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Problem 5.7. Vasya wants to place the numbers from 1 to 6 (each one only once) in the squares such that the following condition is met: if two squares are connected, the number in the higher square is greater. How many ways are there to do this?
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,842 |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to the line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on the line $A L$. Carefully finding the intersection point of thes... | (14;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,843 |
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 | 14,845 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,846 |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA + ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,847 |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,848 |
Problem 9.7. Two parallel lines are drawn through points $A(0 ; 14)$ and $B(0 ; 4)$. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L-... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,850 |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,851 |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
 and \(b\) are such that the equation
\[
a(x-a)^{2}+b(x-b)^{2}=0
\]
has a unique solution. Prove that \(|a|=|b|\).
(N. Agakhanov) | First solution. Let $|b| \neq|a|$. Then $b+a \neq 0$, and the given equation is quadratic: $(a+b) x^{2}-2\left(a^{2}+b^{2}\right) x+\left(a^{3}+b^{3}\right)=0$. In this case, its discriminant $\frac{D}{4}=\left(a^{2}+b^{2}\right)^{2}-(a+b)\left(a^{3}+b^{3}\right)=-a b(a-$ $-b)^{2}$ is not equal to zero, since $a, b$ ar... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,853 |
9.6. Thirty girls - 13 in red dresses and 17 in blue dresses - were dancing in a circle around a Christmas tree. Later, each of them was asked if their right neighbor was in a blue dress. It turned out that those who answered correctly were only the girls standing between girls in dresses of the same color. How many gi... | Answer: 17.
Solution: Consider any girl. The colors of the dresses of her left and right neighbors could have been: blue-blue, blue-red, red-blue, red-red. The girl answered "yes" in exactly the first two cases; therefore, she said "yes" exactly when her left neighbor was wearing a blue dress.
Thus, since exactly 17 ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,854 |
9.7. The perpendicular bisector of side $AC$ of an acute-angled triangle $ABC$ intersects the lines $AB$ and $BC$ at points $B_{1}$ and $B_{2}$, respectively, and the perpendicular bisector of side $AB$ intersects the lines $AC$ and $BC$ at points $C_{1}$ and $C_{2}$, respectively. The circumcircles of triangles $B B_{... | Solution. Let $O$ be the center of the circumcircle of triangle $ABC$. First, we will show that the line $OB$ is tangent to the circumcircle $\omega_b$ of triangle $B B_1 B_2$.
Let $AB < BC$; then the perpendicular bisector of side $AC$ intersects side $BC$ at point $B_2$, and the extension of side $AB$ beyond point $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,855 |
9.8. In the cells of an $8 \times 8$ board, the numbers 1 and -1 are placed (one number per cell). Consider all possible placements of the figure $\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). We will call such a placement unsuccessful if the sum of the numb... | Answer: 36.
Solution: We will show that in each "cross" of five cells on the board, there will be at least one unsuccessful placement. Suppose the opposite; let the numbers in the outer cells of the cross be \(a, b, c, d\), and the number in the central cell be \(e\); denote by \(S\) (M. Antipov) the sum of all these ... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,856 |
7.1. In the thirtieth kingdom, there are only two types of coins: 16 and 27 tugriks. Can you pay for a notebook that costs 1 tugrik and get change? | Answer: Yes, it is possible.
Solution. For example, you can pay with three coins of 27 tugriks and receive change with five coins of 16 tugriks.
Other examples can be provided in a general form:
a) pay with $3+16$ coins of 27 tugriks and receive change with $5+27 n$ coins of 16 tugriks, where $n$ is a natural number... | Yes,itispossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,857 |
7.2. Connect points $A$ and $B$ (see the figure) with a broken line consisting of four segments of equal length so that the following conditions are simultaneously met:
1) the ends of the segments can only be some of the marked points;
2) there should be no marked points inside the segments;
3) adjacent segments shoul... | Answer: One of the options for the construction of the broken line, as shown in Fig. 7.2 a, b, c (with accuracy up to symmetry relative to the line $A B$).
Fig. 7.2 a
, green (green), and yellow (yellow) be denoted as $B, G$, and $Y$ respectively. Since the green color is obtained by mixing two parts of yellow paint and one part of blue, the ... | Blue=27\,^2,\,Green=33\,^2,\,Yellow=16\,^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,859 |
7.4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar? | Answer: In the sixth jar, there are -16 beetles.
Solution. Let there be $x$ beetles in the first jar, then in the second jar there are no fewer than $x+1$ beetles, in the third jar no fewer than $x+2$ beetles, and so on. Thus, in the tenth jar there are no fewer than $x+9$ beetles. Therefore, the total number of beetl... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,860 |
9.1. The numbers $x=999$ and $x=1001$ satisfy the inequality $x^{2}+p x+q<0$. Prove that the discriminant of the quadratic trinomial $x^{2}+p x+q$ is greater than 4. | Solution. The set of solutions of the inequality $x^{2}+\mathrm{px}+\mathrm{q}1001-999=2$. Note that $x \quad=\sqrt{\mathrm{D}}$, i.e., $\mathrm{D}=\left(x_{2}-x_{1}\right)^{2}$. Therefore, $\mathrm{D}>2^{2}=4$. | D>4 | Algebra | proof | Yes | Yes | olympiads | false | 14,862 |
9.3. A caravan of camels 1 km long is moving uniformly across the desert. A rider rode from the end of the caravan to the beginning and returned to the end of the caravan. During this time, the caravan traveled 1 km. What distance did the rider travel if
$ km. | 1+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,863 |
9.4. Two circles of equal radii intersect at points B and C. On the first circle, a point A is chosen. Ray AB intersects the second circle at point D (D ≠ B). On ray DC, a point E is chosen such that DC = CE. Prove that angle DAE is a right angle. | Solution. See fig.
