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Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,365 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,366 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
$ be a non-zero polynomial of degree $n$ with non-negative coefficients such that the function $y=P(x)$ is odd. Can it happen that for different points $A_{1}, A_{2}, \ldots, A_{n}$ on the graph $G: y=P(x)$ the following conditions are satisfied: the tangent to the graph $G$ at point $A_{1}$ passes throu... | # Answer: It cannot.
First solution. We will show that under the given conditions for the polynomial, each subsequent point of tangency lies on the other side of the $O y$ axis compared to the previous one. Let $P(x) = a_{2m+1} x^{2m+1} + a_{2m-1} x^{2m-1} + \ldots + a_1 x$ be the given polynomial, and $Q(x) = a_{2m+1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,374 |
11.3. In the language, there are three letters - Sh, U, and Ya. A word is a sequence of 100 letters, exactly 40 of which are vowels (i.e., U or Ya), and the remaining 60 are the letter Sh. What is the maximum number of words that can be chosen so that for any two chosen words, at least one of the hundred positions cont... | Answer. $2^{40}$.
Solution. Example. Consider all $2^{40}$ words where starting from the 41st position, all letters are Ш, and the first $40-\mathrm{U}$ or Я. This set of words satisfies the condition.
Estimate. To each of our $m$ words, we associate $2^{60}$ words by replacing each letter Ш with У or Я (in all possi... | 2^{40} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,375 |
11.4. In triangle $A B C$, the angle bisectors $A A_{1}$ and $C C_{1}$ intersect at point $I$. A line passing through point $B$ parallel to $A C$ intersects the rays $A A_{1}$ and $C C_{1}$ at points $A_{2}$ and $C_{2}$, respectively. Point $O_{a}$ is the center of the circumcircle of triangle $A C_{1} C_{2}$, and poin... | Solution. We will denote $(X Y Z)$ as the circumcircle of triangle $X Y Z$. Let $P_{a}$ be the center of the circle $\left(B C_{1} C_{2}\right)$, and $P_{c}$ be the center of the circle $\left(B A_{1} A_{2}\right)$. Denote $\angle B A C=2 \alpha$ and $\angle A C B=2 \gamma$. Since $B C_{2} \| A C$, then $\gamma = \angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,376 |
# 2. CONDITION
Three numbers $x, y, z$ satisfy the relation $x^{2}+y^{2}=x y\left(z+\frac{1}{z}\right)$. Prove that at least one of the numbers x or y is equal to the product of the other two numbers. | Solution. According to the condition
$$
\begin{aligned}
& x^{2}-x y z+y^{2}-\frac{x y}{z}=0 \\
& x(x-y z)-y\left(\frac{x}{z}-y\right)=0 \\
& x(x-y z)-\frac{y}{z}(x-y z)=0 \\
& (x-y z)\left(x-\frac{y}{z}\right)=0
\end{aligned}
$$
Therefore, either $x-yz=0$ or $x-\frac{y}{z}=0$, i.e., $y-xz=0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,377 |
# 4. CONDITION
The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$, given that $\mathrm{a}_{1... | Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,379 |
6. Around a circle, 100 integers are written, the sum of which is 1. A chain is defined as several numbers (possibly one) standing consecutively. Find the number of chains, the sum of the numbers in which is positive. | Answer: 4951.
Solution. We will divide all chains (except the chain consisting of all numbers) into pairs that complement each other. If the sum of the numbers in one chain of the pair is $s$, and in the second is $t$, then $s+t=1$. Since $s$ and $t$ are integers, exactly one of them is positive. Therefore, exactly ha... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,380 |
# 3. The sum of two natural addends is 2016. If the last digit of one of them is erased - the second one is obtained. Find these numbers. | Let's denote the second number as A. Then the first number is $10 \mathrm{~A}+\mathrm{B}$, where $\mathrm{B}$ is its last digit. According to the condition, we get ( $10 \mathrm{~A}+\mathrm{B}$ ) $+\mathrm{A}=11 \mathrm{~A}+\mathrm{B}=2016$. Since the single-digit number B is less than 11, the last equality means that ... | 1833+183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,381 |
4. In each of three boxes lie 2016 matches. Two players take turns taking any number of matches from any one box, but only from one. The one who takes the last match wins. Prove that the one who moves first can win, regardless of how his partner plays. | The first move should be to take all the matches from any box. After this, two boxes will remain, and the beginner should, on each subsequent move, take as many matches as the partner took on the previous move, but from the other box. By following this strategy, the first player will equalize the number of matches in t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 17,382 |
1. A 2019-digit number written on the board is such that any number formed by any two adjacent digits (in the order they follow) is divisible by 13. Find the last digit of this number, given that the first digit is 6. | 1. Answer: 2.
Two-digit numbers divisible by 13: $13,26,39,52,65,78,91$. If the first digit of the number is 6, then the second digit must be 5 (forming the number 65, which is divisible by 13), the third digit is 2 (forming 52), and the fourth digit is again 6 (forming 26). Thus, the digits will be arranged in triple... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,383 |
2. On the square, 2019 people - knights and liars - stood in a line. Knights always tell the truth, liars always lie. One of the knights is the king. Everyone except the king said: "The number of liars between me and the king is seven." How many liars are there on the square? List all possible answers. | 2. Answer: any integer from 7 to 14 inclusive.
