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Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,169 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,170 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,171 |
Task No. 1.1
## Condition:
Kirill Konstantinovich's age is 48 years 48 months 48 weeks 48 days 48 hours. How many full years old is Kirill Konstantinovich? | Answer: 53
Exact match of the answer -1 point
## Solution.
48 months is exactly 4 years. 48 weeks is $48 \times 7=336$ days. Together with another 48 days, this totals 384 days, which is 1 year and another 18 or 19 days, depending on whether the year is a leap year. In any case, the remaining days plus another 48 ho... | 53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,172 |
# Task № 1.3
Condition:
Anna Alexandrovna's age is 60 years 60 months 60 weeks 60 days 60 hours. How many full years old is Anna Alexandrovna | Answer: 66
Exact match of the answer -1 point
Solution by analogy with task №1.1.
# | 66 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,174 |
# Task № 1.4
Condition:
Tatyana Timofeevna's age is 72 years 72 months 72 weeks 72 days 72 hours. How many full years old is Tatyana Timofeevna | Answer: 79
Exact match of the answer -1 point
Solution by analogy with task №1.1.
# | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,175 |
# Task № 2.1
Condition:
Dmitry has socks in his wardrobe: 6 pairs of blue socks, 18 pairs of black socks, and 12 pairs of white socks. Dmitry bought some more pairs of black socks and found that now the black socks make up 3/5 of the total number of socks. How many pairs of black socks did Dmitry buy? | Answer: 9
Exact match of the answer -1 point
## Solution.
Dmitry had an equal number of pairs of black and all other pairs of socks, and after the purchase, it turned out that the black pairs make up three parts of all socks, while the other pairs make up two parts. This means that Dmitry bought half of the number o... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,176 |
# Task № 2.3
## Condition:
Dmitry has socks in his wardrobe: 14 pairs of blue socks, 24 pairs of black socks, and 10 pairs of white socks. Dmitry bought some more pairs of black socks and found that now the black socks make up 3/5 of the total number of socks. How many pairs of black socks did Dmitry buy? | Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №2.1.
# | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,177 |
# Task № 3.1
## Condition:
On a sheet of graph paper, there are three $5 \times 5$ squares, as shown in the figure. How many cells are covered by exactly two squares?
 | Answer: 15
Exact match of the answer -1 point
## Solution 1
Direct calculation. We will draw the squares and shade the cells that are covered exactly twice.
Solution 2.
The total area c... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,179 |
# Task № 3.3
## Condition:
On a sheet of graph paper, there are three $5 \times 5$ squares, as shown in the figure. How many cells are covered by exactly two squares?
 | Answer: 13
Exact match of the answer -1 point
Solution by analogy with task №3.1.
# | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,181 |
# Task № 5.2
## Condition:
A car number contains three letters and three digits, for example A123BE. The letters allowed for use are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Katya considers a number lucky if the second letter is a consonant, the first digit... | Answer: 288000
Exact match of the answer -1 point
Solution by analogy with task №5.1.
# | 288000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,183 |
# Task № 5.3
## Condition:
A car number contains three letters and three digits, for example A123BE. The letters allowed for use are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Tanya considers a number lucky if the first letter is a consonant, the second lette... | Answer: 384000
Exact match of the answer -1 point
Solution by analogy with task №5.1.
# | 384000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,184 |
# Task № 5.4
## Condition:
A car number contains three letters and three digits, for example A123BE. The letters allowed for use are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Kira considers a number lucky if the second letter is a vowel, the second digit is ... | Answer: 144000
Exact match of the answer -1 point
Solution by analogy with task №5.1.
# | 144000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,185 |
# Task № 6.2
## Condition:
On the faces of a cube, 6 letters are drawn: A, B, V, G, D, E. The picture shows three images of the cube from different angles. Which letter is drawn on the face opposite the face with the letter D?
![](https://cdn.mathpix.com/cropped/2024_05_06_4cfa7cdc7a8a51b9e752g-23.jpg?height=308&widt... | Solution by analogy with task №6.1.
# | B | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 18,187 |
# Task № 6.3
## Condition:
On the faces of a cube, 6 letters are drawn: A, B, V, G, D, E. The picture shows three images of the cube from different angles. Which letter is drawn on the face opposite the face with the letter $\mathrm{B}$?
![](https://cdn.mathpix.com/cropped/2024_05_06_4cfa7cdc7a8a51b9e752g-24.jpg?heig... | Solution by analogy with task №6.1.
# | \Gamma | Geometry | MCQ | Yes | Yes | olympiads | false | 18,188 |
# Task No. 6.4
## Condition:
On the faces of a cube, 6 letters are drawn: A, B, V, G, D, E. The picture shows three images of the cube from different angles. Which letter is drawn on the face opposite the face with the letter $\mathrm{A}$?
![](https://cdn.mathpix.com/cropped/2024_05_06_4cfa7cdc7a8a51b9e752g-25.jpg?he... | Solution by analogy with problem №6.1.
## Condition:
Anya, Borya, and Vasya took the same test consisting of 6 questions, each of which can be answered with "yes" or "no." The answers are presented in the table:
| Question No. | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Anya | no |... | 3 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 18,189 |
# Task № 7.4
## Condition:
Anya, Borya, and Vasya took the same test consisting of 6 questions, each of which could be answered with "yes" or "no". The answers are presented in the table:
| Question No. | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Anya | no | yes | no | yes | no | y... | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №7.1.
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,190 |
# Task No. 8.1
## Condition:
For the "Handy Hands" club, Anton needs to cut several identical pieces of wire (the length of each piece is a whole number of centimeters). Initially, Anton took a piece of wire 10 meters long and was able to cut only 9 needed pieces from it. Then Anton took a piece 11 meters long, but i... | Answer: 111
Exact match of the answer -1 point
## Solution.
First, note that 9 pieces of 111 cm each make up 999 cm. Therefore, both the first and the second piece are enough for 9 parts, but the second piece is not enough for 10 parts: $10 \times 111=1110>1100$.
We will prove that if the length of the piece is not... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,191 |
1. Variant 1.
Currently, the mother is 24 years and 3 months old, and her daughter is 5 months old. After how many months will the number of years in the mother's age be equal to the number of months in the daughter's age? | Answer: 21.
Solution. Let $x$ be the required number of months. Then we get the equation: $24+(x+3) / 12=x+5$. From this, $x=21$. | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,195 |
Variant 2.
Currently, mom is 23 years and 8 months old, and her daughter is 9 months old. In how many months will the number of years in mom's age be equal to the number of months in her daughter's age? | Answer: 16.
