \%=80 \%$.
Second method. S... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Three mryak are more expensive than five bryak by 10 rubles. And six mryak are more expensive than eight bryak by 31 rubles. By how many rubles are seven mryak more expensive than nine bryak | 3. Answer: by 38 rubles. Solution. By adding 3 bryaks and myraks, we increase the price difference between myraks and bryaks by 21 rubles. This means that one myrak is 7 rubles more expensive than one bryak. Then seven myraks are more expensive than nine bryaks by $31+7=38$ rubles. Criteria: correct solution - 7 points... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13. Find the 2013-th term of the sequence. | Answer: 130
Solution. Let's calculate the first few terms of the sequence. We get: $a_{1}=934 ; a_{2}=16 \times 13=208$; $a_{3}=10 \times 13=130 ; a_{4}=4 \times 13=52 ; a_{5}=7 \times 13=91 ; a_{6}=10 \times 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the seq... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.3. Point $F$ is the midpoint of side $B C$ of square $A B C D$. A perpendicular $A E$ is drawn to segment $D F$. Find the angle $C E F$.
---
Translation provided as requested, maintaining the original format and line breaks. | Answer: $45^{\circ}$.
Solution. Let the line $A E$ intersect the side $C D$ of the square at point $M$ (see Fig. 10.3). Then triangles $A D M$ and $D C F$ are equal (by the leg and acute angle). Therefore, point $M$ is the midpoint of side $C D$. Then triangle $C F M$ is a right and isosceles triangle, so $\angle C M ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. On side $AB$ of triangle $ABC$, point $K$ is marked, and on side $AC$, point $M$ is marked. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find angle $BAC$. | Answer: $60^{\circ}$.
Solution. To both sides of the equality $S_{A K P M}=S_{B P C}$, add the area of triangle $B P K$ (see Fig. 10.5). We get that $S_{A B M}=S_{B C K}$. Therefore, $\frac{1}{2} B C \cdot B K \sin \angle B = \frac{1}{2} A B \cdot A M \sin \angle A$. Then $\frac{B K}{A M} = \frac{A B \sin \angle A}{B ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In an equilateral triangle, 3 points were marked on each side so that they divide the side into 4 equal segments. These points and the vertices of the triangle were painted. How many isosceles triangles exist with vertices at the painted points? | # Answer. 20
Solution. Let the vertices of the triangle be denoted as A, B, C. Suppose points M, N (M closer to A) are taken on side AB, points P, Q (P closer to A) on side AC, and points K, L (K closer to B) on side BC. Then the following cases are possible for isosceles triangles with the vertex at point A: ABC, AKL... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1 From a three-digit number $A$, which does not contain zeros in its notation, a two-digit number $B$ was obtained by writing the sum of the first two digits instead of them (for example, the number 243 turns into 63). Find $A$ if it is known that $A=3 B$. | Answer: 135.
