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9.1. On a circle, 1000 points are marked, each colored in one of $k$ colors. It turns out that among any five pairwise intersecting segments, whose endpoints are 10 different marked points, there will be at least three segments, each of which has endpoints of different colors. For what smallest $k$ is this possible?
(... | Answer. For $k=143$.
Solution. Suppose there are 8 points of one color (say, red) on the circle. Add two more marked points to them, forming a decagon $A_{1} A_{2} \ldots A_{5} B_{1} B_{2} \ldots B_{5}$. Then the segments $A_{1} B_{1}, A_{2} B_{2}, \ldots, A_{5} B_{5}$ intersect each other, and among them, there are t... | 143 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. On the board, the numbers $\sqrt{2}$ and $\sqrt{5}$ are written. It is allowed to add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that it is possible to write the number 1 on the board. | Solution: The simplest way is to provide a sequence of numbers that will lead to the number 1. For example, the following sequence works:
$$
\begin{gathered}
\sqrt{2}+\sqrt{5}, \quad 2 \sqrt{2}+\sqrt{5}, \quad 3 \sqrt{2}+\sqrt{5}, \quad \sqrt{2}-\sqrt{5}, \quad 2 \sqrt{2}-\sqrt{5}, \quad 3 \sqrt{2}-\sqrt{5} \\
(\sqrt{... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
9.5. On a shelf, there are 12 books. In how many ways can 5 of them be chosen if books standing next to each other cannot be chosen? Justify your answer. | Solution: Let's place one more, the thirteenth book, on the right side of the shelf. We will mentally glue each selected book to the book to its right. This is possible since we do not select the 13th book and do not select adjacent books. This will result in 5 glued two-volume sets and 3 separate books. Therefore, the... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.6. Let \(a\) and \(b\) be positive numbers. Find the minimum value of the fraction \(\frac{(a+b)(a+2)(b+2)}{16ab}\). Justify your answer. | Solution: By the inequality between the arithmetic mean and the geometric mean of two numbers, the following three inequalities hold:
$$
a+b \geqslant 2 \sqrt{a b}, \quad a+2 \geqslant 2 \sqrt{2 a}, \quad b+2 \geqslant 2 \sqrt{b}
$$
Multiplying the left and right sides of these three inequalities, we get
$$
(a+b)(a+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. The factory paints cubes in 6 colors (each face in its own color, the set of colors is fixed). How many varieties of cubes can be produced? | Solution: Suppose the procedure of painting the cube goes as follows: an unpainted cube is placed in a machine in a certain fixed position, and then its faces are painted in a certain order: bottom, top, right, left, front, back. First, let's calculate how many ways such a painting can be done. The bottom face can be p... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Matvey decided to start eating properly and every day he ate one bun less and one pear more than the previous day. In total, during the time of proper nutrition, he ate 264 buns and 187 pears. How many days was Matvey on a proper diet? | Answer: 11 days.
Solution: If we "reverse" the sequence of the number of buns, while leaving the pears unchanged, the total number of buns and pears will not change, and the difference between the number of buns and pears eaten each day will become constant. Since $264-187=77=7 \cdot 11$, the correct diet lasted eithe... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 1. Option 1.
Find the value of the expression $101+102-103-104+105+106-107-108+\ldots-619$? | Answer: 100.
Solution. Subtract and add the number 620 at the end of the expression. Then the numbers from 101 to 620 will be divided into $(620-100): 4=130$ quartets. And in each quartet, the sum of the numbers will be equal to -4. Then the value of the expression will be equal to $(101+102-103-104)+(105+106-107-108)... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1
On a sheet, two rectangles are drawn. It is known that both the length and the width of the second rectangle are 3 cm larger than the length and width of the first rectangle, and the area of the second rectangle is 48 cm ${ }^{2}$ larger than the area of the first rectangle. Find the perimeter of the sec... | Answer: 38.
Solution: Let the first rectangle have a length of $a$ cm and a width of $b$ cm. Then the length of the second rectangle is $a+3$ cm, and the width is $b+3$ cm. From the condition, it follows that $(a+3)(b+3)-a b=48$, then $3(a+b)=39, a+b=13$ and $(a+3)+(b+3)=19$. Therefore, the perimeter is 38 cm.
## Var... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
Along a road, 10 lampposts were placed at equal distances, and the distance between the outermost posts was $k$ meters. Along another road, 100 lampposts were placed at the same distances, and the distance between the outermost posts was $m$ meters. Find the ratio $m: k$. | Answer: 11.
Solution: Let the distance between adjacent posts be $x$ meters. Then, in the first case, the distance between the outermost posts is $(10-1) x=9 x=k$ meters. And in the second case, $(100-1) x=99 x=m$ meters. Therefore, the desired ratio is $m: k=(99 x):(9 x)=11$. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1 Sasha wrote the number 765476547654 on the board. He wants to erase several digits so that the remaining digits form the largest possible number divisible by 9. What is this number? | Answer: 7654765464.
