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8.2. The part of the graph of a linear function located in the second coordinate quadrant, together with the coordinate axes, forms a triangle. By what factor will its area change if the slope of the function is doubled and the y-intercept is halved? | Answer: It will decrease by eight times.
Solution. Let the original linear function be defined by the equation $y=k x+b$. From the condition of the problem, it follows that $k>0$ and $b>0$ (see Fig. 8.2). The points of intersection of its graph with the axes are: $A(0 ; b)$ and $B\left(-\frac{b}{k} ; 0\right)$. Since ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. On an $8 \times 8$ chessboard, $k$ rooks and $k$ knights are placed such that no figure attacks any other. What is the largest $k$ for which this is possible? | Answer: 5.
Solution: From the condition, it follows that in one row (column) with a rook, no other figure can stand.
Suppose 6 rooks were placed on the board. Then they stand in 6 rows and 6 columns. Therefore, only 4 unpicked cells will remain (located at the intersection of two empty rows and two empty columns). It... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The numbers $\sqrt{2}$ and $\sqrt{5}$ are written on the board. You can add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that you can write the number 1 on the board. | Solution: For example, we get $\sqrt{5}-\sqrt{2}$, then $\sqrt{5}+\sqrt{2}$ and $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=5-2=3$, then $\sqrt{2} \cdot \sqrt{5}=\sqrt{10}$, then $\sqrt{10}-3$ and $\sqrt{10}+3$ and finally $(\sqrt{10}-3)(\sqrt{10}+3)=10-9=1$.
Criteria. The goal is achieved if the same numbers are used in ... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
3. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest figure... | Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leqslant 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14-$ $43=23$ parts, and each will have between 7 and 12 parts. One figurine is clea... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. The diagonals of the circumscribed trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point O. The radii of the inscribed circles of triangles $A O D, A O B, B O C$ are 6, 2, and $3 / 2$ respectively. Find the radius of the inscribed circle of triangle $C O D$. | Answer: 3
Solution. We will prove a more general statement, that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}$, where $r_{1}, r_{2}, r_{3}$ and $r_{4}$ are the radii of the inscribed circles of triangles $A O D, A O B, B O C$ and $C O D$ respectively.
Let $A B=a, B C=b, C D=c, A D=d, O A=x, O D=y,... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. A beginner gardener planted daisies, buttercups, and marguerites on their plot. When they sprouted, it turned out that there were 5 times more daisies than non-daisies, and 5 times fewer buttercups than non-buttercups. What fraction of the sprouted plants are marguerites? | Answer. Zero. They did not germinate.
Solution. Daisies make up $5 / 6$ of all the flowers, and dandelions make up $1/6$. Therefore, their total number equals the total number of flowers.
Criteria. Only the answer - 0 points. Complete solution - 7 points. | 0 | Other | math-word-problem | Yes | Yes | olympiads | false |
8-3. Vika has been recording her grades since the beginning of the year. At the beginning of the second quarter, she received a five, after which the proportion of fives increased by 0.15. After another grade, the proportion of fives increased by another 0.1. How many more fives does she need to get to increase their p... | Answer: 4.
Solution: Let's say Vika had $n$ grades in the first quarter, of which $k$ were fives. Then, after the first five in the second quarter, the proportion of fives increased by $\frac{k+1}{n+1}-\frac{k}{n}=0.15$. Similarly, after the second five, the increase was $\frac{k+2}{n+2}-\frac{k+1}{n+1}=0.1$. Simplify... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A new series of "Kinder Surprises" - chocolate eggs, each containing a toy car - was delivered to the store. The seller told Pete that there are only five different types of cars in the new series, and it is impossible to determine which car is inside by the appearance of the egg. What is the minimum number of "Kind... | Solution. If Petya buys 10 "Kinder Surprises," in the least favorable situation for him, he will get two cars of each type. If he buys 11 "Kinder Surprises," he will get three cars of one type. Let's prove this. Suppose Petya bought 11 "Kinder Surprises" but did not get three cars of one type. This means the number of ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Two cyclists are training on a circular stadium. In the first two hours, Ivanov lapped Petrov by 3 laps. Then Ivanov increased his speed by 10 km/h, and as a result, after 3 hours from the start, he lapped Petrov by 7 laps. Find the length of the lap. | 4. Answer: 4 km. Solution: let I be Ivanov's speed (initial), Π be Petrov's speed, K be the length of the circle. Then 2I - 2Π = 3K and 2I - 2(Ι - Π) = 3K and 3(Ι - Π) = 3K - 10, express Ι - Π from both equations and equate, we get 14K - 20 = 9K, from which K = 4.
