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# 4. Option 1.
Cyclists Alexei, Vitaly, and Sergey participated in a 10 km race. At the moment when Alexei arrived at the finish line, Vitaly had 1 km left to go. And when Vitaly arrived at the finish line, Sergey had 1 km left to go. By how many kilometers did Alexei lead Sergey at the finish line? | Answer: 1.9.
Solution: Vitaly's speed is 0.9 of Alexey's speed, and Sergey's speed is 0.9 of Vitaly's speed, i.e., 0.81 of Alexey's speed. Therefore, when Alexey has traveled 10 km, Sergey has traveled $10 \cdot 0.81 = 8.1$ km. Thus, Alexey is ahead of Sergey by $10 - 8.1 = 1.9$ km. | 1.9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,366 |
# 5. Variant 1
Baron Munchausen's clock broke at 14:00. Now, for the first 30 seconds of each minute, the clock runs twice as fast as usual, and for the second 30 seconds, it runs twice as slow as usual. The baron has an important meeting at 14:40. What time will his clock show at that moment? | Answer: 14:50
Solution. Note that the first half minute of the clock passes in 15 seconds, and the second half minute passes in 1 minute. This means that 1 minute of real time, according to Baron Münchhausen's clock, lasts 1 minute and 15 seconds, or 1.25 minutes. Then 40 minutes $-40 \cdot 1.25=50$ minutes.
## Varia... | 14:50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,367 |
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 | 14,368 |
# 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 | 14,369 |
# 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 | 14,370 |
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 | 14,371 |
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 | 14,372 |
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 | 14,373 |
Problem 4.4. Cats Leopold, Garfield, Vasiliy, Matilda, and Tom ate two cutlets, two sausages, and one fish in the kitchen. Each of them ate something different. It is known that:
- Leopold, Garfield, and Tom ate 3 different dishes;
- Vasiliy did not eat a cutlet, and Leopold did not eat a sausage;
- Garfield and Matil... | Answer: Garfield and Matilda - cutlets, Basil and Tom - sausages, Leopold - fish.
Solution. Garfield and Matilda ate the same thing, so they either ate a sausage each or a cutlet each.
Case 1. Garfield and Matilda ate a sausage each.
According to the condition, Basil did not eat a cutlet, and he also did not eat a s... | GarfieldMatilda-cutlets,BasilTom-sausages,Leopold-fish | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,374 |
Problem 4.5. Mom and Dad have two children: Kolya and Tanya. Dad is 4 years older than Mom. Kolya is also 4 years older than Tanya and half as old as Dad. How old is each of them if the total age of all family members is 130 years? | Answer: Tanya is 19 years old, Kolya is 23 years old, mom is 42 years old, and dad is 46 years old.
Solution. Let's mentally increase the age of mom and Tanya by 4 years. Then the age of two family members will be equal to dad's age, and another two will have half of dad's age. Then the total age of all family members... | Tanyais19old,Kolyais23old,momis42old,dadis46old | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,375 |
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 | 14,376 |
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 | 14,377 |
Problem 4.8. Vasya has six identical dice, on the faces of each of which the numbers from 1 to 6 are written (each one only once). Vasya rolled all six dice six times in a row. No number appeared twice on the same die.
It is known that the sum of the numbers on the top faces was 21 on the first roll, and 19, 20, 18, a... | Answer: 23.
Solution. Since in six throws, no number appeared twice on any of the dice, each die showed each of the numbers from 1 to 6 exactly once.
Let's calculate the total sum of all numbers that appeared on all dice over the six throws. For one die, this sum is $1+2+3+4+5+6=21$, and for six dice, it is $6 \cdot ... | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,378 |
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 | 14,381 |
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 | 14,382 |
Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "P", as shown in the image, with the number of points on any two touching faces being the same.
 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 | 14,386 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,388 |
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 | 14,392 |
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 | 14,393 |
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 | 14,395 |
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 | 14,396 |
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 | 14,397 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,398 |
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 | 14,399 |
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 | 14,400 |
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 | 14,401 |
8.2 The numbers a, b, c, and d are such that each subsequent number is 1 greater than the previous one. Can it be that $a^{2}+b^{2}=c^{2}+d^{2}$? | 8.2 yes.
For example, these numbers are $-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \frac{3}{2}$.