The one standing directly next to the king is a liar, as there is no one between him and the king. Similarly, the one standing one person away from the king is also a liar, since there is only one person between him and the king. Thus, the first 7 people next to the king ... | 7to14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,384 |
3. If a number is added to the sum of its digits, the result is 328. Find this number. | # 3. Answer: 317
Let's denote the digits of the desired number as a, b, c, and their sum as s. Then abc $+\mathrm{s}=328$. Since $s$ is the sum of three single-digit numbers, s does not exceed 27, so a $=3$. At the same time, b cannot be zero, because otherwise the maximum value of s would be $3+0+9=12$. If $\mathrm{b... | 317 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,385 |
4. There are 36 weights, among which no two have the same mass. Using special scales, you can order any 6 weights by their mass, but you cannot determine their exact mass. How can you determine the 3 heaviest weights in 8 such "weighings"? | 4. Divide 36 weights into 6 groups of 6 weights each. Using the first 6 weighings, we will sort the weights in each group in ascending order. Then, we will select the heaviest weight from each group and sort these weights with the seventh weighing. Let the heaviest of these six weights be in the first of the six initia... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,386 |
5. In a row, there are 10 balloons: blue and green. Can it be that between any two blue balloons there is an even number of balloons, and between any two green balloons - an odd number? | 5. Answer: No, it cannot.
Let's number the balls from left to right. Suppose the ball numbered 1 is green. Then, all balls in even-numbered positions are blue (since between the first green ball and any ball in an even-numbered position, there is an even number of balls). But between any two such blue balls, there is ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,387 |
1. There are candies in five bags. The first has 7, the second has 8, the third has 9, the fourth has 11, and the fifth has 20. Any number of candies can be moved from any bag to any other bag. Can you achieve an equal number of candies in all bags with two moves? What about three? | # Solution.
There are 55 candies in total, and 5 bags, so after the redistribution, each bag should have 11 candies. In the first three bags, there are fewer than 11 candies, so candies need to be added to them, which requires at least three moves. Therefore, it is impossible to equalize the number of candies in the b... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,388 |
5. Petya wants to color several cells of a $6 \times 6$ square so that there are as many vertices as possible that belong to exactly three colored squares. What is the maximum number of such vertices he can achieve? | Answer: 25.
Solution. Each vertex of the grid belongs to one, two, or four squares, and the latter are 25. Therefore, the number of vertices in question is no more than 25. An example is shown in the figure.
Criteria. Example: 4 points. Evaluation: 3
points.
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,398 |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,399 |
Problem 5.7. Vasya wants to place the numbers from 1 to 6 (each one only once) in the squares so that the following condition is met: if two squares are connected, the number in the higher square is greater. How many ways are there to do this?
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,403 |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to the line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on the line $A L$. Carefully finding the intersection point of thes... | (14;2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,404 |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,406 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,407 |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA + ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,408 |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,409 |
Problem 9.7. Two parallel lines are drawn through points $A(0 ; 14)$ and $B(0 ; 4)$. The first line, passing through point $A$, intersects the hyperbola $y=\frac{1}{x}$ at points $K$ and $L$. The second line, passing through point $B$, intersects the hyperbola $y=\frac{1}{x}$ at points $M$ and $N$.
What is $\frac{A L-... | Answer: 3.5.
Solution. Let the slope of the given parallel lines be denoted by $k$. Since the line $K L$ passes through the point ( $0 ; 14$ ), its equation is $y=k x+14$. Similarly, the equation of the line $M N$ is $y=k x+4$.
The abscissas of points $K$ and $L$ (denoted as $x_{K}$ and $x_{L}$, respectively) are the... | 3.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,411 |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,412 |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
^{2}+1=25 m^{2}+40 m+17$ is not divisible by 5. | Answer: $\left\{\begin{array}{c}x=5 m+2, \\ y=5 m^{2}+4 m+1 .\end{array}\right.$ or $\left\{\begin{array}{c}x=5 m+3, \\ y=5 m^{2}+6 m+2 .\end{array}\right.$ where $m \in \mathbb{Z}$.
5 points - the solution is correct, complete, and contains no errors.
4 points - if the answer is incorrect, the solution method is cor... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,414 |
# Problem 11.2 (7 points)
Prove that for any a and $\mathrm{b}$, the equation $\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \mathrm{x}^{2}+2\left(\mathrm{a}^{3}-\mathrm{b}^{3}\right) \mathrm{x}+\left(\mathrm{a}^{4}-\mathrm{b}^{4}\right)=0$ has a solution
# | # Solution:
If $\mathrm{a}^{2}-\mathrm{b}^{2} \neq 0$, then the given equation is quadratic with the discriminant
$\frac{1}{4} D=\left(a^{3}-b^{3}\right)^{2}-\left(a^{2}-b^{2}\right)\left(a^{4}-b^{4}\right)=a^{2} b^{4}-2 a^{3} b^{3}+a^{4} b^{2}=a^{2} b^{2}(a-b)^{2} \geq 0$.
If $\mathrm{a}^{2}-\mathrm{b}^{2}=0$, then... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,415 |
Problem 11.3 (7 points)
Can a cube be cut by a plane so that the cross-section is a pentagon? A regular pentagon? If "yes", construct the section, if "no" - prove it. | Solution:
a) The diagram shows that a pentagon can be obtained in the section (where one vertex coincides with the vertex of the cube, and the other four are midpoints of the edges)
b) A regular pentagon cannot be obtained.
It is easy to prove that there exist two sides of the pentagonal section that lie in opposite... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,416 |
# Problem 11.4 (7 points)
Given 10 weights. It turns out that the total weight of any four weights is greater than the total weight of any three of the remaining ones. Is it true that the total weight of any three weights is greater than the total weight of any two of the remaining ones?