Criteria: correct solution - 7 points; the system is s... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2. How many rectangular trapezoids $A B C D$ exist, where angles $A$ and $B$ are right angles, $A D=2$, $C D=B C$, the sides have integer lengths, and the perimeter is less than 100? | Answer: 5.
Solution. Let $C D=B C=a, A B=b$. Drop a perpendicular from point $D$ to $B C$, and apply the Pythagorean theorem: $(a-2)^{2}+b^{2}=a^{2}$. From this, $b^{2}=4(a-1), a=\frac{b^{2}}{4}+1$. Suppose $A D=2$ is the smaller base of the trapezoid. The perimeter $P=2 a+b+2a$, so $b>2$. Considering that $a=\frac{b^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. What is the maximum number of figures consisting of 4 1x1 squares, as shown in the diagram, that can be cut out from a $6 \times 6$ table, if cutting can only be done along the grid lines?
# | # Solution:
Example:
The diagram shows that 8 figures can be cut out.
Evaluation:
$. | Answer: $x=0$. Solution. The left side of the equation $\geq 1$, and the right side $\leq 1$. Therefore, the equation is equivalent to the system: $\sin x=0, \cos \sqrt{2} x=1$. We have: $x=\pi n, \sqrt{2} x=2 \pi k$ ( $n, k-$ integers). From this, $n=k \cdot \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, the las... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. At a round table, 12 people are sitting. Some of them are knights, who always tell the truth, and the rest are liars, who always lie. Each person declared their left neighbor to be a liar. Can we definitely state how many knights and how many liars are at the table? | 1. Answer: Yes, it is possible.
If a knight is sitting in some place, then he told the truth, and to his left should sit a liar. Conversely, if a liar is sitting in some place, then to his left sits the one who was incorrectly called a liar, that is, a knight. This means that knights and liars alternate around the tab... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.2. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$. | # Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. For various positive numbers $a$ and $b$, it is known that
$$
a^{3}-b^{3}=3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)
$$
By how many times is the larger number greater than the smaller one? | Solution. Let's consider and transform the difference:
$$
\begin{aligned}
& 0=a^{3}-b^{3}-3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}\right)-3\left(2 a b(a-b)-b^{2}(a-b)\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}-6 a b+3 b^{2}\right)= \\
& (a-b)\left(a^{2}-5 a b+4 b^{2}\right)=(a-b)(a-4 b)(a-... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On November 15, a tournament of the game of dodgeball was held. In each game, two teams competed. A win was worth 15 points, a draw 11, and a loss earned no points. Each team played against each other once. At the end of the tournament, the total number of points scored was 1151. How many teams were there? | 6. Answer: 12 teams. Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are earned. Therefore, the number of games was no less than $53(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 te... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. How many positive numbers are there among the first 100 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$ | 2. Note that all members of the sequence, starting from $\sin 1000^{\circ}$, are equal to each other, since the difference between the numbers $10^{k+1}$ and $10^{k}$ for natural $k>2$ is a multiple of 360. Indeed, $10^{k+1}-10^{k}=10^{k}(10-1)=9 \cdot 10^{k}=$ $9 \cdot 4 \cdot 10 \cdot 25 \cdot 10^{k-3} \vdots 360$. B... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive semi-prime numbers that can be semi-prime? | 5. Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assume the opposite. Then we get that the numbers $2n-1, 2n+1, 2n+3$ are prime, and all of them are greate... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the office, each computer was connected by wires to exactly 5 other computers. After some computers were infected by a virus, all wires from the infected computers were disconnected (a total of 26 wires were disconnected). Now, each of the uninfected computers is connected by wires to only 3 others. How many comp... | # 5. Answer: 8.