## | yes | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,402 |
8.4. In an acute-angled triangle $A B C$, altitudes AD, BE, and CF are drawn. It turns out that quadrilateral FBDE is a rhombus. Prove that triangle $A B C$ is equilateral. | 8.4. In an acute triangle \(ABC\), altitudes \(AD\), \(BE\), and \(CF\) are drawn. It turns out that quadrilateral \(FBDE\) is a rhombus. Prove that triangle \(ABC\) is equilateral.
The diagonals of a rhombus are perpendicular, so segment \(DF\), which is perpendicular to altitude \(BE\), will be parallel to side \(AC... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,404 |
8.5. From a square grid, a smaller square was cut out along the lines of the $1 \times 1$ cells. Could 250 cells remain after this? | # 8.5. no.
Let the original square be $n \times n$ cells, and the cut-out square be $m \times m$ cells. Then we get the equation $n^{2}-m^{2}=250$. The numbers $n+m$ and $n-m$ have the same parity, their product $(n+m)(n-m)=n^{2}-m^{2}=250$ is an even number. Therefore, they are both even numbers, but then their produ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,405 |
1. Arrange the numbers from 1 to 101 in a circle so that adjacent numbers differ by 2 or by 5. | Solution. For example, $13,11,9,4,2,7,5,3,1,6,8,10,12, \ldots, 92,94,96,101,99,97,95,100,98,93,91,89, \ldots, 15,13$
Criteria. If the reader has to guess (but can do so) where a certain number is located - no more than 5 points. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,406 |
3. In triangle $ABC$, a circle is inscribed, the points of tangency of the circle with the sides of the triangle are $M, N$, and $K$. The feet of the altitudes of triangle $MNK$ form triangle $PQR$. Prove that triangle $PQR$ is similar to triangle $ABC$. | Solution. Let $M \in B C, N \in C A, K \in A B$ and $P \in K N, Q \in N M, R \in M K$. From the theorem about the angle between a tangent and a chord, it follows that $\angle M N C=\angle M K N=\angle P Q N$. Therefore, $P Q \| A C$. Similarly, $P R \| A B$ and $Q R \| B C$. Therefore, triangles $A B C$ and $P Q R$ are... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,408 |
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 | 14,409 |
5. Given a segment $A B$. Inside it, there are several other smaller segments. These several smaller segments together cover the segment $A B$. Prove that if one half, either the left or the right, is removed from each smaller segment, the remaining halves will cover no less than $1 / 3$ of the length of $A B$. | Solution: Each remaining half of the segment $s$ will be "stretched" three times (homothety with a coefficient of 3 and the center at the midpoint of this half). The result of this homothety obviously contains the entire segment $s$. Therefore, the "triplings," taken together, certainly cover all the original segments,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,410 |
10.1 The equation $(x+a)(x+b)=9$ has a root $a+b$. Prove that then $a b \leqslant 1$. | Solution. According to the condition $(2 a+b)(2 b+a)=9$. Let's perform equivalent transformations:
$$
\begin{aligned}
& 2 a^{2}+2 b^{2}+5 a b=9 \\
& 2(a-b)^{2}+9 a b=9 \\
& a b=1-\frac{2}{9}(a-b)^{2}
\end{aligned}
$$
and the statement of the problem becomes obvious, since $\frac{2}{9}(a-b)^{2}-$ is a non-negative num... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,411 |
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 | 14,412 |
10.3 In quadrilateral $A B C D$, angles $A$ and $C$ are right angles. Perpendiculars from points $B$ and $D$ to diagonal $A C$ intersect at points $M$ and $N$, respectively. Prove that $A M=C N$. | Solution. Since angles $A$ and $C$ are right angles, the quadrilateral $A B C D$ is inscribed in a circle with diameter $B D$. Let point $O$ be its center. Draw a line through $O$ perpendicular to line $A C$, and let it intersect $A C$ at point $P$. Point $P$ is the midpoint of chord $A C$ (a radius perpendicular to a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,413 |
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 | 14,414 |
10.5 Vanya took three Unified State Exams (USE): in Russian, mathematics, and physics. He scored 5 points less in Russian than in physics, and in physics, he scored 9 points less than in mathematics. The golden fish that appeared in Vanya's dream promised to fulfill any number of wishes of the following types:
1) add ... | Solution. As a result of fulfilling each wish, the difference between the scores obtained in any two exams either remains unchanged or changes by 4. This means that this difference always retains the same remainder when divided by 4. The initial differences are 5, 9, and 14; they give remainders of 1, 1, and 2. Therefo... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,415 |
10.6 A certain tetrahedron $DABC$ was cut along the edges $DA, DB,$ and $DC$ and its surface was unfolded onto the plane of the base $ABC$. It turned out that the unfolding forms a triangle. Could this triangle have been a right triangle? Justify your answer. | Solution.
