problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
Problem 6.5. On a line, 5 points $P, Q, R, S, T$ are marked, exactly in that order. It is known that the sum of the distances from $P$ to the other 4 points is 67, and the sum of the distances from $Q$ to the other 4 points is 34. Find the length of the segment $P Q$. | Answer: 11.
Solution. From the condition of the problem, it is known that
$$
P Q+P R+P S+P T=67 \quad \text { and } \quad Q P+Q R+Q S+Q T=34
$$
Let's find the difference of these quantities:
$$
\begin{aligned}
33 & =67-34=(P Q+P R+P S+P T)-(Q P+Q R+Q S+Q T)= \\
& =(P Q-Q P)+(P R-Q R)+(P S-Q S)+(P T-Q T)=0+P Q+P Q+P... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.7. The Amur and Bengal tigers started running in a circle at 12:00, each at their own constant speed. By 14:00, the Amur tiger had run 6 more laps than the Bengal tiger. Then the Amur tiger increased its speed by 10 km/h, and by 15:00, it had run a total of 17 more laps than the Bengal tiger. How many meters ... | Answer: 1250.
Solution. In the first 2 hours, the Amur tiger ran 6 more laps, i.e., in 1 hour, it ran 3 more laps. If it had not increased its speed, in the first 3 hours, it would have run 9 more laps. However, the increase in speed resulted in an additional $17-9=8$ laps in the third hour. Since it increased its spe... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Uncle bought a New Year's gift for each of his nephews, consisting of a candy, an orange, a pastry, a chocolate bar, and a book. If he had bought only candies with the same amount of money, he would have bought 224. He could have bought 112 oranges, 56 pastries, 32 chocolate bars, and 16 books with the same amount o... | 2. Answer. 8. Solution. Let's express the prices of all items in terms of the price of a candy. An orange costs as much as two candies, a pastry - as 4 candies, a chocolate bar - as 224:32 = 7 candies, a book - as 14 candies. The total price of the gift is equal to the price of $1+2+4+7+14=28$ candies, and the number o... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Two brothers sold a flock of sheep, receiving as many rubles for each sheep as there were sheep in the flock. Wishing to divide the proceeds equally, they began to take 10 rubles from the total sum in turns, starting with the older brother. After the older brother took 10 rubles again, the younger brother was left w... | 6. Answer: 2 rubles. Solution. Let there be $n$ sheep in the flock. Then the brothers earned $n^{2}$ rubles. From the condition, it follows that the number of tens in the number $n^{2}$ is odd. Represent the number $n$ as $10 k+m$, where $k-$ is the number of tens, and $m-$ is the number of units in it. Then $n^{2}=100... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2 There were 25 sparrows sitting on two bushes. After 5 sparrows flew from the first bush to the second, and 7 sparrows flew away from the second bush, it turned out that there were twice as many sparrows left on the first bush as on the second. How many sparrows were there on each bush initially? | 7.2 On the first bush, there were 17 sparrows, and on the second, there were 8.
After 7 sparrows flew away, 18 remained. At this point, there were twice as many sparrows on the first bush as on the second. This means there were 12 sparrows on the first bush and 6 on the second. If we return 7 sparrows to the second bu... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. What angle do the hour and minute hands form at twenty minutes past one? (Don't forget to justify your answer). | 7.4 $110^{\circ}$.
At 12:00, the angle between the hour and minute hands is $0^{\circ}$. The minute hand completes a full circle of $360^{\circ}$ in 60 minutes, which is $6^{\circ}$ per minute, and in 20 minutes it will cover $120^{\circ}$. The hour hand moves 12 times slower than the minute hand. Therefore, in 20 min... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.5. Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself (do not forget to justify your answer). | 7.5. The number 198 is unique.
From the condition, we get the relation $11 \cdot(a+b+c)=100 a+10 b+c$, or $10 c+b=89 a$. In this relation, the left side is a number less than 100. If $a$ is greater than 1, then the right side will be a number greater than 100. Therefore, $a=1$, $c=8$, $b=9$. All-Russian School Olympia... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.7. A green chameleon always tells the truth, while a brown chameleon lies and immediately turns green after lying. In a company of 2019 chameleons (green and brown), each in turn answered the question of how many of them are currently green. The answers were the numbers $1,2,3, \ldots, 2019$ (in some order, not neces... | Answer: 1010.
