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10.5. Given an odd number $n>10$. Find the number of ways to arrange the natural numbers $1,2,3, \ldots, n$ in a circle in some order so that each number is a divisor of the sum of the two adjacent numbers. (Ways that differ by rotation or reflection are considered the same.) (D. Khramov) | Answer. Two ways
Solution. Consider an arbitrary arrangement of numbers from 1 to $n$ that satisfies the conditions. Suppose that two even numbers $x$ and $y$ are adjacent, and the next number is $z$. Since $x+z$ is divisible by $y$, the number $z$ is also even. Continuing this movement around the circle, we get that ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. What values can the sum $f_{2}\left(x_{1}\r... | Answer: Only 0.
Solution: Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then
$f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$, since $f_{1}\left(x_{1}\right)=0$. Similarly, we obtain the equalities $f_{3}\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, $2, \ldots, 11$ kg. He also has a bag that will tear if more than 11 kg is placed in it. Archimedes has learned the weights of all the ingots and wants to prove to Hiero that the first ... | Answer. In 2 loads.
Solution. We will show that Archimedes can use the bag only twice. Let him first put in the bag ingots weighing 1, 2, 3, and 5 kg, and then ingots weighing 1, 4, and 6 kg. In both cases, the bag will not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if A... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.1. Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. What values can the sum $f_{2}\left(x_{1}\... | Answer: Only 0.
Solution. Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then
$$
f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1},
$$
since $f_{1}\left(x_{1}\right)=0$. Similarly, we obtain the equalities $f... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. What different values can the digit U take in the puzzle U$\cdot$LAN + U$\cdot$DE = 2020? Justify your answer. (Identical digits are replaced by the same letters, different digits by different letters.) | Answer: two. $\mathrm{V}=2, \mathrm{y}=5$.
Solution: Factor out the common factor: У$\cdot$(ЛАН + ДЭ) $=2020$. Note that У, Л, and Э are not equal to 0.
Factorize the right-hand side: $2020=1 \cdot 2 \cdot 2 \cdot 5 \cdot 101$. Since У is a digit, consider all possible values: $\mathrm{V}=1,2,4,5$.
1) $У=1$. Then ЛА... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.6. Each cell of a $7 \times 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the numbe... | Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones.
In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones.
Th... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Alyosha and Vitya set off from point $N$ to point $M$, the distance between which is 20 km. Unfortunately, they have only one bicycle between them. From $N$, Alyosha sets off on the bicycle, while Vitya starts walking. Alyosha can leave the bicycle at any point along the road and continue on foot. When Vitya reaches... | Answer. 12 km from point $N$.
Solution. Let $x$ (km) be the distance from $N$ to the point where Alyosha leaves the bicycle. Then Alyosha will spend $\frac{x}{15}+\frac{20-x}{4}$ hours on the entire journey, and Vitya will spend $\frac{x}{5}+\frac{20-x}{20}$ hours. By setting up and solving the equation, we find $x=12... | 12 | Algebra | 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 |
Task 1. In a bag, there were cards with numbers from 1 to 20. Vlad drew 6 cards and said that all these cards can be divided into pairs so that the sums of the numbers in each pair are the same. Lena managed to peek at 5 of Vlad's cards: the numbers on them were $2, 4, 9, 17, 19$. What number was on the card that Lena ... | Answer: 12.
Solution. To calculate the answer, one needs to select four numbers out of the given five such that the sum of two of them equals the sum of the other two. By enumeration, it is not difficult to verify that these numbers are $2,4,17,19(2+19=4+17)$.
Thus, the number on the remaining card is $12(2+19=4+17=9... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A row of 11 numbers is written such that the sum of any three consecutive numbers is 18. Additionally, the sum of all the numbers is 64. Find the central number. | Answer: 8.
Solution. Number the numbers from left to right from 1 to 11.
Notice that the sum of the five central numbers (from the fourth to the eighth) is 64 (the sum of all numbers) $-2 \cdot 18$ (the sum of the numbers in the first and last triplets) $=28$.
Then the sixth (central) number is 18 (the sum of the fo... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the numerical value of the expression
$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}+\frac{2}{x y+1}
$$
if it is known that $x$ is not equal to $y$ and the sum of the first two terms is equal to the third. | Answer: 2.
Solution. Let's bring the condition to a common denominator
$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}=\frac{2}{x y+1}
$$
we get
$$
\frac{\left(x^{2}+y^{2}+2\right)(x y+1)-2\left(x^{2}+1\right)\left(y^{2}+1\right)}{\left(x^{2}+1\right)\left(y^{2}+1\right)(x y+1)}=0
$$
expand all brackets in the numerator, c... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. A football team coach loves to experiment with the lineup. During training sessions, he divides 20 available field players into two teams of 10 players each, adds goalkeepers, and arranges a game between the teams. He wants any two field players to end up on different teams at some training session. What is the mi... | # 9.5. 5 Training Sessions.