Option 3.
Currently, the mother is 19 years and 4 months old, and her daughter is 1 month old. After how many months will the number of years in the mother's age be equal to the number of months in the daughter's age?
Answer: 20. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,196 |
2. Variant 1.
In the game "Mathematical Running," nine teams participated (not necessarily equal in the number of participants). On average, there were 7 people in each team. After one team was disqualified, the average number of participants in the remaining teams decreased to 6. How many participants were in the dis... | Answer: 15.
Solution. The total number of participants before disqualification was $7 \cdot 9=63$. After the disqualification of participants, $6 \cdot 8=48$ remained. Therefore, the number of participants in the disqualified team was $63-48=15$. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,197 |
3. Variant 1.
The height $A H$ and the bisector $C L$ of triangle $A B C$ intersect at point $O$. Find the angle $B A C$, if it is known that the difference between the angle $C O H$ and half the angle $A B C$ is $46^{\circ}$. | Answer: 92.
Solution. Let the halves of the angles $A, B, C$ of triangle $ABC$ be denoted by $x, y$, and $z$ respectively. Then, $\angle COH = 90^{\circ} - z$ and $46^{\circ} = 90^{\circ} - z - y$. Since $x + y + z = 90^{\circ}$, we have $x = 44$ and $\angle BAC = 92^{\circ}$. | 92 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,198 |
# 4. Option 1.
Find the number of four-digit numbers where the digit in the units place is exactly 1 more than the digit in the tens place. The number cannot start with zero. | Answer: 810.
Solution. The leading digit of the number can be chosen in 9 ways (any digit except zero). The digit in the hundreds place can be chosen in 10 ways (any digit will do). The digit in the tens place can be any digit from 0 to 8, and the digit in the units place is uniquely determined by the chosen digit in ... | 810 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,199 |
5. Variant 1.
Two square napkins with sizes $1 \times 1$ and $2 \times 2$ were placed on a table such that a corner of the larger napkin fell into the center of the smaller one. What is the maximum area of the table that the napkins can cover? | Answer: 4.75
Solution: The larger napkin covers a quarter of the smaller one. This can be understood by extending the sides of the larger square. Therefore, the area of intersection is 0.25, and the area of the union of the napkins is $1+4-0.25=4.75$ | 4.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,200 |
6. Variant 1.
An artistic film with a duration of 192 minutes consists of four parts. It is known that the duration of any two parts differs by at least 6 minutes. What is the maximum duration that the shortest part can have? Express your answer in minutes. | Answer: 39.
Solution. Let the shortest part be $x$ minutes, then the second (in terms of duration) is no less than $x+6$, the third is no less than $x+12$, and the fourth is no less than $x+18$. Therefore, the entire film lasts no less than $4 x+36$ minutes. Solving the inequality $192 \geq 4 x+36$, we get $x \leq 39$... | 39 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,201 |
7. Variant 1.
In a convex $n$-gon, a diagonal is highlighted. The highlighted diagonal is intersected by exactly 14 other diagonals of this $n$-gon. Find the sum of all possible values of $n$. A vertex of the $n$-gon is not considered an intersection. | Answer: 28.
Solution. Let there be $x$ sides on one side of the diagonal, and $-n-x$ on the other. Then there are $-x-1$ vertices on one side, and $-n-x-1$ on the other. They can be the endpoints of the required diagonals. We get $14=(x-1)(n-x-1)$.
The following cases are possible:
$x-1=14, n-x-1=1$, then $n=17$
$x... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,202 |
# 8. Variant 1.
On the Island of Misfortune, there live truth-tellers, who always tell the truth, and liars, who always lie. One day, 2023 natives, among whom $N$ are liars, stood in a circle, and each said: "Both of my neighbors are liars." How many different values can $N$ take? | Answer: 337.
Solution: Both neighbors of a knight must be liars, and the neighbors of a liar are either two knights or a knight and a liar. Therefore, three liars cannot stand in a row (since in this case, the middle liar would tell the truth). We can divide the entire circle into groups of liars/knights standing in a... | 337 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,203 |
10.1. Which of the numbers is greater $\frac{2^{2021}+1}{2^{2022}+1}$ or $\frac{2^{2022}+1}{2^{2023}+1}$ ? | 10.1. Solution. Let $2^{2021}=n$, then the first expression $\frac{2^{2021}+1}{2^{2022}+1}=\frac{n+1}{2 n+1}$, and the second $\frac{2^{2022}+1}{2^{2023}+1}=\frac{2 n+1}{4 n+1}$. Since the difference between the first and second expressions is positive, the first number is greater than the second. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,204 |
10.2. On the side $A B$ of an equilateral triangle $A B C$, a point $M$ is taken, and on the segment $M C$, on the same side as point $B$, an equilateral triangle $M K C$ is constructed. Prove that the lines $A C$ and $B K$ are parallel. | # 10.2. Solution.
Since $\angle M B C = \angle M K C = 60^{\circ}$, a circle can be drawn through points $M, K, B, C$. Then $\angle K B C = \angle K M C = 60^{\circ}$ (as inscribed angles). T... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,205 |
10.3. The numbers $\frac{1}{a+b}, \frac{1}{a+c}$ and $\frac{1}{b+c}$ form an arithmetic progression. Prove that the numbers $a^{2}, b^{2}$ and $c^{2}$ also form an arithmetic progression. | 10.3. Solution. According to the condition $\frac{1}{a+b}+\frac{1}{b+c}=\frac{2}{a+c}$, i.e., $\frac{a+c+2 b}{(a+b)(b+c)}=\frac{2}{a+c}$. Multiplying both sides of the last equation by $(a+b)(b+c)(a+c)$ and combining like terms, we get $a^{2}+c^{2}=$ $2 b^{2}$, which is what we needed to prove. | ^{2}+^{2}=2b^{2} | Algebra | proof | Yes | Yes | olympiads | false | 18,206 |
10.4. The difference of cubes of two linear functions is a quadratic trinomial. Prove that this trinomial does not have real roots. | 10.4. Solution. Let $f(x)=a x+b$ and $g(x)=c x+d$ be the linear functions in question. Since $f(x)^{3}-g(x)^{3}=(a-c) x^{3}+\cdots$, then $a=c$, otherwise the difference would be a cubic function. At the same time, $b \neq d$. Then $f(x)^{3}-g(x)^{3}=(b-d)\left(f(x)^{2}+f(x) g(x)+g(x)^{2}\right)$. Note that the express... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,207 |
10.5. 72 consecutive natural numbers are divided arbitrarily into 18 groups of 4 numbers each. In each group, the product of the numbers is calculated, and for each of the 18 resulting products, the sum of the digits is calculated. Can all the resulting sums of digits be equal? | # 10.5. Answer. No
Solution. Suppose it is possible and the numbers are divided into 18 quartets in the specified manner. In at least one of the quartets, there is a number divisible by 9, so the sum of the digits of at least one product (and therefore all products) is divisible by 9.
