problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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Problem 4. The median $\mathrm{AA}_{0}$ of triangle $\mathrm{ABC}$ is laid off from point $\mathrm{A}_{0}$ perpendicular to side $\mathrm{BC}$ to the outside of the triangle. Denote the second endpoint of the constructed segment as $\mathrm{A}_{1}$. Similarly, points $\mathrm{B}_{1}$ and $\mathrm{C}_{1}$ are constructe... | Answer._ $\triangle \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$ is equilateral, all angles are $60^{\circ}$.
## Solution.
Since $\triangle \mathrm{ABC}$ is isosceles, $\mathrm{BB}_{0}$ is the perpendicular bisector of the base AC. Therefore, $\mathrm{B}_{1}$ lies on this perpendicular and $\mathrm{CB}_{0} \perp \m... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,761 |
Problem 5. The robbers have 13 gold bars. There are scales that can be used to find the total weight of any two bars. Come up with a way to determine the total weight of all the bars in 8 weighings.
# | # Solution.
We will take the first three ingots and weigh them in pairs: C1+C2, C1+C3, C2+C3, using up three weighings. By adding the results of these weighings and dividing by two, we will find the total weight of these three ingots: ((C1+C2) + (C1+C3) + (C2+C3))/2 = C1+C2+C3. Using the remaining five weighings, we w... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,762 |
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 | 18,763 |
10.2. Can a cube $20 \times 20 \times 20$ be cut into rectangular parallelepipeds such that the length, width, and height of each of them, in some order, are consecutive natural numbers | Answer: No.
Solution: Suppose such a cutting is possible. Since the length, width, and height of the parallelepiped are three consecutive natural numbers, this means that one of these dimensions is a number divisible by 3. Therefore, the volume of each parallelepiped resulting from the cutting will be divisible by 3. ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,764 |
10.3. The numbers $a, b, c$ are the lengths of the sides of a triangle. Prove that if the angle opposite side $b$ in the triangle is greater than $60^{\circ}$, then the equation $a x^{2}+2 b x+c=0$ has two distinct roots. | Solution. Let's write down the cosine theorem: $b^{2}=a^{2}+c^{2}-2 a c \cos \beta$. From the condition, it follows that $\cos \beta a^{2}+c^{2}-a c$. Let's estimate a quarter of the discriminant of the quadratic given in the condition: $D / 4=b^{2}-a c>a^{2}+c^{2}-2 a c=(a-c)^{2} \geq 0$, that is, $D>0$. The statement... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,765 |
10.4. On a circle with center $O$ and diameters $A B$ and $C D$, points $E$ and $F$ are chosen such that $O D$ is the bisector of angle $E O B$ and $A E = C F$ (points $E$ and $F$ are on opposite sides of $C D$). Prove that the chord $E F$ cuts off an isosceles triangle from angle $E D C$. | Solution. Let chord $E F$ intersect radius $O D$ at point $K$. We will prove the equality of angles $E D K$ and $E K D$. By the property of inscribed angles, $\angle E D K = \angle E D C$ is measured by half the degree measure of arc $EAC$, that is, half the sum of the degree measures of arcs $EA$ and $AC$. Angle $EKD$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,766 |
10.5. A circle is inscribed with 103 numbers. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 103 written numbers? | Answer: 42.
Solution. We will show that there will be 3 consecutive numbers, among which there are at least 2 positive ones. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them, there are at least 6 positive numbers. But th... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,767 |
10.1. We added three different two-digit numbers $\overline{x x}, \overline{y y}, \overline{z z}$ and obtained a three-digit number $\overline{x y z}$. Find all possible values of $x, y, z$. The notation $\overline{a b \ldots c}$ means a number where $a$ is the first digit, $b$ is the second digit, $c$ is the last digi... | Answer: $x=1, y=9, z=8$.
Solution. Let's write the equation as $11 x+11 y+11 z=100 x+10 y+$ $+z$. Then $10 z+y=89 x$. If $x=1$, then $10 z+y=89$, and since $z$ and $y$ are digits, we have $z=8, y=9$. If $x \geq 2$, then $10 z+y \geq 178$, which is impossible, as the left side is a two-digit number. | 1,9,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,768 |
10.2. In the bottom left corner of a $7 \times 7$ chessboard, there is a king. In one move, he can move one square to the right, or one square up, or one square diagonally to the right and up. In how many different ways can the king travel to the top right corner of the board, if he is forbidden to visit the central ce... | Answer: 5020.
Solution. We will construct a $7 \times 7$ table, in each cell of which we will write the number equal to the number of allowed paths by which the king can reach this cell from the bottom left corner. First, we fill the left column and the bottom row with ones and write 0 in the central cell (as per the ... | 5020 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,769 |
10.3. The equation $P(x)=0$, where $P(x)=x^{2}+b x+c$, has a unique root, and the equation $P(P(P(x)))=0$ has exactly three distinct roots. Solve the equation $P(P(P(x)))=0$. | Answer: $1 ; 1+\sqrt{2} ; 1-\sqrt{2}$.
Solution. From the condition, it follows that $P(x)=(x-\alpha)^{2}$. The equation $P(P(P(x)))=0$ has the form
$$
\left(\left((x-\alpha)^{2}-\alpha\right)^{2}-\alpha\right)^{2}=0
$$
Thus $\left((x-\alpha)^{2}-\alpha\right)^{2}=\alpha$ and therefore $\alpha>0$. From the last equa... | 1;1+\sqrt{2};1-\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,770 |
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 | 18,771 |
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 | 18,772 |
2. Vasya has coins worth 49 tugriks, and Petya has coins worth 99 tugriks (each has a sufficient number of coins). Vasya owes Petya one tugrik. Will they be able to settle the debt? | Solution. We need to find a solution in non-negative integers for the equation $49 n-99 m=1$. Notice that $99-49 \cdot 2=1$ or $49 \cdot(-2)-99 \cdot(-1)=1(*)$. This solution is not suitable for us due to the negativity of the coefficients. Consider the homogeneous equation $49 x-99 y=0$. It has a series of solutions $... | 97\cdot49-48\cdot99=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,774 |
3. Find the maximum value of the expression $x^{2}+y^{2}+z^{2}$, if $x, y, z$ are integers satisfying the system
$$
\left\{\begin{array}{l}
x y+x+y=20 \\
y z+z+y=6 \\
x z+x+z=2
\end{array}\right.
$$ | Solution. Adding 1 to both sides of each equation in the system, we get:
$$
\left\{\begin{array}{l}
x y + x + y + 1 = 21 \\
y z + z + y + 1 = 7 \\
x z + x + z + 1 = 3
\end{array}\right.