To the solution of problem 10.6
Suppose this is possible. Then the edges of the tetrahedron \(AB, BC\), and \(CA\) will be the midlines of the triangle obtained in the net. All fa... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,416 |
9.6. Given four consecutive natural numbers greater than 100. Prove that from these numbers, one can choose three numbers whose sum can be represented as the product of three different natural numbers greater than 1.
(N. Agakhanov) | Solution. Let $n, n+1, n+2, n+3$ be the given numbers. The sum of the three smallest of them is $3n+3=3(n+1)$, and the sum of the three largest numbers is $3(n+2)$. But at least one of the numbers $n+1$ and $n+2$ is even, that is, equal to the product of the numbers 2 and $k$, where $k>3$. Therefore, the given sum can ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,417 |
9.7. On a rectangular table, several cardboard rectangles are lying. Their sides are parallel to the sides of the table. The sizes of the rectangles can vary, they can overlap, but no two rectangles can have all four vertices in common. Can it happen that every point which is a vertex of a rectangle is a vertex of exac... | Answer. Yes, it can.
Solution. Fig. 1 shows how three pairs of rectangles can be placed so that for each pair all points $A$, $B, C, D, E, F, G, H$ are vertices exactly once. Identical points mark the vertices of one of the rectangles in the pair.
. Drop a perpendicular $D H$ from point $D$ to line $B C$. Since in triangle $B C D$ the angles at vertices $B$ and $C$ are acute, point $H$ lies o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,419 |
9.9. On a board, a convex $n$-gon ($n \geqslant 4$) is drawn. Each of its vertices must be painted either black or white. A diagonal is called bicolored if its ends are painted in different colors. A coloring of the vertices is called good if the $n$-gon can be divided into triangles by bicolored diagonals that do not ... | Answer: $n^{2}-n$.
Solution. Immediately note that coloring all vertices in one color is not a good coloring; such colorings are not considered in the further solution.
Let's call a side of the polygon bicolored if its ends are colored in different colors (i.e., extend the definition of bicolored to sides). We will c... | n^2-n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,420 |
9.10. Petya and Vasya are playing the following game. Petya chooses 100 (not necessarily distinct) non-negative numbers $x_{1}, x_{2}, \ldots$, $x_{100}$, the sum of which is 1. Vasya then pairs them up as he sees fit, calculates the product of the numbers in each pair, and writes the largest of the 50 obtained product... | Answer: $\frac{1}{396}$.
Solution: If Petya chooses the numbers $\frac{1}{2}, \frac{1}{198}, \frac{1}{198}, \ldots, \frac{1}{198}$, then no matter how Vasya pairs these numbers, the number $\frac{1}{198}$ will be paired with $\frac{1}{2}$. Their product will be $\frac{1}{396}$, and the other products will not exceed i... | \frac{1}{396} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,421 |
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 | 14,422 |
2. Can a table of size $n \times n$ be filled with numbers $-1,0,1$ so that the sums in all rows, all columns, and on the main diagonals are different? The main diagonals of the table are the diagonals drawn from the top left corner of the table to the bottom right and from the top right corner of the table to the bott... | Solution. A total of $2 n+2$ sums are considered. Note that each of these sums is an integer, the largest possible sum is $n=1+1+1+\ldots+1$ ( $n$ ones), and the smallest sum is $-n=-1+(-1)+\ldots+(-1)$ ( $n$ minus ones). On the segment $[-n ; n]$, there are exactly $2 n+1$ integers. Then, by the Pigeonhole Principle, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,423 |
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?