Solution: The sum of the digits of the original number is $3 \cdot(7+6+5+4)=66$. According to the divisibility rule for 9, the sum of the erased digits must give a remainder of 3 when divided by 9. It is impossible to select digits with a sum of 3. Digits with a sum of $3+9=12$ can be chosen, and w... | 7654765464 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1
In the garden, there are 26 trees - apple trees and pears. It turned out that among any 18 trees, there is at least one apple tree, and among any 10 trees, there is at least one pear. How many pears are there in the garden? | Answer: 17.
Solution: Since among 18 trees there is at least one apple tree, there are no more than 17 pears. And since among any 10 trees there is at least one pear, there are no more than 9 apple trees. There are 26 trees in total, so there are 9 apple trees and 17 pears.
## Variant 2
In the garden, there are 26 t... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
Petya wrote down all positive numbers that divide some natural number $N$. It turned out that the sum of the two largest written numbers is 3333. Find all such $N$. If there are several numbers, write their sum in the answer. | Answer: 2222.
Solution: Note that one of the listed numbers will be equal to $N$. Since the sum of the two largest listed numbers is odd, these numbers have different parity. This means that the number 2 is a divisor of $N$, so the second number is $N / 2$. According to the condition, $N + N / 2 = 3333$. Therefore, $N... | 2222 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.1. Circular weights weigh 200 grams, square ones weigh 300 grams, and triangular ones weigh 150 grams. 12 weights were placed on the pan balance as shown in the figure. Which pan is heavier and by how many grams?
![](https://cdn.mathpix.com/cropped/2024_05_06_c60160c61c72db74c711g-1.jpg?height=577&width=493&... | Answer: The right side is 250 grams heavier.
Solution. The weight of the left pan in grams is
$$
1 \cdot 300+2 \cdot 150+3 \cdot 200=1200
$$
The weight of the right pan in grams is
$$
1 \cdot 150+2 \cdot 200+3 \cdot 300=1450
$$
Thus, the right pan is 250 grams heavier than the left pan. | 250 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.2. In class 4A, each child has no fewer than 11 classmates and no fewer than 13 female classmates. What is the smallest number of children that can study in this class? | Answer: 26.
Solution. It is not difficult to verify that a class consisting of 12 boys and 14 girls satisfies the condition of the problem. Now let's prove that it cannot be less.
It is clear that there are both boys and girls in the class. Each boy in the class has at least 11 classmates, so there must be at least 1... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Sasha had 47 sticks. Using them all, he formed several letters "B" and "V" as shown in the figure. What is the maximum number of letters "B" that Sasha could have formed?
 | Answer: 8.
Solution. To form the letter "Б", 4 sticks are needed, and to form the letter "В", 5 sticks are needed.
- Sasha could not have formed at least 12 letters "Б" because it would require no less than 48 sticks.
- If Sasha had formed 11 letters "Б", he would have $47-11 \cdot 4=3$ sticks left. This would not be... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.6. Zhenya took a $3 \times 3$ board and placed a column of blue and red cubes on each cell. Then he drew a diagram of the resulting arrangement: he labeled the number of cubes of both colors in each column (the order of the cubes is unknown).
What is the maximum number of blue cubes Zhenya can see if he look... | Answer: 12.
Solution. Let's understand the maximum number of blue cubes Zhenya can see in each of the three rows: left, middle, and right.
Left row. The first column consists of 5 cubes (2 red and 3 blue), so it completely blocks the second column, as well as 5 out of 7 cubes in the last column.
Thus, Zhenya sees al... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, it is allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal? | Answer: 11.
Solution. Suppose $N$ moves were made, after which the number of coins in all stacks became equal.
Let's slightly change the rules. Suppose initially there were not 9, 7, 5, and 10 coins in the stacks, but $N+9, N+7, N+5$, and $N+10$ respectively; and we will perform the moves as follows: instead of addin... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. 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 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles of $... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. A natural number $n$ is called interesting if $2 n$ is a perfect square, and $15 n$ is a perfect cube. Find the smallest interesting number. | Answer: 1800.
Solution. Factorize the number $n$ into prime factors. For a number to be a square, it is necessary that all prime numbers in this factorization appear in even powers, and for a number to be a cube, it is necessary that all prime numbers appear in powers divisible by 3.
Let's look at the power of two th... | 1800 | Number Theory | 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 |
4. Dad is preparing gifts. He distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest gifts, there are 50 candies. How many bags are the candies distributed into? How many candies are in the smallest gift? | Answer: 10 packages, 5 candies.
Solution. Let's number the gifts from the smallest to the largest, from 1 to $n$. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.2 It is known that $10 \%$ of people own no less than $90 \%$ of all the money in the world. For what minimum percentage of all people can it be guaranteed that these people own $95 \%$ of all the money? | Solution. Let $100 S$ be the total amount of money in the world (regardless of the currency), and the total number of people be $100 N$ (where $N$ can be non-integer, which is not important). Then at least $90 S$ of the money is owned by $10 N$ people (let's call this group oligarchs). The remaining $90 N$ people (let'... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4 Let $n$ - be a natural number greater than 10. What digit can stand immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$? Provide all possible answers and prove that there are no others. | Solution. Method 1. $n^{2}+n=(n+0.5)^{2}-0.2510$ ), the digit immediately after the decimal point is no less than 4. That is, it is 4.