$$
Factoring the left-hand sides, we obtain:
$$
\left\{\begin{array}{l}
(x+1)(y+1) = 21 \\
(y+1)(z+1) = 7 \\
(z+1)(x+1) = 3
\end{ar... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,775 |
4. Two circles $\omega_{1}$ and $\omega_{2}$ intersect at points $M$ and $P$. Let $M A$ be a chord of circle $\omega_{1}$ that is tangent to circle $\omega_{2}$ at point $M$, and let $M B$ be a chord of circle $\omega_{2}$ that is tangent to circle $\omega_{1}$ at point $M$. On the line $M P$, a segment $P H = M P$ is ... | Solution. Let $O_{1}, O_{2}, r_{1}, r_{2}$ be the centers and radii of the given circles. Draw perpendiculars from points $O_{1}$ and $O_{2}$ to the chords $A M$ and $B M$ respectively, and denote their point of intersection by $R$. If we prove that $R P \perp M P$, then it follows that $M R = R H$, and we will have pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,776 |
5. The numbers $a, b$ and $c$ satisfy the condition
$$
\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0
$$
Prove that
$$
\frac{a}{(b-c)^{2}}+\frac{b}{(c-a)^{2}}+\frac{c}{(a-b)^{2}}=0
$$ | Solution. Multiply the original equality in turn by the fractions $\frac{1}{b-c}$, $\frac{1}{c-a}$, $\frac{1}{a-b}$ and add them:
$$
\begin{gathered}
\frac{a}{(b-c)^{2}}+\frac{b}{(c-a)^{2}}+\frac{c}{(a-b)^{2}}+\frac{a+c}{(b-c)(a-b)}+\frac{a+b}{(b-c)(c-a)}+ \\
+\frac{b+c}{(c-a)(a-b)}=0 \\
\frac{a+c}{(b-c)(a-b)}+\frac{a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,777 |
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 | 18,778 |
7.3. While Cheburashka eats two portions of ice cream, Winnie-the-Pooh manages to eat five such portions, and while Winnie-the-Pooh eats three portions, Karlson eats seven. Working together, Cheburashka and Karlson ate 82 portions. How many portions did Winnie-the-Pooh eat during this time? Justify your answer. | Solution: Suppose Winnie-the-Pooh ate exactly 15 portions of ice cream. Since $15=3 \cdot 5$, Cheburashka ate $3 \cdot 2=6$ portions, and Karlson ate $-5 \cdot 7=35$ portions. Together, Karlson and Cheburashka ate $6+35=41$ portions, which is half of the 82 portions given in the problem. Therefore, the feast actually l... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,780 |
7.4. Along a straight path, fir trees grow at intervals of 10 m. A magpie has settled on each fir tree. Magpies can fly from one fir tree to another, and if one magpie flies from one fir tree to another, then another magpie must necessarily fly from some fir tree to another by the same distance, but in the opposite dir... | Justify the answer.
Solution: a) They can. For example, like this: the magpie from the first tree on the left flies 30 m to the right, and the magpie from the first tree on the right flies 30 m to the left. The magpie from the second tree on the left flies 20 m to the right, and the magpie from the second tree on the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,781 |
7.5. Once in a swamp there were 6 perch, 7 pike, and 8 shuk. Since there was no food in the swamp, the perch started eating the pike, the pike started eating the shuk, and the shuk started eating the perch, with the condition that no fish could eat another fish that had previously eaten an odd number of other fish. Aft... | Solution: Suppose the pike remained. Then the pikeperch ate exactly 7 pikes, so one of the pikeperch must have eaten an odd number of pikes. This means that this pikeperch could not have been eaten. Contradiction.
Similarly, we get a contradiction if we assume that the perch remained: in this case, the pikes ate an od... | \pikeperch\remained | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,782 |
7.6. In a square garden, there were 49 trees (7 rows of 7 trees each - see the diagram). The gardener gave the worker the following instruction: "Leave only five rows of trees, with four trees in each, and cut down the rest." After the work was done, the garden was almost empty: instead of 20 trees (as the gardener wan... | Solution: For example, as shown in the figure.
For the solution to problem 7.6
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct example | 7 points |
| In... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,783 |
1. The number 400 was divided into four parts such that if 1 is added to the first part, 2 is subtracted from the second part, the third part is multiplied by 3, and the fourth part is divided by 4, all the results will be equal. Into what parts was the number 400 divided? | Answer: $62, 65, 21$ and 252.
Solution. Let $4 x$ be the last (fourth) part of the number. After dividing it by 4, the result, equal to $x$, coincides with the third part multiplied by 3, so the third part of the number is $-\frac{x}{3}$. If the first part is increased by 1, and the second part is decreased by 2, thei... | 62,65,21,252 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,784 |
2. In the school, there are boys and girls. The average age of the boys differs from the average age of the girls, but the average of these two numbers coincides with the average age of all students. Who is more in the school - boys or girls? | Answer: equally.
Solution. Let $m$ be the number of boys and $d$ be the number of girls in the school, and let the sum of the ages of all boys be $M$, and the sum of the ages of all girls be $D$. Then the average age of all boys is $\frac{M}{m}$, the average age of all girls is $\frac{D}{d}$, and the average age of al... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,785 | |
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 | 18,786 |
5. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$. | Answer: $A C=13$.
Solution. (Fig. 3). First, we prove that $M I=M A$ (trident lemma).
Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is equa... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,787 |
3. Solve the inequality: $\left(x^{2}+2 x+3\right)\left(2 x^{4}-4 x^{2}+3\right)<2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3.