The common chord $\mathrm{BC}$ of equal circles subtends equal arcs, so the inscribed angles BDC and BAC, which subtend these arcs, are equal. But then triangle $\mathrm{A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,864 |
9.5. M is a four-digit number composed of non-zero digits, N is the number written with the same digits in reverse order. Prove that if $\mathrm{M}+\mathrm{N}$ is divisible by 101, then the sum of the two outer digits of the number $\mathrm{M}$ is equal to the sum of the two middle digits of the number M. | Solution. Let $\mathrm{M}=\overline{\mathrm{abcd}}(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}-$ digits), i.e., $\mathrm{M}=1000 \mathrm{a}+100 \mathrm{~b}+10 \mathrm{c}+\mathrm{d}$. Then $\mathrm{N}=1000 \mathrm{~d}+100 \mathrm{c}+10 \mathrm{b}+\mathrm{a}$ and $\mathrm{M}+\mathrm{N}=1001 \mathrm{a}+110 \mathrm{b}+1... | Number Theory | proof | Yes | Yes | olympiads | false | 14,865 | |
7.1 In the glass, there was a solution in which water made up $99 \%$. The glass with the solution was weighed, and the weight turned out to be 500 gr. After that, some of the water evaporated, so that in the end, the proportion of water was 98\%. What will be the weight of the glass with the resulting solution, if the... | Answer: 400 g.
Indication: Initially, the water was $0.99 \cdot(500-300)=198$ (g), and the substance was $200-198=2$ (g). After the water evaporated, 2 g of the substance make up $100 \%-98 \%=2 \%$ of the solution, so the entire solution weighs 100 g, and together with the glass, it weighs 400 g. | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,866 |
7.3. A cube is placed on a table. Can 8 numbers; $1,2, \ldots, 8$, be placed at its vertices so that the following two conditions are met: 1) if the two numbers at the ends of each vertical edge are added, then all four sums are the same; 2) the sum of the numbers on the top face is equal to the sum of the numbers on t... | Answer: Yes. Instruction Let's provide an example. Place the numbers $1,7,6,4$ at the vertices of the square ABCD of the lower base, and above them, at the vertices of the square $A_{1} B_{1} C_{1} D_{1}$, place the numbers $8,2,3,5$. Then the sums at the vertices of both squares are 18 each. On each of the vertical ed... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,867 |
7.4. Do there exist such two two-digit numbers that if 20 is added to the first one and 15 is subtracted from the second one, the resulting numbers will remain two-digit, and their product will be equal to the product of the original numbers? | Answer: they exist. Hint: An example can be given $a=16, b=27$. Indeed, $(16+20) \cdot(27-15)=16 \cdot 27=432$ (The analysis of the solution and all possible pairs are considered in problem 8.4). | 16\cdot27=432 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,868 |
11.1. A literature teacher decided to find out from the students of class 11 "A" how many people from their class are absent. He received the following answers: Peter: "More than one." Victor: "More than two." Tatyana: "More than three." Charles: "More than four." Polina: "Less than four." Shurik: "Less than three." Ho... | Solution: Note that if Charles is right, then not only he, but also Peter, Victor, and Tatyana told the truth - a total of 4 students. This is a contradiction. Therefore, Charles lied, and no more than four people are absent. Let's assume Tatyana is right. Then exactly four people are absent, and in this case, she, Pet... | 2,3,or4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,869 |
11.2. On the board, 10 real numbers are written; the sum of any three of them is greater than seven. Can it happen that
a) the sum of any seven of them is less than sixty;
b) the sum of any five of them is less than twenty-four?
Justify your answers. | Solution: a) Let the numbers be $a_{1}, a_{2}, \ldots, a_{10}$. Let $S$ be their sum. Then
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}>7 \\
a_{2}+a_{3}+a_{4}>7 \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
a_{9}+a_{10}+a_{1}>7 \\
a_{10}+a_{1}+a_{2}>7
\end{array}\right.
$$
If we add the left and right... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 14,870 |
11.3. On the edge $A A_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge length $3 \sqrt{2}$, a point K is marked. In space, a point $T$ is marked such that $T B=7$ and $T C=\sqrt{67}$. What are the smallest and largest values that the length of the segment TK can have? Justify your answer.
 g(y)=a x y+b x+c y+1
$$
holds, where $a, b, c$ are constants, and $x$ and $y$ are any real numbers. Prove that $a=b c$. | Solution: Substitute $y=0$ into the identity and we get that $g(0) f(x)=b x+1$. Similarly, substituting $x=0$, we get that $f(0) g(y)=c y+1$. Finally, substituting $x=y=0$, we get that $f(0) g(0)=1$. In particular, $f(0) \neq 0$ and $g(0) \neq 0$. Then
$$
\begin{gathered}
f(x)=\frac{b x+1}{g(0)}, \quad g(y)=\frac{c y+... | Algebra | proof | Yes | Yes | olympiads | false | 14,872 | |
11.5. Given an acute angle O. On one of its sides, we take a point $A_{1}$ and drop a perpendicular $A_{1} A_{2}$ to the other side of the angle. Then from point $A_{2}$, we drop a perpendicular to $O A_{1}$, obtaining point $A_{3}$, and so on. How can we use a compass and a straightedge to construct a segment equal to... | Solution: Let the measure of angle $O$ be $\alpha$, and the length of segment $A_{1} A_{2}=m-$ see the figure. From the similarity of triangles $O A_{1} A_{2}$ and $A_{1} A_{2} A_{3}$, we find that $\angle A_{1} A_{2} A_{3}=\alpha$, and $A_{2} A_{3}=m \cos \alpha$. Next, from the similarity of triangles $O A_{2} A_{3}$... | \frac{}{1-\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,873 |
11.6. In a competition held in a single round-robin format, each team plays against every other team exactly once. At the end of a single round-robin football tournament among 16 teams, it turned out that the team "Joker" won against all the teams that, in the final standings, are placed higher than "Joker" (by points ... | # Solution:
Method 1. We will call the teams that outperformed the "Joker" in the tournament table strong, and the teams that lagged behind - weak. Note that the position of the "Joker" will not change if we replay all matches where a strong team met a weak team and award the victory to the strong team. Therefore, we ... | 6thplace | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,874 |
10.1. New smartwatches cost 2019 rubles. Namzil has $\left(500^{2}+4 \cdot 500+3\right) \cdot 498^{2}-500^{2}$. $503 \cdot 497$ rubles. Will he have enough money to buy the smartwatches? | Answer: No.