Method 2. The required digit is the last digit in the number $\left[10 \sqrt{n^{2}+n}\right]=$ $\left[\sqrt{100 n^{2}+100 n}\right]$. Note that $100 n^{2}+100 n100 n^{2}+80 n+16=(10 n+... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On the island of knights and liars (liars always lie, knights always tell the truth), each resident supports exactly one football team. In a survey, all residents of the island participated. To the question "Do you support 'Rostov'?", 40% of the residents answered "Yes". To a similar question about 'Zенit', 30% answ... | Solution. Let $x \%$ of the island's inhabitants be liars. Then $(100-x) \%$ are knights. Since each knight answered affirmatively to exactly one of the questions, and each liar to three, we have $(100-x)+3 x=40+30+50$, from which $x=10$. Since none of the island's inhabitants said they support CSKA, all liars support ... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Let quadrilateral $A B C D$ be inscribed. The rays $A B$ and $D C$ intersect at point K. It turns out that points $B, D$, and the midpoints $M$ and $N$ of segments $A C$ and $K C$ lie on the same circle. What values can the angle $A D C$ take?
. Similarly... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) An apprentice chef took two buckets of unpeeled potatoes and cleaned everything in an hour. In the process, $25 \%$ of the potatoes went into peels. How long did it take for him to accumulate exactly one bucket of peeled potatoes? | Solution. Since a quarter of the potatoes went into peels, then in one hour, three quarters of two buckets of cleaned potatoes were obtained. This means that in one hour, the trainee cook got $\frac{3}{2}$ buckets, and exactly one bucket in 40 minutes.
Answer. 40 minutes.
## Comments on Evaluation.
Correct answer on... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." That child said to their right neigh... | Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. A group of boys sitting next to each other is followed by a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so their number... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. There are more than 20 but fewer than 30 people in the class, and everyone has a different birthday. Petya said: "There are twice as many people older than me as there are younger than me in the class." Katya said: "There are three times fewer people older than me as there are younger than me in the class." How many... | Answer: 25.
Solution: From Petya's words, it is clear that without him, the number of students in the class is a multiple of 3 (2 parts + 1 part). That is, together with him, there can be 22, 25, or 28 people in the class. Similarly, from Katya's words, it is clear that without her, the number of students in the class... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Three three-digit numbers, in the notation of which all digits except zero participate, add up to 1665. In each number, the first digit was swapped with the last digit. This resulted in three new three-digit numbers. What is the sum of the new numbers | Solution: The sum of the last digits of the three original numbers is 5, 15, or 25. However, 5 and 25 are excluded, as they cannot be represented as the sum of three different digits (from 1 to 9), so 15 remains. Therefore, the sum of the middle digits is also 15, and the sum of the first digits is 15. It then becomes ... | 1665 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Petya cut an $8 \times 8$ square along the cell boundaries into parts of equal perimeter. It turned out that not all parts were equal. What is the maximum number of parts he could have obtained?
# | # Answer: 21.
Solution: Let $S$ be the smallest area among the figures obtained by Petya.
If $S=1$, then this figure is a $1 \times 1$ square and its perimeter is 4. Then all of Petya's figures have a perimeter of 4, meaning they are all $1 \times 1$ squares. But this means that all figures are equal, which contradic... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are 4 stacks of coins. The first stack has 9 coins, the second has 7, the third has 5, and the fourth has 10. In one move, you are allowed to add one coin to three different stacks. What is the minimum number of moves required to make the number of coins in all stacks equal?
# | # Answer: in 11 moves.
Solution 1. Consider the differences between the number of coins in each of the other stacks and the number of coins in stack No. 3. In one move, either each of these differences increases by 1 (if all stacks except 3 are chosen for adding coins), or exactly 1 of these differences decreases by 1... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.4. Find the largest natural number, all digits of which are different, and the product of these digits is a square of a natural number. | Answer: 986431. Solution: Obviously, among these digits there is no zero. Further, we have $1 \cdot 2 \cdot \ldots \cdot 9=2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$. Therefore, we need to remove the digits 5 and 7, and also need to make the odd power of two even. This means we need to remove the digit 2: obviously, we... | 986431 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the pop ensemble "Sunshine," everyone plays either the violin or the double bass. The average age of those who play the violin is 22 years, and those who play the double bass is 45 years. Igor changed his instrument and started playing the violin instead of the double bass. As a result, the average age of those w... | Solution. Let $x$ be the number of people who play the contrabass, excluding Igor, and $y$ be the number of people who play the violin (also excluding Igor).