$$
\begin{aligned}
& \left(x^{2}+2 x+3\right)\left(2 x^{4}-4 x^{2}+3\right)<2 \\
& \left((x+1)^{2}+2\right)\left(2\left(x^{2}-1\right)^{2}+1\right)<2
\end{aligned}
$$
Since $(x+1)^{2}+2 \geq 2$ and $2\left(x^{2}-1\right)^{2}+1 \geq 1$, then $\left(x^{2}+2 x+3\right)\left(2 x^{4}-4 x^{2}+3\right) \geq 2$. Therefor... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,788 |
4. In triangle $\mathrm{ABC}$ inscribed in a circle, $\mathrm{AB}<\mathrm{AC}$. A point $\mathrm{D}$ is marked on side $\mathrm{AC}$ such that $\mathrm{AD}=\mathrm{AB}$. Prove that the perpendicular bisector of segment DC bisects the arc BC that does not contain point A. | 4. Extend side $\mathrm{BD}$ to intersect the circumscribed circle at point E. Since $\angle \mathrm{ADB} = \angle \mathrm{CDE}$ (vertical angles); $\angle \mathrm{ABE} = \angle \mathrm{ACE}$ (inscribed angles subtending the same arc), by the first similarity criterion $\Delta \mathrm{DAB} \sim \Delta \mathrm{DEC}$. Th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,789 |
5. There are 250 light bulbs arranged in a circle. Initially, all the light bulbs are on. It is allowed to either switch any 4 consecutive bulbs (from on to off or vice versa), or take 5 consecutive bulbs and switch all except the middle one. Is it possible to turn off all the bulbs using these operations? | 5. Let the bulbs be of two colors: red and blue, and the colors of the bulbs will alternate. Then we will have 125 bulbs of each color. We will only consider the red bulbs. Note that any switching operation affects exactly 2 red bulbs. We will count the number of red bulbs that are on. Initially, there are 125. If the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,790 |
7.1. Pasha wrote down an equality consisting of integers and arithmetic operation signs in his notebook. Then, in the expression on the left side of the equality, he encrypted each digit and operation sign with a letter, replacing identical digits or signs with the same letters and different ones with different letters... | Solution. There are many ways to obtain the required equality. We will present some of them (for a complete solution to the problem, only one is needed):
$$
\begin{gathered}
0+1 \times 7 \times 289 \\
2065+5-47 \\
1963+3 \times 20
\end{gathered}
$$ | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,791 |
7.2. Given a quadrilateral $A B C D$. Its diagonals $A C$ and $B D$ intersect inside the quadrilateral. These diagonals divide the angles of the quadrilateral into two smaller parts, i.e., 8 angles are formed, two at each vertex of the quadrilateral. Can 3 of these 8 angles be obtuse? | Solution. Let the diagonals of the quadrilateral intersect at point $O$. At point $O$, two pairs of vertical angles meet. Two of these angles are not less than $90^{\circ}$, so even if the triangles they form are obtuse, the obtuse angle is not at the vertex of the quadrilateral. In the other two triangles, there are n... | Theycannot | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,792 |
7.3. There are 55 boxes standing in a row, numbered in order from 1 to 55. Each box contains no more than 10 balls, and the number of balls in any two adjacent boxes differs by exactly 1. It is known that the boxes numbered $1,4,7,10, \ldots, 55$ contain a total of 181 balls. What is the minimum total number of balls t... | Solution. Since the number of balls in two adjacent boxes differs by 1, the parity of this number always differs. Therefore, throughout the entire row of boxes, the parity alternates - the number of balls in boxes with different parity numbers also has different parity. We will divide the boxes into pairs: 1 with 4, 7 ... | 487 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,793 |
7.4. All natural numbers from 1 to 100 inclusive are written on the board. Vasya chooses a pair of numbers on the board, the greatest common divisor of which is greater than one, and erases one of them. What is the smallest number of numbers Vasya can leave with such actions? | Solution. Evaluation. In order to erase a number, Vasya must find two numbers in whose prime factorization there will be at least one identical prime number. However, the number one (which has no prime numbers at all) and prime numbers greater than 50 (for which there is no second number that divides by it) cannot be e... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,794 |
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 | 18,795 |
9.1. Find the smallest six-digit number that is a multiple of 11, where the sum of the first and fourth digits is equal to the sum of the second and fifth digits and is equal to the sum of the third and sixth digits. | Solution. Let the desired number have the form $\overline{1000 x y}$, where $x, y$ are some digits. Then, by the condition, the sum of the first and fourth digits is 1, from which $x=y=1$. But the number 100011 is not divisible by 11. Therefore, we will look for a number of the form $\overline{1001 x y}$. Then $x=y=2$,... | 100122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,796 |
9.2. Point I is the center of the inscribed circle of triangle $ABC$. The circumcircle of triangle $BIC$ is tangent to the line $AB$. Prove that this circle is also tangent to the line $AC$. | Solution. Immediately note that $B I$ and $C I$ are the angle bisectors of triangle $A B C$. Let point $O$ be the center of the circumcircle of triangle $B I C$. From the tangency condition, it follows that $\angle O B A=90^{\circ}$. Let $\angle I B A=\angle I B=\beta$, then $\angle O B C=90^{\circ}-2 \beta$. Triangle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,797 |
9.3. Petya named two non-zero real numbers \(x\) and \(y\) to Vasya and asked him to compute two new numbers: \(x + \frac{1}{y^2}\) and \(y^2 + \frac{1}{x}\). Vasya got confused and calculated the values of the expressions \(x^2 + \frac{1}{y}\) and \(y + \frac{1}{x^2}\). Nevertheless, the results turned out to be the s... | Solution. Suppose that $x \neq y$. According to the condition, we have the system of equations
$$
\left\{\begin{array}{l}
x+\frac{1}{y^{2}}=y+\frac{1}{x^{2}} \\
y^{2}+\frac{1}{x}=x^{2}+\frac{1}{y}
\end{array}\right.
$$
Rewrite the first equation of the system as $(x-y)(x+y)=x^{2}-y^{2}=\frac{1}{x}-\frac{1}{y}=\frac{y... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,798 |
9.4. Given the set $S=\{1,2,3, \ldots, 2022\}$ and the set $A=\left\{a_{1}, a_{2}, a_{3}, \ldots, a_{2022}\right\}$, which is a permutation of the set $S$. It is known that for any $1 \leqslant n, m \leqslant 2022$ the expression $a_{n}+a_{m}$ is divisible by the GCD $(n ; m)$. Find the number of possible sets $A$. | Solution. Let $n$ and $m$ be equal and equal to some odd number. Then $2 a_{n} \vdots \gcd(n ; n)=n$, which means $a_{n} \vdots n$. For $n \geqslant 1013$, this is only true if $a_{n}=n$ for $n=\{1013,1015, \ldots, 2021\}$. Let $n$ and $m$ be even. Then $a_{n}$ and $a_{m}$ must have the same parity, which means all num... | Two\sets | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,799 |
9.5. Tom Sawyer's fence consists of $n$ boards. Ben Rogers and Billy Fisher take turns painting one board each, with Ben painting in red and Billy in blue. Ben starts. If anyone paints two adjacent boards, they will immediately be sent to swim, and the remaining one will get the core of the apple; if the fence is compl... | Solution. Let's number the fence boards from 1 to $n$ in order from one end of the fence to the other.
a) Billy is sufficient to paint the board opposite to the one painted by Ben (i.e., if Ben paints the board with number $k$, then Billy paints the board with number $n+1-k$).