Therefore, the product of the nu... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,208 |
10.6. On the board, the expression $\cos x$ is written. It is allowed to add or multiply several expressions written on the board (an expression can be used multiple times) and write the new obtained expression on the board. Is it possible to get an expression that takes the value 0 when $x=\pi$ after several actions?
... | Answer. Yes, it is possible.
Solution. The first action is to append $\cos ^{2} x$, the second is to append $\cos ^{2} x+\cos x$. Since $\cos \pi=-1$, the value of the last expression at $x=\pi$ is 0.
Comment. An answer without presenting the required expression - 0 points.
Only presenting any correct required expre... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,209 |
10.7. On the sides of a convex quadrilateral $A B C D$, rectangles are constructed outward. It turns out that all vertices of these rectangles, different from points $A, B, C, D$, lie on one circle. Prove that the quadrilateral $A B C D$ is cyclic. | Solution. $\quad$ Let $A B X Y$ be one of the given rectangles, and $O-$ be the center of the circle on which the eight vertices from the problem's condition lie (see Fig. 3). Then $O$ lies on the perpendicular bisector $\ell$ of the segment $X Y$. But $\ell$ coincides with the perpendicular bisector of the segment $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,210 |
10.8. On the board, $n$ quadratic trinomials of the form $* x^{2}+* x+*$ (instead of coefficients, asterisks are written) are written. Is it possible for some $n>100$ to replace the $3 n$ asterisks with some $3 n$ consecutive natural numbers (in some order) so that each of the $n$ given trinomials has two distinct inte... | Answer. No, it cannot.
Solution. The solution will consist of three steps (A, B, C). A) We will prove the following lemma.
Lemma. Suppose for some natural numbers $a, b$, the quadratic trinomial $a x^{2}+b x+c$ has integer roots. Then $b$ and $c$ are divisible by $a$.
Proof. By Vieta's theorem, $b / a = -\left(x_{1}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,211 |
10.9. Let's call a polygon good if it has a pair of parallel sides. A certain regular polygon was cut by non-intersecting (at interior points) diagonals into several polygons, each having the same odd number of sides. Can it happen that at least one of these polygons is good?
(I. Bogdanov) | Answer. No, it cannot.
Solution. We will prove the following simple lemma.
Lemma. If an $n$-gon is divided by non-intersecting diagonals into $(d+2)$-gons, the number of which is $t$, then $n=td+2$.
Proof. Induction on $t$; the base case for $t=1$ is obvious.
For the inductive step. Assuming that the statement is t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,212 |
10.10. Petya thought of two polynomials $f(x)$ and $g(x)$, each of the form $a x^{2} + b x + c$ (i.e., the degree of each polynomial does not exceed 2). On each turn, Vasya names a number $t$, and Petya tells him (at his discretion) one of the values $f(t)$ or $g(t)$ (without specifying which one he reported). After $n... | Answer. When $n=8$.
Solution. We will call a polynomial of the form $a x^{2}+b x+c$ simply a polynomial, and the graph of such a polynomial - simply a graph. We will use the following well-known lemma.
Lemma. Through any three points $\left(a_{i}, b_{i}\right)(i=1,2,3)$ with different abscissas, there passes exactly ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,213 |
Task 1. Little Red Riding Hood decided to visit her grandmother, whose cottage was 1 km away from her house. The wolf did not encounter her that day, so she walked the same route to and from. On flat sections, her speed was 4 km/h, uphill - 3 km/h, and downhill - 6 km/h. How long was she on the way? | Answer: Half an hour.
Solution. Consider any inclined section of the path of length $s$. Climbing it, Red Riding Hood will spend $s / 3$ time, descending $-s / 6$, in total -- $s\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{s}{2}$. Therefore, her average speed on this section is $2 s: \frac{s}{2}=4$. Thus, the average sp... | Half\an\ | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,214 |
Problem 2. Petya claims that two spinners are more expensive than five ice creams, Vasya - that three spinners are more expensive than eight ice creams. It is known that only one of them is right. Is it true that 7 spinners are more expensive than 19 ice creams? | Answer. Incorrect.
Solution. Let the price of the spinner be $s$, and the price of the ice cream be $\mathrm{m}$. The first statement means that $s>\frac{5 m}{2}=\frac{15 m}{6}$, and the second statement means that $s>\frac{8 m}{3}=\frac{16 m}{6}$. If the second condition were true, then the first condition would also... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,215 |
Problem 3. In a convex quadrilateral, the lengths of the diagonals are 2 and 4 cm. Find the area of the quadrilateral, knowing that the lengths of the segments connecting the midpoints of opposite sides are equal. | Answer: 4 cm$^2$.
Solution: The segment connecting the midpoints of adjacent sides of a quadrilateral is parallel to its diagonal (as the midline of the corresponding triangle). Therefore, quadrilateral $P Q R T$ is a parallelogram. By the condition, the diagonals of this parallelogram are equal, so it is a rectangle.... | 4^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,216 |
Problem 4. Three lines intersect to form 12 angles, and $n$ of them turn out to be equal. What is the maximum possible value of $n$? | Answer: 6.
Solution: Three lines limit a certain triangle. If this triangle is equilateral, then out of twelve angles, six are $60^{\circ}$, and the other six are $120^{\circ}$.
Can any external angle of the triangle be equal to its internal angle? It is equal to the sum of the non-adjacent internal angles, so it is ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,217 |
Problem 5. Consider four consecutive numbers $n, n+1, n+2, n+3$. For which $n$ is the LCM of the first three numbers greater than the LCM of the last three? | Answer. Any odd number not less than 5.