Note that four training sessions are insufficient. Let's take 10 football players who played on the same team during the first training session. During the second training session, at least five of them will be on the same team again. During the third training session, at least three of the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.2. Fishermen caught several carp and pike. Each caught as many carp as all the others caught pike. How many fishermen were there if the total number of carp caught was 10 times the number of pike? Justify your answer. | # Solution:
Method 1. Each fisherman caught as many carp and pike together as the total number of pike caught. Summing the catches of all fishermen, we get that the total catch of all fishermen (in terms of the number of fish) is equal to the total number of pike caught, multiplied by the number of fishermen. On the o... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. We consider all possible pairs of quadratic equations $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0$ such that each equation has two distinct roots. Is it true that the expression $\frac{1}{x_{1} x_{3}} + \frac{1}{x_{1} x_{4}} + \frac{1}{x_{2} x_{3}} + \frac{1}{x_{2} x_{4}}$, where the numbers $x_{1}, x_{2}$ are... | Solution: According to Vieta's theorem
$$
x_{1} x_{2}=-\left(x_{3}+x_{4}\right)=q \text { and } x_{3} x_{4}=-\left(x_{1}+x_{2}\right)=p
$$
Then
$$
\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}=\frac{x_{3}+x_{4}}{x_{1} x_{3} x_{4}}=-\frac{q}{p x_{1}}
$$
Similarly
$$
\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}=-\frac{... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
10.3. In the distant times of stagnation in the Soviet Union, 15 and 20 kopeck coins were in circulation. Schoolboy Valera had a certain amount of money only in such coins. Moreover, the number of 20 kopeck coins was greater than the number of 15 kopeck coins. Valera spent one-fifth of all his money, paying two coins f... | Solution: One fifth of Valera's capital could be either 30, 35, or 40 kopecks. Then, after buying the ticket, he should have had 120, 140, or 160 kopecks left, and the cost of the lunch was either 60, 70, or 80 kopecks. The maximum value of three coins is 60 kopecks, so the last two scenarios are impossible. Therefore,... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The city center is a rectangle measuring $5 \times 8$ km, consisting of 40 blocks, each $1 \times 1$ km, with boundaries formed by streets that create 54 intersections. What is the minimum number of police officers needed to be placed at the intersections so that any intersection can be reached by at least one polic... | Solution. Evaluation. Consider the intersections on the boundary. There are 26 in total. Each police officer can control no more than 5 intersections (if he is on the boundary, then exactly 5, if he is inside the city, then no more than 3 on each side and no more than 5 in the corner). Therefore, at least 6 police offi... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. The hunter told a friend that he saw a wolf with a one-meter tail in the forest. That friend told another friend that a wolf with a two-meter tail had been seen in the forest. Passing on the news further, ordinary people doubled the length of the tail, while cowards tripled it. As a result, the 10th channel report... | Solution: Note that when information is transmitted by ordinary people, the length of the tail is multiplied by 2, and when transmitted by cowardly people, it is multiplied by 3. Therefore, the number of twos in the product equals the number of ordinary people (and the number of threes equals the number of cowardly peo... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. Lisa wrote a quadratic equation. Artem erased its free term, so the equation now looks like $\operatorname{\operatorname {mak}} 2 x^{2}+20 x+\ldots=0$. Lisa doesn't remember what number Artem erased, but she remembers that the equation has exactly one real root. What is this root? | # Answer: -5
Solution. A quadratic equation has one root if and only if its discriminant is 0. And if the discriminant is 0, then the root is calculated using the formula $x_{1}=-b /(2 a)=-20 / 4=-5$. | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. A paper rectangle $4 \times 8$ was folded along the diagonal as shown in the figure. What is the area of the triangle that is covered twice?
 | Answer: 10.
Solution. By the Pythagorean theorem, $A C=\sqrt{A B^{2}+B C^{2}}=4 \sqrt{5}$. Triangle $A E C$ is isosceles, and if we drop the height $E H$ in it, then triangle $A E H$ will be similar to triangle $A C D$. Therefore, $A H / E H=A D / D C=2 . A H=2 \sqrt{5}$, since the height in an isosceles triangle is a... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. Real $x, y, z$ are such that $x y + x z + y z + x + y + z = -3, x^{2} + y^{2} + z^{2} = 5$. What is $x + y + z$? | Answer: -1
Solution. Add twice the first equation to the second, we get $(x+y+z)^{2}+2(x+y+z)=-1$. Therefore, $(x+y+z+1)^{2}=0, x+y+z=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. A $7 \times 7$ checkered board was assembled using three types of figures (see the image), not necessarily all. How many figures, composed of four cells, could have been used?
 | Answer: only one.
Solution. We will prove that only one figure consisting of four cells can be used. We will color the cells of the board as shown in Fig. 11.5a: Each of the given figures can cover no more than one shaded cell, therefore, the number of figures must be no less than 16.
Since 16 three-cell figures cove... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In the basket, there are fruits (no less than five). If you randomly pick three fruits, there will definitely be an apple among them. If you randomly pick four fruits, there will definitely be a pear among them. What fruits can be picked and in what quantities if you randomly pick five fruits? | Solution. From the condition of the problem, it follows that the "non-apples" in the box are no more than two fruits (otherwise, you could pull out 3 fruits, none of which would be apples). Similarly, "non-pears" are no more than three fruits (otherwise, you could pull out 4 fruits, none of which would be pears). Thus,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation: $\cos ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)+\sin ^{4}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)=\sin ^{-2}\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. | # Solution.
Let's make the substitution $t=\left(\sqrt{\frac{\pi^{2}}{4}-|x|}\right)$. Clearly, $0<t \leq \frac{\pi}{2}$.
$\cos ^{4} t+\sin ^{4} t=\sin ^{-2} t$.
$\cos ^{4} t+\sin ^{4} t=\left(\cos ^{2} t+\sin ^{2} t\right)^{2}-2 \sin ^{2} t \cdot \cos ^{2} t=1-2 \sin ^{2} t \cdot \cos ^{2} t \leq 1 . \frac{1}{\sin ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the height of the pyramid if, by cutting it only along the lateral edges and unfolding the lateral faces onto the plane of the base, outside of it, you get a square with a side of 18. If it is impossible to get such a square, explain why.
# | # Solution:
The figure shows such a net. If we flip the pyramid and place it on the face $A B C$, it is easy to calculate the volume of the pyramid $V=\frac{1}{3} \cdot \frac{1}{2} \cdot 9 \c... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.2. On the board, nine quadratic trinomials are written: $x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \ldots, x^{2}+a_{9} x+b_{9}$. It is known that the sequences $a_{1}, a_{2}, \ldots, a_{9}$ and $b_{1}, b_{2}, \ldots, b_{9}$ are arithmetic progressions. It turned out that the sum of all nine trinomials has at least o... | 10.2. Answer. 4.