. Prove that the equation $a x^{2}+2 \sqrt{2} b x+c=0$ has two roots. | Solution. By the condition $b=a q, c=a q^{2}(a \neq 0$ and $q \neq 0)$, and the original equation will take the form
$$
a x^{2}+2 \sqrt{2} a q x+a q^{2}=0 \text { or } x^{2}+2 \sqrt{2} q x+q^{2}=0
$$
- the discriminant of the last equation $D=8 q^{2}-4 q^{2}=4 q^{2}>0$, so the equation has two distinct roots. | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,427 |
10.2. In a row, 21 numbers are written sequentially: from 1999 to 2019 inclusive. Enthusiastic numerologists Vova and Dima performed the following ritual: first, Vova erased several consecutive numbers, then Dima erased several consecutive numbers, and finally, Vova erased several consecutive numbers (at each step, the... | Solution. Let $n$ be the sum of the numbers erased by Dima, then $4n$ is the sum of the numbers erased by Vova, $5n$ is the sum of the numbers erased by both, and $42189 - 5n$ is the number that remains unerased (42189 is the sum of all numbers from 1999 to 2019). According to the problem, $1999 \leq 42189 - 5n \leq 20... | 1999or2019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,428 |
10.4. All angles of the pentagon $A B C D E$ are equal. Prove that the perpendicular bisectors of segments $A B$ and $C D$ intersect on the bisector of angle $E$. | Solution. $A B C D E$ is a pentagon with equal angles, $O M, O N$ are the perpendicular bisectors of $A B$ and $C D$ (with $M, N$ being their feet, and $O$ being their intersection point). Extend $E A$ and $C B$ to intersect at point $F$. Extend $E D$ and $B C$ to intersect at point $G$. The sum of the angles of $A B C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,430 |
10.5. Rational numbers $a$ and $b$ satisfy the equation $a^{3} b+a b^{3}+2 a^{2} b^{2}+2 a+2 b+1=0$. Prove that $\sqrt{1-a b}$ is a rational number. | Solution. $a^{3} b+a b^{3}+2 a^{2} b^{2}+2 a+2 b+1=(a b-1)(a+b)^{2}+(a+b+1)^{2}$, from which it follows that $\sqrt{1-a b}=\left|\frac{a+b+1}{a+b}\right|$: the sum, quotient, and absolute values of rationals are rational. | \sqrt{1-}=|\frac{+b+1}{+b}| | Algebra | proof | Yes | Yes | olympiads | false | 14,431 |
10.1. Find the sum $\sin x + \sin y + \sin z$, given that $\sin x = \tan y$, $\sin y = \tan z$, $\sin z = \tan x$. | Answer: 0.
First solution. From $\sin x = \tan y$, we get $\sin x \cos y = \sin y$. Therefore, $|\sin x| \cdot |\cos y| = |\sin y|$. This means $|\sin x| \geq |\sin y|$, and the inequality becomes an equality only if either $\sin y = \sin x = 0$, or $|\cos y| = 1$ (which again implies $\sin y = \sin x = 0$). Similarly... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,432 |
9.1. There are 2019 vases arranged in a circle. Each vase contains white and red roses. Masha wants to move one flower from each vase to the next one in a clockwise direction so that the number of both white and red roses in each vase differs from the original. Can she do this? | Answer. No.
Solution. Let's say Masha moved a white rose from vase $V_{i}$ to the next vase $V_{i+1}$, then a red rose must be moved from the previous vase $V_{i-1}$ (considering $V_{0}=V_{2020}$) to vase $V_{i}$, otherwise the number of red roses in it would not change. Similarly, if Masha moved a red rose from a vas... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,435 |
9.3. Will the equation $x^{2019}+2 x^{2018}+3 x^{2017}+\cdots+2019 x+2020=0$ have integer roots | Answer: No.
Solution. If there is an integer root $a$, then $ab^{2019}, 4 b^{2016}>3 b^{2017}, \ldots, 2020>2019 b$.
If $b \geq 2$, then $b^{2019} \geq 2 b^{2018}, 3 b^{2017}>4 b^{2016}, \ldots, 2019 b>2020$.
Therefore, there are no integer roots. | No | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,436 |
9.4. $M M_{1}$ and $P P_{1}$ are the angle bisectors of triangle $M N P$. The lengths of the perpendiculars dropped from vertex $N$ to the lines $M M_{1}$ and $P P_{1}$ are equal. Prove that triangle $MNP$ is isosceles. | Solution. 1st method. Let $N D$ and $N E$ be the perpendiculars dropped from vertex $N$ to the lines $M M_{1}$ and $P P_{1}$. Extend the perpendiculars $N E$ and $N D$ until they intersect the line $M P$ (the points of intersection are $T$ and $S$, respectively). Triangles $NPT$ and $NMS$ are isosceles (the bisectors $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,437 |
9.5. From the natural numbers from 1 to 1239, 384 different numbers were selected such that the difference between any two of them is not equal to 4, 5, or 9. Is the number $625$ selected? | Answer. Yes.