Solution. Consider the triplet of numbers $n, n+1, n+2$. Any two adjacent numbers are coprime. If the numbers $n$ and $n+2$ have a common divisor, then it is a divisor of their difference, 2. Thus, if the number $n$ is odd, then all three numbers $n, n+1, n+2$ are pairwise copri... | Any\odd\\not\\than\5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,218 |
1. Find the sum of the coefficients of the polynomial obtained after expanding the brackets and combining like terms in the expression
$\left(2 x^{2021}-x^{2020}+x^{2019}\right)^{11}-29$. | Solution. First, note that the sum of the coefficients of any polynomial in canonical form $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}$ is $P(1)$. Next, note that after raising the bracket to the 11th power, the polynomial $Q(x)=\left(2 x^{2021}-x^{2020}+x^{2019}\right)^{11}-29$ will take a canonical form (s... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,219 |
2. Can the number 2019 be represented as the sum of 90 natural numbers with the same digit sum?
翻译完成，如果您需要进一步的帮助，请告诉我。 | Solution. Each addend has the same sum of digits, so their remainders when divided by 9 are the same (the sum of the digits of a number gives the same remainder when divided by 9 as the number itself). The sum of 90 identical remainders is divisible by 9, which means the sum of any 90 natural numbers with the same sum ... | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,220 |
4. A semicircle is inscribed in triangle $ABC$ such that its diameter lies on side $BC$, and the arc touches sides $AB$ and $AC$ at points $C_{1}$ and $B_{1}$, respectively. Prove that
$$
\frac{A C_{1}}{C_{1} B} \cdot \frac{B H}{H C} \cdot \frac{C B_{1}}{B_{1} A}=1
$$
where $H$ is the foot of the altitude dropped fro... | # Solution.
Let $MN$ be the diameter of the given semicircle. First, note that the quadrilateral $C_1AB_1O$ is cyclic, since the sum of its opposite angles is $180^\circ$. Let $\Omega$ be the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,222 |
5. Find all triples of natural numbers for which the following condition is satisfied: the product of any two of them, increased by one, is divisible by the remaining number. | Solution. Only the triplet $(1 ; 1 ; 1)$ works. First, note that among these numbers there cannot be any even numbers: if $a$ is even, then $bc+1$ is odd. Now let's prove that these numbers must be coprime. If this is not the case, assume $\text{GCD}(a, b)=m>1$, then $ac+1$ gives a remainder of 1 when divided by $m$ an... | (1;1;1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,223 |
# 1. Option 1
A cyclist is riding on a track at a constant speed. It is known that at $11:22$ he had traveled 1.4 times the distance he had traveled by $11:08$. When did he start? | Format the answer as follows: $15: 45$.
Answer. 10:33.
Solution. Let the cyclist have traveled $x$ km by 11:08, then by 11:22 he had traveled $1.4 x$ km. Therefore, in 14 minutes (from 11:08 to 11:22), he traveled $0.4 x$ km. To travel $x$ km, it took him two and a half times longer, that is, 35 minutes.
## Variant ... | 10:33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,224 |
# 2. Option 1
Masha wrote the number 547654765476 on a piece of paper. She erased several digits so that the resulting number is the largest possible multiple of 9. What is this number? | Answer: 5476547646.
Solution: The sum of the digits of the original number is $3 \cdot(7+6+5+4)=66$. From the divisibility rule by 9, it follows that the sum of the erased digits must give a remainder of 3 when divided by 9. It is impossible to select digits with a sum of 3. Digits with a sum of $3+9=12$ can be chosen... | 5476547646 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,225 |
# 3. Variant 1
Vasya cut out a triangle from cardboard and numbered its vertices with the digits $1,2,3$. It turned out that if Vasya's triangle is rotated 12 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, it will return to its original position. If Vasya's triangle is rota... | Answer: 4.
Solution: Since after 12 turns around the first vertex of the triangle, it returns to its original position, it makes one or several full rotations of $360^{\circ}$. Therefore, the angle at the first vertex is not less than $30^{\circ}$. Reasoning similarly, we get that the angle at the second vertex is not... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,226 |
# 4. Variant 1
A number with the sum of its digits equal to 2021 was divided by 7, resulting in a number that is written only with the digit 7. How many digits 7 can be in it? If there are multiple answers, indicate their sum. | Answer: 503.
Solution. By multiplying $777 \ldots 77$ and 7 in a column, we get 54...439. Let $x$ be the number of sevens in the original number, then the number of fours in the resulting product is $x-2$. Its sum of digits is $4(x-2)+5+3+9=4x+9$, but on the other hand, it is equal to 2021, so $x=\frac{2021-9}{4}=$ $\... | 503 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,227 |
# 5. Option 1
Unit cubes were used to build a large cube. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 4 neighbors is 132. Find the number of cubes that have exactly 5 neighbors. | Answer: 726.
Solution: Cubes that have exactly 4 neighbors touch exactly one edge of the large cube. There are 12 edges in total, so 11 such cubes touch each edge. Therefore, the number of cubes that are strictly inside each face is $11 \cdot 11 = 121$. Since there are 6 faces, the answer is: $121 \cdot 6 = 726$.
## ... | 726 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,228 |
# 6. Option 1
Dasha poured 9 grams of feed into the aquarium for the fish. In the first minute, they ate half of the feed, in the second minute - a third of the remaining feed, in the third minute - a quarter of the remaining feed, and so on, in the ninth minute - a tenth of the remaining feed. How many grams of feed ... | Answer: 0.9.
Solution: After the first minute, $\frac{1}{2}$ of the initial feed remains, after the second minute, $\frac{1}{2} \cdot \frac{2}{3}$, and so on. After the 9th minute, $\frac{1}{2} \cdot \frac{2}{3} \cdots \cdots \frac{9}{10} = \frac{1}{10}$ of the initial amount of feed remains, i.e., 0.9 grams.
## Vari... | 0.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,229 |
# 7. Variant 1
In the garden, there are 46 trees - apple trees and pears. It turned out that among any 28 trees, there is at least one apple tree, and among any 20 trees, there is at least one pear. How many pears are there in the garden? | Answer: 27.
Solution: Since among 28 trees there is at least one apple tree, the number of pears is no more than 27. Since among any 20 trees there is at least one pear, the number of apple trees is no more than 19. There are 46 trees in total, so there are 19 apple trees and 27 pears.
## Variant 2
In the garden, th... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,230 |
# 8. Variant 1
In trapezoid $A B C D(A D \| B C)$, the bisectors of angles $D A B$ and $A B C$ intersect on side $C D$. Find $A B$, if $A D=5, B C=2$. | Answer: 7.
Solution.
Mark point $L$ on side $AB$ such that $LB = BC$. Let $K$ be the intersection point of the angle bisectors of $\angle DAB$ and $\angle ABC$. Then triangles $LBK$ and $BC... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,231 |
5.1. Vanya thought of a number and added the sum of its digits. The result was 2021. What number could Vanya have thought of? | Answer: 2014 or 1996
Solution. $2014+7=2021, 1996+25=2021$.