# | # Solution:
Proof: Let's order the weights of our weights in non-decreasing order: $m_{1} \leq m_{2} \leq \ldots \leq m_{10}$. Then, from the condition that the total weight of any four weights is greater than the total weight of any three of the remaining, it follows that $m_{1}+m_{2}+m_{3}+m_{4} > m_{8}+m_{9}+m_{10}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,417 |
# Problem 11.5 (7 points)
Solve the system of equations $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x-y+\sqrt{\frac{x-y}{x+y}}=\frac{20}{x+y}\end{array}\right.$
# | # Solution:
Consider two cases:
1) Let $x+y>0$, then $x-y \geq 0 ; \sqrt{(x+y)^{2}}=x+y$.
By eliminating the denominator, solve the second equation of the system, $x^{2}-y^{2}+\sqrt{x^{2}-y^{2}}=20$. Let $\sqrt{x^{2}-y^{2}}=t, t \geq 0, t^{2}+t=20 \Rightarrow t=-5, t=4$. Find the solutions of the system $\left\{\beg... | (5;-3),(5;3),(-\frac{\sqrt{118}}{2};\frac{3\sqrt{2}}{2}),(-\frac{\sqrt{118}}{2};-\frac{3\sqrt{2}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,418 |
1. Given positive numbers $x>y$. Let $z=\frac{x+y}{2}$ and $t=\sqrt{x y}$. It turns out that $(x-y)=3(z-t)$. Find
$x / y$. | Answer: 25 (in the second variant 9). Note that $z-t=(\sqrt{} x-\sqrt{} y)^{2} / 2$. Denoting $a=\sqrt{} x, b=\sqrt{} y$, we get that $a^{2}-b^{2}=3(a-b)^{2} / 2$, which transforms into $\kappa(a-b)(2 a+2 b-3 a+3 b)=(a-b)(5 b-a)=0$. By the condition $a \neq b$, therefore $a=5 b$, i.e. $x / y=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,419 |
2. Schoolchildren from grades 6 to 11 participate in the olympiad. The chairman of the jury knows that exactly 1000 schoolchildren will come, but does not know their distribution by grades. He has exactly 500 sheets of paper. Two copies of the conditions (one on the left, one on the right) are printed on one sheet; dif... | Answer: He is right. Without loss of generality, we can assume the number of participants in each class is ordered as follows: $6,7,8,9,10,11$. The first team will print all the conditions for the 6th grade, and the same number for the 7th grade. The second team will print all the remaining conditions for the 7th grade... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,420 |
3. Sasha writes down 30 different natural numbers $a_{1}, a_{2}, \ldots, a_{30}$. Olya computes the GCD or LCM of numbers $a_{1}$ and $a_{2}$, then the GCD or LCM of the obtained result and number $a_{3}$, and so on, until she computes the GCD or LCM of the previous result and number $a_{30}$. In the process, she must ... | Answer: No, it cannot. Suppose one of the last two numbers chosen by Sasha is greater than 300. Then Olya can calculate the LCM in the last two steps, and the result will be greater than 300. Now let one of the last two numbers be less than 300. Then Olya can calculate the GCD in the last two steps, and the result will... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,421 |
4. In a regular 20-gon, four consecutive vertices $A, B, C$ and $D$ are marked. Inside it, a point $E$ is chosen such that $A E=D E$ and $\angle B E C=2 \angle C E D$. Find the angle $A E B$. | Answer: $39^{\circ}$ (in the 2nd variant: $36^{\circ}$).
Note that $ABCD$ is an isosceles trapezoid with angles $\angle ABC = \angle DBC = 180^{\circ} \cdot 18 / 20 = 162^{\circ}$. Point $E$ lies on the perpendicular bisector of the base $AC$, and therefore, triangle $BEC$ is isosceles. Draw the height $EH$ in it, and... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,422 |
5. The company specializes in manufacturing "boards with a hole": this is a $300 \times 300$ grid board, in which a hole in the form of a rectangle, not extending to the board's edge, is cut out by cells. Each such board comes with a tag indicating the maximum number of non-attacking rooks that can be placed on this bo... | Answer: 400 rooks (in the $2-nd$ variant 440 rooks).
Let the hole have dimensions $a \times b$. Note that all cells of the board with the hole are covered by $300-b$ columns and 300 - a rows. If $a, b \geqslant 100$, then all cells are covered by no more than 400 lines, and it is impossible to place more than 400 rook... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,423 |
1. According to a sociological survey conducted in 7 "Y" grade, it was found that $20 \%$ of students interested in mathematics are also interested in physics, and $25 \%$ of students interested in physics are also interested in mathematics. Only Pete and Vasya are not interested in either of these subjects. How many p... | # Solution
The number of people in 7 "I" excluding Petya and Vasya exceeds 18 but is less than 28. Let $n$ be the number of students interested in both mathematics and physics. Then, the total number of students interested in mathematics is $5n$, and the total number of students interested in physics is $4n$. Therefor... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,424 |
2. Along the school corridor, there is a New Year's garland consisting of red and blue bulbs. Next to each red bulb, there is definitely a blue one. What is the maximum number of red bulbs that can be in this garland if there are 50 bulbs in total?
# | # Solution
Let's calculate the minimum number of blue bulbs that can be in the garland. We can assume that the first bulb is red. Since there must be a blue bulb next to each red bulb, three red bulbs cannot go in a row. Therefore, among any three consecutive bulbs, at least one bulb must be blue. Then, among the firs... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,425 |
3. Is it possible to arrange 4 ones, 3 twos, and 3 threes in the circles of a five-pointed star (see figure) so that the sums of the four numbers on each of the five lines are equal?
 | # Solution
Suppose the numbers have been arranged in the required manner. Let $S$ be the sum of the numbers on each line, then the sum of the numbers on all five lines is $5 S$. Since each circle lies at the intersection of two lines, in this calculation, the number written in each circle is counted twice. Therefore, ... | No,itisnotpossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,426 |
4. Prove that if any two digits are denoted by the letters $X$ and $A$, then the six-digit number $X A X A X A$ is divisible by 7 without a remainder.