Solution. Consider two chameleons who spoke in a row. One of them was brown at the moment of speaking; indeed, if both were green, the number of green chameleons would not have changed after the first one spoke, and the second one would have named the same number as the first. We can divide all the chame... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.8. In an acute-angled triangle $A B C$, the bisector $B L$ is drawn. The circumcircle of triangle $A B L$ intersects side $B C$ at point $D$. It turns out that point $S$, symmetric to point $C$ with respect to the line $D L$, lies on side $A B$ and does not coincide with its endpoints. What values can $\angle A B C$ ... | Answer: $60^{\circ}$.
First solution. From the symmetry, triangles $C L D$ and $S L D$ are equal, so $D S = D C$, $\angle C D L = \angle S D L$, and $\angle D L C = \angle D L S$. Since the quadrilateral $A L D B$ is inscribed in a circle, we have $\angle B A L = \angle L D C$ (see Fig. 1). The chords $A L$ and $D L$ ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1.
Tourists Vitya and Pasha are walking from city A to city B at equal speeds, while tourists Katya and Masha are walking from city B to city A at equal speeds. Vitya met Masha at 12:00, Pasha met Masha at 15:00, and Vitya met Katya at 14:00. How many hours after noon did Pasha meet Katya? | Answer: 5.
Solution: The distance between Masha and Katya and their speeds do not change, and the speeds of Vitya and Pasha are equal. Vitya met Katya 2 hours after Masha, so Pasha will also meet Katya 2 hours after Masha, i.e., at 5:00 PM - 5 hours after noon. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 4. Variant 1
If from the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ we subtract the discriminant of the quadratic polynomial $g(x)=$ $(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$. | Answer: 6.
Solution. We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=4 f(-2)$.
## Variant 2
If the discriminant of the quadratic trinomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic trinomial $g(x)=$ $(a+1) x^{2}+2(b+2) x+c+4$, the result is 28. Find ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 16. What is the greatest value that the largest number on the board can take | Answer: 108.
Solution: The sum of the given numbers is $9 \cdot 16=144$. Since all the numbers are distinct, the sum of the 8 smallest of them is no less than $1+2+\cdots+8=36$. Therefore, the largest number cannot be greater than $144-36=108$. This is possible: $(1+2+\cdots+8+108): 9=16$.
Variant 2 Petya wrote 9 dif... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7. Option 1.
Unit cubes were used to assemble a large parallelepiped with sides greater than 3. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 429. Find the number of cubes that have exactly 4 ... | Answer: 108.
Solution: Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$, and $c-2$ is greater than 1 and their product equals the product of the three prime numbers 3, 11, and 13, t... | 108 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1. Point $I$ is the center of the circle inscribed in triangle $A B C$ with sides $B C=6$, $C A=8, A B=10$. Line $B I$ intersects side $A C$ at point $K$. Let $K H$ be the perpendicular dropped from point $K$ to side $A B$. Find the distance from point $I$ to line $K H$. | Answer: 2.
Solution.
By the converse of the Pythagorean theorem, angle $C$ is a right angle. Then, triangles $B K C$ and $B K H$ are congruent by the hypotenuse and an acute angle. Therefore... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 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 $1 \... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Find all positive solutions of the equation
$$
x^{101} + 100^{99} = x^{99} + 100^{101}
$$
Do not forget to justify your answer. | Solution. The equation is reduced to the form
$$
x^{99}\left(x^{2}-1\right)=100^{99}\left(100^{2}-1\right)
$$
Obviously, the solution is $x=100$. For $0<x<1$, the left side is less than the right side. For $x>1$, the left side is an increasing function, so the required value is taken no more than once. Therefore, the... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1 The sum of 100 numbers is 1000. The largest of these numbers was doubled, and some other number was decreased by 10. After these actions, the sum of all the numbers did not change. Find the smallest of the original numbers. | 9.1 The sum of 100 numbers is 1000. The largest of these numbers was doubled, and some other number was decreased by 10. After these actions, the sum of all the numbers did not change. Find the smallest of the original numbers.