Comment. Any correct answer - 7 points. | 2014or1996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,232 |
5.2. Masha eats a bowl of porridge in 12 minutes, and Bear eats it twice as fast. How long will it take them to eat six bowls of porridge? | Answer: 24 min.
Solution: In 12 minutes, Masha eats one bowl of porridge, and Bear eats 2 bowls. In total, 3 bowls of porridge are eaten in 12 minutes. Six bowls of porridge will be eaten in twice the time, i.e., $12 * 2=24$ min.
Comment: A correct answer without justification - 3 points. | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,233 |
5.5. The hedgehogs collected 65 mushrooms and divided them so that each hedgehog got at least one mushroom, but no two hedgehogs had the same number of mushrooms. What is the maximum number of hedgehogs that could be | Answer: 10
Solution: If there were 11 hedgehogs, then together they would have collected no less than $1+2+3+\ldots+10+11=66$ mushrooms, which exceeds the total number of mushrooms collected. Therefore, there were fewer than 11 hedgehogs. We will show that there could have been 10 hedgehogs. Suppose the first found 1 ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,236 |
10.2. Pete and Vasya take turns writing natural numbers on the board, not exceeding 2018 (writing a number that is already on the board is prohibited); Pete starts. If after a player's move, there are three numbers on the board that form an arithmetic progression, this player wins.
. If no one has won by this point, Vasya wins on the next move - he just needs to find two written numbers of the same parity and write their arithmetic mean (which is an integer).
Moreover, note that if three inte... | Vasya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,238 |
10.4. Let $O$ be the center of the circumcircle $\Omega$ of an acute-angled triangle $ABC$. On the arc $AC$ of this circle, not containing point $B$, a point $P$ is taken. On the segment $BC$, a point $X$ is chosen such that $PX \perp AC$. Prove that the center of the circumcircle of triangle $BXP$ lies on the circumci... | Solution. Let $G$ be the center of the circle $\gamma$ circumscribed around triangle $B X P$ (see Fig. 5). Then $\angle B G P = \widehat{B X P} = 2 \angle C X P$ (since angle $C X P$ is acute). Since $G B = G P$ and $O B = O P$, triangles $G O B$ and $G O P$ are equal by three sides, hence $\angle B G O = \angle O G P ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,239 |
10.5. Given an odd number $n>10$. Find the number of ways to arrange the natural numbers $1,2,3, \ldots, n$ in a circle in some order so that each number is a divisor of the sum of the two adjacent numbers. (Ways that differ by rotation or reflection are considered the same.) (D. Khramov) | Answer. Two ways
Solution. Consider an arbitrary arrangement of numbers from 1 to $n$ that satisfies the conditions. Suppose that two even numbers $x$ and $y$ are adjacent, and the next number is $z$. Since $x+z$ is divisible by $y$, the number $z$ is also even. Continuing this movement around the circle, we get that ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,240 |
10.5. Non-zero numbers $a, b, c$ are such that $a x^{2}+b x+c>c x$ for any $x$. Prove that $c x^{2}-b x+a>c x-b$ for any $x$. (M. Murashkin) | Solution. Since for all $x$ the inequality $P(x)=$ $=a x^{2}+(b-c) x+c>0$ holds, the discriminant of the quadratic polynomial $P(x)$ is negative: $D=(b-c)^{2}-4 a c=b^{2}+c^{2}-2 b c-4 a c<0$. This means that the leading coefficient of the quadratic polynomial $Q(x)=c x^{2}-(b+c) x+(a+b)$ is positive, and its discrimin... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,241 |
10.6. Lines tangent to the circle $\omega$ at points $B$ and $D$ intersect at point $P$. A line passing through $P$ intersects the circle at points $A$ and $C$. A line parallel to $BD$ is drawn through an arbitrary point on the segment $AC$. Prove that it divides the lengths of the broken lines $ABC$ and $ADC$ in the s... | Solution. Triangles $\quad P B A$ and $P C B$ are similar because $\angle B P C$ is common, and $\angle P B A=\angle P C B=\frac{1}{2} \stackrel{A B}{ }$. Therefore, $\frac{B A}{B C}=\frac{P B}{P C}$. Similarly, from the similarity of triangles $P D A$ and $P C D$, it follows that $\frac{D A}{D C}=\frac{P D}{P C}$. Sin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,242 |
10.7. Do there exist three pairwise distinct non-zero integers, the sum of which is zero, and the sum of their thirteenth powers is the square of some natural number?
(V. Senderov) | Answer. They exist.
Solution. For a natural number $t$, the triplet of numbers $3t, -t, -2t$ satisfies all conditions, except possibly the last one. To make the sum $(3t)^{13} + (-t)^{13} + (-2t)^{13} = t^{13}(3^{13} - 1 - 2^{13})$ a perfect square, it suffices to set, for example, $t = 3^{13} - 1 - 2^{13}$.
Thus, th... | 3t,-,-2t | Number Theory | proof | Yes | Yes | olympiads | false | 18,243 |
10.8. Let's call a staircase of height $n$ a figure consisting of all cells of the square $n \times n$ that lie no higher than the diagonal (the figure below shows a staircase of height 4).
In... | Answer: $2^{n-1}$.
Solution. In each column of the staircase, mark one top cell; call their union the top layer. No two of the $n$ cells of this layer can lie in the same rectangle of the partition, so in any partition of the staircase, there are at least $n$ rectangles. On the other hand, the minimum total area of $n... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,244 |
9.1. Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. What values can the sum $f_{2}\left(x_{1}\r... | Answer: Only 0.
Solution: Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then
$f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$, since $f_{1}\left(x_{1}\right)=0$. Similarly, we obtain the equalities $f_{3}\... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,245 |
9.2. Given an isosceles triangle \(ABC\) with \(AB = BC\). In the circumcircle \(\Omega\) of triangle \(ABC\), a diameter \(CC'\) is drawn. A line passing through point \(C'\) parallel to \(BC\) intersects segments \(AB\) and \(AC\) at points \(M\) and \(P\) respectively. Prove that \(M\) is the midpoint of segment \(C... | Solution. Since $C C^{\prime}$ is the diameter of $\Omega$, we have $\angle C^{\prime} A C=90^{\circ}$. Since $\quad M P \| B C, \quad$ we get $\angle M P A=\angle B C A=\angle B A C$ (see Fig. 1). Therefore, triangle $A M P$ is isosceles, and its height $M D$ is also a median. Since $A D=D P$ and $A C^{\prime} \| D M$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,246 |
9.3. Petya chose several consecutive natural numbers and wrote each one either in red or blue pencil (both colors are present). Can the sum of the least common multiple of all red numbers and the least common multiple of all blue numbers be a power of two?