Let $\mathrm{m}$ be the number of infected computers, and $\mathrm{n}$ be the number of uninfected computers. Then, before the infection, there were $5(\mathrm{~m}+\mathrm{n}) / 2$ cables, and after the disconnection, there were $3 \mathrm{n} / 2$ cables (from which it follows that $\mathrm{n}$ is even... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the table, there are candies of three types: caramels, toffees, and lollipops. It is known that there are 8 fewer caramels than all the other candies, and there are 14 fewer toffees than all the other candies. How many lollipops are on the table? Be sure to explain your answer. | # Solution.
Method 1. Since there are 8 fewer caramels than other candies, there are 4 fewer caramels than half of the candies. Since there are 14 fewer toffees than all other candies, there are 7 fewer toffees than half of the candies. Thus, if we remove all caramels and toffees, 4 + 7 = 11 candies will remain. Since... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given an angle of $13^{0}$. How to obtain an angle of $11^{0}$? | 3. One possible option: lay off the angle of $13^{0}$, 13 times, then the difference between the straight angle and the obtained angle will give the required angle $\left(180^{\circ}-13 \cdot 13^{0}=11^{\circ}\right)$ | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. CONDITION
Given a right triangle $A B C$ with legs $A C=3$ and $B C=4$. Construct triangle $A_{1} B_{1} C_{1}$ by sequentially moving point $A$ a certain distance parallel to segment $B C$ (point $A_{1}$ ), then point $B-$ parallel to segment $A_{1} C$ (point $B_{1}$ ), and finally point $C$ - parallel to segment... | Solution. When a vertex of a triangle is moved parallel to its base, the area of the triangle does not change. Therefore, we sequentially obtain the equality of the areas of triangles $A B C, A_{1} B C, A_{1} B_{1} C$, and finally, $A_{1} B_{1} C_{1}$. Thus, $B_{1} C_{1}=2 S / A_{1} B_{1}=12$.
Answer: 12. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 5. CONDITION
Vladislav Vladimirovich, taking less than 100 rubles, went for a walk. Entering any cafe and having at that moment $m$ rubles $n$ kopecks, he spent $n$ rubles $m$ kopecks ( $m$ and $n$ - natural numbers). What is the maximum number of cafes Vladislav Vladimirovich could visit? | Solution. Method one. Let Vladislav Vladimirovich have $a$ rubles $b$ kopecks upon entering the first cafe. It is clear that $b \leqslant a$. Then upon exiting, he will have $a-b-1$ rubles and $b-a+100$ kopecks. Let $a-b=t \leqslant 99$. Thus, Vladislav Vladimirovich now has $t-1$ rubles and $100-t$ kopecks. The condit... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Option 1.
In the Ivanov family, both the mother and the father, and their three children, were born on April 1st. When the first child was born, the parents' combined age was 45 years. The third child in the family was born a year ago, when the sum of the ages of all family members was 70 years. How old is the midd... | Answer: 5.
Solution.
If the first child is older than the second child by $x$ years, and the middle child is older than the third child by $y$ years, then $70-45=3(x+y)+y$, because the age of each parent and the eldest child increased by $(x+y)$ years by the time the third child was born, and the age of the second ch... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. The children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will have as many mushrooms as all the others combined. How many children went to pick mushrooms | Answer: 6 children.
Solution: Let Anya give half of her mushrooms to Vitya. Now all the children have the same number of mushrooms (this means that Vitya did not have any mushrooms of his own). For Sanya to now get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushr... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Andrey, Boris, Vasily, Gennady, and Dmitry played table tennis in pairs such that every two of them played with every other pair exactly once. There were no draws in the tennis matches. It is known that Andrey lost exactly 12 times, and Boris lost exactly 6 times. How many times did Gennady win?