From the condition, it follows that the total age of those who play the contrabass was initially equal to: $(x+1) \cdot 45$, and when Igor changed the instrument... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. 3.1. Find the number of four-digit numbers composed of the digits $1,2,3,4,5,6,7$ (each digit is used no more than once) that are divisible by 15. | Answer: 36.
Solution.
For a number to be divisible by 15, it must be divisible by 3 and by 5. Therefore, according to the divisibility rule for 5, the last digit can only be 5. The sum of the digits of the number must be divisible by 3. For this, the sum of the first three digits must give a remainder of 1 when divid... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. 4.1. In a right triangle $ABC$ (right angle at $C$), the bisector $BK$ is drawn. Point $L$ on side $BC$ is such that $\angle CKL = \angle ABC / 2$. Find $KB$, if $AB = 18, BL = 8$. | Answer: 12.
## Solution.
Note that $\angle L K B=\angle C K B-\angle C K L=\angle C A B+\angle A B K-\angle C K L$ (the last equality holds because $\angle C K B$ is an exterior angle of triangle $A B K$). Since $\angle C K L=\angle A B C / 2=\angle A B K$, we have that $\angle L K B=\angle C A B$. From the fact that... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. 5.1. Find the largest natural number in which all digits are different and any two adjacent digits differ by 6 or 7. | Answer: 60718293.
## Solution.
We will map each digit from 0 to 9 to a vertex in a graph and connect the vertices with an edge if the corresponding digits differ by 6 or 7.
We see that the ... | 60718293 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. 6.1. The figure "lame rook" can move to an adjacent cell in one move. On a $20 \times 20$ board, crosses were placed in all cells that the "lame rook" can reach from the top-left corner in exactly 10 moves. How many cells were marked with a cross? | Answer: 36.
## Solution.
Let's set up a coordinate system so that the top-left cell has coordinates $(0,0)$. In one move, the sum of the rook's coordinates changes by 1. In 10 moves, the sum of the coordinates cannot exceed 10, and it will become an even number.
. It turned out that the distance between $a$ and $b$ is half the distance between $b$ and $c$. Find $M$.
, $ab+bc+ac=-84$ (2), $abc=M$ (3). According to the condition, $2(b-a)=c-b$ or $3b=c+2a$. Considering (1), $c... | 160 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. 8.1. How many increasing arithmetic progressions of 22 different natural numbers exist, in which all numbers are no greater than 1000? | Answer: 23312.
## Solution.
Consider the 22nd term of each such progression, it will have the form $a_{22}=a_{1}+21d$. This means that $a_{1}$ and $a_{22}$ will have the same remainders when divided by 21. Each pair of numbers not exceeding 1000, giving the same remainders when divided by 21, defines one of the requi... | 23312 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Indicate the smallest number ending in 37 with the sum of its digits equal to 37 and divisible by 37. | Answer: 99937.
Solution. The number is the smaller, the fewer digits are required to write it. Two digits - these are the last two digits, their sum is 10. Therefore, the sum of the other digits is 27, and there are at least three of them, since the largest digit is 9. Thus, the number 99937 satisfies three of the fou... | 99937 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Cyclists Petya, Vlad, and Timur simultaneously started a warm-up race on a circular cycling track. Their speeds are 27 km/h, 30 km/h, and 32 km/h, respectively. After what shortest time will they all be at the same point on the track again? (The length of the cycling track is 400 meters.) | Answer: 24 min.
Solution. Vlad rides 3 km/h faster than Petya. Therefore, he will overtake him by one lap (i.e., ride 400 m more) in $\frac{0.4}{3}$ h $=8$ min. Timur rides 2 km/h faster than Vlad. Therefore, he will overtake him by one lap in $\frac{0.4}{2}$ h $=12$ min. The least common multiple of the numbers 8 and... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the rebus
$$
\mathbf{K}\mathbf{O}>\mathbf{H}>\mathbf{A}>\mathbf{B}>\mathbf{U}>\mathbf{P}>\mathbf{y}>\mathbf{C}
$$
different letters represent different digits. How many solutions does the rebus have? | Answer: 0.
Solution. From the rebus, it follows that $\mathbf{P}>\mathbf{O}>\mathbf{P}$. This cannot be! Evaluation. 7 points for the correct solution. | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. Karlson counts 1000 buns baked by Fräulein Bock: «one, two, three, ..., nine hundred ninety eight, nine hundred ninety nine, thousand». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer: 2611.
Solution. One word will be required to pronounce 37 numbers: $1,2,3,4,5,6,7$, $8,9,10,11,12,13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200,300,400$, $500,600,700,800,900,1000$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required f... | 2611 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. A semicircle with diameter $A B$ and center at point $O$ is divided by points $C$ and $D$ into three parts 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 inte... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Seryozha decided to start eating properly and every day he ate one fewer chocolate cookie and one more sugar-free cookie than the previous day. In total, during the time of proper nutrition, he ate 264 chocolate cookies and 187 sugar-free cookies. How many days was Seryozha on a proper diet? | Answer: 11 days.