Otvet: 10.
$\boldsymbol{P e s h e n i e : ~ P u s t ь ~} M$ - the largest number, $t$ - so... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be? | 9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be equal to?
Answer: 0.
Solution: Let $\left(x_{1}, y_{1}\right)$ be the coordinates of the vertex of one para... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the box, there are balls of seven colors. One tenth of the balls are red, one eighth are orange, and one third are yellow. There are 9 more green balls than red ones, and 10 more blue balls than orange ones. There are 8 blue balls in the box. The remaining balls are purple. What is the smallest possible number of... | Answer: 25 balls.
Solution. Let the total number of balls be $x$, and the number of violet balls be $y$. Then
$$
\frac{x}{10}+\frac{x}{8}+\frac{x}{3}+\frac{x}{10}+9+\frac{x}{8}+10+8+y=x
$$
from which $\frac{47 x}{60}+27+y=x$, that is, $y=\frac{13 x}{60}-27$, and the smallest value of $y$ is achieved at the smallest ... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Vanya and Petya decided to mow the lawn for a football field. Vanya could do it alone in 5 hours, and Petya in 6 hours. They started at 11 o'clock, and stopped mowing simultaneously when their parents called them, but Petya took a one-hour lunch break, and Vanya's lunch break lasted two hours. One tenth of the lawn ... | Answer: At 15 o'clock.
Solution: Let $t$ be the time from 11 o'clock to the end, then Petya worked $(t-1)$ hours, and Vanya worked $(t-2)$ hours. In one hour, Petya completed $\frac{1}{6}$ of the work, and Vanya completed $\frac{1}{5}$ of the work. We get the equation: $\frac{1}{6} \cdot(t-1)+\frac{1}{5} \cdot (t-2)=\... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Masha and Alina are playing on a $5 \times 5$ board. Masha can place one chip in some cells. After that, Alina covers all these cells with L-shaped pieces consisting of three cells (non-overlapping and not extending beyond the boundaries of the square, L-shaped pieces can only be placed along the grid lines). If Ali... | Answer: 9 chips.
Solution. Example. Masha can place chips in the cells indicated in the figure (a). Then, Alina will need nine corners, as one corner cannot cover more than one cell with a chip. However, nine corners without overlapping cannot be placed on the board, since $27>25$.
Estimate. If Masha places fewer tha... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A company produces a lemon drink by diluting lemon juice with water. Initially, the company produced a drink containing $15\%$ lemon juice. After some time, the CEO ordered to reduce the lemon juice content to $10\%$. By what percentage will the amount of lemon drink produced increase with the same volumes of lemon ... | Answer: By $50 \%$.
Solution. Method 1. The content of lemon juice in the drink after the CEO's directive decreased by one and a half times. This means that from the same lemons, one and a half times more lemon drink can be made. In other words, the amount of lemon drink produced will increase by one and a half times ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. All natural numbers, the sum of the digits in the representation of which is divisible by 5, are listed in ascending order: $5,14,19,23,28,32, \ldots$ What is the smallest positive difference between consecutive numbers in this sequence? Provide an example and explain why it cannot be smaller.
---
The smallest pos... | Answer. The smallest difference is 1, for example, between the numbers 49999 and 50000.
Solution. The difference cannot be less than 1, as we are talking about the difference between different natural numbers.
Comment. How to guess the solution.
It is clear that if two adjacent numbers differ only in the units place... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On a standard graph paper, an angle is drawn (see figure). Find its measure without using measuring instruments. Justify your answer. | Answer: $45^{\circ}$.
Solution. Connect the two "extreme" points with a segment (as shown in the figure). The resulting triangle is isosceles because two of its sides, $A B$ and $B C$, are diagonals of three-cell rectangles. Diagonal $A B$ divides the angle of the rectangle at vertex $B$ into two angles that complemen... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task № 2
The captain's assistant, observing the loading of the ship, smoked one pipe after another from the start of the loading. When $2 / 3$ of the number of loaded containers became equal to $4/9$ of the number of unloaded containers, and the ship's bells struck noon, the old seafarer began to light another pipe.... | Answer: The assistant smoked 5 pipes.