Solution.
Lemma. Among any 13 consecutive natural numbers, one can choose no more than four such that no two of them differ by 4, 5, or 9.
Proof of the lemma. Divide the 13 numbers $a, a+1, \ldots, a+12$ into 9 groups (of one or two numbers) and write the groups in a circle in the following order: $\{a+... | Yes | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,438 |
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 | 14,439 |
4. (7 points) Six mathematicians went fishing. Together they caught 100 fish, and each caught a different number. After fishing, they noticed that any of them could distribute all their fish to the other five fishermen so that the remaining five would have an equal number of fish. Prove that one fisherman can go home w... | Solution. After one fisherman distributes his fish, the others should each have $100: 5=20$ fish. This means that each of them caught no more than 20 fish.
Suppose, for example, that Fisherman Pyotr Petrovich has exactly 20 fish. When another mathematician distributes his fish, Pyotr Petrovich does not receive anythin... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 14,442 |
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 | 14,443 |
1. Is it possible to place six different integers at the vertices and midpoints of the sides of a triangle so that each number at a vertex is equal to the sum of the numbers at the other two vertices and the number at the midpoint of the opposite side? | Answer: Yes, it is possible.
Solution: For example: A(0), B(3), C(1), $\mathrm{A}_{1}(-4), \mathrm{B}_{1}(2), \mathrm{C}_{1}(-2)$. Here $\mathrm{A}_{1}$ is the midpoint of side $\mathrm{BC}, \mathrm{B}_{1}$ is the midpoint of side $\mathrm{AC}, \mathrm{C}_{1}$ is the midpoint of side AB.
Grading Criteria: Correct exa... | Yes,itispossible | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,444 |
2. Dima, Sasha, Kolya, and Gleb participated in an olympiad and took the first four places. A year later, their classmates managed to recall only three facts: "Dima took first place or Gleb took third," "Gleb took second place or Kolya took first," "Sasha took third place or Dima took second." Who performed better - Di... | Answer: Dima.
Solution. If Dima took first place, then from the last statement it follows that Sasha took third place, which means Dima performed better. If it is not true that Dima is in first place, then Gleg must be in third place. Then from the second statement, we get that Kolya took first place. Sasha could not ... | Dima | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,445 |
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 | 14,446 |
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 | 14,447 |
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 | 14,448 |
8.2. There are 100 natural numbers. If any one of them is removed, the sum of the remaining 99 will be even. Prove that all the numbers are even. | 8.2. Suppose that among the numbers there is an odd one. All numbers cannot be odd, since the sum of 99 odd numbers is odd. Therefore, among the numbers there is an even one. But when removing an even and an odd number, the sum of the remaining numbers will have a different parity, and all sums cannot be even. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,450 |
8.3. A domino tile covers exactly two squares of a chessboard. Is it possible to lay out a set of dominoes (28 pieces) on this board so that none of the tiles can be moved from their position in the plane of the board? (moving off the edge of the board is not allowed). | 8.3. Answer: it is possible.
An example of such an arrangement is shown in the figure. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,451 |
8.4. In the cells of a $15 \times 15$ table, the numbers $1, 2, \ldots, 15$ are written. In each row and each column, each of these numbers appears exactly once. In cells symmetric with respect to the main diagonal of the table (running from the top left to the bottom right), the same numbers are written. Prove that al... | 8.4. The numbers located outside the main diagonal,
are paired into identical numbers located in symmetric cells. This means that outside the main diagonal, each number appears an even numbe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,452 |
8.5. In triangle $A B C$, the bisector $B L$ is drawn, and a point $K$ is chosen on its extension beyond point $L$ such that $L K=A B$. It turns out that $A K \| B C$. Prove that $A B>B C$. | 8.5. Let $\angle A B C=2 x$. Then $\angle A B L=\angle C B L=x$, and $\angle C B L=\angle A K B=x$ (as alternate interior angles when $A K$ and $B C$ are parallel and $B K$ is the transversal). Therefore, triangle $B A K$ is isosceles, hence $A K=A B$. Since by the condition $L K=A B$, triangle $A K L$ is isosceles, fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,453 |
1. One day, Alice found herself in a magical forest where two amazing twins, Tra-la-la and Tru-la-la, lived. Tra-la-la always lied on Mondays, Tuesdays, and Wednesdays, but told the truth all other days of the week. Tru-la-la always lied on Thursdays, Fridays, and Saturdays, but told the truth all other days of the wee... | Answer: only on Wednesday.