Let $P_{i}(x)=x^{2}+a_{i} x+b_{i}, P(x)=P_{1}(x)+\ldots+P_{9}(x)$. Notice that $P_{i}(x)+P_{10-i}(x)=2 x^{2}+\left(a_{i}+a_{10-i}\right) x+\left(b_{i}+b_{10-i}\right)=$ $=2 P_{5}(x)$. Therefore, $P(x)=9 P_{5}(x)$, and the condition is equivalent to $P_{5}(x)$ having at least one root.
Let $x_{0}$ be ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Anya has a chocolate bar of size 5 x 6 squares. It contains 6 squares, forming a $2 \times 3$ rectangle, in which there are nuts (the rectangle can be positioned either vertically or horizontally). Anya does not know exactly where the nuts are. She wants to eat the smallest number of squares, but in such a way that ... | 2. Answer: 5.
Solution. An example of which slices to eat.
Evaluation. Let's number the slices as shown in the figure.
| 1 | 1 | 2 | 2 | 3 | 3 |
| :--- | :--- | :--- | :--- | :--- | :--- |
... | 5 | Logic and Puzzles | 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. How many emeralds are there in total in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Mom is walking around a lake with a stroller and completes a full lap around the lake in 12 minutes. Vanya rides a scooter on the same path in the same direction and meets (overtakes) mom every 12 minutes. At what intervals will Vanya meet mom if he rides at the same speed but in the opposite direction? | Answer: Every 4 minutes.
Solution. Since Mom completes a full lap around the lake in 12 minutes and meets Vanya every 12 minutes, in 12 minutes Vanya rides around the lake exactly 2 times, while Mom completes one lap. Therefore, Vanya's speed is twice that of Mom's. From this, it follows that when Vanya was riding in ... | 4 | Algebra | 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 is 1 m
The third islander is 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 ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. There is a set of 2021 numbers. Moreover, it is known that if each number in the set is replaced by the sum of the others, the same set will be obtained. Prove that the set contains a zero. | Solution. Let the sum of the numbers in the set be $M$, then the number $a$ in the set is replaced by the number $b=M-a$. Summing these equations for all $a$:
$$
b_{1}+\ldots+b_{2021}=2021 M-\left(a_{1}+\ldots+a_{2021}\right)
$$
from which $M=0$, since $b_{1}+\ldots+b_{2021}=a_{1}+\ldots+a_{2021}=M$. Therefore, for a... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
# 5. Variant 1
It is known that $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{1}{5}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{4}{5}}$. Find $\cos (\alpha-\beta)+\cos (\beta-$ $\gamma)+\cos (\gamma-\alpha)$ | Answer: -1.
Solution. Consider the expression $(\cos \alpha+\cos \beta+\cos \gamma)^{2}+(\sin \alpha+\sin \beta+\sin \gamma)^{2}$ and expand the brackets: $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2(\cos \alpha \cdot \cos \beta+\cos \alpha \cdot \cos \gamma+\cos \beta \cdot \cos \gamma)+\sin ^{2} \alpha+\sin ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 6. Option 1
Initially, there were 20 balls of three colors in the box: white, blue, and red. If we double the number of blue balls, the probability of drawing a white ball will be $\frac{1}{25}$ less than it was initially. If we remove all the white balls, the probability of drawing a blue ball will be $\frac{1}{16}... | Answer: 4.
Solution: Let there be $a$ white balls, $b$ blue balls, and $c$ red balls in the box. We can set up the following equations:
$$
\begin{gathered}
a+b+c=20 \\
\frac{a}{20}=\frac{a}{20+b}+\frac{1}{25} \\
\frac{b}{20}+\frac{1}{16}=\frac{b}{20-a}
\end{gathered}
$$
Transform the second equation: $\frac{a b}{20(... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
At the base of the quadrilateral pyramid $S A B C D$ lies a square $A B C D, S A$ - the height of the pyramid. Let $M$ and $N$ be the midpoints of the edges $S C$ and $A D$. What is the maximum value that the area of triangle $B S A$ can have if $M N=3 ?$ | Answer: 9.
Solution: Let $O$ be the center of the square $ABCD$. Then $MO$ is the midline of the triangle $SAC$, so $SA = 2MO$. Similarly, $ON$ is the midline of the triangle $BDA$, so $AB = 2ON$. Therefore, $SA^2 + AB^2 = 4(MO^2 + ON^2) = MN^2 = 36$. Let $SA = x, AB = y$. From the formula $S = 0.5 \cdot SA \cdot AB =... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.1. In the village, seven people live. Some of them are liars (always lie), and the rest are knights (always tell the truth). Each of them said about each of the others whether they are a knight or a liar. Out of the 42 answers received, 24 were “He is a liar.” What is the smallest number of knights that can live in t... | Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of knight-liar, the phrase "He is a liar" will be said twice. Since this phr... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. In a row, there are 27 matchboxes, each containing a certain number of matches. It is known that in any four consecutive boxes, the total is 25 matches, and in all of them, the total is 165. How many matches are in the eighth box? | Answer: 10
Solution. In the first 24 boxes, there are a total of $6 \cdot 25=150$ matches. In the last three boxes, there are 15 matches. Therefore, the 4th from the end (or 24th from the start) has 10 matches. Then, in boxes $24, 23, 22, 21$, there are 25 matches in total, meaning in boxes $21, 22, 23$, there are 15 ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A $7 \times 7$ board has a chessboard coloring. In one move, you can choose any $m \times n$ rectangle of cells and repaint all its cells to the opposite color (black cells become white, white cells become black). What is the minimum number of moves required to make the board monochromatic?