(O. Dmitriev, R. Zhenodarov) | Answer. No, it cannot.
Solution. Suppose the opposite. Consider the powers of two by which the written numbers are divisible; let $2^{k}$ be the largest of them. If at least two of the written numbers are divisible by $2^{k}$, then two adjacent such numbers will differ by $2^{k}$. Therefore, one of them will be divisi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,247 |
9.4. King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, $2, \ldots, 11$ kg. He also has a bag that will tear if more than 11 kg is placed in it. Archimedes has learned the weights of all the ingots and wants to prove to Hiero that the first ... | Answer. In 2 loads.
Solution. We will show that Archimedes can use the bag only twice. Let him first put in the bag ingots weighing 1, 2, 3, and 5 kg, and then ingots weighing 1, 4, and 6 kg. In both cases, the bag will not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if A... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,248 |
10.1. Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. For each trinomial $f_{i}(x)$, one root was chosen and denoted by $x_{i}$. What values can the sum $f_{2}\left(x_{1}\... | Answer: Only 0.
Solution. Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then
$$
f_{2}\left(x_{1}\right)=a x_{1}^{2}+b x_{1}+c_{2}=\left(a x_{1}^{2}+b x_{1}+c_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1},
$$
since $f_{1}\left(x_{1}\right)=0$. Similarly, we obtain the equalities $f... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,249 |
10.2. Petya chose 10 consecutive natural numbers and wrote each one either in red or blue pencil (both colors are present). Can the sum of the least common multiple of all red numbers and the least common multiple of all blue numbers end in 2016?
(o. Dmitriev, R. Zhenodarov) | Answer: No, it cannot.
Solution: Assume the opposite. Note that a number ending in 2016 is necessarily divisible by 16.
Among the ten of Petya's numbers, there is either one or two numbers that are divisible by 8. In the first case, one of the obtained least common multiples (LCM) is divisible by 8, and the other is ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,250 |
10.3. On side $A B$ of the convex quadrilateral $A B C D$, points $K$ and $L$ are taken (point $K$ lies between $A$ and $L$), and on side $C D$, points $M$ and $N$ are taken (point $M$ lies between $C$ and $N$). It is known that $A K=K N=D N$ and $B L=B C=C M$. Prove that if $B C N K$ is a cyclic quadrilateral, then $A... | Solution. In the case $A B \| C D$, we have $B C=K N$, so $A K=B L=C M=D N$. Therefore, the quadrilateral $L M D A$ is obtained from $B C N K$ by a parallel translation by the vector $\overrightarrow{B L}$.
 | Answer. $6 \cdot 50^{2}-5 \cdot 50+1=14751$ pairs.
Solution. Let the side length of the table be $2 n = 100$ (so $n=50$) and number the rows from top to bottom and the columns from left to right with numbers from 1 to $2 n$.
In each row, there can be from 0 to $2 n$ black cells. Since the number of black cells in all... | 14751 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,252 |
9.1. What different values can the digit U take in the puzzle U$\cdot$LAN + U$\cdot$DE = 2020? Justify your answer. (Identical digits are replaced by the same letters, different digits by different letters.) | Answer: two. $\mathrm{V}=2, \mathrm{y}=5$.
Solution: Factor out the common factor: У$\cdot$(ЛАН + ДЭ) $=2020$. Note that У, Л, and Э are not equal to 0.
Factorize the right-hand side: $2020=1 \cdot 2 \cdot 2 \cdot 5 \cdot 101$. Since У is a digit, consider all possible values: $\mathrm{V}=1,2,4,5$.
1) $У=1$. Then ЛА... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,253 |
9.2. A student-entrepreneur bought several packs of masks from a pharmacy and sold them to classmates at a higher price, making a profit of 1000 rubles. With all the money earned, he again bought masks from the pharmacy (at the same price as the first time) and sold them to classmates (at the same price as the first ti... | Answer: 2000 rubles.
Solution 1: Let the package of masks in the pharmacy cost x rubles, and the student sold the masks for y rubles, and bought a packages of masks the first time. Then, according to the condition, $a(y-x)=1000$.
The revenue amounted to ay rubles, so the second time the student was able to buy $\frac... | 2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,254 |
9.3. In triangle $M P Q$, a line parallel to side $M Q$ intersects side $M P$, median $M M_{1}$, and side $P Q$ at points $D, E$, and $F$ respectively. It is known that $D E=5$, and $E F=7$. What is the length of $M Q$? | Answer: 17.
Solution: Draw a line through point $E$ parallel to $P Q$ ( $K$ and $L$ - the points of intersection of this line with sides $M P$ and $M Q$ respectively). Since $M M_{1}$ is a median, then $L E = E K$, and since $D F \parallel M Q$, it follows that $D E$ is the midline of triangle $M K L$. Therefore, $M L... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,255 |
9.4. On the coordinate plane, the graphs of a linear and a quadratic function are plotted as shown in the figure (one of the intersection points is at the vertex of the parabola), and the line $y=k x+b$ passes through the point ( $-1 ; 2020$ ), while the coefficients $a$ and $c$ of the parabola $y=(x-c)^{2}$ are intege... | Answer: two. $k=-404 ;-1010$.
Solution: Note that $k<0$. Since the line passes through the point (-1, $2020)$, we have $2020=-k+b$, from which $b=2020+k$. Then the equation of the line can be written as $y=kx+2020+k$.
Let's find the points of intersection of the line with the coordinate axes.
Point $A: y_{A}=k \cdot... | -404,-1010 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,256 |
9.5. A group of children, after the municipal stage of the mathematics competition, formed a circle, discussing the solved problems. It turned out that there are exactly 20 future prizewinners and exactly 25 future winners of the municipal stage, such that each of them has at least one neighbor who is a participant who... | Solution. If among the children standing in a circle, several participants stand in a row, none of whom are winners or prizewinners (we will simply call them participants), we will expel all of them except one. Obviously, this will not affect the conditions of the problem. Now, each participant has only prizewinners or... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,257 |
1. The number 2019 has an interesting property: the sum of $2+019$ is exactly 10 times less than the sum of $201+9$. Find all four-digit numbers that have this property. | Answer: 2019, 3028, 4037, 5046, 6055, 7064, 8073, 9082.