Solution: The last digit of the numbers $B$ and $A=3B$ is the same, so this digit is 5. Moreover, $A$ is divisible by 3, which means $B$ is also divisible by 3 (the sums of the digits are the same). Therefore, $B$ is one of the numbers $15, 45, 75$. By checking, we find that the number $B=45$ satisfies th... | 135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. The lines containing the bisectors of the exterior angles of a triangle with angles of 42 and 59 degrees intersected pairwise and formed a new triangle. Find the degree measure of its largest angle. | Answer: 69
Solution. Let the vertices of the given triangle be $A, B, C$, and its angles be $\alpha, \beta$, and $\gamma$, respectively. Additionally, let $A_{1}, B_{1}$, and $C_{1}$ be the points of intersection of the external angle bisectors of angles $B$ and $C$, $A$ and $C$, and $A$ and $B$, respectively. Then, $... | 69 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. Petya wrote a natural number $A$ on the board. If it is multiplied by 8, the result is the square of a natural number. How many three-digit numbers $B$ exist such that $A \cdot B$ is also a square of a natural number? | # Answer: 15
Solution. If $8 A$ is the square of a natural number, then any prime number greater than 2 must appear in $A$ with an even exponent, and the number 2 must appear with an odd exponent. Therefore, in $B$, any prime number greater than 2 must also appear with an even exponent, and the number 2 must appear wi... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. How many six-digit numbers exist that consist only of the digits 1 and 2, given that each of them appears? | Answer: 62
Solution. There are a total of $64=2^{6}$ six-digit numbers consisting of the digits 1 and 2, since there are 6 positions and for each position, there are two options to place a digit. However, two numbers - the one consisting only of ones and the one consisting only of twos - do not satisfy the condition, ... | 62 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. Coins come in denominations of 50 kopecks, 1 ruble, 2 rubles, 5 rubles, 10 rubles. $V$ has several coins in his wallet. It is known that no matter which 20 coins are taken out of the wallet, there will be at least one 1-ruble coin, at least one 2-ruble coin, and at least one 5-ruble coin. What is the maximum numbe... | Answer: 28
Solution. Example: 9 coins of 1 ruble, 9 coins of 2 rubles, 9 coins of 5 rubles, and 1 coin of 10 rubles. Note that the wallet contains a total of $9+9+1=19$ coins that are not 1 ruble, so among any 20 coins, there will definitely be a 1-ruble coin. Similarly, this can be verified for all other denomination... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. The real number a is such that $2a - 1/a = 3$. What is the value of $16a^4 + 1/a^4$? | Answer: 161
Solution. Squaring the equality from the condition, we get $4 a^{2}-4+1 / a^{2}=9$, that is, $4 a^{2}+1 / a^{2}=13$. Squaring again, we obtain $16 a^{4}+8+1 / a^{4}=169$. Therefore, $16 a^{4}+1 / a^{4}=161$. | 161 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8-1. Kolya made a figure from four identical blocks as shown in the picture. What is the surface area of this figure? Express your answer in square centimeters.
 | Answer: 64.
Solution. The surface area of one block is 18 cm². Out of this area, 2 cm² is "lost" at the joints with other blocks, leaving a total area of $18-2=16$ cm². Since there are 4 blocks, the answer is $4 \cdot 16=64$ cm². | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-2. On a distant planet in the mangrove forests, there lives a population of frogs. The number of frogs born each year is one more than the difference between the number of frogs born in the previous two years (the larger number minus the smaller number).
For example, if 5 frogs were born last year and 2 in the year ... | Answer: 2033.
Solution: This is a fairly simple problem where the only thing you need to do is correctly understand the condition and apply it the necessary number of times. The ability to correctly understand the condition is an important skill in solving problems!
By applying the algorithm described in the conditio... | 2033 | Other | math-word-problem | Yes | Yes | olympiads | false |
8-3. Olya bought three gifts and packed them in three rectangular boxes: blue, red, and green. She tried to place these gifts in different ways: one on the table, and two on top of each other on the floor. Some distances are given in the diagram. Find the height of the table $h$. Express your answer in centimeters.
![... | Answer: 91.
Solution: Let the height of the blue rectangle be $b$, the height of the red rectangle be $r$, and the height of the green rectangle be $g$. Then, according to the condition, $h+b-g=111, h+r-b=80, h+g-r=82$. Adding all these equations, we get $3h=273$, from which $h=91$. | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-4. There are 50 parking spaces on a car park, numbered from 1 to 50. Currently, all parking spaces are empty. Two cars, a black one and a pink one, have arrived at the parking lot. How many ways are there to park these cars such that there is at least one empty parking space between them?
If the black and pink cars ... | Answer: 2352.
Solution I. Carefully consider the cases of the placement of the black car. If it parks in spot number 1 or 50, the pink car can park in any of the 48 spots (numbered from 3 to 50 or from 1 to 48, respectively). If the black car parks in a spot numbered from 2 to 49, the pink car has only 47 options (all... | 2352 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-6. In a kindergarten, 150 children are standing in a circle. Each child is looking at the teacher standing in the center of the circle. Some children are wearing blue jackets, and the rest are wearing red ones. There are 12 children in blue jackets whose left neighbor is in a red jacket. How many children have a left... | Answer: 126.