Solution 1. The total number of cookies eaten each day is the same. In total, Seryozha ate $264+187=451=11 \cdot 41$ cookies. Since the proper diet lasted more than one day and he ate more than one cookie each day, either he ate for 11 days, 41 cookies each day, or for 41 days, 11 cookies each day. Bu... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.6. 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 |
2. Three bear cubs were dividing three pieces of cheese weighing 10 g, 12 g, and 15 g. A fox came to help them. She can simultaneously bite and eat 1 g of cheese from any two pieces. Can the fox leave the bear cubs equal pieces of cheese? | Answer: She can.
Solution. Let's provide one of the possible examples of how the fox could do it. For convenience, let's record the results of the fox's "work" in a table.
| 10 | 12 | 15 |
| :---: | :---: | :---: |
| 9 | 12 | 14 |
| 8 | 12 | 13 |
| 7 | 12 | 12 |
| 7 | 11 | 11 |
| 7 | 10 | 10 |
| 7 | 9 | 9 |
| 7 | 8 |... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In the underwater kingdom, there live octopuses with seven and eight legs. Those with 7 legs always lie, while those with 8 legs always tell the truth. One day, a conversation took place between three octopuses.
Green octopus: "We have 21 legs together." Blue octopus (to the green one): "You're lying!" Red octopus:... | Answer. 1) Could not.
2) The green octopus has 7 legs, the blue one has 8 legs, and the red one has 7 legs.
Solution.
1) If the green octopus had told the truth, then each octopus would have 7 legs. This means that the green octopus, according to the condition, should have lied. We get a contradiction, so the green ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Private Petrov took a bucket of unpeeled potatoes and cleaned them in 1 hour. In the process, $25 \%$ of the potatoes went into peels. How long did it take for him to have half a bucket of peeled potatoes? | Answer. In 40 minutes.
Solution. Since a quarter of the potatoes went into peels, Petrov received three quarters of a bucket of peeled potatoes in 1 hour. This means that a quarter of a bucket of peeled potatoes Petrov received in 20 minutes, and half a bucket - in 40 minutes.
## Grading Criteria.
- Correct solution... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The figure shown on the right is made of matches (the side of a small square is one match). The area of the entire shaded figure is 300 square centimeters. Find the total length of all the matches used.
. What did the total weight of the tomatoes become? | Solution. The weight of the dry matter in tomatoes is $1 \%$ of 1 ton - 10 kg. After drying, the weight of the dry matter does not change and constitutes $5 \%(100 \%-$ $(99 \%-4 \%))$ of the total weight of the tomatoes. In other words, the total weight is 200 kg.
Remark. Perhaps, the student can be given 2 - 3 point... | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 11.2. (7 points)
The master makes a whole number of parts in one hour, more than 5, and the apprentice makes 2 parts less. The master completes the order in a whole number of hours, and two apprentices together - one hour faster. How many parts does the order consist of? | Answer: 24.
Solution: Let $x$ be the number of parts in the order; $y$ (parts per hour) be the master's productivity $(y>5)$; $y-2$ (parts per hour) be the apprentice's productivity; $2 y$ - 4 (parts per hour) be the productivity of two apprentices.
Then $\frac{x}{y}$ is the time for the master to complete the order,... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 11.3. (7 points)
The base of the pyramid with vertex $P$ is a quadrilateral $A B C D$, where the sum of angles $A$ and $D$ is five times smaller than the sum of angles $B$ and $C$. Find the angle between the planes of the faces $P A B$ and $P C D$, if both are perpendicular to the base. | Answer: $60^{\circ}$.
Solution: Let the planes $A P B$ and $C P D$ intersect along the line $P X$ (point $X$ lies in the plane $A B C$, see the figure).
Since $(A P B) \perp(A B C)$ and $(C ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 11.5. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars al... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Petya came up with a four-digit number, in which all the digits are different. It is known that the sum of the first three digits of this number is divisible by 9 and the sum of the last three digits of this number is divisible by 9. What values can the sum of all the digits of this number take? Find all possible va... | 2. Answer. The sum of the digits is 18.
According to the condition, the sum of the first three digits is divisible by 9, and the sum of the last three is also divisible by 9. Therefore, the difference between the first and the last digit is divisible by 9. Since all digits are different, the first digit is 9, and the ... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $ABC$, $AB = 3 \cdot BC$, point $M$ is the midpoint of side $AB$, and $BD$ is the angle bisector. Find the angle $MDB$.
---
Here is the translation of the given text, maintaining the original formatting and structure. | 4. Answer. $90^{\circ}$.
Let $X$ be the midpoint of segment $M B$, and draw segment $D X$. By the property of the angle bisector, we have
$$
A D: D C = A B: B C = 3: 1
$$
On the other hand,... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. CONDITION
In what minimum number of colors should natural numbers be painted so that any two numbers, the difference between which is 3, 4, or 6, are of different colors? | Solution. Let the number $n=1$ be color $A$, then the numbers 4, 5, and 7 must be painted in another color. Let $n=4$ be color $B$, then from $7-4=3$, it follows that the number $n=7$ is of the third color $C$. Therefore, at least 3 colors are required. The coloring $A A A B B B C C C A A A B B B \ldots$ is the desired... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by o... | Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. Thus, the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y = x - n ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. After watching the movie, viewers rated it one by one with an integer score from 0 to 10. At any given time, the movie's rating was calculated as the sum of all the given scores divided by their number. At some point in time $T$, the rating became an integer, and then with each new voting viewer, it decreased by ... | # Answer: 5.