Om vem: Gennady won... | Solution. The first pair can be formed in $5 \times 4: 2=10$ ways, the second pair can be formed in $3 \times 2: 2=3$ ways. In total, we get $10 \times 3: 2=15$ games. Andrei played in 4 pairs, and they played with 3 pairs. Therefore, Andrei played $4 \times 3=12$ times. According to the problem, he lost 12 times, whic... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Masha wrote a three-digit number on the board, and Vera wrote the same number next to it, but she swapped the last two digits. After that, Polina added the obtained numbers and got a four-digit sum, the first three digits of which are 195. What is the last digit of this sum? (The answer needs to be justified.)
... | Solution. Let Masha write the number $100 x+10 y+z$. Then Vera wrote the number $100 x+10 z+y$, and the sum of these numbers is $200 x+11 y+11 z$. For $x \leqslant 8$ this expression does not exceed 1798, and therefore cannot start with 195. Thus, $x=9$. Then $11(y+z)$ is a three-digit number starting with 15. Among th... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5.1. Points $A, B, C, D, E, F$ on the diagram satisfy the following conditions:
- points $A, C, F$ lie on the same line;
- $A B=D E=F C$
- $\angle A B C=\angle D E C=\angle F C E$
- $\angle B A C=\angle E D C=\angle C F E$
- $A F=21, C E=13$.
Find the length of segment $B D$.
, there are no fewer balls than in the previous one. It is also known that ther... | Answer: 1 red ball, 3 blue balls.
Solution. Among all the boxes, there can be:
- a maximum of 2 boxes with one ball (one with a red ball, the other with a blue ball),
- a maximum of 3 boxes with two balls (one with two red balls, another with one red and one blue, and the third with two blue balls),
- a maximum of 4 ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The distance between points A and B is 90 km. At 9:00, a bus left point A for point B at a speed of 60 km/h. Starting from 9:00, every 15 minutes, buses leave point B towards point A at a speed of 80 km/h. The bus that left point A, after traveling 45 km, reduces its speed to 20 km/h due to a breakdown and continues... | 2. A bus leaving point A will travel 45 km in $\frac{45}{60}$ hours. During this time, a bus leaving point B at 9:00 will travel $\frac{45}{60} \cdot 80=60$ km and will be closer to point A than the bus leaving point A. A bus leaving point B at 9:15 will travel $\left(\frac{45}{60}-\frac{15}{60}\right) \cdot 80=40$ km ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $\mathrm{k}$ transfers. At the same time, the number of air routes should not exceed four. What is the smallest $\mathrm{k}$ for which this is... | 5. $\mathrm{k}=2$. At least two transfers will be required. From an arbitrary city A, one can reach no more than four cities without a transfer, and with one transfer - no more than $4 \times 3=12$ cities. That is, if using no more than one transfer, one can fly to no more than 16 other cities, but 19 are required. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. Semyon was solving the quadratic equation $4 x^{2}+b x+c=0$ and found that its two roots are the numbers $\operatorname{tg} \alpha$ and $3 \operatorname{ctg} \alpha$ for some $\alpha$. Find $c$. | Answer: 12
Solution. By Vieta's theorem, $c / 4$ equals the product of the roots. Considering that $\operatorname{tg} \alpha \cdot \operatorname{ctg} \alpha=1$, we get $c / 4=3$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
# 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 |
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 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 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 |
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 |
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 |
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 |
# 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 |
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 |
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 |
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 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 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 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
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 |