# | # Solution
$\overline{X A X A X A}=\overline{X A} \cdot 10000+\overline{X A} \cdot 100+\overline{X A}=\overline{X A} \cdot 10101$. Since $10101=7 \cdot 1443$, then ХАХАХА is divisible by 7 without a remainder. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,427 |
5. A rectangular sheet of paper $ABCD$ was folded as shown in the figure. Find the ratio $DK: AB$, if $C_{1}$ is the midpoint of $AD$.
# | # Solution
$A C_{1}=D C_{1}=1 / 2 A D$. Triangles $B C_{1} K$ and $B C K$ are equal, so $B C_{1}=B C$. In the right triangle $A B C_{1}$, the leg $A C_{1}$ is half the hypotenuse $B C_{1}$, which means $\angle A B C_{1}=30^{\circ}$. Therefore, $\angle A C_{1} B=60^{\circ}, \angle D C_{1} K=180^{\circ}-60^{\circ}-90^{\... | 1:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,428 |
7.1. Riding along a forest road, Ivan-tsarevich met a bear, a fox, and a wolf. The bear always tells the truth, the fox always lies, and the wolf alternates between truth and lies, always starting with the truth. Each of the animals said two sentences to Ivan-tsarevich. The first: "You will save the horse. But you will... | Solution: The first parts of the statements are: "you will save the horse," "you will remain unharmed," "you will remain intact." Among them, two should be true and one false, so Ivan survives, but his horse does not. This means that both parts of the first statement are false, and the first speaker was the fox. Simila... | Thefirststatementbelongstothefox,thetothewolf,thethirdtothebear.Ivanwillsurvive,butwithouthishorse | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,429 |
7.2. Let \( x \) and \( y \) be integers such that \(\frac{1+y}{x-y}=x\). Prove that the numbers \( x \) and \( y \) differ by 1. | Solution: We will transform the equation (for $x \neq y$) in an equivalent manner
$$
\begin{aligned}
& \frac{1+y}{x-y}=x \Leftrightarrow 1+y=x(x-y) \Leftrightarrow 1+y=x^{2}-x y \Leftrightarrow \\
& \Leftrightarrow 1-x^{2}=-y-x y \Leftrightarrow(1-x)(1+x)=-y(1+x)
\end{aligned}
$$
Since the number $1+x$ is not equal t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,430 |
7.3. In a rectangle $10 \times 7$, a central rectangle $1 \times 6$ cm has been cut out, as shown in the figure. Cut the resulting figure into two parts so that they can be assembled into an $8 \times 8$ square. Show how to cut and then assemble. | Solution: For example, as shown in the figure.
To the solution of problem 7.3
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correctly shown how to cut and how... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,431 |
7.4. Seryozha wrote a five-digit natural number on the board. It turned out that in this number, among any two adjacent digits, the one on the right is greater. Vasya multiplied this number by 9 and calculated the sum of the digits of the resulting product. Find all possible values that this sum can take. Justify your ... | Solution: Let Sergey write the number $x=\overline{a b c d e}$. Then $10 x=\overline{a b c d e 0}$. Notice that $9 x=10 x-x$ and perform the subtraction "in columns":
Due to the condition $a$... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,432 |
7.5. Seryozha and Vasya set out towards each other at the same time: the first from point $C$, and the second from point $B$, the distance between which is 21 km. Seryozha's dog Bobik starts moving from point $C$ together with Seryozha and runs towards Vasya. Upon meeting him, without taking a breath, he rushes back to... | Solution: The approach speed of the pedestrians is $3+4=7$ km/h, and they will meet after $21: 7=3$ hours from the start of their movement. Continuously running, Bobik will cover $3 \cdot 11=33$ km during this time.
Answer: 33 km.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct an... | 33 | Other | math-word-problem | Yes | Yes | olympiads | false | 17,433 |
7.6. Let $a, b, c$ be natural numbers, and the product $ab$ is divisible by $5c$, the product $c$ is divisible by $13a$, and the product $ca$ is divisible by $31b$. Find the smallest possible value of the product $abc$. Justify your answer. | Solution: From the condition that $ab$ is divisible by $5c$, it follows that at least one of the numbers $a$ and $b$ is divisible by 5. If this number is $a$, then from the condition that $bc$ is divisible by $13a$, it follows that one of the numbers $b$ or $c$ is also divisible by 5; if this number is $b$, then one of... | 4060225 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,434 |
3. Find such an integer $\mathrm{n}$ that for any integer $\mathrm{k} \neq 2013$, the number $n-k^{2014}$ is divisible by $2013-k$.
Answer: $2013^{2014}$. | # Solution
Let $t=2013-k$, then $\frac{n-k^{2014}}{2013-k}=\frac{n-(2013-t)^{2014}}{t}$. If we raise the number $2013-t$ to the power of 2014, we get a sum of terms, each of which contains a factor $t^{m}$ (for some $m=1, \ldots, 2014$) except for $2013^{2014}$, then we get $\frac{n-(2013-t)^{2014}}{t}=\frac{n-2013^{2... | 2013^{2014} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,435 |
4. The numbers $a>0, b$, and $c$ are such that the quadratic trinomial $f(x)=a x^{2}+b x+c$ has two roots greater than 1. Prove that the quadratic trinomial $g(x)=3 a x^{2}-(c+2013) x+b+2015$ will have two distinct roots. | # Solution
Let $x_{1}$ and $x_{2}$ be the roots of the function $f(x)$, then by Vieta's formulas we have:
$\left\{\begin{array}{c}x_{1}+x_{2}=-\frac{b}{a} \\ x_{1} x_{2}=\frac{c}{a}\end{array}\right.$, and since $x_{1}>1, x_{2}>1$ and $a>0$, we get $\left\{\begin{array}{c}-b>2 a \\ c>a\end{array}\right.$. Therefore,
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,436 |
5. What is the maximum number of cells that can be painted on a $6 \times 6$ board so that it is impossible to select four painted cells such that the centers of these cells form a rectangle with sides parallel to the sides of the board.