Solution. Let's represent the desired number in the form $1000a + 100b + 10c + d$, where $a, b, c, d$ are digits, and $a \neq 0$. Since the number $100a + 10b + c + d$ must be divisible by 10, either $c + d = 0$ or $c + d = 10$. The first case is only possible wh... | 2019,3028,4037,5046,6055,7064,8073,9082 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,258 |
3. Prove that for all $x>0$ the inequality $1 / x+4 x^{2} \geq 3$ holds. | Solution. Since $y+1 / y \geq 2$ for all $y>0$, then
$$
\frac{1}{x}+4 x+4 x^{2}-4 x+1-1=2\left(\frac{1}{2 x}+2 x\right)+(2 x-1)^{2}-1 \geq 4-1=3
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 18,260 |
4. Consider the family of parabolas of the form $y=-x^{2}+p x+q$, the vertices of which lie on the graph $y=x^{2}$. Prove that all curves of this family pass through a common point. | Solution. Since the point ( $a, a^{2}$ ) on the parabola $y=x^{2}$ is the vertex of the parabola $y=-x^{2}+p x+q=-\left(x-\frac{p}{2}\right)^{2}+\frac{p^{2}}{4}+q, \quad$ then $\quad p=2 a$. Therefore, $a^{2}=-a^{2}+2 a^{2}+q$ and $q=0$. The parabola $y=-x^{2}+p x$ intersects the $O x$ axis at points 0 and $p$, if $p \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,261 |
5. A cube with an edge length of 7 was cut into unit cubes and the cube located at the center of the large cube was removed. Can the remaining figure be assembled from blocks of $1 \times 1 \times 2$? | Answer: No.
Solution. We will color the unit cubes in a "checkerboard pattern": in the bottom layer $7 \times 7 \times 1$, the corner cubes will be painted black, the two adjacent cubes along the edge - white, the adjacent ones to them - black again, and so on; in each subsequent layer, we will paint the cubes in colo... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,262 |
1. Find the smallest 10-digit number, the sum of whose digits is greater than that of any smaller number. | Answer: 1999999999.
Solution. Among 9-digit numbers, the largest sum of digits is for the number 999999 999, which is 81. Since the sought 10-digit number is greater than 999999 999, we need to find the smallest number with a sum of digits no less than 82. If the first digit of this number is 1, then the sum of the re... | 1999999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,263 |
2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 9 times fewer points than all the others combined. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament? | # Answer: 15 chess players.
Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. They drew the other half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,264 |
3. Prove that any odd composite number can be represented as the sum of three or more consecutive odd positive addends. How many such representations exist for the number $2021?$ | Answer: exactly one way, $2021=5+7+9+\ldots+89$.
Solution. Let $n=a \cdot b$ be a composite number, and $1<a \leqslant b<n$. Try to represent $n$ as a sum of $m$ consecutive odd terms:
$$
n=(2 k+1)+(2 k+3)+\ldots+(2 k+2 m-1)=m \cdot(2 k+m) .
$$
Since $n=a \cdot b$, one of the solutions to the equation $a \cdot b=m \... | 2021=5+7+9+\ldots+89 | Number Theory | proof | Yes | Yes | olympiads | false | 18,265 |
4. Let $x, y$ and $z$ be real numbers. Find the minimum and maximum value of the expression $f=\cos x \sin y+\cos y \sin z+\cos z \sin x$. | Answer: $\max f=\frac{3}{2}, \min f=-\frac{3}{2}$.
Solution. To find the maximum and minimum values, we estimate the expression $|f|$, using the property of absolute value: the absolute value of the sum of several numbers does not exceed the sum of the absolute values of these numbers. Therefore, $|f| \leqslant|\cos x... | \maxf=\frac{3}{2},\f=-\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,266 |
5. Side $A B$ of triangle $A B C$ is greater than side $B C$, and angle $B$ is $40^{\circ}$. A point $P$ is taken on side $A B$ such that $B P = B C$. The bisector $B M$ intersects the circumcircle of triangle $A B C$ at point $T$. Find the angle $M P T$. | Answer: $20^{\circ}$.
Solution. (Fig. 3.) In the quadrilateral $A P M T$, the angle at vertex $A$ is measured by half the arc $T C B$. Triangles $P M B$ and $C M B$ are equal by two sides and the angle between them, so $\angle P M B = \angle C M B = \angle A M T$. The angle $\angle A M T$ is measured by half the sum o... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,267 |
11.1. Does there exist a natural number $n$, greater than 1, such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number? | Answer: Yes, it exists.
Solution. For example, $n=2^{8}=256$.
Indeed, $\sqrt{n \sqrt{n \sqrt{n}}}=\sqrt{n \sqrt{n \cdot n^{\frac{1}{2}}}}=\sqrt{n \sqrt{n^{\frac{3}{2}}}}=\sqrt{n \cdot n^{\frac{3}{4}}}=\sqrt{n^{\frac{7}{4}}}=n^{\frac{7}{8}}$. Then, for $n=2^{8}$, the value of this expression is $\left(2^{8}\right)^{\f... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,268 |
11.2. Are there such integers $p$ and $q$ that for any integer values of $x$ the expression $x^{2}+p x+q$ is divisible by $3 ?$ | Answer: No, they do not exist.
Solution. Suppose such $p$ and $q$ exist. Then:
1) if $x=0$, then $x^{2}+p x+q=q$ is divisible by 3;
2) if $x=1$, then $x^{2}+p x+q=1+p+q$ is divisible by 3;
3) if $x=-1$, then $x^{2}+p x+q=1-p+q$ is divisible by 3.
We can reason in different ways.
First method. From 1), 2), and 3), ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,269 |
11.3. In square $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively. Segments $A E$ and $B F$ intersect at point $G$. Which is larger: the area of triangle $A G F$ or the area of quadrilateral GECF? | Answer: $S_{A F G}>S_{C E G F}$.
Solution. Let the area of triangle $A G F$ be denoted by $S_{1}$, and the area of quadrilateral $G E C F$ by $S_{2}$ (see Fig. 11.3a, b). Let the area of the square be $S$, then $S_{1}+S_{2}+S_{A B E}+S_{A D F}=S$.
Considering that $S_{A B E}=S_{A D F}=\frac{1}{4} S$, we get: $S_{1}+S... | S_{AFG}>S_{CEGF} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,270 |
11.5. Each lateral face of the pyramid is a right triangle, where the right angle is adjacent to the base of the pyramid. A height is drawn in the pyramid. Can it lie inside the pyramid? | Answer: No, it cannot.
Solution. Let the base of the pyramid $S A_{1} \ldots A_{n}$ be the polygon $A_{1} \ldots A_{n}$ (see Fig. $11.5 \mathrm{a}$, b). There are two possible cases.
1) Adjacent angles in two adjacent lateral faces are right angles.