b) We will show that Billy has such a st... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,800 |
5.1. Evgeny is tiling the floor of his living room, which measures 12 by 16 meters. He plans to place 1 m $\times$ 1 m square tiles along the perimeter of the room, and the rest of the floor will be covered with 2 m $\times$ 2 m square tiles. How many tiles will he need in total? | Solution. Let's step back 1 meter from the living room's boundaries. We will be left with a rectangle of size $10 \times 14$, which needs to be covered with $2 \times 2$ tiles. The area of the rectangle is 140, and the area of one tile is 4, so we will need $140: 4=35$ large tiles. The area of the boundary strip is $12... | 87 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,801 |
5.2. Ivan and Irina Ivanov, Mikhail and Maria Mishin, and Petr and Polina Petrov want to cross a river. There is a two-person boat. Ivan, Mikhail, and Polina can row. A man cannot be left alone with another man's wife on the shore or in the boat. If a family is completely on the other side, they immediately leave. How ... | $$
\begin{aligned}
& \text { (1) i, m, p, P } \xrightarrow{\text { I, M }}(2) \\
& \text { (1) i, m, p, P } \stackrel{\text { I }}{\leftarrow}(2) \mathrm{M} \\
& \text { (1) i, I, m } \xrightarrow{\text { p, P }}(2) \mathrm{M} \\
& \text { (1) i, I, m } \stackrel{\text { M }}{\leftarrow}(2) \\
& \text { (1) } \mathrm{i... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,802 |
5.3. In ten boxes lie balls: $1,3,5, \ldots, 19$ balls. Two friends, Peter and Basil, take turns taking one ball from any box. Peter takes the first ball. The one who, after their move, makes the number of balls in any two boxes the same (possibly zero) loses. Who among them can win regardless of the opponent's moves? | Solution. Peter will win if he makes any permissible moves, that is, such moves after which there will not be a pair of boxes with the same number of balls. Let's prove this. Consider the situation (let's call it critical) when the boxes contain different numbers of balls, but any taking of a ball will lead to a situat... | Peterwillwin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,803 |
5.4. Daniil and Pasha were driving in a car along a straight road at a constant speed. While Daniil was driving, Pasha had nothing to do and was looking at the kilometer markers. Pasha noticed that exactly at noon they passed a marker with the number $X Y$ (where $X, Y$ are some digits), at $12:42$ - a marker with the ... | Solution. Method 1. In 42 minutes from 12:00 to 12:42, the car traveled no more than 100 km, and in the remaining 18 minutes, it traveled even less (in $42 / 18=7 / 3$) times. Therefore, by 13:00, the car had not reached the sign with the number 200. Then $X=1$. In this case, regardless of the number $Y$, the car trave... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,804 |
5.5. On a circle, some 11 red points are marked, and inside the circle - some 11 yellow points. Prove that if no yellow point lies on a segment connecting two red points, then one can choose 5 red points such that they form a pentagon with no more than 4 yellow points inside it. | Solution. Let's number the red points sequentially, moving clockwise, with natural numbers from 1 to 11. And consider three such pentagons (only the numbers of their vertices - red points are indicated): 1) $1,2,3,4,5$, 2) $1,5,6,7,8$, $1,8,9,10,11$. It is easy to see that the interior areas of the pentagons do not hav... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,805 |
6.1. Can a figure with a perimeter of less than 25 cells be formed from four five-cell crosses and four four-cell corners? The figures cannot overlap. | Solution. For example, like this:
To the solution of problem 6.1
Answer. Yes, it is possible. | Yes,itispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,806 |
6.2. Mathematicians Andrey, Boris, and Viktor were solving problems from an olympiad. First, Andrey solved several problems, then Boris solved a third of the remaining problems. After this, a third of the problems remained unsolved, which Viktor completed. What part of all the problems did Andrey solve? | Solution. When Andrey solved several problems, Boris then solved a third of the remainder, and Victor, accordingly, two-thirds of the remainder. If this, according to the condition, is a third of all the problems, multiplying Victor's problems by three we get that twice the remainder after Andrey is the total number of... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,807 |
6.3. Three horses participated in the race. A betting office accepts bets with the following odds: for the first horse to win $4: 1$ (i.e., if the first horse wins, the player gets back the money they bet on it and four times as much; otherwise, the player loses all the money they bet), for the second horse $-3: 1$, fo... | Solution. To succeed, Pinocchio can act as follows: leave one coin for himself, and from the remaining 19 coins, place 4 coins on the first horse, 5 coins on the second horse, and 10 coins on the third horse. In the event of success for each horse (taking into account the coefficient), he will receive 20 coins, and tog... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,808 |
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 | 18,809 |
6.5. 2023 Minecraft players gathered and divided into two servers. Every minute, someone was upset that there were more players on their server than on the other, and switched to the other server. Over 2023 minutes, each player changed servers only once. How many people could have initially gathered on the first server... | Solution. Let's call the servers A and B. If any of the servers is empty at the beginning, then after 1012 transitions, it will become the largest, and all those who have not changed servers will be on the other one. That is, both servers are not empty at the beginning. Note that at the moment when a player who started... | 1011or1012 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,810 |
6.6 All natural numbers from 1 to 50 inclusive are written on the board. Vasya chooses a pair of numbers on the board, the greatest common divisor of which is greater than one, and erases one of them. What is the smallest number of numbers Vasya can leave with such actions? | Solution. Evaluation. Note that we will not be able to choose the number 1 with any other number as a pair so that their GCD is greater than one, so one will definitely remain on the board. Also, note that the numbers $29,31,37,41,43$, and 47 will definitely remain on the board. They are prime and greater than half of ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,811 |
# 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
$$
We will show that there are no other roots. Since the left side of the equation is defined only for $x \geqslant 1$, it is sufficient to consider the case $x>1$. Using the valid inequalities in this case $x^{4}>x^... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,812 |
11.2. Given a tetrahedron. Is it necessarily true that there exist four parallel planes, each passing through one vertex of the tetrahedron, such that the distances between any two adjacent planes are the same? | Solution. Let a tetrahedron $ABCD$ be given. Mark two points $K, L$ on the edge $AD$ such that $AK: KL: LD = 1: 1: 1$. Note that segments $KB$ and $CL$ do not lie on the same line. Then we can draw planes $\alpha_{A}, \alpha_{B}, \alpha_{C}, \alpha_{D}$ through vertices $A, K, L, D$ parallel to $KB$ and $CL$. It remain... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,813 |
11.3. Given distinct prime numbers $p, q, r$. The product $p q r$ is divisible by $p+q+r$. Prove that $(p-1)(q-1)(r-1)+1$ is a square of a natural number. | Solution. The divisors of the number $p q r$ are the numbers $1, p, q, r, p r, p q, q r$ and $p q r$. The number $p+q+r$ is greater than the first four of them and less than the last one, so it is one of the numbers $p r, p q, q r$. Without loss of generality, $p+q+r=p q$, that is, $-p-q+p q=r$. Then