Solution. Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. This means the sum of the scores was $n x$. Let the next viewer give $y$ points. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which $y =... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. For what least natural $n$ do there exist such natural $a$ and $b$ that
$$
\text { GCD }(a, b)=999 \text { and } \text{LCM}(a, b)=n!
$$
(here $n!=1 \cdot 2 \cdot \ldots \cdot n)$ ? | 2. For what smallest natural $n$ do there exist such natural $a$ and $b$ that
$$
\text { GCD }(a, b)=999 \text { and } \text{LCM}(a, b)=n!
$$
(Here $n!=1 \cdot 2 \cdot \ldots \cdot n) ?$
ANSWER. $n=37$.
SOLUTION. Since $\text{LCM}(a, b)$ is divisible by $\text{GCD}(a, b)=999=27 \cdot 37$, and the number 37 is prime... | 37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades? | 5. Vova has 19 math grades in his journal, all twos and threes, and the first four grades are twos. It turned out that among the quartets of consecutive grades, all 16 possible combinations of four twos and threes are present. What are Vova's last four grades?
ANSWER. 3222.
SOLUTION. Note that there are exactly 16 qu... | 3222 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The lateral sides $K L$ and $M N$ of trapezoid $K L M N$ are equal to 15 and 12, respectively, and the base $L M=3$. The bisector of angle $N K L$ passes through the midpoint of side $M N$. Find the area of the trapezoid. Answer: 80. | # Solution.
Let $Q$ be the midpoint of segment $M N$. Extend the bisector of angle $N K L$ to intersect line $L M$ at point $P$. $\angle N K Q = \angle Q K L$.
$\angle N K Q = \angle K P L$ ... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. The working day at the enterprise lasted 8 hours. During this time, the labor productivity was as planned for the first six hours, and then it decreased by $25 \%$. The director (in agreement with the labor collective) extended the shift by one hour. As a result, it turned out that again the first six hours were wor... | 1. Answer: by 8 percent. Solution. Let's take 1 for the planned labor productivity (the volume of work performed per hour). Then before the shift extension, workers completed $6+1.5=7.5$ units of work per shift. And after the extension, $8+2.1=8.1$ units. Thus, the overall productivity per shift became $8.1: 7.5 \times... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Each cell of a $5 \times 5$ square contains a number. The numbers in any row and any column form an arithmetic progression. The numbers in the corners of the square are $1, 25, 17, 81$. What number is in the central cell? Don't forget to justify your answer.
| 1 | | | | 25 |
| :--- | :--- | :--- | :--- | :--- |
... | 2. Answer: 31. Solution. In any arithmetic progression $a_{5}=a_{1}+4 d, a_{3}=a_{1}+2 d \Rightarrow$ $a_{3}=\left(a_{1}+a_{5}\right) / 2$. Therefore, in the third cell of the first column stands the number 41, in the third cell of the last column stands the number 21, and in the middle cell of the second row (i.e., in... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Triangle $A B C$ is similar to the triangle formed by its altitudes. Two sides of triangle $A B C$ are 4 cm and 9 cm. Find the third side. | 4. Answer. 6 cm. Solution. Let $A B=9, B C=4, A C=x, S$ - the area of triangle $A B C$. Then the heights of the triangle are $2 S / 4, 2 S / 9, 2 S / x$. By the triangle inequality, side $A C$ can be either the largest or the middle in length. Applying the triangle inequality to the triangle formed by the heights, we s... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural number that has exactly 12 different natural divisors, the largest prime divisor of which is the number 101, and the last digit is zero. | Solution. Let $n$ be the desired number. According to the condition, $n: 101, n: 2, n: 5$. Consider the number $m=2 \cdot 5 \cdot 101$, note that it has exactly 8 different natural divisors $(1$, $2,5,101,2 \cdot 5, \ldots, 2 \cdot 5 \cdot 101$ ), so $n>m$. Since $n$ is the smallest natural number satisfying the condit... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. How many four-digit numbers exist where the digit in the thousands place is greater than the digit in the hundreds place? | Solution. The digit in the thousands place can take one of 9 possible values: $1,2,3, \ldots, 9$ (we cannot take 0, since the number is four-digit). For each of these options, we can specify the corresponding number of options for the hundreds digit: $1,2,3$, ..., 9. That is, in total $1+2+\ldots+9=45$ options. The oth... | 4500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. What is the minimum number of chips that need to be taken so that in any arrangement of them on the cells of a chessboard, there would necessarily be 4 chips standing in a row horizontally? | Solution. Consider one row. To ensure that there are 4 chips in it, there must be no less than 7 chips (6 chips can be placed in two groups of 3 chips). If there are a total of $6 \cdot 8+1=49$ chips, then by the pigeonhole principle, there will definitely be a row with no fewer than 7 chips. On the other hand, it is i... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. We consider all possible $100-$-digit numbers, where each digit is equal to 1 or 2. For each, the remainder when divided by 1024 is calculated. How many different remainders are there among these? | 5. Solution: Subtract the number $11 \ldots 1$ (one hundred ones) from all the numbers. The numbers under consideration have turned into numbers consisting of zeros and ones from 0 to $11 \ldots 1$. We will show that if two of these numbers have differences in the last 10 digits, then they have different remainders whe... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. In a supermarket, there are two types of fruit sets for sale. The first type of set consists of 3 apples and 15 oranges and costs 360 rubles. The second type of set consists of 20 apples and 5 oranges and costs 500 rubles. Fruits are sold only in sets, and sets cannot be divided. Seryozha came to the supermarket a... | Solution: Let Sergey buy $x$ sets of the first type and $y$ sets of the second type. Then he bought $3x + 20y$ apples and $15x + 5y$ oranges. From the problem's condition, it follows that these numbers are equal, that is, $15y = 12x$ or $5y = 4x$. The smaller $y$ is, the smaller $x$ is, and the smaller the amount paid.... | 3800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. Six different natural numbers from 6 to 11 are placed on the faces of a cube. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 36, and the second time it was 33. What number is written on the face opposite the one with the number 10? Justify your answer. | Solution: The sum of the numbers on all 6 faces of the die is 51. Therefore, the sum of the numbers on the top and bottom faces during the first roll is 15, and during the second roll, it is 18. The number 15 can be obtained (as the sum of two different integers from the interval [6; 11]) in two ways: 15 = 9 + 6 = 8 + ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. On a line, several points were marked. Then, between each pair of neighboring points, one more point was marked, and this operation was repeated once more. As a result, 101 points were obtained. How many points were marked initially? | 9.1. Answer: 26. Solution. Let there initially be $k$ points marked. Then $k-1$ more points were added to them (one between the first and second, second and third, $\ldots, k-1$-th and $k$-th marked points), and then another $(k+(k-1))-1=$ $2 k-2$ points. In total, the number of points became $4 k-3$. Solving the equat... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. The average height of 11 football players on the team is 182 cm. During the match, the referee sent off one player, and the average height of the remaining players became 181 cm. What is the height of the player who was sent off? | Answer: 192 cm. Solution. Let $S$ be the sum that results from adding the heights of the 10 remaining football players. Then $\frac{S}{10}=181$ and $\frac{S+x}{11}=182$, where $x$ is the height of the removed football player. From this, $S=1810$ and $x=182 \cdot 11-S=2002-1810=192$. | 192 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. Petya wrote down all natural numbers from 1 to $n$ in a row on the board and counted the total number of digits written. It turned out to be 777. What is $n$? | Answer: 295. Solution: Since a total of 777 digits have been written, $n$ must be a three-digit number: indeed, in the case of a two-digit $n$, no more than $9+2 \cdot 90=189$ digits would have been written, and in the case of a four-digit (or more) number, more than $9+2 \cdot 90+3 \cdot 900=2889$ digits would have be... | 295 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. In a bag for a bingo game, there are 90 barrels with numbers from 1 to 90. What is the minimum number of barrels that need to be drawn at random from the bag to guarantee getting a barrel with a number divisible by 3 or 5 (or both 3 and 5)? | Answer: 49. Solution. We will call numbers that are divisible by 3 or 5 desirable. The number of numbers from 1 to 90 that are divisible by 3 is 30 ( =90: 3 ), and those divisible by 5 is 18 ( =90:5). If we add $30+18=48$, then all numbers that are divisible only by 3 and only by 5 will be counted once, and numbers tha... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. How many six-digit natural numbers exist, each of which has adjacent digits with different parity | Answer: 28125. Solution. If the first (most significant) digit is even, then it can be chosen in four ways $(2,4,6,8)$, and all subsequent ones can be chosen in five ways (possible candidates for the second, fourth, and sixth digits are $1,3,5,7,9$, and for the third and fifth - $0,2,4,6,8$). In the end, by the rule of... | 28125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. Points $A, B, C, D$ are located at the vertices of rectangle $ABCD$, its sides and diagonals $AC$ and $BD$ are roads. The first car traveled the route $B \rightarrow$ $C \rightarrow A \rightarrow D$ in one hour, and the second car traveled the route $D \rightarrow B \rightarrow C \rightarrow A$ in one hour. After ... | Answer: In 40 minutes.
Solution: The diagonals of a rectangle have the same length and are longer than any of its sides. In one hour, together the two cars would travel three times the length of side $B C$ and three times the diagonal, since one car travels two sides equal to $B C$ and one diagonal in an hour, while t... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A straight road passes not far from Mount Fuji. The driver of a car noticed the mountain 60 km to the north, and an hour later - 45 km to the west. What is the shortest distance the car passed from Mount Fuji? | Answer: 36 km.