## Solution.
Let $x$ be the part of containers that were loaded by noon, and $y$ be the remaining part of containers. Then from the conditions we get: $\left\{\begin{array}{l}\frac{2}{3} x=\frac{4}{9} y \\ x+y=1\end{array} \Rightarrow x=\frac{2}{5}, \quad y=\frac{3}{5}\right.$.
... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4
Find the area of the figure defined by the inequality
$$
|x|+|y|+|x-y| \leq \mathbf{2}
$$ | Answer: The area of the figure is 3 sq.units.
## Solution.
It is easy to see that if the point $(\boldsymbol{x}, \boldsymbol{y})$ satisfies the original inequality $|\boldsymbol{x}|+|\boldsymbol{y}|+|\boldsymbol{x} \boldsymbol{y}| \leq 2$, then the point $(-x,-\boldsymbol{y})$ will also satisfy this inequality, since... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) In a bag, there are 15 balls (see the figure). Color each ball in one of three colors: blue, green, or red - so that two of the statements are true, and one is false:
- there is one more blue ball than red balls;
- there are an equal number of red and green balls;
- there are 5 more blue balls than green... | # Answer.
7 blue balls, 6 red balls, 2 green balls.
Solution. We will prove that the second statement cannot be true. Indeed, if the first and second statements are true, then if we remove ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Four girls are singing songs, accompanying each other. Each time, one of them plays the piano while the other three sing. In the evening, they counted that Anya sang 8 songs, Tanya - 6 songs, Olya - 3 songs, and Katya - 7 songs. How many times did Tanya accompany? Justify your answer. | Answer. Twice.
Solution. If we add up the specified number of songs sung, each song will be counted 3 times (from the perspective of each of the three singing girls). Thus, we can find out how many songs were sung in total: $(8+6+3+7): 3=8$. It is known that Tanya sang 6 out of 8 songs, so she accompanied $8-6=2$ time... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Numbers $1,2,3,4,5,6,7$ and 8 are placed at the vertices of a cube, such that the sum of any three numbers belonging to any face is not less than 10. Find the minimum possible sum of four numbers belonging to one face. | Answer: 16. Evaluation. Consider an arbitrary face. If the largest number written at the vertex of this face is no more than 5, then the sum of the remaining numbers is no more than $4+3+2=9$. Therefore, the largest number written at the vertex of any face is at least 6, and the minimum possible sum of four numbers bel... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a two-digit number, where the first digit is greater than the second. Its digits were rearranged, and a new two-digit number was obtained. The difference between the first and the second number turned out to be a perfect square. How many possible two-digit numbers satisfy this condition? | 1. Answer: 13 numbers.
Let the first number be $10a + b$. Then the second number is $10b + a$, where $a > b > 0$. The difference is $9(a - b)$, which implies that the difference $a - b$ is a perfect square. The difference can be 1 or 4. For the first case, we have 8 values for the larger number $(21, 32, \ldots, 98)$,... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Between the digits of the number 987654321, signs + should be placed so that the sum is 99. In how many ways can this be achieved? | 3. Answer: in two ways.
It is clear that there cannot be three-digit addends, and there must be at least one two-digit addend, since the sum of all digits is 45. Let there be one two-digit addend, and we group the digits \(a+1\) and \(a\). Then the two-digit addend is \(10(a+1) + a = 11a + 10\), and the sum of the rem... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. On a plane, there are 10 different points. We considered the midpoints of all segments connecting all pairs of points. What is the smallest number of midpoints that could have arisen? | 4. Answer: 17 midpoints.
Consider the case when the points lie on a single line. Introduce coordinates on it, and let the points be $a_{1}<\cdots<a_{10}$. Consider the midpoints of segments involving $a_{1}$. There are nine of them, and they form an increasing sequence $\frac{a_{1}+a_{2}}{2}<\frac{a_{1}+a_{3}}{2}<\cdo... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There is a $10 \times 10$ grid for playing "Battleship". In this case, the game follows unusual rules, and only ships of size $1 \times 2$ are allowed, which can be placed both horizontally and vertically. What is the maximum number of these ships that can be placed on such a field, if the ships must not extend beyo... | 5. Answer: 13.