# | # Solution
Let $n$ be the number of cells that can be painted. Denote by $x_{1}, x_{2}, \ldots, x_{6}$ the number of painted cells in the 1st, 2nd, ..., 6th columns, respectively. Then $x_{1}+x_{2}+\ldots+x_{6}=n$. Let's count the number of pairs of cells in the columns that can form the sides of the desired rectangle... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,437 |
6. Can space be divided into cubes with integer edges so that among all their edges, there are no more than 324 identical ones? Justify your answer
# | # Solution
Step 1: Construct a composite cube from 27 smaller cubes with edges equal to 1.
Step 2: Surround this composite cube with a side length of 3 with 26 more cubes of the same size, forming a composite cube with a side length of \(3^2 = 9\).
Continue the process.
 = 6d(6d-1)$ matches. Since each match awards one point, the total number of p... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,441 |
5. Angle $A$ of the rhombus $A B C D$ is $60^{\circ}$. A line passing through point $C$ intersects segment $A B$ at point $M$ and line $A D$ - at point $N$. Prove that the angle between lines $M D$ and $N B$ is $60^{\circ}$. - | Solution. Let $K$ be the intersection point of the lines $M D$ and $N B$ (Fig. 4). Note that triangle $A B D$ is equilateral, $\angle B A D=\angle A B D=\angle B D A=60^{\circ}$. From the similarity of triangles $B M C$ and $A M N$, as well as $M A N$ and $C D N$, it follows that
$$
\frac{M B}{A B}=\frac{M C}{C N}, \q... | 60 | Geometry | proof | Yes | Yes | olympiads | false | 17,442 |
10.6. Given an acute-angled triangle $A B C$. On the extensions of its altitudes $B B_{1}$ and $C C_{1}$ beyond points $B_{1}$ and $C_{1}$, points $P$ and $Q$ are chosen respectively such that angle $P A Q$ is a right angle. Let $A F$ be the altitude of triangle $A P Q$. Prove that angle $B F C$ is a right angle.
(A. ... | 10.6. Points $B_{1}$ and $C_{1}$ lie on the circle constructed with $B C$ as its diameter. To solve the problem, it is sufficient to prove that $F$ also lies on this circle, that is, it is sufficient to prove that the quadrilateral $C B_{1} F C_{1}$ is cyclic.
Since $\angle A B_{1} P = \angle A F P = 90^{\circ}$, poin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,444 |
10.7. For natural numbers $a>b>1$, define the sequence $x_{1}, x_{2}, \ldots$ by the formula $x_{n}=\frac{a^{n}-1}{b^{n}-1}$. Find the smallest $d$ such that this sequence does not contain $d$ consecutive terms that are prime numbers, for any $a$ and $b$.
(V. Senderov) | 10.7. Answer. 2.
For $a=4, b=2$ we have $\frac{a^{1}-1}{b^{1}-1}=3, \frac{a^{2}-1}{b^{2}-1}=5$. It remains to show that more than two consecutive prime numbers will not occur.
We will prove a stronger statement than required: for $n \geqslant 2$ at least one of the numbers $\frac{a^{n}-1}{b^{n}-1}, \frac{a^{n+1}-1}{b... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,445 |
10.8. A $2010 \times 2010$ square grid is cut into three-cell corners. Prove that it is possible to mark one cell in each corner such that in each row and each column there are an equal number of marked cells.
(I. Bogdanov, O. Podlipsky) | 10.8. Let $n=2010 / 3=670$. We will call a line a row or a column; we will number the rows from top to bottom, and the columns from left to right. Note that to satisfy the conditions of the problem, it is sufficient that for any $k$, the left $k$ columns and the top $k$ rows contain $k n$ marked cells each.
We say tha... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,446 |
1. On the board, three quadratic equations are written:
$$
\begin{gathered}
2020 x^{2}+b x+2021=0 \\
2019 x^{2}+b x+2020=0 \\
x^{2}+b x+2019=0
\end{gathered}
$$
Find the product of the roots of all the equations written on the board, given that each of them has two real roots. | Solution. According to the theorem converse to Vieta's theorem, the product of the roots of the first equation is $-\frac{2021}{2020}$, the product of the roots of the second equation is $\frac{2020}{2019}$, and the third is 2019. Therefore, the product of the roots of all equations is $\frac{2021}{2020} \cdot \frac{20... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,447 |
2. Masha wrote one non-zero digit on each of three cards (all digits are different). Dasha formed all possible three-digit numbers from them. Can the sum of these numbers be equal to 2021? | Solution. Let the digits on the cards be $a, b$, and $c$. Since they are all distinct, Dasha formed six numbers. Since each digit appears in the hundreds, tens, and units places twice, their sum is $200(a+b+c)+$ $20(a+b+c)+2(a+b+c)=222(a+b+c)$. Then the equation $222(a+b+c)=2021$ cannot hold, since 222 is divisible by ... | itcannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,448 |
4. In triangle $A B C$, the bisector $B E$ and the median $A D$ are perpendicular and have the same length, equal to 60. Find the sides of triangle $A B C$.
 | Solution. Let $P$ be the intersection point of segments $B E$ and $A D$. Triangle $A B D$ is isosceles because its bisector $B P$ is also an altitude. Therefore, $A P = P D = 30 ; B C = 2 B D = 2 A B$.