1. Russia, Germany, and France decided to build the Nord Stream 2 gas pipeline, 1200 km long, agreeing to finance this project equally. In the end, Russia built 650 kilometers of the pipeline, Germany built 550 kilometers of the pipeline, and France contributed its share in money. Germany received 1.2 billion euros fro... | Solution. Each country was supposed to build 400 kilometers of the gas pipeline. Thus, Russia built 250 km of the pipeline for France, and Germany built 150 km of the pipeline. Therefore, the money from France should be distributed between Russia and Germany in a ratio of 5:3. Hence, Russia will receive 1.2 billion $\t... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The numbers $1,2,3,4,5,6,7,8,9$ are written into the cells of a $3 \times 3$ table. After that, all possible sums of numbers standing in adjacent (by side) 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 |
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 |
10.5. 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 |
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 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4 The pirates, led by John Silver on Treasure Island, found Billy Bones' chest, which contained 40 coins worth 1 ducat each and 40 coins worth 5 ducats each. John Silver has not yet decided how to divide this money among all the pirates (he does not want to take anything for himself). For what maximum number of pira... | Solution. We will show that if there are no more than 11 pirates, Silver can divide the coins as he wishes. Indeed, let the $i$-th pirate need to receive $S_{i}$ coins. $S_{i}=5 x_{i}+a_{i}$, for some integers $x_{i}$ and $a_{i}$ (where $a_{i}$ is the remainder of the division of $S_{i}$ by 5). Note that the sum of all... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. Uncle Chernomor assigns 9 or 10 of his thirty-three bogatyrs (knights) to duty each evening. What is the smallest number of days after which it can happen that all bogatyrs have been on duty the same number of times? | # 10.5. 7.
Let \( m \) and \( n \) be the number of days when 9 and 10 heroes were on duty, respectively. Let \( k \) be the number of days each hero was on duty. Then \( 9m + 10n = 33k \).... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Let all numbers $x, y, z$ be non-zero. Find all values that the expression
$$
\left(\frac{x}{|y|}-\frac{|x|}{y}\right) \cdot\left(\frac{y}{|z|}-\frac{|y|}{z}\right) \cdot\left(\frac{z}{|x|}-\frac{|z|}{x}\right)
$$
can take. | Solution: By the Pigeonhole Principle, among the numbers $x, y$, and $z$, there will be two numbers of the same sign. Then the corresponding bracket will be equal to 0, and the entire product will also be equal to 0.
Answer: Only the number 0.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. All natural numbers from 1 to 20 were divided into pairs, and the numbers in each pair were added. What is the maximum number of the ten resulting sums that can be divisible by 11? Justify your answer. | Solution: The number 11 is the only number in the set that is divisible by 11, so adding it to any other number will disrupt divisibility by 11. Therefore, all 10 sums cannot be divisible by 11. One example where nine sums are divisible by 11 is as follows:
$(1,10),(2,20),(3,19),(4,18),(5,17),(6,16),(7,15),(8,14),(9,1... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Seven friends entered a cafe and ordered 3 small cups of coffee, 8 medium cups, and 10 large cups. The volume of a small cup is half the volume of a medium cup, and the volume of a large cup is three times the volume of a small cup. How should the friends divide the cups of coffee among themselves so that everyone d... | Solution: Let's call the amount of coffee in a small cup a "norm". Then, in total, we have $3+8 \times 2+10 \times 3=49$ norms. Since there are seven friends, each should get 7 norms.
We divide as follows: 1 small + 2 large - 3 people; 2 medium + 1 large - 4 people.
Criteria. A correct example - 7 points. No addition... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a football tournament, 12 teams participated. By September, they had played several games, and no two teams had played each other more than once. It is known that the first team played exactly 11 games. Three teams played 9 games each. One team played 5 games. Four teams played 4 games each. Two teams played only... | Answer: 5 games.
Solution. Let the first team K1 play 11 games - i.e., once with everyone. Teams K2 and K3 played 1 game each - these are games with team K1.
There are 9 teams left (K4-K12). Three of these teams (K4, K5, K6) played 9 games each. One of these games was with K1. And 8 with all teams K4-K12 (except them... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Given a convex quadrilateral ABCD. Point $M$ is the midpoint of side BC, and point $N$ is the midpoint of side CD. Segments AM, AN, and $MN$ divide the quadrilateral into four triangles, the areas of which, written in some order, are consecutive natural numbers. What is the maximum possible area of triangle $\mathrm... | Answer: 6.