Answer: in 6 moves. | Solution. Consider segments of length equal to the side of a cell, separating pairs of cells adjacent to the side of the board. Along each side of the board, there are 6 such segments, totaling $6 \cdot 4=24$. Each of these segments separates cells that initially have different colors, so each segment must end up on th... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. There is a ruler 10 cm long without divisions. What is the smallest number of intermediate divisions that need to be made on the ruler so that segments of length 1 cm, 2 cm, 3 cm, ..., 10 cm can be laid off, applying the ruler in each case only once. | Answer: 4
Solution: First, let's prove that three divisions are not enough. There are a total of 10 segments with endpoints at five points. Therefore, if three divisions are made, each length from 1 to 10 should be obtained exactly once. If any division is made at a non-integer distance from the left end, there will b... | 4 | 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 |
10.2 A group of friends went for a morning run around a lake. During the run, one by one they realized they had miscalculated their strength, and switched from running to walking. One of the friends calculated that he had run one-eighth of the total distance that the entire group had run, and walked one-tenth of the to... | Solution 1: Let the person who ran cover $x$ part of the road, then $0<x<1$, and he walked $(1-x)$ part of the way. If there were $n$ people in total, then according to the condition, the total distance covered by the group (expressed in terms of parts) is on one side $n$, and on the other side $8 x + 10(1-x) = 10 - 2 ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The midpoints of adjacent sides of a rectangle with a perimeter of 32 were connected by segments. The same operation was performed on the resulting quadrilateral: the midpoints of adjacent sides were connected by segments (see figure). How many times in total does one need to perform such an operation so that the pe... | Answer: 11.
Solution: After two operations, a quadrilateral is obtained, the sides of which are the midlines of triangles with bases parallel to the sides of the original rectangle. Therefore, this quadrilateral is a rectangle, and each of its sides is half the length of the corresponding side of the original rectangl... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. All 25 students in class 7A participated in a quiz consisting of three rounds. In each round, each participant scored a certain number of points. It is known that in each round, as well as in the total of all three rounds, all participants scored a different number of points.
Student Kolya from 7A was thi... | Answer: 10.
Solution. In the first round, 2 classmates overtook Kolya, in the second - 3, in the third - 4. Then, in the sum of all three rounds, he could be overtaken by no more than $2+3+4=9$ classmates, i.e., in the sum of the three rounds, he could not end up lower than 10th place.
Now let's provide an example of... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. What is the minimum number of unit-radius circles required to completely cover a triangle with sides $2 ; 3 ; 4$? | Answer: three circles. Solution. Let $A C=4, A B=2, B C=3$ and let $C_{1}, A_{1}$ and $B_{1}$ be the midpoints of sides $A B, B C$ and $A C$ respectively. Note that angle $B$ is obtuse, since $A C^{2}>A B^{2}+B C^{2}$. Therefore, points $B$ and $B_{1}$ lie inside the circle of radius 1 with center at point $O$ - the mi... | 3 | 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 |
3. After teacher Mary Ivanovna moved Vovochka from the first row to the second, Vanechka from the second row to the third, and Mashenka from the third row to the first, the average age of students sitting in the first row increased by one week, those sitting in the second row increased by two weeks, and those sitting i... | 3. Let there be x people in the third row. Since the average age is the sum of the ages divided by the number of people, after the rearrangement, the total age of the children in the first row increased by 12 weeks, in the second row by 24 weeks, and in the third row by -4x weeks. Since the total sum of the ages of all... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. While waiting for customers, the watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). The seller recorded on a piece of paper the weights of the weights he used. What is ... | Answer: 6.
Solution. With one or two weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, any of the given watermelons can be weighed. Indeed, $2=1+1, 4=3+1$, $6=5+1, 8=7+1, 11=10+1, 12=9+3, 13=10+3, 14=9+5, 15=10+5, 16=9+7, 17$ $=10+7, 18=9+9, 19=10+9, 20=10+10$. Thus, six different numbers could have been recorded.
... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the notebook, all irreducible fractions with the numerator 15 are written down, but which are greater than $\frac{1}{16}$ and less than $\frac{1}{15}$. How many such fractions are written in the notebook? | Answer: 9 fractions.
## Solution:
We are looking for all suitable irreducible fractions of the form $\frac{15}{n}$. Since $\frac{1}{16}<\frac{15}{n}<\frac{1}{15}$, then $15 \cdot 15<n<15 \cdot 16$ or $225<n<240$ (with the fraction $\frac{15}{n}$ being irreducible, meaning $n$ is not divisible by 3 or 5). It is not di... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In each cell of a $2 \times 2$ table, a number was written, and all the numbers are different. It turned out that the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second colum... | # Answer: 0.
## Solution:
| $a$ | $b$ |
| :--- | :--- |
| $c$ | $d$ |
Let's denote the numbers in the table as shown on the left. According to the condition, $a+b=c+d$, $ac=bd$. Then $a=c+d-b$, substitute this into the product equality: $(c+d-b)c=bd$. Expand the brackets and move everything to the left: $c^2 + cd - ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The organizers of a mathematics olympiad decided to photograph 60 participants. It is known that no more than 30 participants can fit in a single photograph, however, any two students must appear together in at least one photograph. What is the minimum number of photographs needed to achieve this? | # Answer: 6.
## Solution:
Example with 6 photos: divide 60 participants into 4 groups of 15 people (groups $A, B, B$, Г). Take 6 photos of all possible pairs of groups: $A+D, A+B, A+\Gamma, B+B, B+\Gamma, B+\Gamma$ - in each photo, there will be 30 people, and it is easy to see that in this way any two people will be... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8.2. (7 points)
Twelve people are carrying 12 loaves of bread. Each man carries 2 loaves, each woman carries half a loaf, and each child carries a quarter of a loaf. How many men, women, and children were there? | Answer: 5 men, one woman, and 6 children.