Solution I. Let each of the children say whether the color of their jacket is the same as that of the child to their left. We need to find out how many children will say "The same."
Let's find out how many children said "Different." Note that if we go around the circle, the jacket colors of such children... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-8. The numbers $a, b$, and $c$ (not necessarily integers) are such that
$$
a+b+c=0 \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=100
$$
What is $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ ? | Answer: -103.
Solution. We have
$$
100=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{-b-c}{b}+\frac{-a-c}{c}+\frac{-a-b}{a}=-3-\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)
$$
from which $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}=-103$.
## Variants of tasks | -103 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangular grid of size 1 x 60. In how many ways can it be cut into grid rectangles of size 1 x 3 and 1 x 4? | 3. Answer: 45665.
Solution. Let $x$ be the number of rectangles of size 1 x $3, y$ be the number of rectangles of size $1 \times 4$. Then $3 x+4 y=60$, where $x, y$ are non-negative integers. The value $\mathrm{x}=\frac{60-4 \mathrm{y}}{3}=20-\frac{4 \mathrm{y}}{3}$ is an integer if $\mathrm{y}=3 \mathrm{t}(\mathrm{t}... | 45665 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the board, there are 2010 non-zero numbers $\mathrm{a}_{1}$, $\mathrm{a}_{2}, \ldots \mathrm{a}_{2010}$ and the products of all pairs of adjacent numbers: $\mathrm{a}_{1} \cdot \mathrm{a}_{2}, \mathrm{a}_{2} \cdot \mathrm{a}_{3}, \ldots \mathrm{a}_{2009} \cdot \mathrm{a}_{2010}$. What is the maximum number of neg... | 4. Answer: 3014.
Solution. Consider the triples of numbers ( $\left.\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{1} \cdot \mathrm{a}_{2}\right)$ ), ( $\left.\mathrm{a}_{3}, \mathrm{a}_{4}, \mathrm{a}_{3} \cdot \mathrm{a}_{4}\right), \ldots,\left(\mathrm{a}_{2009}, \mathrm{a}_{2010}, \mathrm{a}_{2009} \cdot \mathrm{a}_{... | 3014 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the required trio of people. How man... | Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.3. The height $CH$, dropped from the vertex of the right angle of triangle $ABC$, bisects the bisector $BL$ of this triangle. Find the angle $BAC$.
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The provided text has been translated from Russian to English, maintaining the original formatting and structure. | Answer: $30^{\circ}$.
Solution. Let $C H$ and $B L$ intersect at point $K$ (see Fig. 8.3). Then $C K$ is the median of the right triangle $B C L$, drawn to the hypotenuse, so $C K = 0.5 B L = B K$. Therefore, $\angle K C B = \angle K B C = \angle K B H$. Since the sum of these three angles is $90^{\circ}$ (from triang... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. On the Island of Liars and Knights, a circular arrangement is called correct if each person standing in the circle can say that among their two neighbors, there is a representative of their tribe. Once, 2019 natives formed a correct circular arrangement. A liar approached them and said: "Now we can also form a cor... | Answer: 1346.
Solution. We will prove that a correct arrangement in a circle is possible if and only if the number of knights is at least twice the number of liars.
Indeed, from the problem's condition, it follows that in such an arrangement, each liar has two knights as neighbors, and among the neighbors of any knig... | 1346 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. For a natural number $N$, all its divisors were listed, then the sum of digits for each of these divisors was calculated. It turned out that among these sums, all numbers from 1 to 9 were found. Find the smallest value of $N$. | Answer: 288.
Solution. Note that the number 288 has divisors $1,2,3,4,32,6,16,8,9$. Therefore, this number satisfies the condition of the problem. We will prove that there is no smaller number that satisfies the condition.
Indeed, since $N$ must have a divisor with the sum of digits 9, $N$ is divisible by 9. Now cons... | 288 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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