Solution: Let $a=500$. Then the original expression is $\left(a^{2}+4 \cdot a+3\right) \cdot(a-2)^{2}-a^{2}$. $(a+3) \cdot(a-3)=\left(a^{2}+4 \cdot a+3\right) \cdot\left(a^{2}-4 a+4\right)-a^{2} \cdot\left(a^{2}-9\right)=a^{4}-4 a^{3}+4 a^{2}+4 a^{3}-16 a^{2}+$ $16 a+3 a^{2}-12 a+12-a^{4}+9 a^{2}=4 a+12$. ... | 2012<2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,875 |
10.2. Is the number $4^{2019}+6^{2020}+3^{4040}$ prime? | Answer: No.
Solution. Transform the expression: $4^{2019}+6^{2020}+3^{4040}=\left(2^{2}\right)^{2019}+2^{2020} \cdot 3^{2020}+3^{2 \cdot 2020}=$ $\left(2^{2019}\right)^{2}+2 \cdot 2^{2019} \cdot 3^{2020}+\left(3^{2020}\right)^{2}=\left(2^{2019}+3^{2020}\right)^{2}$. Therefore, the original number is the square of the ... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,876 |
10.3. In trapezoid $A B C D$, the bases $B C=3$ and $A D=9$, angle $\angle B A D=30$ degrees, and $\angle A D C=60$ degrees. A line through point $D$ divides the trapezoid into two equal areas. Find the length of the segment of this line that is inside the trapezoid. | Answer: $\sqrt{39}$.
Solution: Let $B C=x$, then $A D=3 x$. Drop a perpendicular from point $B$ to the base $A D$, denoted as $B H$. Then the area of triangle $A B D$ is $S_{A B D}=\frac{1}{2} \cdot A D \cdot B H=\frac{3}{2} x h$, and $S_{B C D}=\frac{1}{2} \cdot B C \cdot B H=\frac{1}{2} x h$, i.e., $S_{A B D}=3 \cdo... | \sqrt{39} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,877 |
10.4. A cube of size $5 \times 5 \times 5$ cells is divided into $1 \times 1 \times 1$ cubes, each of which contains a grasshopper. At some moment, each grasshopper moves to an adjacent cube (up, down, left, or right, provided that the cube exists). Will there be a cube that does not contain any grasshopper? | Answer: Yes, it will remain.
Solution: Let's color the cube in a checkerboard pattern. Since the total number of cells in the cube is \(5 \times 5 \times 5 = 125\), which is odd, the number of black and white cells cannot be equal. Let's assume, for definiteness, that there are more black cells. Then, the number of gr... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,878 |
10.5. In a unit square, 76 points are marked arbitrarily. Prove that there exists a circle of radius $\frac{1}{7}$ that can cover at least 4 points. | Solution: Divide the square into 25 equal squares. We will prove that there exists at least one square containing 4 points. Assume the opposite, then in any square there are no more than 3 points, and the total number of points is no more than 75 - a contradiction.
Describe a circle around the found square. The diamet... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 14,879 |
# Task № 3.2
Condition:
Mashenka beautifully writes the names on New Year's gifts. She wrote "MAMA" in 12 minutes, and the name "MILA" in 8 minutes. She spends the same amount of time on identical letters, and possibly different times on different letters. How long will it take her to write the name "LILI"? | Express the answer in minutes.
# | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,880 | |
# Task № 3.4
## Condition:
Grandma is embroidering her grandchildren's names on their towels. She embroidered the name "LANA" in 15 minutes, and the name "ALLA" in 12 minutes. She spends the same amount of time on the same letters, and possibly different times on different letters. How long will it take her to embroi... | Express the answer in minutes.
# | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,881 | |
# Task No. 8.1
## Condition:
Postman Pechkin thinks that his watch is 10 minutes slow, but in reality, it is 20 minutes fast. He agreed with Uncle Fyodor to meet at the post office at 12:00, and then called and said that he wouldn't make it on time and would be 17 minutes late. At what time will Pechkin actually arri... | Write the answer in the format HH:MM.
# | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,882 |
# Task No. 8.4
## Condition:
Hemul thinks that his watch is fast by 10 minutes, but in reality, it is slow by 15 minutes. He agreed with Snusmumrik to meet by the river at 14:00, and then decided to come 5 minutes earlier. What time will Hemul actually arrive at the meeting? | Write the answer in the format HH:MM.