Solution: Let $x$ be the number of men, $y$ be the number of women, and $z$ be the number of children; $x, y, z$ are natural numbers.
Then $x+y+z=12$ and $2 x+\frac{y}{2}+\frac{z}{4}=12$. From the last equation, it follows that $8 x+2 y+z=48$.
Transform the last equation: $7... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Solve the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
Answer: 2 | 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$. Acting similarly, we get that $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this e... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. The math teacher agreed with the eleven students who came to the elective that he would leave the classroom, and the students would agree among themselves who would be a liar (always lie) and who would be a knight (always tell the truth). When the teacher returned to the class, he asked each student to say about ... | Answer: 7.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of "knight-liar," the phrase "He is a liar" will be said twice. Since this p... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.4. At competitions, an athlete's performance is evaluated by 7 judges, each of whom gives a score (an integer from 0 to 10). To obtain the final score, the best and worst scores from the judges are discarded, and the arithmetic mean is calculated. If the average score were calculated based on all seven scores, the a... | # Answer. 5.
Solution. Suppose there are no fewer than six dancers. Let $A, a, S_{A}$ be the best score, the worst score, and the sum of all non-discarded scores of the winner, respectively, and $B, b, S_{B}$ be the same for the last athlete. Instead of averages, we can arrange the dancers by the sum of all scores or ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. In a regular pentagon $A B C D E$, a point $F$ is marked on side $A B$, and a point $G$ is marked on side $B C$ such that $F G=G D$. Find the angle $C D G$, if the angle $F D E$ is $60^{\circ}$. | Answer: $6^{\circ}$.
Solution. The angles of a regular pentagon are each $108^{\circ}, \angle E D A = 36^{\circ}, \angle F D E = 60^{\circ}$. Therefore, $\angle A D F = 24^{\circ}$.
Additio... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. During a physical education class, all students in the class lined up in a row. It turned out that boys and girls alternated in the row. It is known that exactly $52 \%$ of the students are girls. Find the number of boys in the class. Justify your answer. | Solution: Since there are more than half girls, the first and last student in the row are girls. Let's remove the last girl from the row - there will be an equal number of boys and girls left. In this row, all the boys in the class make up $48 \%$. There are as many girls, another $48 \%$. The remaining $100-2 \cdot 48... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the equations $x^{2}+a x+b=0$ and $x^{3}+b x+a=0$ have a common root and $a>b>0$. Find it. | Answer: -1.
Solution. Multiply the first equation by $x$ and subtract the second from it. The common root of the original equations will also be a root of the resulting equation
$$
\left(x^{3}+a x^{2}+b x\right)-\left(x^{3}+b x+a\right)=0 \quad \Longleftrightarrow \quad a\left(x^{2}-1\right)=0
$$
The last equation h... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Pasha is playing a computer game. The game takes place on an infinite grid. Each cell contains either a treasure or a natural number. The number indicates the distance to the nearest treasure in cells (if it takes $A$ steps vertically and $B$ steps horizontally to reach the treasure, the cell contains the number $... | Solution. a) Note that if Pasha finds a certain number $K$ in a cell, then at least one treasure must be located in the cells along the perimeter of a square rotated by $45^{\circ}$ relative to the grid lines, where the side of the square is exactly $K+1$ cells. At the same time, there cannot be any treasures inside su... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.4. Ivan on a tractor and Petr in a "Mercedes" left point A for point B. Petr arrived at point B, waited for 10 minutes, and called Ivan to find out that Ivan had only covered a third of the distance and was currently passing by a cafe. Petr drove to meet him. Not noticing Ivan, he reached the cafe and spent half an h... | Solution. From the fact that Ivan had traveled a third of the distance by the time Peter had traveled the whole distance and waited for 10 minutes, we can conclude that by the time Ivan reaches point B, Peter could have traveled the route three times and waited for half an hour. Instead, he traveled the route $1+2 / 3+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 11.1. Solve the equation
$$
x^{4}+2 x \sqrt{x-1}+3 x^{2}-8 x+4=0
$$ | Solution. Let's check that $x=1$ is a root. Indeed,
$$
1^{4}+2 \sqrt{1-1}+3 \cdot 1^{2}-8+4=1+3-8+4=0
$$
We will show that there are no other roots. Since the left side of the equation is defined only for $x \geqslant 1$, it is sufficient to consider the case $x>1$. Using the valid inequalities in this case $x^{4}>x^... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. In how many ways can all natural numbers from 1 to $2 n$ be arranged in a circle so that each number is a divisor of the sum of its two neighboring numbers? (Ways that differ only by rotation or symmetry are considered the same) | Solution. Note that if the numbers in the circle do not alternate in parity, then some two even numbers are adjacent. An even number is a divisor of the sum of the even number adjacent to it and the second adjacent number, which means the latter is also even. Continuing this reasoning around the circle, we get that all... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
36. A pair of natural numbers $a>b$ is called good if the least common multiple (LCM) of these numbers is divisible by their difference. Among all natural divisors of the number $n$, exactly one good pair was found. What can $n$ be? | 36. Answer: $n=2$. Two odd numbers cannot form a good pair. Therefore, the number $n$ has an even divisor, and thus it is even. But then it has two good pairs of divisors: $(2,1)$ and $(n, n / 2)$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find all integers $x, y$ for which $x+y, 2x+3y$ and $3x+y$ are perfect squares. | Answer: $x=y=0$. Let $x+y=a^{2}, 2 x+3 y=b^{2}$ and $3 x+y=c^{2}$. Note that $2 b^{2}+c^{2}=7 a^{2}$. We will prove that this equation has a unique solution: $a=b=c=0$. Indeed, let $(a, b, c)$ be a solution with the minimal sum $a^{2}+b^{2}+c^{2}$, in which not all numbers $a, b, c$ are equal to 0. Then $2 b^{2}+c^{2}:... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28. When preparing a district olympiad, each jury member participated in no more than 10 discussions. Discussions can be large or small. In a small discussion, 7 jury members participate, each sending exactly one email to each of the other 6. In a large discussion, 15 jury members participate, each sending exactly one ... | 28. Answer: The secretary participated in 6 small discussions and 2 large ones.