Solution. Estimation. Let $n, n+1, n+2, n+3$ be the areas of the four triangles. Then the area of quadrilateral $ABCD$ is $4n+6$. $MN$ is the midline of triangle $BCD$, so $S_{BCD} = 4S_{MCN}$, but $S_{MCN} \geq n$, hence $S_{BCD} \geq 4n$. Then $S_{ABD} = S_{ABCD} - S_{BCD} \leq 6$.
Example. If $ABCD$ is ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Three jumps of a two-headed dragon are equal to 5 jumps of a three-headed one. But in the time it takes for the two-headed dragon to make 4 jumps, the three-headed one makes 7 jumps. Which one runs faster? Justify your answer.
# | # Answer. Three-headed.
Solution. Consider the time it takes for a two-headed dragon to make 3*4=12 jumps. In this time, a three-headed dragon makes $3 * 7=21$ jumps. Since 12=4*3, 12 jumps of the two-headed dragon are equal to 4*5=20 jumps of the three-headed dragon. Thus, in the same amount of time, the three-headed... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Three lines intersect at one point 0. Outside these lines, a point M is taken and perpendiculars are dropped from it to them. The points $\mathrm{H}_{1}, \mathrm{H}_{2}$ and $\mathrm{H}_{3}$ are the bases of these perpendiculars. Find the ratio of the length of the segment OM to the radius of the circle circumscribe... | Solution
$2, \prime \prime$
First, consider two intersecting... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. On the board, there are 2017 digits. From these, several numbers were formed, the sums of the digits of these numbers were calculated, and then the sum of all the numbers was subtracted by the sum of the sums of their digits. The resulting number was broken down into digits, and the above operation was repea... | Solution. Since the difference between a number and the sum of its digits is divisible by 9, the first operation will result in a number that is a multiple of 9. Moreover, if we take the sum of several numbers and subtract the sum of the digits of these numbers, the result will also be a multiple of 9. Continuing the c... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. The numbers from 1 to 20 are arranged in a circle. We will paint a number blue if it is divisible without a remainder by the number to its left. Otherwise, we will paint it red. What is the maximum number of blue numbers that could be in the circle? | # Solution.
Evaluation. It is obvious that numbers cannot be blue if the number to their left is greater than or equal to 11. That is, no more than 10 numbers can be blue.
Example. As an example, both any correct arrangement and a correct algorithm are counted. An example of a correct algorithm.
1) write down the nu... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.4. It is known that $a b c=1$. Calculate the sum
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}
$$ | Solution: Note that
$$
\frac{1}{1+a+a b}=\frac{1}{a b c+a+a b}=\frac{1}{a(1+b+b c)}=\frac{a b c}{a(1+b+b c)}=\frac{b c}{1+b+b c}
$$
Similarly, by replacing 1 with the number $a b c$, we have
$$
\frac{1}{1+c+c a}=\frac{a b}{1+a+a b}=\frac{a b^{2} c}{1+b+b c}=\frac{b}{1+b+b c} .
$$
Then
$$
\frac{1}{1+a+a b}+\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. On a certain segment, its endpoints and three internal points were marked. It turned out that all pairwise distances between the five marked points are different and are expressed in whole centimeters. What is the smallest possible length of the segment? Justify your answer. | Solution: There are 5 points, so there are 10 pairwise distances. If all of them are expressed as positive whole numbers of centimeters and are distinct, at least one of them is not less than 10. Therefore, the length of the segment is not less than 10. Suppose the length of the segment is exactly 10. Then the pairwise... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 4.1. Condition:
In front of the elevator stand people weighing 50, 51, 55, 57, 58, 59, 60, 63, 75, and 140 kg. The elevator's load capacity is 180 kg. What is the minimum number of trips needed to get everyone up? | Answer: 4 (or 7)