Solution: The mountain and the observation points are at the vertices of a right-angled triangle with legs of 60 and 45. The hypotenuse in this case is 75. By equating the areas of the triangle calculated in different ways, we get that the product of the legs is equal to the product of the hypotenuse an... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. This year, the son and daughter are so many years old that the product of their ages is 7 times less than the father's age. And in three years, the product of their ages will already be equal to the father's age. Find the father's age. | Answer: 21 years
Solution. Let $n$ be the son's age, $m$ be the daughter's age, then the father's age is $7 m n$. In three years, the son's age will be $n+3$, the daughter's age will be $m+3$, and the father's age will be $7 \mathrm{mn}+3$, and according to the condition, the equation $(n+3)(m+3)=7 m n+3$ will be sati... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A right triangle $A B C$ (cathetus $B C$ is greater than cathetus $A C$) is inscribed in a circle. A point $D$ is chosen on side $B C$ such that $B D=A C$, and point $M$ is the midpoint of the arc $A C B$. Find the angle $C D M$. | Answer: $45^{\circ}$.
Solution. Note that $A B$ is the diameter of the circumscribed circle. Connect point $M$ with points $A, B, C$, and $D$. Since arcs $A M$ and $B M$ are equal, the chords $A M$ and $B M$ that subtend them are also equal. Segments $B D$ and $A C$ are equal by the condition. Finally, angles $M B C$ ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. When a certain number is divided by 13 and 15 with a remainder, the same incomplete quotients are obtained. Find the largest such number. | Answer: 90. Suppose the number a satisfies the condition of the problem. Then the equalities $a=13 d+b=15 d+c$ hold, where $d$ is the quotient from the condition of the problem. From this, $2 d=b-c \leq 12$, i.e., $d \leq 6$, and the remainder $b$ from dividing the number by 13 is no more than 12. For $d=6$ and $b=12$,... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. Oleg drew an empty $50 \times 50$ table and wrote a non-zero number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers written... | Answer: 1250 works.
Solution. First, we show that there are at least 1250 irrational numbers in the table. Suppose along the left side of the table, there are $x$ irrational and $50-x$ rational numbers. Then along the top side, there are $50-x$ irrational and $x$ rational numbers. Since the product of a non-zero ratio... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101 ?$ | Answer. It exists.
Solution. For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$.
Remark. A number $\overline{a b c d}$ with distinct non-zero digits satisfies the condition if and only if $a+d=b+c$.
Comment. Any correct example with verification that it fits - 7 points.
Any correct example wi... | 1234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are 9 cards with numbers $1,2,3,4,5,6,7,8$ and 9. What is the maximum number of these cards that can be laid out in some order in a row so that on any two adjacent cards, one of the numbers is divisible by the other? | Answer: 8.
Solution: Note that it is impossible to lay out all 9 cards in a row as required. This follows from the fact that each of the cards with numbers 5 and 7 can only have one neighbor card with the number 1. Therefore, both cards 5 and 7 must be at the edges, and the card with the number 1 must be adjacent to e... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. Petya bought one cupcake, two muffins, and three bagels, Anya bought three cupcakes and a bagel, and Kolya bought six muffins. They all paid the same amount of money for their purchases. Lena bought two cupcakes and two bagels. How many muffins could she have bought for the same amount she spent? | Answer: 5 cupcakes.
Solution: The total cost of Petya and Anya's purchases is equal to the cost of two of Kolya's purchases. If we denote P, K, and B as the costs of a cake, a cupcake, and a bagel, respectively, we get the equation: $(P + 2K + 3L) + (3P + 5) = 12K$, from which it follows that $4P + 4B = 10K$, or $2P +... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. What is the minimum number of 3-cell corners that need to be painted in a $6 \times 6$ square so that no more corners can be painted? (Painted corners must not overlap.) | Answer: 6.
Solution. Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted. Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest number that contains only the digits 2 and 3 in equal quantities, and is divisible by 2 and 3. | Answer: 223332.
Sketch of the solution. If a number is divisible by 3, then the sum of its digits is divisible by 3, which means the number of twos is a multiple of three, and thus, there must be at least three twos. Therefore, the desired number is a six-digit number. Since it is divisible by 2, it ends in 2. The num... | 223332 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A rectangle with sides $6 \mathrm{~cm}$ and $3 \mathrm{~cm}$ was cut into three rectangles of equal perimeter. What can the perimeter of these rectangles be? Find all possible answers. | Answer: 14 cm, 10 cm, 10.5 cm.
Sketch of the solution.
Let's cut a rectangle with sides $a$ and $b$. Cases 1 and 2. Cut the rectangle into three equal rectangles with two cuts parallel to side $a$.
![](https://cdn.mathpix.com/cropped/2024_05_06_f259ce87565f6e7a499dg-2.jpg?height=208&width=373&top_left_y=1224&top_lef... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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