In a small discussion, 7 jury members participate, each sending exactly one email to each of the 6 others, so in the end, 42 emails are sent in a small discussion. Similarly, participants in a large discussion send 210 emails.
Let a total... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. There are $2 n$ cards, each with a number from 1 to $n$ (each number appears on exactly two cards). The cards are lying on the table face down. A set of $n$ cards is called good if each number appears exactly once. Baron Munchausen claims that he can point out 80 sets of $n$ cards, at least one of which is guarantee... | Answer: $n=7$. We will present an algorithm for how to indicate $2^{n-1}$ sets on $2n$ cards, one of which is suitable. (In our case, this is $2^{6}=64<80$ sets.) Imagine that identical cards are connected by (invisible to us for now) red edges. We will arbitrarily pair the cards with blue edges. The red-blue graph is ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
24. A polynomial of degree 10 has three distinct roots. What is the maximum number of zero coefficients it can have?
(A. Khryabrov) | 24. Answer: Answer: 9 zero coefficients. For example, the polynomial $x^{10}-x^{8}$ has roots $0,1,-1$. If a polynomial has only one non-zero coefficient, it is of the form $a x^{10}$, and therefore has exactly one root. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the bank of the river stand 10 sheikhs, each with a harem of 100 wives. Also at the bank stands an $n$-person yacht. By law, a woman should not be on the same bank, on the yacht, or even at a transfer point with a man if her husband is not present. What is the smallest $n$ for which all the sheikhs and their wive... | Answer: 10. Example: first, all 1000 wives move quietly, then one returns and 10 sheikhs leave. Finally, another returns and picks up the first. Evaluation: let the number of places not exceed 9. Consider the moment when the first sheikh appears on the other shore - or several at once, but not all, for they could not h... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Given a prime number $p$. All natural numbers from 1 to $p$ are written in a row in ascending order. Find all $p$ for which this row can be divided into several blocks of consecutive numbers so that the sums of the numbers in all blocks are equal. | Answer: $p=3$. Let $k-$ be the number of blocks, $S-$ be the sum in each block. Since $p(p+1) / 2=k S$, either $k$ or $S$ is divisible by $p$. Clearly, $k<p$, so $S$ is a multiple of $p$. Let the leftmost group consist of numbers from 1 to $m$. Then $m(m+1) / 2$ is a multiple of $p$, from which $m \geqslant p-1$, i.e.,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Given $n$ distinct natural numbers, any two of which can be obtained from each other by permuting the digits (zero cannot be placed in the first position). For what largest $n$ can all these numbers be divisible by the smallest of them? | Answer: 9. It is clear that there cannot be more than nine numbers.
We will use a known property of the period of a purely periodic rational fraction $\alpha=$ $a / b<1$ with coprime $(a, b)$: the length of the period is the smallest natural $t$ for which $\left(10^{t}-1\right) \vdots b$, and the period $T$ itself is ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
52. Olya wrote fractions of the form $1 / n$ on cards, where $n-$ are all possible divisors of the number $6^{100}$ (including one and the number itself). She arranged these cards in some order. After that, she wrote the number on the first card on the board, then the sum of the numbers on the first and second cards, t... | 52. Answer: two denominators. Let's represent all fractions in the form $a_{n} / 6^{100}$, then $a_{1}, a_{2}, \ldots$ are again all divisors of the number $6^{100}$, each appearing once. Let the partial sums be denoted by $S_{n} / 6^{100}$. Then the denominator of the irreducible representation of the partial sum depe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. A grasshopper starts moving in the top-left cell of a $10 \times 10$ square. It can jump one cell down or to the right. Additionally, the grasshopper can fly from the bottom cell of any column to the top cell of the same column, and from the rightmost cell of any row to the leftmost cell of the same row. Prove that... | 13. Consider the diagonal running from the bottom-left corner to the top-right corner. We will paint all 10 cells on this diagonal red. Note that from any red cell, without making any jumps, you can only move to cells that are to the right of it, below it, or to the right and below it. Therefore, it is impossible to mo... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false |
66. An isosceles triangle \(ABC\) with a perimeter of 12 is inscribed in a circle \(\omega\). Points \(P\) and \(Q\) are the midpoints of the arcs \(ABC\) and \(ACB\) respectively. The tangent to the circle \(\omega\) at point \(A\) intersects the ray \(PQ\) at point \(R\). It turns out that the midpoint of segment \(A... | 66. Answer: 4. Let $I_{A}, I_{B}, I_{C}$ be the centers of the excircles of triangle $ABC$, touching sides $BC, CA$, and $AB$ respectively. Then the lines $A I_{A}, B I_{B}, C I_{C}$ will be the angle bisectors of triangle $ABC$, and the lines $I_{B} I_{C}, I_{C} I_{A}, I_{A} I_{B}$ will be its external angle bisectors... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
49. What is the maximum number of solutions that the equation $\max \left\{a_{1} x+b_{1}, \ldots, a_{10} x+b_{10}\right\}=0$ can have, if $a_{1}, \ldots, a_{10}, b_{1}, \ldots, b_{10}$ are real numbers, and all $a_{i}$ are not equal to 0? | 49. Answer: 2 solutions. For example, 5 functions $-x-1$ and 5 functions $x-1$.