One of the examples of placing 13 ships is shown in the figure:
| | | | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | |
| | | | | | | | | | |... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Let $n$ - be a natural number. What digit stands immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$? | Answer: 4.
Solution. We will prove that $(n+0.4)^{2} < 0.8$. The last inequality is true for any natural $n$.
2) $n^{2}+n < (n+0.5)^{2} \Leftrightarrow n < n+0.25$, which is obvious.
Evaluation Criteria.
“+” A complete and justified solution is provided
“Ғ” The correct answer is provided, but only one of the two re... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. The tsar has eight sons, and they are all fools. Every night the tsar sends three of them to guard the golden apples from the firebird. The tsareviches cannot catch the firebird, blame each other for it, and therefore no two of them agree to go on guard duty a second time. For how many nights can this continue at... | Answer: 8 nights.
Solution. Evaluation. Consider any of the sons. In each guard, he is with two brothers, so after his three appearances, there will be one brother with whom he has not been on guard and will not be able to go, as there will be no third for them. This situation is "symmetric," meaning if son A has not ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Arkady, Boris, and Vasili decided to run the same distance, consisting of several laps. Arkady ran each lap 2 minutes faster than Boris, and Boris ran 3 minutes faster than Vasili, and all of them ran at a constant speed. When Arkady finished the distance, Boris had one lap left to run, and Vasili had two laps le... | Answer: 6.
Solution: Let the distance be $n$ laps, and Arkady takes $t$ minutes per lap. Then in $n t$ minutes, Arkady runs the entire distance. In this time, Boris runs one lap less, and Vasily runs two laps less. Therefore, $n t=(n-1)(t+2)$ and $n t=(n-2)(t+5)$. From this, it follows that $2 n=t+2$ and $5 n=2 t+10$,... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. On each of ten cards, a real number is written. For each non-empty set of these cards, the sum of all numbers written on the cards in this set was found. It is known that not all of the obtained sums are integers. What is the maximum possible number of integer sums that could have resulted? | Answer: 511.
Solution: It is clear that at least one of the cards has a non-integer number, otherwise the sum of the numbers in any set would be an integer. We select one of these cards and call it $a$, and the number on it $-x$. For each set $C$ that does not contain $a$ (including the empty set), we associate the se... | 511 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. In the store, Sasha bought two pens, three notebooks, and one pencil and paid 33 rubles, Dima bought one pen, one notebook, and two pencils and paid 20 rubles. How much did Tanya pay for four pens, five notebooks, and five pencils? | Answer: 73 rubles.
Let $a, b, c$ be the price of a pen, a notebook, and a pencil, respectively. Then, according to the condition, we have $2a + 3b + c = 33$ and $a + b + 2c = 20$. Multiplying the second relation by 2 and adding it to the first equation, we get the answer. | 73 | Algebra | 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 numb... | 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 $1 \t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
 cells are written down in a notebook. What is the smallest number of different numbers that could have been written in the notebook? | # Solution:
Consider the number in the central cell of the table. Next to it are 4 different neighbors - they give 4 different sums with the central number, so there are already at least 4 different sums written down. An example where there are exactly 4 of them exists (one is shown on the right, the sums are $8,9,10,... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A palindrome is a natural number that reads the same from left to right as it does from right to left. Does there exist a five-digit palindrome that is equal to the sum of two four-digit palindromes? | Answer: For example, $6006+5005=11011$.
Other examples are possible: 7777+4444=12221.
Criteria. Answer without an example: 0 points. | 11011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Which three-digit numbers are more numerous: those in which all digits have the same parity, or those in which adjacent digits have different parity | Answer: equally
Instructions. Method 1. We will consider the numbers of interest by tens. In each ten, the hundreds and units digits are the same (they are of the same parity). There are exactly 10 tens: five of each parity, and thus, from this ten, five numbers will be added to each type. Therefore, the numbers of ea... | 225 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On a line, seven points A, B, C, D, E, F, G are marked in the given order. It turned out that $A G=23 \mathrm{~cm}, B F=17 \mathrm{~cm}$ and $\mathrm{CE}=9 \mathrm{~cm}$. Find the sum of the lengths of all segments with endpoints at these points. | Answer: 224 cm.