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result directly. | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,883 | |
1. Dima wrote down three consecutive natural numbers, where the first (the smallest) is divisible by 5, the second is divisible by 4, and the third is divisible by 3. What numbers could Dima have written down? (Provide at least one example.) | # Solution:
For example, the following sequence works: 55, 56, 57 (these are the three smallest natural numbers with the required property). There are different ways to arrive at this solution. Firstly, a logical enumeration works: the first number must be divisible by 5 and also odd (so that the next number is divisi... | 55,56,57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,884 |
2. Usain runs one lap around the school stadium at a constant speed, while photographers Arina and Marina are positioned around the track. For the first 4 seconds after the start, Usain was closer to Arina, then for 21 seconds he was closer to Marina, and then until the finish, he was closer to Arina again. How long do... | # Solution:
It is not hard to see that regardless of Arina and Marina's positions, the entire circle of the school stadium is divided into two equal parts - half of the circle is closer to Arina and the other half is closer to Marina (this is half of the shorter arc between Arina and Marina and half of the longer arc ... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,885 |
3. A football player can pass the ball to another player if there are no other players on the line segment connecting them. Is it possible to arrange 6 football players on the field so that each can pass the ball to exactly 4 other players?
# | # Solution:
It is possible. An example is shown in the diagram on the right.
## Criteria:
A correct example of player placement - 7 points, no correct example - 0 points (an incorrect exam... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,886 |
4. Avoska and Neboska collected 64 nuts and laid them out in a row so that each pair of adjacent nuts differed by 1 gram. Prove that then they can divide all the nuts between themselves so that they get an equal number of nuts and an equal amount of mass.
# | # Solution:
Let's divide all the nuts into quartets of consecutive nuts. Consider any of these quartets. Let the mass of the first nut in this quartet be \( m \), then the following mass variants for the four nuts (in grams) are possible:
(1) \( m, (m+1), (m+2), (m+3) \) or \( m, (m-1), (m-2), (m-3) \);
(2) \( m, (m... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 14,887 |
5. On the board are all natural numbers from 1 to 100: 1, 2, 3, 4, ..., 99, 100. Each student who enters the classroom circles the numbers that are not divisible by any of the other written numbers, and then erases all the circled numbers (obviously, the first student to enter will erase only the number 1). When Karl e... | Answer: 64 and 96.
## Solution:
In the next step, after erasing the ones, all prime numbers will be erased (they have no divisors other than 1 and themselves). Next, the numbers that have exactly two prime factors (not necessarily distinct) in their factorization will be erased, followed by the numbers that have exac... | 6496 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,888 |
10.5. Thirty girls -13 in red dresses and 17 in blue dresses - were dancing in a circle around a Christmas tree. Later, each of them was asked if their right neighbor was in a blue dress. It turned out that those who answered correctly were only the girls standing between girls in dresses of the same color. How many gi... | # Answer: 17.
Solution. Consider any girl. The colors of the dresses of her left and right neighbors could have been: blue-blue, blue-red, red-blue, red-red. The girl answered "yes" in exactly the first two cases; therefore, she said "yes" exactly when her left neighbor was wearing a blue dress.
Thus, since exactly 1... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,889 |
10.6. Natural numbers $a, b$ and $c$, where $c \geqslant 2$, are such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Prove that at least one of the numbers $a+c, b+c$ is composite.
(V. Senderov) | Solution. It is sufficient to show that at least one of the two numbers $d_{a}=\operatorname{GCD}(a, c)$ and $d_{b}=\operatorname{GCD}(b, c)$ is greater than 1. Indeed, if, for example, $d_{a}>1$, then $a+c$ is divisible by $d_{a}$ and $a+c>d_{a}$, which means that $a+c$ is a composite number.
From the equality $\frac... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,890 |
10.8. On a circle of length 2013, 2013 points are marked, dividing it into equal arcs. A chip is placed at each marked point. We define the distance between two points as the length of the shorter arc between them. For what largest $n$ can the chips be rearranged so that there is again one chip at each marked point, an... | Answer. $n=670$.
Solution. Let's number the points and the chips placed on them in a clockwise direction with consecutive non-negative integers from 0 to 2012. Consider an arbitrary permutation and the chips with numbers 0, 671, and 1342, initially located at the vertices of an equilateral triangle. The pairwise dista... | 670 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,892 |
Problem 1. A paper rectangle $3 \times 7$ was cut into squares $1 \times 1$. Each square, except those that stood at the corners of the rectangle, was cut along both diagonals. How many small triangles were obtained? | Answer: 68.
Solution. Note that the total number of squares cut along the diagonals is $3 \cdot 7-4=17$. Each of them is cut into 4 small triangles. Therefore, there will be $4 \cdot 17$ - 68 small triangles. | 68 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,893 |
Problem 2. Alla thought of a three-digit number, in which there is no digit 0, and all digits are different. Bella wrote down the number in which the same digits are in reverse order. Galia subtracted the smaller number from the larger one. What digit stands in the tens place of the resulting difference?
# | # Answer: 9
Solution. Since we are subtracting a smaller number from a larger one, the digit in the hundreds place of the first number is greater than that of the second. Then, in the units place, conversely, the digit of the first number is smaller than that of the second. Therefore, when subtracting, we will have to... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,894 |
Problem 3. Masha has 4 pieces of red plasticine, 3 pieces of blue plasticine, and 5 pieces of yellow plasticine. First, she divided each non-red piece of plasticine in half, and then she divided each non-yellow piece of plasticine in half. How many pieces of plasticine did Masha get | Answer: 30 pieces of plasticine.
Solution. After Masha's first action, the number of blue and yellow pieces of plasticine doubles. They become 6 and 10, respectively. After Masha's second action, the number of red and blue pieces of plasticine doubles. They become 8 and 12, respectively. Then the total number of piece... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,895 |
Problem 4. A square area was paved with square tiles (all tiles are the same). A total of 20 tiles adjoin the four sides of the area. How many tiles were used in total?
 | Answer: 36 tiles.