Suppose this equation has three roots: $u<v<w$. At point $v$, one of the linear functions $a_{i} x+b_{i}$ is zero. On the other hand, its values at points $u$ and $w$ do not exceed zero. However, such a linear function can only be a consta... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
52. In the table, there are 25 columns and 300 rows, and Kostya painted all its cells in three colors. Then Lesha, looking at the table, names one of the three colors for each row and marks all the cells of this color in that row. (If there are no cells of the specified color in the row, he marks nothing in it.) After ... | 52. Answer: Two columns. To leave at least two columns, Lesha must for each row name a color that does not appear in the first two cells of that row. With this strategy, the first two columns will not be crossed out.
Note now that $300=C_{25^{2}}$. Using this observation, we can associate each row with its own pair of... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
30. What is the minimum number of colors needed to color the cells of a $5 \times 5$ square so that among any three consecutive cells in a row, column, or diagonal, there are no cells of the same color?
(M. Antipov) | 30. Answer: in five colors.
Consider a cross of five cells, the central cell of which coincides with the central cell of the square. By the condition, three cells in its column have different colors. Similarly, the colors of the three cells in its row are also different. Finally, any two cells at the "ends" of the cro... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
36. Many city residents engage in dancing, many in mathematics, and at least one in both. Those who engage only in dancing are exactly $p+1$ times more than those who engage only in mathematics, where $p-$ is some prime number. If you square the number of all mathematicians, you get the number of all dancers. How many ... | 36. Answer: 1 person is engaged in both dancing and mathematics.
Let $a$ people be engaged only in mathematics, and $b \geqslant 1$ people be engaged in both dancing and mathematics. Then, according to the condition, $(a+b)^{2}=(p+1) a+b$. Subtract $a+b$ from both sides: $(a+b)^{2}-(a+b)=p a$. Factor out the common te... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
66. The sum
$$
\begin{gathered}
2 \\
3 \cdot 6
\end{gathered}+\begin{gathered}
2 \cdot 5 \\
3 \cdot 6 \cdot 9
\end{gathered}+\ldots+\begin{gathered}
2 \cdot 5 \cdot \ldots \cdot 2015 \\
3 \cdot 6 \cdot \ldots \cdot 2019
\end{gathered}
$$
was written as a decimal fraction. Find the first digit after the decimal point. | 66. Answer: the first digit after the decimal point is 5.
To start, let's simplify the given sum. Each term can be written as a difference
$$
\begin{aligned}
\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}=\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1) \cdot(3 k+3)}{3 \cdot 6 \cdot ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. From the edge of a large square sheet, a small square was cut off, as shown in the figure, and as a result, the perimeter of the sheet increased by $10 \%$. By what percent did the area of the sheet decrease? | 5. Answer: by $4 \%$.
Let the side of the larger square be denoted by $a$, and the side of the smaller square by $b$. As a result of cutting out the smaller square from the perimeter of the sheet, one segment of length $b$ disappears, and three such segments appear instead, meaning the perimeter increases by $2 b$. Th... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Sasha went to bed at 10 PM and set the alarm clock (with hands and a 12-hour dial) for 7 AM. During the night, at some point, the alarm clock, which had been working properly, broke, and its hands started moving in the opposite direction (at the same speed). Nevertheless, the alarm rang exactly at the scheduled time... | 1. Answer: the alarm clock broke at 1 o'clock at night.
Let's imagine that the minute hand on the alarm clock is missing, and the hour hand, at the moment when the alarm clock broke, split into two halves, one of which (as stated in the condition) started moving in the opposite direction, while the other continued its... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the inequality $2021 \cdot \sqrt[202]{x^{2020}}-1 \geq 2020 x$ for $x \geq 0$. (10 points) | Solution. Transform the inequality into the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021} \leq \sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1
\end{aligned}
$$
But by the r... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation $2021 x=2022 \cdot \sqrt[2022]{x^{2021}}-1 .(10$ points $)$ | Solution. $x \geq 0$. Transform the equation to the form:
But ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation $2021 \cdot \sqrt[202]{x^{2020}}-1=2020 x$ for $x \geq 0 \cdot$ (10 points) | Solution. Transform the equation to the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021}=\sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1 \\
& \frac{(2020 \text { instances) }}{2021}=\sqrt[202]{x^{2020}}
\end{aligned}
$$
But by the relation for the arithmetic mean and the geometric mean
 | Solution. Transform the equation to the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021}=\sqrt[202]{x^{2020}}, \text { from which } \\
& \frac{\begin{array}{l}
x+x+\ldots+x+1 \\
(2020 \text { terms) }
\end{array}}{2021}=\sqrt[202]{x^{2020}} .
\end{aligned}
$$
But by the inequality between the arithmetic mean and the ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. An athlete with a mass of 78.75 kg is testing a net used by firefighters to save people. The net sags by 100 cm when the athlete jumps from a height of 15 m. Assuming the net behaves elastically like a spring, calculate how much it will sag when a person with a mass of 45 kg jumps from a height of 29 m.
Given:
$m_... | Solution. The mechanical system "Earth-athlete-net" $x_{2}-?$ can be considered closed. According to the law of conservation of energy, when the athlete jumps, his potential energy should completely transform into the energy of the elastic deformation of the net: $m_{2} g\left(h_{2}+x_{2}\right)=\frac{k x_{2}^{2}}{2} ;... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $9^{x}+4 \cdot 3^{x+1}=13$. | Solution: $3^{2 x}+4 \cdot 3^{x+1}=13, 3^{2 x}+12 \cdot 3^{x}-13=0$,
$\left(3^{x}\right)_{1,2}=\frac{-12 \pm \sqrt{144+52}}{2}=\frac{-12 \pm 14}{2}=\left[\begin{array}{l}1, \\ -13 .- \text { n.s. }\end{array} 3^{x}=1 \Rightarrow x=0\right.$. Answer: $x=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=3 \sin x+4 \cos x$. | Solution: $f(x)=3 \sin x+4 \cos x=\sqrt{3^{2}+4^{2}} \sin \left(x+\operatorname{arctg} \frac{4}{3}\right)=5 \sin \left(x+\operatorname{arctg} \frac{4}{3}\right)$.