Solution. The statement "I lie on Sundays" is false for either of the brothers, so the first one was definitely lying. This means his first statement is also false. Thus, the first brother does not lie on Saturdays, so the first one is Tralala. Then the second one is Trulala, and on the day ... | Wednesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,454 |
2. Does there exist an integer $\mathrm{n}$ such that $\mathrm{n}^{2022}-2 \cdot \mathrm{n}^{2021}+3 \cdot \mathrm{n}^{2019}=2020$? | Answer: No.
Solution 1. Consider the possible remainders of $n$ when divided by 3. If $n$ is divisible by 3 (remainder 0), then $(n^{2022}-2 \cdot n^{2021}+3 \cdot n^{2019})$ is also divisible by 3 (i.e., the remainder is 0). If $n$ gives a remainder of 1 when divided by 3, then $n^{2022}, n^{2021}$, and $n^{2019}$ al... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,455 |
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 | 14,456 |
4. In an acute-angled triangle $\mathrm{ABC}$, the altitude $\mathrm{CH}$ was drawn, and it turned out that $\mathrm{CH} = \mathrm{AB}$. Inside triangle $\mathrm{CHB}$, there is a point $\mathrm{T}$ such that triangle $\mathrm{BTH}$ is a right isosceles triangle with hypotenuse $\mathrm{HB}$. Prove that angle $\mathrm{... | Solution. Since triangle BTH is a right isosceles triangle, then $\mathrm{TH}=\mathrm{TB}$ and $\angle \mathrm{THB}=\angle \mathrm{TBH}=45^{\circ}$. Therefore, $\angle \mathrm{CHT}=90^{\circ}-\angle \mathrm{THB}=45^{\circ}$. Triangles ABT and CHT are equal by two sides and the angle between them ( $\mathrm{AB}=\mathrm{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,457 |
5. From the natural numbers from 1 to 99, 50 different numbers were chosen such that no two of them add up to 100 and no two of them add up to 99. Which numbers were chosen? | Answer: numbers from 50 to 99.
Solution. Let's divide the numbers into 49 pairs $(1,99),(2,98), \ldots,(49,51)$, and the number 50 will remain unpaired. In each pair, the sum of the numbers is 100, which means no more than 1 number is chosen from each pair. There are 49 pairs, and a total of 50 numbers are chosen, so ... | \from\50\to\99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,458 |
10.2. Plot on the coordinate plane the set of points whose coordinates satisfy the equation $4 x^{2} y^{2}=4 x y+3$. | Solution. The given equation can be reduced to an equivalent one: $(2 x y-1)^{2}=4 \Leftrightarrow 2 x y=1 \pm 2$. Therefore, the set of points in question consists of two hyperbolas $y=\frac{3 / 2}{x}$ and $y=\frac{-1 / 2}{x}$ (each quadrant will have one branch of the corresponding hyperbola). | \frac{3/2}{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,460 |
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 | 14,462 |
10.5. Given a triangle $A B C$, where $\angle A=40^{\circ}, \angle C=20^{\circ}$. Prove that the length of the angle bisector drawn from vertex $B$ is $A C-B C$. | Solution. Let $B M$ be the bisector of angle $B$. Extend $\triangle A B C$ to an isosceles triangle $A D C$ by extending segment $C B$ beyond point $B$ such that $C D = C A$ (obviously, $C B < C A$, since angle $B$ is greater than angle $A$). We have: $\angle D B A = 20^{\circ} + 40^{\circ} = 60^{\circ}$ by the propert... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,463 |
1. Five consecutive natural numbers are written in a row. Is it possible that the sum of the digits of the first number is 52, and that of the fifth number is 20? | Answer: possibly.