By the property of the bisector of a triangle, $\frac{C E}{A E} = \frac{B C}{A B} \Rightarrow \frac{C E}{A E} = 2 \Ri... | 15\sqrt{13},30\sqrt{13},45\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,450 |
5. Let $k$ be a natural number. It is known that among 29 consecutive numbers $30 k+1$, $30 k+2, \ldots, $30 k+29$ there are exactly 7 primes. Prove that the first and last of them are primes. | Solution. Let's cross out from this sequence the numbers that are multiples of 2, 3, or 5. There will be 8 numbers left: $30k+1, 30k+7, 30k+11, 30k+13, 30k+17, 30k+19, 30k+23, 30k+29$. Suppose that among them there is a composite number. We will prove that this number is a multiple of 7. The first seven of these number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,451 |
1. The equation $x^{2}-a x+2022=0$ has 2 positive integer roots. What is the smallest value of $a$ for which this is possible? | Solution. According to the theorem converse to Vieta's theorem, we have: $x_{1}+x_{2}=a$, $x_{1} \cdot x_{2}=2022$. Note that the product of the roots can be factored into two factors in 4 ways: $1 \cdot 2022, 2 \cdot 1011, 3 \cdot 674, 6 \cdot 337$. The smallest sum of the factors is $6+337=343$. Answer: 343. | 343 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,452 |
2. There are 25 coins, 12 of which are counterfeit and differ in weight by exactly 1 g from the genuine ones. All coins weigh an integer number of grams. Some may be lighter than the genuine ones, while others may be heavier. There are balance scales without weights, with a needle that shows the difference in weight. W... | Solution. We will prove that one weighing is sufficient. Set aside the coin under investigation, and place the rest on the scales, 12 coins on each side. If the scales show a difference in an even number of grams, then the coin is genuine; if in an odd number, then it is counterfeit. Indeed, if an odd number of counter... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,453 |
3. The numbers $p, q$ and $p q+1$ are prime. Prove that $(2 p+q)(p+2 q)$ is divisible by four. | Solution. Since $p q+1$ is the largest of the three given prime numbers, it is odd, which means the product $p q$ is even. Therefore, at least one of the numbers $p, q$ is equal to 2. Let's assume for definiteness that $p=2$, then $p+2 q=2(1+q)$ is divisible by 4 if $q>p$, since $1+q$ is even. If, however, $q=p=2$, the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,454 |
4. Find the hypotenuse of a right triangle if the height drawn to it is 1 cm, and one of the angles of the triangle is $15^{\circ}$. If the answer is not an integer, round it to the tenths. | Solution. Let's call the original triangle $ABC$. Let $CH$ be the height drawn to the hypotenuse; $\angle C=90^{\circ}$, and $\angle A=15^{\circ}$. Draw the median $CM$. It is clear that $CM=MA=MB$, so triangle $CMA$ is isosceles $\left(CM=MA \text{ and } \angle MCA=\angle MAC=15^{\circ}\right)$. Note that $\angle BMC=... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,455 |
5. How many even six-digit numbers exist, in the notation of which identical digits do not stand next to each other | Solution. Let $N_{k}(k>1)$ denote the number of even $k$-digit numbers in which identical digits do not stand next to each other. Similarly, define $N_{1}$, but exclude the number 0, so $N_{1}=4$. We can directly calculate $N_{2}=41$. Let $k>2$. Notice that from $k$-digit numbers satisfying the condition, we can form s... | 265721 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,456 |
2. Is it possible to arrange the digits $2,0,1,9$ (one digit per cell) in a $3 \times 3$ square so that each row, each column, and both diagonals contain three different digits? Provide an example or explain why it is not possible. | Solution. Answer: no. Consider the corner and central cells of a $3 \times 3$ square (a total of 5 cells). Any two of these cells lie either in the same column, or in the same row, or on the same diagonal. Since there are 5 cells in total, and there are 4 different digits to fill them, by the pigeonhole principle, ther... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,458 |
4. At a round table sit knights, who always tell the truth, and liars, who always lie. There might be a demon at the table: the demon lies only, but everyone considers him a knight. When those sitting at the table were asked, 10 of them said: "My left neighbor is a liar," the rest said "My left neighbor is a knight." C... | Solution. There is no demon at the table. Let's prove this.