$$
(p-1)(q-1)=p q... | r^2 | Number Theory | proof | Yes | Yes | olympiads | false | 18,814 |
11.4. Nyusha has 2022 coins, and Barash has 2023. Nyusha and Barash toss all their coins simultaneously and count how many heads each of them gets. The one who gets more heads wins, and in case of a tie, Nyusha wins. $C$ What is the probability that Nyusha wins? | Solution. Let's assume Barash has temporarily forgotten about his last coin. If Barash has more, he has already won; if Nyusha has more, he has already lost. The chances of these scenarios are equal since they have the same number of coins. If a tie occurs, the forgotten coin will decide the outcome, with each having a... | 0.5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,815 |
11.5. Krosh, having found the roots of the equation $x^{2}-x-1=0$, invented the "Krosh's golden system of numeration," a system of numeration with base $\varphi=\frac{1+\sqrt{5}}{2}$. Losyash noted that, in fact, Krosh finds decompositions of positive numbers into sums of certain different integer powers of the number ... | Solution. a) Note that $\varphi$, as a root of $\varphi^{2}-\varphi-1=0$, gives $1=\varphi^{-1}+\varphi^{-2}=$ $0.11_{\varphi}$. Sequentially using $\varphi^{-k}=\varphi^{-k-1}+\varphi^{-k-2}$ for all $k$, we get $1=1,0_{\varphi}=0,11_{\varphi}=0,1011_{\varphi}=0,101011_{\varphi}=0,10101011_{\varphi}=\ldots$, that is, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,816 |
8.1. Arrange the natural numbers from 1 to 9 in a $3 \times 3$ table so that each number is used exactly once, and the sum of the numbers in each row and each column is a prime number. | Solution. Example of number placement
| 6 | 8 | 9 |
| :--- | :--- | :--- |
| 1 | 7 | 3 |
| 4 | 2 | 5 | | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,817 |
8.2. Can the numbers $a, b, c$ (not necessarily integers) in some order coincide with the numbers $a+1, b^{2}+2, c^{3}+3$? | Solution. Let $M$ be the largest of the numbers $a, b, c$. Note that $M$ is positive, since $b^{2}+2$ is positive. Clearly, $M+1>M$. We will show that the numbers $M^{2}+2$ and $M^{3}+3$ are also greater than $M$. If $M \geqslant 1$, then $M^{2}+2>M^{2} \geqslant M$ and similarly $M^{3}+3>M^{3} \geqslant M$, and if $M<... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,818 |
8.3. Inside the square $A B C D$, a point $O$ is chosen. On the segments $A O, B O, C O, D O$, squares $O A A_{1} A_{2}, O B B_{1} B_{2}, O C C_{1} C_{2}, O D D_{1} D_{2}$ are constructed (all vertices are named in the order of clockwise traversal). Prove that $A_{2} B_{2} C_{2} D_{2}$ is a square. | Solution. Method 1.
Consider triangles $A O B$ and $A_{2} O B_{2}$. By construction, $A O = A_{2} O$ and $B O = B_{2} O$. If angles $A O A_{2}$ and $B O B_{2}$ have a common part, then $\angle A O B = 90^{\circ} - \angle A O B_{2} = \angle A_{2} O B_{2}$, if not, $\angle A O B = 90^{\circ} + \angle A O B_{2} = \angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,819 |
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 | 18,820 |
8.5. A positive rational number is written on the board. For any already written numbers \(a\) and \(b\) (including identical ones), it is allowed to write the numbers \(a + 2b\), \(ab^2\), and \(a / b^2\) on the board. Is it always possible (possibly in several steps):
a) to write the number 1 if initially one odd na... | Solution. First, let's make one simple
Observation. If the number $x$ is written on the board, then for any odd $n$, it is possible to write $n x$.
For this, it is enough to write the numbers
$$
x+2 x=3 x, 3 x+2 x=5 x, \ldots,(2 k-1) x+2 x=(2 k+1) x
$$
until we get $n x$. Now let's move on to solving the problem.
... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,821 |
10.1. The numbers $x$ and $y$ satisfy the inequalities $x^{3}>y^{2}$ and $y^{3}>x^{2}$. Prove that $x+y>2$. | Solution. The numbers $x$ and $y$ are positive, since the cube of each is greater than the non-negative number - the square of the other variable. Therefore, we can multiply the left and right parts of the inequalities. We get
$$
x^{3} y^{3}>x^{2} y^{2} \Leftrightarrow x y>1 \Leftrightarrow y>\frac{1}{x}
$$
Then
$$
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,822 |
10.2. When talking about a round-robin chess tournament (where each participant played one game against each other), the commentator said the following: "15 people participated in the tournament; the tournament winner scored twice as many points as the participant who finished last. The other 13 participants scored the... | Solution. In total, $\frac{15 \cdot 14}{2}=105$ games were played in the tournament; therefore, the participants scored exactly 105 points in total. Let the last-place finisher have scored $x$ points, then the winner scored $2 x$ points. Let each of the other 13 participants have scored $y$ points (The numbers $x$ and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,823 |
10.3. In a rectangular trapezoid $ABCD$ on the larger lateral side $CD$, there is a point $T$ such that the circles with diameters $CT$ and $TD$ are tangent to the lateral side $AB$ each. Denote the points of tangency of side $AB$ with the circles as $X$ and $Y$. Can it happen that $AY = BX$? Justify your answer. | Solution. Let $M$ be the center of the circle with diameter $C T$, and $N$ be the center of the circle with diameter $D T$.
Then $M C = M T = M X$ and $N D = N T = N Y$. Moreover, $M X \perp A B$ and $N Y \perp A B$, as radii drawn to the point of tangency of the circle and the line. Drop a perpendicular from point $T... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,824 |
10.4. A $7 \times 7$ square was cut without remainder (along the grid lines) into three-cell corners and small squares of size $2 \times 2$. Prove that exactly one small square was obtained. | Solution. We will color the cells of the square in 4 colors (see the figure). Note that two cells of the same color cannot fall into any one of the figures into which the square is cut. Then, since we have 16 red cells, the total number of corners and squares with side 2 is at least 16. Let the number of squares be \( ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,825 |
10.5. In a horse race, three horses participated. A betting office accepts bets with the following odds: for the first horse to win 4:1 (i.e., if the first horse wins, the player gets back the amount they bet and four times more; otherwise, the player loses all the money they bet), for the second horse -3:1, and for th... | Solution. a) If Pinocchio bets 11 gold coins on the first horse, 13 on the second, and 26 on the third, he will always receive at least 2 more gold coins. Indeed, if the first horse wins, he will receive 55 gold coins (11 - return of the bet and 44 winnings), if the second - 52 gold coins (of which 13 is the return of ... | )11,13,26\\b) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,826 |
66. It is known that the quadratic trinomial
$$
(b+c) x^{2}+(a+c) x+(a+b)
$$
has no roots. Prove that $4 a c-b^{2} \leqslant 3 a(a+b+c)$. | 66. By replacing the signs of $a, b$, and $c$ if necessary, we can assume that $a > 0$. Then, $P(x) + (a x^2 + b x + c) = (a + b + c)(x^2 + x + 1)$, from which it follows that $a + b + c \geqslant 0$ (since the right-hand side is bounded from below), and the sum of the minima of the left-hand sides is no greater than t... | 4-b^2\leqslant3() | Algebra | proof | Yes | Yes | olympiads | false | 18,827 |
67. A regular hexagon is divided into equal rhombi with sides parallel to the sides of the hexagon. Directions are assigned to three sides of the hexagon, none of which are adjacent, in the order of traversing the hexagon counterclockwise. Then, an arrow is placed on each side of the rhombi, pointing in the same direct... | 67. Let there be a cycle in the graph, and let it pass through the horizontal segment a1. Take the rhombus adjacent to side a1 and mark the parallel side a2 in it. Take the rhombus adjacent to side a2 and mark the parallel side a3 in it. And so on.