Let, for example, $\angle S A_{2} A_{1}=\angle S A_{2} A_{3}=90^{\c... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,272 |
11.6. Each cell of a $7 \times 8$ table (7 rows and 8 columns) is painted in one of three colors: red, yellow, or green. In each row, the number of red cells is not less than the number of yellow cells and not less than the number of green cells, and in each column, the number of yellow cells is not less than the numbe... | Answer: 8.
Solution. 1) In each row of the table, there are no fewer red cells than yellow ones, so in the entire table, there are no fewer red cells than yellow ones.
In each column of the table, there are no fewer yellow cells than red ones, so in the entire table, there are no fewer yellow cells than red ones.
Th... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,273 |
11.1. A natural number greater than 1000000 gives the same remainder when divided by 40 and by 625. What digit can stand in the thousands place of this number? (N. Agakhanov, K. Sukhov) | Answer: 0 or 5.
Solution. Let $n$ be the given number, and $t$ be its remainder when divided by 40 and 625. Then the number $n-t$ is divisible by 40 and 625, which means it is divisible by $\operatorname{LCM}(40; 625)=5000$. Therefore, the difference $n-t$ ends in either 5000 or 0000. The remainder $t<40$. Thus, the d... | 0or5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,274 |
11.2. Non-zero numbers $x$ and $y$ satisfy the inequalities $x^{4}-y^{4}>x$ and $y^{4}-x^{4}>y$. What sign can the product $xy$ have (list all possibilities)
(N. Agakhanov) | Answer. Plus sign.
First solution. Adding the given inequalities, we get: $x+y>y^{4}$ and $y^{4}-y>x^{4}$, and we multiply them (this is permissible since their right-hand sides are positive). We have: $x y\left(1-x^{3}-y^{3}\right)>0$. Since $x1>0$. Therefore, $x y$ is positive.
Second solution. We will prove that $... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,275 |
11.3. On the $O x$ axis, points $0,1,2, \ldots, 100$ were marked and graphs of 200 different quadratic functions were drawn, each of which passes through two of the marked points and touches the line $y=-1$. For each pair of graphs, Oleg wrote on the board a number equal to the number of common points of these graphs. ... | Answer: Could not.
Solution. Each of our 200 polynomials corresponds to two integer points $a$ and $b$ on the $O x$ axis. Without loss of generality, we will assume that $a < b$. The width of the polynomial is $w=b-a$, and the axis is $c=\frac{a+b}{2}$. If the width is $w>0$ and the axis is $c$, then it is written as ... | 39698 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,276 |
11.4. A triangular pyramid $S A B C$ is inscribed in a sphere $\Omega$. Prove that the spheres symmetric to $\Omega$ with respect to the lines $S A, S B, S C$ and the plane $A B C$ have a common point. A sphere symmetric to a given one with respect to a line $\ell$ is a sphere of the same radius, the center of which is... | The first solution. Let $R$ be the radius of the circumscribed sphere $\Omega$ of the tetrahedron $SABC$, and let $O$ be its center. Mark the point $O'$, symmetric to the point $O$ with respect to the plane $(ABC)$, and the point $P$ such that $\overrightarrow{O'P} = \overrightarrow{OS}$ (in the case when the points $S... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,277 |
11.5. In Flower City, there live $99^{2}$ dwarfs. Some of the dwarfs are knights (always tell the truth), while the rest are liars (always lie). The houses in the city are located in the cells of a $99 \times 99$ square (a total of $99^{2}$ houses, arranged in 99 vertical and 99 horizontal streets). Each house is inhab... | Answer: 75.
Solution: Example. Let's show that if $k=74$, we cannot guarantee finding the house of Znayka. Place Znayka and the liar Neznayka in houses with numbers $(50 ; 49)$ and $(49 ; 50)$, respectively. We will show that it might be such that from the answers of the residents, we cannot uniquely determine which o... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,278 |
1. There are 5 identical-looking boxes with masses of 10, 11, 12, 14, and 17 kg, as well as electronic scales that show the exact mass of the weighed items (any number of boxes can be placed on the scales). Can the weight of each box be determined in 3 weighings? | Answer: Yes
Solution 1. We will weigh 2 boxes at a time. Note that all pairwise sums of masses are distinct, so the result of the weighing can uniquely determine which pair of masses was weighed $(21=10+11, 22=10+12, 23=11+12, 24=10+14, 25=11+14, 26=12+14, 27=10+17, 28=11+17, 29=12+17, 31=14+17)$. Let's denote the box... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,279 |
2. Alyosha and Vitya set off from point $N$ to point $M$, the distance between which is 20 km. Unfortunately, they have only one bicycle between them. From $N$, Alyosha sets off on the bicycle, while Vitya starts walking. Alyosha can leave the bicycle at any point along the road and continue on foot. When Vitya reaches... | Answer. 12 km from point $N$.
Solution. Let $x$ (km) be the distance from $N$ to the point where Alyosha leaves the bicycle. Then Alyosha will spend $\frac{x}{15}+\frac{20-x}{4}$ hours on the entire journey, and Vitya will spend $\frac{x}{5}+\frac{20-x}{20}$ hours. By setting up and solving the equation, we find $x=12... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,280 |
3. In an equilateral triangle $\mathrm{ABC}$, the height $\mathrm{BH}$ is drawn. On the line $\mathrm{BH}$, a point $\mathrm{D}$ is marked such that $\mathrm{BD}=\mathrm{AB}$. Find $\angle \mathrm{CAD}$. | Answer: $15^{\circ}$ or $75^{\circ}$.
Solution. Obviously, $\triangle \mathrm{ADH}=\Delta \mathrm{CDH}$ (right-angled, by two legs), hence $\mathrm{AD}=\mathrm{CD}$. Therefore, $\triangle \mathrm{BDA}=\triangle \mathrm{BDC}$ (by three sides) $\Rightarrow \angle \mathrm{DAB}=\angle \mathrm{DCB}, \angle \mathrm{BDA}=\an... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,281 |
5. On the table, there are 7 clean sheets of paper. Every minute, Vasya chooses any 4 of them and draws one star on each of the chosen sheets. Vasya wants to end up with a different number of stars on each sheet (that is, there should be no two sheets with the same number of stars drawn on them). What is the minimum to... | Answer: 28
Solution: Estimation. We will prove that to achieve the desired result, the number of moves must be at least 7. If 5 or fewer moves are made, then on each sheet there are no more than 5 stars, meaning there are only 6 different variants of the number of stars (from 0 to 5), while there are 7 sheets. By the ... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,282 |
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