First, consider the case when there is no demon at the table. The phrase "My left neighbor is a liar," regardless of who says it (a knight or a liar), indicates that among the speaker and their left neighbor, there is one knight and one liar. For the phrase "... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,460 |
5. In the cash register of the commercial director Vasily, money appeared for the first time in a long time - 2019 coins. Unfortunately, it is known that one of them is counterfeit, differing in weight from the genuine ones. Enraged workers employed by Vasily are demanding immediate payment of their salaries, and only ... | Solution. Answer: no. If each weighing results in equality, then all the coins used in it are genuine, and thus should be given away immediately. After this, a smaller odd number of coins will remain. Eventually, the fake coin will be the last one left, and Vasily will never learn its weight. | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,461 |
1. Can one of the angles of a triangle ( $x$ ) satisfy the equation: $\cos \frac{x}{2}=\cos \frac{x}{3} ?$ | Solution. Let's find the possible values of $x$ from the given equation. For this, we will transform it:
$\cos \frac{x}{2}-\cos \frac{x}{3}=-2 \sin \frac{5 x}{12} \sin \frac{x}{12}=0$
Then the possible cases are:
$\sin \frac{5 x}{12}=0$ or $\frac{5 x}{12}=n \pi, n \in \boldsymbol{Z}$ and the smallest positive value ... | cannot | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,462 |
2. From the set $\{10 ; 11 ; 12 ; \ldots ; 19\}$, 5 different numbers were chosen, and from the set $\{90 ; 91 ; 92 ; \ldots ; 99\}$, 5 different numbers were also chosen. It turned out that the difference between any two of the ten chosen numbers is not divisible by 10. Find the sum of all 10 chosen numbers. | Solution. Obviously, all the selected numbers have different last digits. Therefore, the sum can be found by separately adding the tens and units of these numbers. Among the tens, there are five 10s and five 90s, so the sum of the tens is $5 \cdot 10 + 5 \cdot 90 = 500$. And among the units, each digit appears exactly ... | 545 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,463 |
3. Does there exist a natural number $\boldsymbol{n}$ for which both numbers $\frac{2 n-5}{9}$ and $\frac{n-2}{15}$ are integers | Solution. Suppose, for the sake of contradiction, that such an $n$ exists, then denote the obtained integers as $m=\frac{2 n-5}{9}$ and $k=\frac{n-2}{15}$. From the first equation, $2 n=9 m+5$, and from the second, $n=15 k+2$. Then $2 n=9 m+5=30 k+4$ or $9 m+1=30 k$, which contradicts divisibility by 3.
Answer: Such a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,464 |
4. The diagonals $A C$ and $B D$ of a convex quadrilateral $A B C D$ are perpendicular and intersect at point $\boldsymbol{O}$. Circles $S_{1}, S_{2}, S_{3}, S_{4}$ with centers at points $O_{1}, O_{2}, O_{3}, O_{4}$ and radii $r_{1}, r_{2}, r_{3}, r_{4}$ are inscribed in triangles $A O B, B O C, C O D, D O A$, respect... | Solution. a) We use the known formula for the radius of the inscribed circle for a right triangle with legs $a, b$ and hypotenuse $c: r=\frac{1}{2}(a+b-c)$ (Fig. 1).
$$
\begin{gathered}
2\left(r_{1}+r_{2}+r_{3}+r_{4}\right) \leq 2(O A+O B+O C+O D)-(A B+B C+C D+D A)= \\
=2(A C+B D)-(A B+B C+C D+D A)= \\
=2(A C+B D)-\sq... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,465 |
5. The function $\mathbf{y}=f(x)$ satisfies the following conditions:
a) $f(4)=2$;
b) $f(n+1)=\frac{1}{f(0)+f(1)}+\frac{1}{f(1)+f(2)}+\ldots+\frac{1}{f(n)+f(n+1)}, n \geq 0$.
## Find the value of $f(2022)$. | Solution. For any $\quad n \geq 0 \quad f(n+2)-f(n+1)=\frac{1}{f(n+2)+f(n+1)}, \quad$ from which $f^{2}(n+2)=f^{2}(n+1)+1$. Using the condition $f(4)=2$, we get:
$$
\begin{gathered}
4=f^{2}(4)=f^{2}(3)+1 \Rightarrow f(3)=\sqrt{3} \\
3=f^{2}(3)=f^{2}(2)+1 \Rightarrow f(2)=\sqrt{2} \\
2=f^{2}(2)=f^{2}(1)+1 \Rightarrow f... | \sqrt{2022} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,466 |
9.1 In a $7 \times 7$ grid, several cells were marked, with at least one cell marked in each row and each column. It turned out that in any row of the grid, an odd number of cells were marked, and in any column, the number of marked cells is divisible by 3. What is the smallest and largest number of cells in the grid t... | Solution: Lower bound: In each column, either 3 or 6 cells are marked, so there are at least $3 \cdot 7=21$ marked cells in total.
Upper bound: Summing the number of marked cells by rows, we get that the total number of marked cells is odd, while summing by columns gives that it is divisible by 3. Since in each column... | 2139 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,467 |
9.2 In the fishing, 11 experienced fishermen and $n$ children participated. Together they caught $n^{2}+$ $5 n+22$ fish, with all experienced fishermen catching the same amount, and all children catching the same amount, but each 10 less than an experienced fisherman. Who was there more of at the fishing - experienced ... | Solution: Let each child catch $m$ fish. Then $n m+11(m+10)=n^{2}+5 n+22$. From this, $(n+11) m=n^{2}+3 n-88$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+5 n-88=$ $(n+11)(n-6)-22$, so 22 is divisible by $n+11$. The only divisor of 22 greater than 11 is 22 itself, so $n+11=22, n=11$. Therefore, the... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,468 |
9.4 Is there a rectangular box, all three dimensions of which (height, width, and depth) are expressed as irrational numbers, while the surface area and volume are integers? | Solution 1: Consider a box with sides $\sqrt{2}-1, \sqrt{2}-1,3+2 \sqrt{2}$. Its volume is $(\sqrt{2}-1)^{2}(3+2 \sqrt{2})=(3-2 \sqrt{2})(3+2 \sqrt{2})=1$, and the surface area is $2(\sqrt{2}-1)^{2}+4(\sqrt{2}-1)(3+2 \sqrt{2})=2(3-2 \sqrt{2})+4(\sqrt{2}+1)=10$.