We will perform the same construction in the other direction: Take the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,828 |
68. Positive irrational numbers $\alpha$ and $\beta$ are such that for all $x>0$ the equality $[\alpha[\beta x]]=[\beta[\alpha x]]$ holds. Prove that $\alpha=\beta$. (J. Lagarias, T. Marayama, D. Richman) | 68. Let's introduce a renotation: we will consider that we are given two such irrational parameters $\alpha$ and $\beta$ that for all $x>0$ the equality $\left[{ }_{\alpha}^{1}\left[\begin{array}{l}1 \\ \beta\end{array}\right]\right]=\left[{ }_{\beta}^{1}\left[\begin{array}{l}1 \\ \alpha\end{array} x\right]\right]$ hol... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,829 |
35. There are several parallel rails. On these rails, there are 30 freight and 20 passenger cars, each car standing on two adjacent rails. A car is called movable if both rails it stands on are not occupied by other cars (in the provided diagram, there are no movable cars). Among the passenger cars, exactly 10 are mova... | 35. Ten stationary passenger cars occupy no more than 20 rails. In total, there are 21 stationary freight cars. If each of them occupies at least one of these rails, then there will be two stationary freight cars occupying the same rail. Otherwise, there will be a stationary car $B$ that does not occupy the same rails ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,830 |
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 | 18,831 |
37. In an isosceles triangle \(ABC\), the bisector \(AK\) is drawn. A point \(D\) is chosen on the base \(AC\) such that \(BC = CD\). Prove that it is possible to form an isosceles triangle with the lateral side \(BK\) and the base \(BD\). | 37. In triangle $A B C$, let's draw the bisector $C L$. Points $B$ and $D$ are symmetric with respect to this bisector, so $B L=D L$. Furthermore, from the isosceles property, it follows that $B K=B L$. Thus, $B K L-$ is the desired triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,832 |
40. At the vertices of a convex 2020-gon, numbers are placed such that among any three consecutive vertices, there is both a vertex with the number 7 and a vertex with the number 6. On each segment connecting two vertices, the product of the numbers at these two vertices is written. Andrey calculated the sum of the num... | 40. Answer: 1010. This answer is achieved if the numbers 7 and 6 alternate.
We can assume that all numbers are equal to 6 or 7. Indeed, if a number is different from 6 and 7, then among its neighbors, one is 6 and one is 7, and the same is true for the neighbors two steps away. Then, replacing this number with 7 does ... | 1010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,835 |
41. On the board, there is a natural number $M$. Olya chooses a natural number $N$ and appends the number 1 to the right of $M$, then the number 2, and so on, up to the number $N$. Prove that Olya can choose the number $N$ in such a way that the resulting number is divisible by 77. | 41. It is not hard to check that if to a natural number $k$ you append two consecutive three-digit numbers on the right, then its remainder when divided by 77 will increase by 1. Thus, the numbers written by Olya for $N=102,104, \ldots, 998$ give consecutive remainders when divided by 77. There are no fewer than 77 of ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,836 |
1. On a circle, 2021 points are marked. Kostya marks a point, then marks the next point to the right, then he marks the point to the right of the last marked point, skipping one, then the point to the right of the last marked point, skipping two, and so on. On which move will a point be marked for the second time? | Answer: on the 67th move. The question is about the smallest natural $b$ for which $(a+1)+(a+2)+\cdots+(b-1)+b$ is divisible by $2021=43 \cdot 47$. This sum is equal to $b(b+1) / 2-a(a+1) / 2=$ $(b-a)(b+a+1) / 2$ and it is not hard to see that the smallest $b=(b-a) / 2+(b+a) / 2$ is achieved with the factorization $(66... | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,837 |
2. Misha has a $100 \times 100$ chessboard and a bag with 199 rooks. In one move, he can either place one rook from the bag on the bottom-left cell, or remove two rooks standing on the same cell, place one of them on a cell adjacent to the right or above, and put the other back in the bag. Misha wants to arrange 100 no... | Yes, any. By induction, it is easy to establish that to place a rook on a cell $(x, y)$ (if we consider that the bottom-left corner has coordinates $(1,1)$), a reserve of $x+y-1$ rooks is required. In the arrangement $B$ of rooks thought up by Misha, there is a cell $(100, y_{100})$ in the rightmost column - to place a... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,838 |
4. On a plane, $n$ points with distinct abscissas are marked. Through each pair of points, a parabola, the graph of a quadratic trinomial with a leading coefficient of one, is drawn. A parabola is called good if there are no marked points on it or above it, except for the two points through which it is drawn. What is t... | Answer: $n-1$ parabolas. Example: parabolas passing through $n$ points on the X-axis. Evaluation: connect two marked points with an edge if the parabola passing through them is good. Suppose that two good parabolas pass through a given point $A$. To the right of point $A$, one of the parabolas lies above the other, so ... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,839 |
20. The continuation of the bisector $B L$ of triangle $A B C$ intersects its circumscribed circle at point $K$. The bisector of the external angle $B$ intersects the extension of segment $C A$ beyond point $A$ at point $N$. Prove that if $B K=B N$, then segment $L N$ is equal to the diameter of the circumscribed circl... | 20. Let $\angle B L N=\varphi$. Then
$$
\angle B C K=\angle B C A+\angle A C K={ }_{2}^{1}(\breve{A B}+\breve{A K})={ }_{2}^{1}(\breve{A B}+\breve{C K})=\varphi
$$
Note that $\angle L B N=90^{\circ}$, as the angle between the external and internal bisectors of a triangle. Then $B K=B N=B L \operatorname{tg} \varphi$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,842 |
21. Fedia listed all natural divisors of the number $n$, lying in the interval $\left[2^{100}, 3^{100}\right]$. Could it happen that he listed exactly 100 numbers, and exactly 35 of them are cubes of natural numbers? | 21. From the: cannot.
Lemma: If $n$ is divisible by $a^{3}$ and $b^{3}$, then $n$ is divisible by $a^{2} b$ and $a b^{2}$.
The proof of the lemma is completely analogous to the lemma from problem 13.