Instructions. There are five pairs of segments that sum up to AG and AG itself. There are three pairs of segments that sum up to BF and BF itself. There is one pair of segments that sum up to CE and CE itself. Therefore, the total length of all segments is $6 \cdot 23 + 4 \cdot 17 + 2 \cdot 9 = 138 + 6... | 224 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. From a paper square $8 \times 8$, p seven-cell corners were cut out. It turned out that it was impossible to cut out any more such corners. For what smallest $n$ is this possible? A seven-cell corner is obtained by cutting out a $3 \times 3$ square (in cells) from a $4 \times 4$ square. | Answer: $\mathrm{n}=3$.
Instructions. Example. The figures show two examples of placing three corners so that no more can be cut out.
Cells belonging to the same corner are marked with the s... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. From a three-digit number, the sum of its digits was subtracted. The same operation was performed on the resulting number, and so on, 100 times. Prove that the result will be zero. (6 points) | Solution. Since $\overline{a b c}-(a+b+c)=9 \cdot(11 a+b)$, the first difference is divisible by 9. The sum of its digits is divisible by 9, which means the second, and similarly, all other differences will be divisible by 9.
The sum of the digits of a three-digit number divisible by 9 can be 9, 18, or 27. Therefore, ... | 0 | Number Theory | proof | Yes | Yes | olympiads | false |
10.2. Given a parallelogram $A B C D$, where $A B<A C<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$; moreover, segments $A D$ and $C E$ intersect. It turns out that $\angle A B F=\angle D C E$. Find th... | Answer: $60^{\circ}$.
Solution. Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C>A C$, we have $\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A$; thus, $A^{\prime}$ lies on the extension of segment $D C$ beyond poi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with? (N. Agakhanov) | Answer: Up to 8 zeros.
Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. Let $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$. Find all values that the number $x+y$ can take, and prove that no other values are possible. | Solution: Consider the function $f(x)=x+\sqrt{x^{2}+1}$. Since for any real $x$ we have $\sqrt{x^{2}+1}>\sqrt{x^{2}}=|x| \geqslant -x$, this function is positive everywhere. Let $a>0$. Solving the equation $f(x)=a$, we find that the value $a$ is taken by the function at the unique point $x=\frac{a^{2}-1}{2 a}$. Our equ... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
1. Option 1. To a confectionery factory for cake packaging, 5 rolls of ribbon, each 60 m long, were delivered. How many cuts need to be made to get pieces of ribbon, each 1 m 50 cm long? | Answer: 195.
Solution: From one roll, 40 pieces of ribbon, each 1 m 50 cm long, can be obtained. For this, 39 cuts are needed. Therefore, a total of $5 \cdot 39=195$ cuts are required. | 195 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 2. Option 1
Given a rectangular grid. We will call two cells adjacent if they share a side. Let's count the number of cells that have exactly four adjacent cells. It turned out to be 23. How many cells have exactly three adjacent cells? | Answer: 48.
Solution. Let $a$ and $b$ be the sides of the rectangle. The total number of cells that have exactly 4 neighbors by side is $(a-2)(b-2)$, and on the other hand, there are 23 such cells. Since 23 is a prime number, the numbers $a-2$ and $b-2$ are equal to the numbers 1 and 23 in some order. The number of ce... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 3. Variant 1
Three rectangles A, B, and C are drawn on the sheet (see figure).
Rectangles A and B have the same width, while rectangles B and C have the same length (width - top to bottom... | Answer: 20.
Solution: Let rectangle A have a length of $a$ cm and a width of $b$ cm. If the length is increased by 3 cm, the area increases by $3 b$. Therefore, $3 b=12, b=4$. The area of rectangle B is larger than that of A by $24 \mathrm{~cm}^{2}$, so the length of rectangle B is $24: 3=8$ cm. Therefore, the length ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 4. Variant 1
The train was moving from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 15 minutes. After that, the driver had to increase the train's speed by 4 times to arrive at point B on schedule. How many minutes does the train travel from poi... | Answer: 40.
Solution. Let's represent the train's path as a segment divided into 8 parts. Denote the intervals it travels in equal time with arcs.