Answer: The maximum value is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $9^{x-1}+3^{x+2}=90$. | Solution: $9^{x-1}+3^{x+2}=90,\left(3^{x-1}\right)^{2}+27 \cdot 3^{x-1}=90$, $3^{x-1}=\frac{-27 \pm \sqrt{3^{6}+4 \cdot 3^{2} \cdot 10}}{2}=\frac{-27 \pm 3 \sqrt{121}}{2}=\frac{-27 \pm 33}{2}=\left[\begin{array}{l}3, \\ -30 .\end{array} \quad x-1=1, x=2\right.$ Answer: $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{x-3}{2 x+1}}+2=3 \sqrt{\frac{2 x+1}{x-3}}$. | Solution: $t=\frac{x-3}{2 x+1}>0 ; \sqrt{t}+2=\frac{3}{\sqrt{t}}, \sqrt{t}+2 \sqrt{t}-3=0, \sqrt{t}=\left[\begin{array}{l}1 \\ -3 \text {, but this is not valid }\end{array}\right.$
$\frac{x-3}{2 x+1}=1 ; x-3=2 x+1, x=-4, \sqrt{\frac{-7}{-7}}+2=3 \sqrt{\frac{-7}{-7}}$. Answer: $x=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{2 x+2}{x+2}}-\sqrt{\frac{x+2}{2 x+2}}=\frac{7}{12}$. | Solution: $\sqrt{\frac{2 x+2}{x+2}}-\sqrt{\frac{x+2}{2 x+2}}=\frac{7}{12} ; t=\frac{2 x+2}{x+2}>0, \sqrt{t}-\sqrt{\frac{1}{t}}=\frac{7}{12}$,
$\sqrt{t}=\frac{7 \pm \sqrt{49+4 \cdot 144}}{24}=\frac{7 \pm 25}{24}=\left[\begin{array}{l}\frac{32}{24}=\frac{4}{3} \\ -\frac{18}{24}=-\frac{3}{4}\end{array}, \frac{2 x+2}{x+2}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=6 \sin x+8 \cos x$. | Solution: $f(x)=6 \sin x+8 \cos x=\sqrt{6^{2}+8^{2}} \sin \left(x+\operatorname{arctg} \frac{8}{6}\right)$.
Answer: The maximum value is 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the greatest and least values of the function $y=x^{3}-3 x^{2}+5$ on the interval $[1 ; 3]$. | Solution: $y=x^{3}-3 x^{2}+5,[1 ; 3] ; y^{\prime}=3 x^{2}-6 x=3 x(x-2) \Rightarrow x=0 ; 2$
$y(1)=3 ; y(3)=5 ; y(2)=1$. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate $\sqrt{4+\sqrt{12}}-\sqrt{4-\sqrt{12}}$. | Solution: $\sqrt{4+\sqrt{12}}-\sqrt{4-\sqrt{12}}=A ; A^{2}=8-2 \sqrt{16-12}=4 ; \quad A= \pm 2$ Answer: $A=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate $\sqrt{3+\sqrt{8}}-\sqrt{3-\sqrt{8}}$. | Solution: $\sqrt{3+\sqrt{8}}-\sqrt{3-\sqrt{8}}=A ; A^{2}=6-2 \sqrt{9-8}=4, \quad A= \pm 2, \quad A=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $3 \cdot 9^{x}+2 \cdot 3^{x}=1$. | Solution: $3 \cdot 9^{x}+2 \cdot 3^{x}=1,3 \cdot\left(3^{x}\right)^{2}+2 \cdot 3^{x}-1=0,3^{x}=\frac{-2 \pm \sqrt{4+12}}{6}=\left[\begin{array}{l}\frac{1}{3}, \\ -1-\text { not valid }\end{array} \quad x=-1\right.$.
Answer: $x=-1$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Two vertices of a square with an area of $256 \mathrm{~cm}^{2}$ lie on a circle, while the other two vertices lie on a tangent to this circle. Find the radius of the circle. | Solution: $S_{A B C D}=256 \, \text{cm}^{2} \Rightarrow a=16 \, \text{cm}=x$,
$E F=2 R-x, F O=R-E F=R-(2 R-x)=x-R$,
$\triangle F C O: F O^{2}=R^{2}-F C^{2} \Rightarrow(x-R)^{2}=R^{2}-8^{2}$,
$x^{2}-2 R x+R^{2}=R^{2}-8^{2}, 2 R x=16^{2}+8^{2} \Rightarrow R=\frac{16^{2}+8^{2}}{2 \cdot 16}=10$. Answer:
$R=10$.
![](ht... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{x+3}{3 x-5}}+1=2 \sqrt{\frac{3 x-5}{x+3}}$. | Solution:
$$
\sqrt{\frac{x+3}{3 x-5}}+1=2 \sqrt{\frac{3 x-5}{x+3}}
$$
$\frac{x+3}{3 x-5}>0$
$t=\frac{x+3}{3 x-5}, \sqrt{t}+1=\frac{2}{\sqrt{t}}, \frac{t+\sqrt{t}-2}{\sqrt{t}}=0, \sqrt{t}=\frac{-1 \pm \sqrt{1+8}}{2}=\frac{-1 \pm 3}{2}=\left[\begin{array}{l}-2 \\ 1\end{array}\right.$
$t=1, \frac{x+3}{3 x-5}=1,3 x-5=x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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