Let's return to the problem. Arrange the listed divisors in a row in ascending order. We will prove that between any ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,843 |
22. Quadratic trinomials $f(x)$ and $g(x)$ are such that
\[
\begin{aligned}
& f(2) \\
& g(2)
\end{aligned}=\begin{aligned}
& f(3) \\
& g(3)
\end{aligned}=2
\]
Find $f(1)$, given that $g(1)=2, f(5)=7$ and $g(5)=2$.
(A. Golev) | 22. Answer: $f(1)=5$.
Consider the quadratic trinomial $h(x)=2 g(x)-f(x)$. The conditions state that the numbers 2 and 3 are roots of this quadratic trinomial. Therefore,
$$
2 g(x)-f(x)=a(x-2)(x-3)
$$
where $a$ is the leading coefficient of this trinomial.
Now substitute $x=5$ into this equation and we find that
$... | f(1)=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,844 |
23. Given a positive number $c$. In space, 99 points are marked in such a way that for each of the marked points, the distances to the two nearest marked points differ by at least a factor of $c$. What is the largest $c$ for which this is possible? | 23. Answer: $c=\frac{1+\sqrt{5}}{2}$.
First, let's prove that $c \leqslant \frac{1+\sqrt{5}}{2}$.
From the condition, it follows that for any marked point, the distance to any other marked point (except the nearest one) exceeds the distance to the nearest one by at least a factor of $c$.
Suppose there are three such... | \frac{1+\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,845 |
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 | 18,847 |
4. On the table, there are 100 weights of different masses. A weight is called successful if its mass is equal to the sum of the masses of some two other weights on the table. What is the smallest number of successful weights that would allow us to assert with certainty that the masses of some two weights differ by at ... | Answer: 87. Let there be 87 lucky numbers. If one of the two addends of a lucky number $a=b+c$ is itself lucky, for example, $b=d+e$, then $a=c+d+e \geqslant 3 \min (c, d, e)$, which means there will be two numbers differing by at least a factor of three. If, however, all the addends of the eighty-seven lucky numbers a... | 87 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,848 |
56. Andrey has 100 stones of different weights, and although he can distinguish the stones by their appearance, he does not know the exact weight of each stone or how they are ordered by weight. Andrey can place exactly 10 stones on the table in the evening, and at night, the house spirit will arrange them in ascending... | 56. Answer: will be able to.
Conduct the "Barabashka test" with each set of ten stones.
If Barabashka is not present, then any two stones in each test will always lie in the same order (the lighter one to the left). Let's check that if Barabashka is present, this property will be violated. Suppose in some set of ten ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,849 |
57. Find all natural numbers such that the sum of the squares of their five smallest divisors is a perfect square. | 57. Answer: such numbers do not exist.
If the sought number $n$ is odd, then all its divisors would be odd, so the sum of the squares of any five of its divisors gives a remainder of 5 modulo 8 and therefore cannot be a perfect square. If the number $n$ is not divisible by three, then this sum of squares gives a remai... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,850 |
58. On the base $AD$ of trapezoid $ABCD$, a point $E$ is marked. It is known that $\angle CAD = \angle ADC = \angle ABE = \angle DBE$. Prove that triangle $BCE$ is isosceles. | 58. Let $\alpha$ denote the angle $C A D$ and the angles equal to it. Let $C^{\prime}$ be the point symmetric to point $C$ with respect to the line $A D$. Then $A C D C^{\prime}$ is a rhombus. In triangle $A D C^{\prime}: 180^{\circ}=\angle A C^{\prime} D+2 \alpha=\angle A C^{\prime} D+\angle A B D$, so the quadrilater... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,851 |
60. From point $O$, rays $\ell, \ell_{1}, \ell_{2}$ emanate, the angle between $\ell$ and $\ell_{2}$ is acute, and ray $\ell_{1}$ lies inside this angle. On ray $\ell$ lies a fixed point $F$ and an arbitrary point $L$. Through points $F$ and $L$ pass a circle tangent to ray $\ell_{1}$ at point $L_{1}$, and a circle tan... | 60. Note that $O F \cdot O L=O L_{1}^{2}=O L_{2}^{2}$. Therefore, the perpendicular bisector of the segment $L_{1} L_{2}$ coincides with the bisector of the angle $L_{1} O L_{2}$. Hence, any circle passing through the points $L_{1}$ and $L_{2}$ is symmetric with respect to this bisector. This means that the circle pass... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,853 |
61. In a social network, each user has no more than ten friends (the friendship relation is symmetric). The network is connected: if, upon learning interesting news, a user starts sending it to their friends, who in turn send it to their friends, and so on, eventually all users will learn the news. Prove that the netwo... | 61. A social network can be represented as a graph where people are vertices and the "friendship" relationship is the edges. It is sufficient to consider the case where this graph is a tree. According to the conditions of the problem, a group with 1 to 100 members will be called small, and a group with 100 to 900 membe... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,854 |
62. In the exam, there are 25 topics, each with 8 prepared questions. A ticket includes 4 questions on different topics. Is it possible to prepare 50 tickets such that each question appears in them exactly once and for any two topics, there is a ticket that includes questions from both of them? | 62. Answer: it is possible. We will identify topics with pairs of residues $(x, y)$ modulo 5. Each ticket will be understood as a quadruple of topics, and we need to present 50 quadruples such that each topic is in exactly 8 quadruples and each pair of topics is in exactly one quadruple.
Working with pairs of residues... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,855 |
34. Solution 1. Let there be $N$ students in the school before the New Year, among which there were $M$ boys, constituting $k$ percent of the total. This means that $M = \frac{k}{100} N$, or $100 M = k N$.
After the New Year, the number of boys became $M+1$, and the total number of students became $N+3$. If now the bo... | Solution 2. Let $m$ and $d$ be the number of boys and girls in the school before New Year. Let $\frac{100 m}{m+d}=x$, where $x$ is an integer by the problem's condition.
First, consider the case where the proportion of boys has not changed, i.e., $\frac{100(m+1)}{m+d+3}=x$. Then
$$
\frac{100 m}{m+d}=\frac{100(m+1)}{m... | +<200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,856 |
68. Let's consider the mathematicians' acquaintance graph. By condition, this graph is chordal, i.e., any cycle of 4 or more vertices contains a chord (a pair of adjacent vertices not adjacent in the cycle). We will prove that for a chordal graph $G$, the expression $f(G):=c_{1}-c_{2}+c_{3}-\ldots$, where $c_{i}$ is th... | Solution 1. Assume the contrary and consider $G$ as the smallest counterexample in terms of the number of vertices. It is clear that the graph $G$ contains more than one vertex and is connected (otherwise, one of the connected components would be a smaller counterexample). Remove an arbitrary vertex $v$ from the graph ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,860 |
56. Misha came to a country with $n$ cities, and every two cities are directly connected by a road. He plans to start from some city and drive through several cities without visiting any city twice. Each time Misha drives on a road, the president destroys $k$ roads leading from the city to which this road leads. (If th... | 56. Answer: Misha will be able to visit $n-k$ cities if $k \leqslant n-3$, and 2 cities otherwise (these answers coincide when $k=n-2$).