Since the train stopped for 15 minutes, it... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
An apple, three pears, and two bananas together weigh 920 g; two apples, four bananas, and five pears together weigh 1 kg 710 g. How many grams does a pear weigh? | Answer: 130
Solution: From the first condition, it follows that two apples, four bananas, and six pears together weigh 1 kg 840 g. Therefore, a pear weighs $1840-1710=130$ g. | 130 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7. Option 1
Diligent Masha wrote down in a row all natural numbers from 372 to 506 inclusive. Then she calculated two sums: first, the sum of all odd numbers in this row, and then the sum of all even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get? | Answer: 439.
Solution. Let's divide all numbers from 372 to 506, except 506, into pairs such that each even number corresponds to the next odd number (this is possible because the last number is 505). In each pair, the odd number will be 1 greater than the even one. There will be (505 - 371) : 2 = 67 pairs in the rang... | 439 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 11 sticks, and laid out a pattern inside the triangle such that the triangle was divided into smaller triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle wi... | Answer: 198.
Solution (1st method). All the sticks can be divided into those that go horizontally, at an angle to the right, and at an angle to the left:
In the first such row, there is 1 s... | 198 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. Answer. $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$. | Solution.
Let $O$ be the common center of the given circles. From the condition, it follows that $BO$ and $CO$ are the bisectors of angles $ABC$ and $BCD$ ($O$ is the incenter) and, moreover... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.6. On the board, three natural numbers were written: two ten-digit numbers $a$ and $b$, as well as their sum $a+b$. What is the maximum number of odd digits that could have been written on the board?
(I. Bogdanov, P. Kozhevnikov) | Answer: 30.
Solution: Note that the number $a+b$ has no more than 11 digits, so in total, no more than 31 digits are written on the board. At the same time, all three numbers $a, b, a+b$ cannot be odd simultaneously. Therefore, one of their last three digits is even, which means that there are no more than 30 odd digi... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. From two settlements, which are 60 km apart, two cars set off in the same direction at the same time. The first car, traveling at a speed of 90 km/h, caught up with the second car after three hours. What is the speed of the second car? | Solution. The closing speed of the cars is $60: 3=20(\kappa m / h)$, so the speed of the second car is $90-20=70(km / h)$.
Answer: 70 km $/$ h.
Comment. The correct answer was obtained through correct reasoning - 7 points. The closing speed was correctly found, but the wrong answer was given - 3 points. | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. On a grid sheet of size $100 \times 100$, several non-overlapping cardboard isosceles right triangles with a leg of 1 were placed; each triangle occupies exactly half of one of the cells. It turned out that each unit segment of the grid (including boundary segments) is covered by exactly one leg of a triangle. Fi... | Answer. $49 \cdot 50=2450$ cells
Solution. Let $n=50$. We will call a triangle upper if it is located above the line containing its horizontal leg, and lower otherwise. Number the horizontal lines of the grid from bottom to top with numbers from 0 to $2n$.
Denote by $u_{k}$ (respectively $d_{k}$) the number of segmen... | 2450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.3. The numbers $a, b, c$ satisfy the relation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$. Find $(a+b)(b+c)(a+c)$. | Answer: 0. Solution: Move $\frac{1}{a}$ to the right side, we get $\frac{b+c}{b c}=\frac{-(b+c)}{a(a+b+c)}$. If $b+c \neq 0$, then we will have (multiplying by the denominator)
$$
a^{2}+a b+a c+b c=0 \Leftrightarrow a(a+b)+c(a+b)=0 \Leftrightarrow(a+b)(a+c)=0
$$
Thus, in any case $(a+b)(b+c)(a+c)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. Kolya drew 10 segments and marked all their intersection points in red. Counting the red points, he noticed the following property: on each segment, there are three red points. a) Provide an example of the arrangement of 10 segments with this property. b) What is the maximum number of red points for 10 segments wi... | Answer: b) 15. Solution. a) See the example in the figure. As another example, you can take two copies of the right part of the figure from problem 7.5. b) The example given in the figure shows that 15 red points can be obtained. Let's prove that this is the maximum possible number. Number all 10 segments and write dow... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima laid out a structure in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.