Solution. For example, the numbers that fit are: 889999, 890000, 890001, 890002, 890003. | 889999,890000,890001,890002,890003 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,464 |
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 | 14,465 |
4. $A L$ is the bisector of triangle $A B C, K$ is a point on side $A C$ such that $C K=C L$.
Line $K L$ and the bisector of angle $B$ intersect at point $P$. Prove that $A P=P L$. | Solution. Let $\angle B A C=2 a, \angle A B C=2 b$, then $\angle B C A=180^{\circ}-(2 a+2 b)$.
In the isosceles triangle $L C K: \angle C L K=\angle C K L=\left(180^{\circ}-\angle B C A\right) / 2=a+b$. $\angle L K C=\angle L A K+\angle A L K$, as the exterior angle in triangle $A L K$, from which $\angle A L K=b$.
T... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,466 |
5. In a chess tournament, 20 chess players participated. Each pair of participants played exactly one game against each other. A win was awarded 1 point, a draw was awarded $1 / 2$ point, and a loss was awarded 0 points. In the end, all chess players scored a different number of points. Prove that there is a participan... | Solution.
Assume that the tournament winner won fewer than 10 games. Then the maximum number of points he could have scored is $9 \cdot 1 + 10 \cdot 0.5 = 14$ (a maximum of 9 wins and 10 draws). Then the runner-up could have scored a maximum of 13.5 points, the third place 13 points, ..., the 19th place 5 points, and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,467 |
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 | 14,470 |
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 | 14,471 |
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 | 14,472 |
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 | 14,474 |
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 | 14,475 |
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 | 14,477 |
3. There are 9 coins of different weights. Detector L5, in one operation, examines 5 coins and indicates the lightest of them. How can you find the four heaviest coins in five operations? | Solution. After each operation, we set aside the light coin found during this operation and do not use it again. In the end, we will have 5 coins that were light. The remaining 4 are the ones we are looking for, as none of them can be the result of an operation.
## Evaluation Criteria.
The correct method is indicated... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,478 |
4. Vasya must plant 6 trees in a row. Each planted tree is either a birch or an oak. The planting must satisfy the condition: next to each tree of one type, there must be a tree of the other type. In how many different ways can Vasya perform the planting? Two ways are different if there is a spot where different types ... | Solution. Note that by changing the type of each planted tree to another, we get another way. We will call two such ways complementary. Thus, it is enough to count the number of such pairs. Which tree to plant at the edge is completely determined by its neighbor. Therefore, it is enough to determine the number of ways ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,479 |
5. Darья Ivanovna and Mariya Petrovna bought identical boxes of tea bags. It is known that one tea bag is enough for 3 cups or 4 cups of tea. Darья Ivanovna drank 74 cups of tea, and Mariya
Petrovna 105 cups of tea. Prove that they either made a mistake in counting the cups of tea or the number of tea bags in the boxes... | Solution. Since $105-74=26$, Maria Petrovna drank 26 more cups than Daria Ivanovna, and since $4-3=1$, she used 26 more tea bags than Daria Ivanovna. Therefore, there must be at least 26 tea bags in the box, but then Daria Ivanovna should have drunk at least $3 \times 26=78$ cups, which is more than the 74 she drank, w... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,480 |
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 | 14,481 |
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 | 14,483 |
5. Find all pairs of natural numbers $n$ and $m$, for which
$$
(n+1)(2n+1)=2m^2
$$ | Answer: there are none.
Solution. Since $2 m^{2}=(n+1)(2 n+1)>2 n^{2}, m>n$. On the other hand, $2(n+1)^{2}>(n+1)(2 n+1)=2 m^{2}$ and $m<n+1$. It turns out that $n<m<n+1$. This is impossible.
Remark. Other solutions are possible. For example, if we consider the equation as a quadratic in $n$, we get the discriminant ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,484 |
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 | 14,485 |
9.2. For what values of $q$ does the quadratic equation $x^{2}-12 x+q=0$ have two distinct roots, one of which is the square of the other? | Answer: -64 or 27.
Solution. Let the roots of the equation be $a$ and $a^{2}$. By Vieta's theorem, $a + a^{2} = 12$ and $a \cdot a^{2} = q$. Solving the quadratic equation for $a$, we find that $a = -4$ or $a = 3$. Since $q = a^{3}$, then $q = -64$ or $q = 27$.
## Criteria.
7 points. Correct solution with the correc... | -64or27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,486 |
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