Solution. Let's call the tiles adjacent to the top side of the platform "top tiles". Similarly, we define "right", "left", and "bottom" tiles. Notice that if we add the number of top tiles, the number of bottom tiles, the number of right tiles, and the number of left tiles, we will count the corner t... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,896 |
Problem 5. Merlin decided to weigh King Arthur on enchanted scales that always err by the same weight in the same direction. When Merlin weighed Arthur, they showed a weight of 19 stones. Then Merlin weighed the royal horse and got a weight of 101 stones. Finally, Merlin weighed Arthur on the horse, and the scales show... | Answer: 13 stones.
Solution. Note that if we add 19 stones and 101 stones, we get the combined weight of Arthur and the horse, to which the scale error has been added (or subtracted) twice. Meanwhile, 114 stones is the combined weight of Arthur and the horse, to which the scale error has been added only once. Therefor... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,897 |
Problem 6. Peter has 5 rabbit cages (the cages are in a row). It is known that there is at least one rabbit in each cage. We will call two rabbits neighbors if they sit either in the same cage or in adjacent ones. It turned out that each rabbit has either 3 or 7 neighbors. How many rabbits are sitting in the central ca... | Answer: 4 rabbits.
Solution. Let's number the cells from 1 to 5 from left to right.
Notice that the neighbors of the rabbit in the first cell are all the rabbits living in the first two cells. The neighbors of the rabbit in the second cell are all the rabbits living in the first three cells. The third cell cannot be ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,898 |
Problem 7. In the queue for the school cafeteria, 16 schoolchildren are standing in such a way that boys and girls alternate. (The first is a boy, followed by a girl, then a boy again, and so on.) Any boy who is followed by a girl in the queue can swap places with her. After some time, it turned out that all the girls ... | Answer: 36 exchanges.
Solution. Note that in the end, each boy will make one exchange with each girl who is in line after him. That is, the first boy will make 8 exchanges, the second - 7, the third - 6, and so on. Then the total number of exchanges is
$$
8+7+6+5+4+3+2+1=36.
$$
It is not hard to understand that if e... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,899 |
Problem 8. Seryozha placed numbers from 1 to 8 in the circles so that each number, except one, was used exactly once. It turned out that the sums of the numbers on each of the five lines are equal. Which number did Seryozha not use?
}=\frac{1}{n}-\frac{1}{n+1}$. Then the sum equals $1-\frac{1}{2021}=\frac{2020}{2021}$. | \frac{2020}{2021} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,902 |
7.4. How to cut a $5 \times 5$ square with straight lines so that the resulting pieces can be used to form 50 equal squares? It is not allowed to leave unused pieces or overlap them. | Solution. First, cut the $5 \times 5$ square into 25 squares of $1 \times 1$, then cut each of the resulting squares along the diagonals into 4 triangles, from which, by attaching the longer sides of two triangles to each other, 2 squares can be formed:
^{2}-4\left(2 \mathrm{x}^{2}+\mathrm{x}-1\right)=\mathrm{x}^{2}-4 \mathrm{x}+4=(\mathrm{x... | -x-1-2x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,905 |
11.1. Each of 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than $2 ”, \ldots$, the tenth said: “My number is greater than 10”. After that, all ten, sp... | Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My num... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,907 |
11.2. It is known that each of the trinomials $x^{2}+a x+b$ and $x^{2}+a x+b+1$ has at least one root, and all roots of these trinomials are integers. Prove that the trinomial $x^{2}+a x+b+2$ has no roots. | The first solution. Let $D_{1}, D_{2}, D_{3}$ be the discriminants of these three polynomials, respectively. The first two equations have only integer roots, so $D_{1}=m^{2}, D_{2}=n^{2}$, where the numbers $m$ and $n$ can be considered non-negative integers. Subtracting the second equation from the first, we get $4=m^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,908 |
11.3. Let's call the distance between two cells on a checkerboard the minimum number of moves a chess king can make to get from one to the other. Find the maximum number of cells that can be marked on a $100 \times 100$ board such that there are no two marked cells with a distance of 15 between them.
(I. Bogdanov) | Answer. $55^{2}=3025$ cells.
Solution. Divide the board into 9 squares $30 \times 30$, 6 rectangles $10 \times 30$, and one square $10 \times 10$ (see Fig. 5). In each $30 \times 30$ square, the cells are divided into $15^{2}$ groups of four such that the distance between any two cells in the same group is 15 (each gr... | 3025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,909 |
11.4. An infinite sequence of non-zero numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that for all natural $n \geqslant 2018$ the number $a_{n+1}$ is the smallest root of the polynomial
$$
P_{n}(x)=x^{2 n}+a_{1} x^{2 n-2}+a_{2} x^{2 n-4}+\ldots+a_{n} .
$$
Prove that there exists an $N$ such that in the infinite sequen... | Solution. Let $n \geqslant 2018$. Notice that $P_{n}(a)=P_{n}(-a)$ for all $a$. Therefore, since $P_{n}(x)$ has a non-zero root, it also has a negative root, from which it follows that $a_{n+1}<0$.
Furthermore, since $P_{n+1}(x)=x^{2} P_{n}(x)+a_{n+1}$, we have
$$
P_{n+1}\left(a_{n+1}\right)=a_{n+1}^{2} P_{n}\left(a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,910 |
11.5. In the tetrahedron \(ABCD\), the altitudes \(BE\) and \(CF\) are drawn. The plane \(\alpha\) is perpendicular to the edge \(AD\) and passes through its midpoint. It is known that the points \(A, C, D\), and \(E\) lie on the same circle, and the points \(A, B, D\), and \(F\) also lie on the same circle. Prove that... | Solution. Line $C F$ is perpendicular to plane $A B D$, so $C F \perp A D$. Similarly, $B E \perp A D$. Therefore, lines $C F$ and $B E$ are parallel to plane $\alpha$ or lie in it. Points $B$, $C$, $E$, and $F$ lie on the sphere $\omega$ circumscribed around the tetrahedron $A B C D$.