In the case $k \geqslant n-2$, while Misha is traveling from the first city to the second, the President will destroy all the other $n-2$ roads leading from the second city, and Mish... | n-k | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,861 |
57. Let's color the vertices of a 2018-gon in two colors such that any two adjacent vertices are of different colors. If the sum of the angles at the vertices of one color is equal to the sum of the angles at the vertices of the other color, we will call such a 2018-gon interesting. In a convex 2019-gon, one vertex is ... | 57. Let's denote the adjacent vertices of the 2019-gon by the letters $A, B, C$ and $D$ (clockwise). Suppose vertices $B$ and $C$ are not marked. Consider the angles of the 2019-gon, taken every other one counterclockwise, starting from $A$ and ending at $D$. Denote their sum by $u$. Consider the remaining angles, and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,862 |
58. Along a circle, there are $n$ coins, each lying heads or tails up. If two adjacent coins lie the same way (both heads or both tails), it is allowed to flip both. How many arrangements of coins are there that cannot be obtained from each other by applying such operations? | 58. Answer: 2 variants for odd $n$ and $n+1$ variants for even $n$.
Notice that the operation is reversible. For odd $n$, we will show that all coins can be oriented in the same direction. Divide the coins into groups of consecutive coins with the same orientation. If there are at least two groups, they alternate, so ... | 2variantsforoddnn+1variantsforevenn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,863 |
59. The coefficients of the polynomial $f(x)$ are integers, in absolute value not exceeding 5000000. Moreover, each of the equations
$$
f(x)=x, \quad f(x)=2 x, \quad \ldots, \quad f(x)=20 x
$$
has an integer root. Prove that $f(0)=0$.
(M. Antipov) | 59. We will prove that $f(0)$ is divisible by all prime numbers less than 20, from which it will follow that $f(0)=0$, since this product is too large.
Indeed, suppose $f(0)$ is not divisible by some $p$. Then $x \equiv 0(\bmod p)$ cannot be a solution to any of the equations in the condition (the left side is not div... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,864 |
60. Given a cyclic quadrilateral $A B C D$. A line perpendicular to $B D$ intersects segments $A B$, $B C$ and rays $D A$, $D C$ at points $P, Q, R, S$ respectively. It is known that $P R=Q S$. Prove that the midpoint of segment $P Q$ is equidistant from points $A$ and $C$. | 60. Let $T$ be the midpoint of segment $P Q$. Suppose point $A^{\prime}$ is symmetric to point $A$ with respect to point $T$. Triangles $S A^{\prime} Q$ and $R A P$ are symmetric and, therefore, equal. Then
$$
\angle S A^{\prime} Q=\angle R A P=180^{\circ}-\angle B A D=\angle Q C D
$$
Thus, point $A^{\prime}$ lies on... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,865 |
61. Let \(a, b, c, d > 0\). Prove the inequality
\[
a^{4} + b^{4} + c^{4} + d^{4} - 4abcd \geq 4(a - b)^{2} \sqrt{abcd}
\]
(S. Gorsky, F. Petrov)
 | 61. We will use the obvious inequality many times
$$
x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2}
$$
Let's denote the left side of the inequality to be proved by $L$. Notice that
$$
2 L=\left(a^{2}-c^{2}\right)^{2}+\left(c^{2}-b^{2}\right)^{2}+\left(a^{2}-d^{2}\right)^{2}+\left(d^{2}-b^{2}\right)^{2}+2(a c-b d)^{2}+2(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,866 |
62. A checker moves from the bottom left corner of a $100 \times 100$ board to the top right corner, moving one cell to the right or one cell up on each step. Let $a$ be the number of paths in which the checker makes exactly 70 steps below the diagonal running from the bottom left corner to the top right corner, and $b... | 62. Answer: $a=b$.
For each $k, 0 \leqslant k \leqslant n$, let $M_{k}$ denote the set of routes the checker can take from the bottom-left to the top-right corner of an $n \times n$ board, making exactly $2 k$ steps below the main diagonal. There is a remarkable fact: for a fixed $n$, all sets $M_{k}$ contain the same... | b | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,867 |
1. Sasha thought of three natural numbers. He wrote down all the common divisors of the first two numbers, then wrote down all the common divisors of the second and third, and finally wrote down all the common divisors of the first and third. In total, 18 numbers were written down. Four of them are $-3,3,4,4$. Find the... | Answer: 1, 1, 1, 2, 2, 2, 3, 4, 6, 6, 6, 12, 12, 12. From the condition, it follows that all three thought-of numbers are multiples of 3 and 4, which means they are multiples of 12. This already gives 6 common divisors for each pair of numbers, so there are no other common divisors. | 1,1,1,2,2,2,3,4,6,6,6,12,12,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,868 |
5. Olya and Dima are playing a game. First, Olya lays out $2 n$ cards with numbers from 1 to $2 n$ in a row in some order. Then they take turns, starting with Dima. On each turn, a player takes one of the outer cards. What is the maximum number of consecutive numbers that Olya can collect, regardless of Dima's actions? | Answer: $n$ numbers. Olya should lay out the cards as follows: $1, n+1,2, n+2, \ldots, n-1,2 n-1, n, 2 n$. Olya will take cards from the same side as Dima. Then she will take (from the left) several numbers from $n+1$ in the larger direction, and (from the right) several numbers from $n$ in the smaller direction. | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,869 |
24. The numbers $a, b, c$ satisfy the equations
$$
a b+a+b=c, \quad b c+b+c=a, \quad c a+c+a=b
$$
Find all possible values of $a$. | 24. Answer: $0,-1,-2$.
Add 1 to both sides of all three equations and factorize the left sides. Let $x=a+1, y=b+1, z=c+1$, we get the system of equations
$$
x y=z, \quad y z=x, \quad x z=y .
$$
We can proceed similarly to the first solution, but we can also do it completely differently. Multiply all three equations:... | 0,-1,-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,871 |
25. The sequence $\left(x_{n}\right)$ is defined by the conditions: $x_{1}=1$, and for each natural number $n$, the number $x_{n+1}$ is the largest number that can be obtained by rearranging the digits of the number $x_{n}+1$. Find the smallest $n$ for which the decimal representation of the number $x_{n}$ has exactly ... | 25. Answer: the smallest such number $n$ is $18298225=9(1+2+3+\ldots+2016)+1$.
To understand how this sequence is constructed, let's write out the first few terms. Initially, the sequence contains single-digit numbers:
$$
1, \quad 2, \quad 3, \quad \ldots, \quad 9 .
$$
Next come the two-digit numbers:
![](https://c... | 18298225 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,872 |
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