## Solution.
In one trip, the elevator can move no more than three people, as the minimum possible weight of four people will be no less than $50+51+55+57=213>180$. Note that no one will be able to go up with the person weighing 140 kg, so a separate trip will be required for his ascent. For the rema... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4.2. Condition:
In front of the elevator stand people weighing 150, 60, 70, 71, 72, 100, 101, 102, and 103 kg. The elevator's load capacity is 200 kg. What is the minimum number of trips needed to get everyone up? | Answer: 5 (or 9)
## Solution
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+70+71=201>200$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4.3. Condition:
In front of the elevator stand people weighing 150, 62, 63, 66, 70, 75, 79, 84, 95, 96, and 99 kg. The elevator's load capacity is 190 kg. What is the minimum number of trips needed to get everyone up? | # Answer: 6 (or 11)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $62+63+66=191>190$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remai... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4.4. Condition:
In front of the elevator stand people weighing $130,60,61,65,68,70,79,81,83,87,90,91$ and 95 kg. The elevator's load capacity is 175 kg. What is the minimum number of trips needed to get everyone up? | # Answer: 7 (or 13)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+61+65=186>175$. Note that no one will be able to go up with the person weighing 135 kg, so a separate trip will be required for his ascent. Six trips wil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 5.1. Condition:
In the warehouse, there are 8 cabinets, each containing 4 boxes, each with 10 mobile phones. The warehouse, each cabinet, and each box are locked. The manager has been tasked with retrieving 52 mobile phones. What is the minimum number of keys the manager should take with them? | # Answer: 9
Solution. To retrieve 52 phones, at least 6 boxes need to be opened. To open 6 boxes, no fewer than 2 cabinets need to be opened. Additionally, 1 key to the warehouse is required. In total, $6+2+1=9$ keys need to be taken by the manager. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 8.2. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander -1 m
The third islander -1 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then from his statement, it follows that the third one is also a knight; by the principle of exclusion, the second and fourth must be liars. The fourth said that... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 8.3. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 1 meter away from me."
- The second from the left said: "My fellow t... | # Answer:
The third islander -1 m; 3 m; 4 m.
The fourth islander -2 m.
## Solution.
Let's number the islanders from right to left. Suppose the first one is a knight. Then, from his statement, it follows that the second one is also a knight. However, the second one said that his fellow tribesman is two meters away f... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9-1. Postman Pechkin calculated that he walked half the distance (at a speed of 5 km/h) and only a third of the time he was cycling (at a speed of 12 km/h). Did he make a mistake in his calculations? | Answer. Mistaken.
Solution. Let's denote the entire distance Pechkin traveled as $2 S$ km. Then, on foot, he covered a distance of $S$ km and spent $S / 5 = 0.2 S$ (hours) on it. According to the problem, this constituted $2 / 3$ of the total time spent, meaning the entire journey took $0.2 S : 2 / 3 = 0.3 S$ (hours),... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-2. The school volleyball team played several matches. After they won another match, the share of victories increased by $1 / 6$. To increase the share of victories by another 1/6, the volleyball players had to win two more consecutive matches. What is the minimum number of victories the team needs to achieve to incre... | Answer: 6.
Solution: Let the team initially play $n$ matches, of which $k$ were won. Then, after the next win, the share of victories increased by $\frac{k+1}{n+1}-\frac{k}{n}=\frac{1}{6}$. Similarly, after two more wins, the increase was $\frac{k+3}{n+3}-\frac{k+1}{n+1}=\frac{1}{6}$. Simplifying each equation, we get... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. 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 |
5.2. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 11 girls in thi... | # Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 11 = 5: x$, from which we get $m x = 55$. The number 55 has only one divisor greater than 11, which is 55. Therefore, $... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.4. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 17 girls in thi... | Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls got be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 17 = 5: x$, from which $m x = 85$. The number 85 has only one divisor greater than 17, which is 85. Therefore, $m=85, x=1$, th... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved two seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Wh... | # Answer: 2
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 2, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decrease... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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