Also, since $\angle B E C=90^{\c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 14,911 |
1. Solve the equation $1-(2-(3-(\ldots 2010-(2011-(2012-x)) \ldots)))=1006$. | # Answer. $x=2012$
Solution. Opening the brackets, we get $1-2+3-4+\ldots+2011-2012+x=$ 1006; $-1006+x=1006 ; x=2012$. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,912 |
2. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part. | Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,913 |
3. One of the angles of the trapezoid is $60^{\circ}$. Find the ratio of its bases, if it is known that a circle can be inscribed in this trapezoid and a circle can be circumscribed around this trapezoid. | Answer. $1: 3$
## Solution.
A
Since $ABCD$ is a cyclic quadrilateral, it is isosceles, i.e., $AB = CD$. Since $\angle BAD = 60^\circ$, then $\angle ABC = 120^\circ$. The center of the insc... | 1:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,914 |
4. Solve the numerical riddle: TETA+BETA=GAMMA. (Different letters - different digits.)
# | # Answer. $4940+5940=10880$
Solution. Since $A+A$ ends in $A$, then $A=0$. Since $\Gamma$ is the result of carrying to the next place, then $\Gamma=1$. Since $A+A$ ends in $A$, then $A=0$. This means there is no carry to the tens place, i.e., $T+T$ ends in $M$, and thus $M$ is even. There is also no carry to the hundr... | 4940+5940=10880 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,915 |
5. Does there exist a natural number $\mathrm{n}$ such that the number $\mathrm{n}^{2012}-1$ is some power of two | Answer. No, it does not exist.
Solution. Transform: $n^{2012}-1=\left(n^{1006}\right)^{2}-1=\left(n^{1006}-1\right)\left(n^{1006}+1\right)$. Suppose that this number is a power of two, then each of the two resulting factors is also a power of two, and these factors differ by 2. This is only possible in one case, if $n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,916 |
6. In five 15-liter buckets, there are 1, 2, 3, 4, and 5 liters of water respectively. It is allowed to triple the amount of water in any container by pouring water from one other container (if there is not enough water to triple the amount, then it is not allowed to pour from this bucket). What is the maximum amount o... | # Answer
The answer depends on the interpretation of the condition -- whether it is allowed to pour NOT all the contents of the bucket (essentially -- whether it is possible to measure OUT ONE LITER of water)
A) If it is not allowed, then the answer is 9 liters
B) If it is allowed, then the answer is 12 liters
## S... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,917 |
11.1. While walking in the park, Seryozha and Misha stumbled upon a meadow surrounded by lindens. Seryozha walked around the meadow, counting the trees. Misha did the same but started from a different tree (although he went in the same direction). The tree that was the $20-\mathrm{th}$ for Seryozha was the $7-\mathrm{t... | # Answer: 100.
Solution. First method. Let there be $n$ trees growing around the glade. We will calculate in two ways the number of intervals between the two trees mentioned in the problem's condition. During Sergei's walk: $20-7=13$. During Misha's walk: $7+(n-94)=n-87$. Therefore, $n-87=13$, which means $n=100$.
Se... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,918 |
11.2. In triangle $A B C$, angle $C$ is $75^{\circ}$, and angle $B$ is $60^{\circ}$. The vertex $M$ of the isosceles right triangle $B C M$ with hypotenuse $B C$ is located inside triangle $A B C$. Find angle $M A C$. | Answer: $30^{\circ}$.
First method. From the problem statement, it follows that $\angle B A C=45^{\circ}$. Draw a circle with center $M$ and radius $M B=M C$ (see Fig. 11.2). Since $\angle B M C=90^{\circ}$, the larger arc $B C$ of this circle is the locus of points from which the chord $B C$ is seen at an angle of $4... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,919 |
11.3. For a quadratic trinomial $f(x)$ and some real numbers $l, t$ and $v$, the following equalities are satisfied: $f(l)=$ $t+v, f(t)=l+v, f(v)=l+t$. Prove that among the numbers $l, t$ and $v$ there are equal ones. | Solution. Let $f(x)=a x^{2}+b x+c(a \neq 0)$. From the condition of the problem, the following equalities follow:
$$
\left\{\begin{aligned}
a l^{2}+b l+c & =t+v \\
a t^{2}+b t+c & =v+l \\
a v^{2}+b v+c & =l+t
\end{aligned}\right.
$$
Subtracting equation (2) from equation (1), and then equation (3) from equation (1), ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,920 |
11.4. On the computer screen is the number 12. Every second, the number on the screen is either multiplied or divided by 2 or 3. The result of the operation appears on the screen in place of the written number. Exactly one minute later, a number appears on the screen. Could it have been the number 54? | Answer: No, it could not.
Solution. Note that $12=2^{2} \cdot 3^{1}$, which means the total exponent of the factors (twos and threes) is 3. Regardless of the operation performed, each change in the number results in a change of the total exponent by 1. A total of 60 such changes should occur. Therefore, after 60 secon... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,921 |
11.5. Given a regular triangular pyramid $S A B C$, the edge of the base of which is equal to 1. Medians of the lateral faces are drawn from vertices $A$ and $B$ of the base $A B C$, which do not have common points. It is known that the edges of a certain cube lie on the lines containing these medians. Find the length ... | Answer: $\frac{\sqrt{6}}{2}$.
Solution. The specified medians $A D$ and $B E$ of the lateral faces $A S B$ and $B S C$ lie on skew lines, and the skew edges of the cube are mutually perpendicular. Therefore, the condition for the existence of a cube with edges on the specified lines is equivalent to the perpendiculari... | \frac{\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,922 |
11.6. On a circle, 20 points are marked. How many triples of chords with endpoints at these points exist such that each chord intersects each other (possibly at the endpoints)? | Answer: 156180.
Solution. The ends of the sought chords can be 3, 4, 5, or 6 points. Let's consider these cases.
1) The ends of the chords are 3 points (see Fig. 11.6a). They can be chosen in $C_{20}^{3}$ ways. Each triplet of points can be connected by chords in a unique way.
![](https://cdn.mathpix.com/cropped/202... | 156180 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,923 |
8.1. "Just you wait!" roared the wolf, noticing the hare 30 m away, and rushed after him when the hare had 333 m left to reach his hiding place. Will the wolf catch the hare if the wolf runs 600 m per minute and the hare runs 550 m per minute? Justify your answer. | # 8.1. Answer. Will catch up.
Solution. The closing speed of the wolf and the hare is 50 m per minute. So, the wolf needs $\frac{3}{5}$ minutes to catch the hare. But in this time, the hare will run 330 m, which means it will just barely not make it to the shelter. | Willcatchup | Other | math-word-problem | Yes | Yes | olympiads | false | 14,924 |
8.2. In a triangle, one of the medians is perpendicular to one of the angle bisectors. Prove that one of the sides of this triangle is twice as long as another. | 8.2. Solution. Note that the given median and bisector cannot come from the same vertex, as otherwise the angle at this vertex would be greater than $180^{\circ}$. Now let the bisector $A D$ and median $C E$ of triangle $A B C$ intersect at point $F$. Then $A F$ is the bisector and altitude in triangle $A C E$, which m... | AB=2AC | Geometry | proof | Yes | Yes | olympiads | false | 14,925 |
8.3. Prove that the number 2022 cannot be represented as the difference of squares of two numbers. | 8.3. Solution. The square of any integer, when divided by 4, gives a remainder of either 0 or 1. Then the difference of the squares of two integers, when divided by 4, can give a remainder of 0, 1, or 3. The number $2022=4 \cdot 505+2$ gives a remainder of 2 when divided by 4. Therefore, it cannot be the difference of ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,926 |
8.4. Can the numbers from 1 to 9 be placed in the cells of a $3 \times 3$ square so that the sum of any two numbers in adjacent cells (cells sharing a side) is a prime number? | # 8.4. Answer. No.
Solution. Let $a, b, c, d, e, f, g, h, k$ be the desired arrangement.
| $a$ | $b$ | $c$ |
| :--- | :--- | :--- |
| $d$ | $e$ | $f$ |
| $g$ | $h$ | $k$ |
Then, from the simplicity of the sums $a+b, b+c, \ldots, a+d, \ldots, f+k$, it follows that they are odd (each sum is greater than 2). Therefore,... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 14,927 |
8.5. One hundred athletes are lined up. Each of them is wearing a red or blue sports suit. If an athlete is wearing a red suit, then the athlete standing nine people away from him is wearing a blue suit. Prove that no more than 50 athletes are wearing red suits. | 8.5. Solution. Let's number the athletes in the row and consider the first 20
According to the condition, in each such pair, there is no more than one athlete in a red costume. Therefore, amo... | 50 | Combinatorics | proof | Yes | Yes | olympiads | false | 14,928 |
11.5. An immortal flea jumps along integer points on the number line, starting from point 0. The length of the first jump is 3, the second is 5, the third is 9, and so on (the length of the $k$-th jump is $2^{k}+1$). The direction of the jump (right or left) is chosen by the flea. Can it happen that the flea eventually... | Answer. Yes.
Solution. We will show how the flea can jump, sequentially landing in points $0,1,2,3, \ldots$ (each time - in several jumps). For this, it is sufficient to show how, having landed in point $n$, it can jump to point $n+1$ in several jumps.
Suppose that before landing in point $n$, the flea has made $k-1$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 14,929 |
11.6. Real numbers \(a, b, c, d\), each with a modulus greater than one, satisfy the relation
\[
a b c + a b d + a c d + b c d + a + b + c + d = 0
\]
Prove that
\[
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} + \frac{1}{d-1} > 0
\]
(K. Ivanov) | Solution. Let $x=\frac{a+1}{a-1}, y=\frac{b+1}{b-1}, z=\frac{c+1}{c-1}, t=$ $=\frac{d+1}{d-1}$. Since the absolute values of $a, b, c, d$ are greater than one, the numbers $x, y, z, t$ are positive (and not equal to 1).
The given relation can be rewritten as
$$
(a+1)(b+1)(c+1)(d+1)=(a-1)(b-1)(c-1)(d-1)
$$
or $x y z ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,930 |
11.7. In a scalene triangle $ABC$ inscribed in a circle $\omega$, the tangent to this circle at point $C$ intersects the line $AB$ at point $D$. Let $I$ be the center of the circle inscribed in triangle $ABC$. The lines $AI$ and $BI$ intersect the bisector of angle $CDB$ at points $Q$ and $P$ respectively. Let $M$ be t... | The first solution. Let, without loss of generality, point $D$ lie on ray $B A$. Let the angle bisectors $A I$ and $B I$ of the triangle intersect $\omega$ again at points $A^{\prime}$ and $B^{\prime}$, respectively. Finally, let $L$ be the midpoint of arc $A C B$ (see Fig. 4).
$.
Solution. Introduce a coordinate system on the plane so that the centers of the cells, and only they, have integer coordinates. We will say that a cell has the same coordinates as its center. We will call a rectangle $a \times b$ vertical or horizontal if its side ... | \alpha=1/(2^{2}-2+b^{2}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,932 |
1. (7 points) During a sale, Petr bought trousers with a $40 \%$ discount and a shirt with a $20 \%$ discount. The next day, Ivan bought the same trousers and shirt without any discounts. Could Ivan have paid one and a half times more than Petr? Justify your answer. | Answer: Could.
Solution: Let the trousers cost $x$ rubles and the shirt cost $y$ rubles without any discount. Then, Peter paid $0.6 x + 0.8 y$ rubles, and Ivan paid $x + y$ rubles. We get the equation $1.5 \cdot (0.6 x + 0.8 y) = x + y$, from which $x = 2 y$. Thus, if the trousers cost twice as much as the shirt, Ivan... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,933 |
2. (7 points) Provide an example of a number $x$ for which the equality $\sin 2017 x - \operatorname{tg} 2016 x = \cos 2015 x$ holds. Justify your answer. | Answer. For example, $\frac{\pi}{4}$.
Solution. Since $\frac{2016 \pi}{4}=504 \pi=252 \cdot 2 \pi$ is a multiple of the period, we have
$$
\begin{gathered}
\sin \frac{2017 \pi}{4}=\sin \left(252 \cdot 2 \pi+\frac{\pi}{4}\right)=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2} \\
\cos \frac{2015 \pi}{4}=\cos \left(252 \cdot 2 \p... | \frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,934 |
3. (7 points) Rubik made a net of a $3 \times 3 \times 3$ cube and marked two points on it - see the image. What will be the distance between these points after Rubik glues the net into a cube?
. The given points are two opposite vertices of the cube $2 \times 2 \times 2$. In a cube $2 \times 2 \times 2$, the diagonal has a length of $2 \sqrt{3}$.
 Do there exist such three real numbers that if they are placed in one order as the coefficients of a quadratic trinomial, it will have two distinct positive roots, and if placed in another order, it will have two distinct negative roots? | Answer. No.
Solution. Let the quadratic polynomial $a x^{2}+b x+c$ have two negative roots $x_{1}$ and $x_{2}$. Then $b / a=-\left(x_{1}+x_{2}\right)>0$ and $c / a=x_{1} x_{2}>0$, which means that the numbers $b$ and $c$ have the same sign as the number $a$. Suppose, by some permutation of the coefficients, we obtaine... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,936 |
5. (7 points) From the midpoint of each side of an acute-angled triangle with area $S$, perpendiculars are drawn to the other two sides. Find the area of the hexagon bounded by these perpendiculars. | Answer: $\frac{S}{2}$.
## Solution.
 | \frac{S}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,937 |
6. (7 points) If the number $A$ is written on the board, you can add any of its divisors, except 1 and $A$ itself. Can you get 1234321 from $A=4$? Answer: Yes. | Solution. Adding a divisor $n$ to a number means adding $n$ to a number of the form $k n$. The result will be a number of the form $(k+1)n$. Note that the number 1234321 is divisible by 11. Then to the number $A=4=2 \cdot 2$, we will add 2 until we get the number $2 \cdot 11$: $2 \cdot 2 \rightarrow 2 \cdot 3 \rightarr... | 1234321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 14,938 |
1. Prove that the expression $x^{5}-4 x^{4} y-5 y^{2} x^{3}+20 y^{3} x^{2}+4 y^{4} x-16 y^{5}$ is not equal to 77 for any integer values of $x$ and $y$. | Solution: Let's factorize the given expression:
$$
\begin{aligned}
& x^{5}-4 x^{4} y-5 y^{2} x^{3}+20 y^{3} x^{2}+4 y^{4} x-16 y^{5}=x^{4}(x-4 y)-5 x^{2} y^{2}(x-4 y)+4 y^{4}(x-4 y)= \\
& =(x-4 y)\left(x^{4}-5 x^{2} y^{2}+4 y^{4}\right)=(x-4 y)(x+2 y)(x-y)(x+y)
\end{aligned}
$$
We need to check that the factors are p... | proof | Algebra | proof | Yes | Yes | olympiads | false | 14,939 |
2. Point $\mathbf{E}$ is the midpoint of side AB of parallelogram ABCD. On segment DE, there is a point F such that $\mathrm{AD}=\mathbf{B F}$. Find the measure of angle CFD. | Solution: Extend $\mathrm{DE}$ to intersect line $\mathrm{BC}$ at point $\mathrm{K}$ (see the figure). Since $\mathrm{BK} \| \mathrm{AD}$, we have $\angle \mathrm{KBE} = \angle \mathrm{DAE}$. Additionally, $\angle \mathrm{KEB} = \angle \mathrm{DEA}$ and $\mathrm{AE} = \mathrm{BE}$, so triangles $\mathrm{BKE}$ and $\mat... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 14,940 |
3. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter Г on the board, at least one colored cell is found? | Solution: Consider a $2 \times 3$ rectangle. In it, obviously, a minimum number of cells need to be colored. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown in the figure.
... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 14,941 |
4. Let $f(x)=x^{2}+b x+c$. It is known that the discriminant of the quadratic equation $f(x)=0$ is 2020. How many roots does the equation $f(x-2020)+f(x)=0$ have? | Solution: Since the distance between the roots is $\mathrm{D}$, the equations $\mathrm{f}(\mathrm{x})=0$ and $\mathrm{f}(\mathrm{x}-\mathrm{D})=0$ have a common root. If we draw the graphs of these functions, they will intersect the x-axis at the same point (this root). The vertical line passing through this point is t... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 14,942 |
8.1. Can the numbers from 1 to 2022 be arranged in a circle so that each number is divisible by the difference of its neighbors?
Solution. No, because an odd number cannot be divisible by an even number, so odd numbers must be placed in pairs, but there are only 1011 odd numbers, making such an arrangement impossible. | Answer: no.
## Criteria:
7 points - complete solution;
3 points - noted that odd numbers must stand in pairs, further progress is absent;
1 point - noted that next to an odd number stand numbers of different parity or, that from 1 to 2022 there are 1011 odd numbers. | no | Number Theory | proof | Yes | Yes | olympiads | false | 14,944 |
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