Solution 2: Consider the polynomial $f(x)=x^{3}-30 x^{2}+... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,469 |
9.5 What is the largest number of non-overlapping groups into which all integers from 1 to 20 can be divided so that the sum of the numbers in each group is a perfect square? | Solution: A group consisting of a single number can only be formed by 4 squares. The remaining 16 numbers must be divided into groups of at least two. Therefore, there will be no more than 12 groups in total. Let's check that exactly 12 groups are not possible. Indeed, in such a case, the numbers 1, 4, 9, 16 would form... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,470 |
11.5. From the integers from 0 to 1000, 101 numbers were chosen. Prove that among the absolute values of their pairwise differences, there are ten different numbers not exceeding 100.
(I. Bogdanov) | Solution. Let $a_{0}<a_{1}<\ldots<a_{100}$ be the chosen numbers, ordered in ascending order. The sum of ten differences $a_{10}-a_{0}, a_{20}-a_{10}, \ldots, a_{100}-a_{90}$ is $a_{100}-a_{0} \leqslant 1000$, so one of these differences does not exceed 100. Let this difference be $a_{10 i+10}-a_{10 i};$ then
$$
0<a_{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,471 |
11.6. Positive numbers $a, b, c$ and $d$ satisfy the condition $2(a + b + c + d) \geqslant a b c d$. Prove that $a^{2} + b^{2} + c^{2} + d^{2} \geqslant a b c d$.
(A. Khryabrov) | Solution. There are two cases possible.
Case 1. Suppose that $a b c d \geqslant 16$. Then, by the inequality between the quadratic mean and the arithmetic mean, we have
$$
\begin{aligned}
a^{2}+b^{2}+c^{2}+ & d^{2} \geqslant 4\left(\frac{a+b+c+d}{4}\right)^{2} \geqslant \\
& \geqslant 4\left(\frac{a b c d}{8}\right)^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 17,472 |
11.7. The Master of the Mint wants to issue coins of 12 denominations (each - a natural number of rubles) so that any amount from 1 to 6543 rubles can be paid without change, using no more than 8 coins. Will he be able to do this? (When paying an amount, several coins of the same denomination can be used.)
(0. Podlips... | Answer: Yes.
Solution. Note that $9^{4}=6561>6543$. We will show that it is possible to choose 12 denominations such that any amount from 1 to 6560 rubles can be paid without change using no more than 8 coins.
First, let's show how to issue coins of three denominations so that any amount from 1 to 8 rubles can be pai... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,473 |
11.8. In triangle $A B C$, a circle $\omega$ is inscribed with its center at point $I$. A circle $\Gamma$ is circumscribed around triangle $A I B$. Circles $\omega$ and $\Gamma$ intersect at points $X$ and $Y$. The common tangents to circles $\omega$ and $\Gamma$ intersect at point $Z$. Prove that the circumcircles of ... | Solution. Let the circumcircle of triangle $ABC$ be denoted by $\Omega$. Suppose the bisector $CI$ intersects $\Omega$ again at point $S$. Then, as is known, $SA=SB=SI$, which means point $S$ is the center of circle $\Gamma$. By symmetry, point $Z$ lies on the line $SC$.
Let the common tangents to circles $\omega$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,474 |
2. Masha invited three bear cubs and a tiger cub to her place to solve math problems. After solving the problems, Masha counted the pastries in the kitchen and noticed that two were missing. Masha has a balance scale without weights, on which she can place pastries, bear cubs, and the tiger cub. All pastries weigh the ... | Solution. Denote the bear cubs as M1, M2, M3. First, compare the bear cubs M1 and M2. There are two possible outcomes.
1) One of the bear cubs turned out to be heavier, for example, $\mathrm{M} 1>\mathrm{M} 2$. Then, for the second weighing, compare M1 and M3 + pastry. If M1 < M3 + pastry, then M1 ate both pastries.
2... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,476 |
3. Find the numerical value of the expression
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{2}{a b+1}
$$
if it is known that $a$ is not equal to $b$ and the sum of the first two terms is equal to the third. | Answer: 2.
Solution. Let's bring the condition to a common denominator
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}=\frac{2}{a b+1}
$$
we get
$$
\frac{\left(a^{2}+b^{2}+2\right)(a b+1)-2\left(a^{2}+1\right)\left(b^{2}+1\right)}{\left(a^{2}+1\right)\left(b^{2}+1\right)(a b+1)}=0
$$
expand all brackets in the numerator, c... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,477 |
4. In the multi-story fraction
$$
D+\frac{1}{P+\frac{1}{0+\frac{1}{b+\frac{1}{b}}}}
$$
the letters were replaced with non-zero digits. What is the smallest result that can be obtained? (As usual, different letters correspond to different digits). | Answer. $1+\frac{1}{9+\frac{1}{2+\frac{1}{8+\frac{1}{3}}}}=1 \frac{53}{502}$.
Solution. Let the entire fraction be $T$ and write
$$
\mathrm{T}=\text { D }+\frac{1}{X}, \quad X=\mathrm{P}+\frac{1}{Y}, \quad Y=\mathrm{O}+\frac{1}{\mathrm{Z}}, \quad Z=\mathrm{B}+\frac{1}{\mathrm{~b}}
$$
Notice that all numbers $\mathrm... | 1\frac{53}{502} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,478 |
5. After the topic "triangles," all students were given paper triangles and scissors to solve the next problem. Mischievous Vasya, not listening to the teacher, did the following:
he folded his triangle along a straight line, made a straight cut with scissors, and got three pieces, unfolded the folded parts. Vasya ran ... | Answer: Yes, they can.
Solution. Consider a triangle $ABC$ where $\angle C=90^{\circ}, \angle A=60^{\circ}$. Draw its angle bisector $AD$ and the perpendicular $DE$ to the leg $BC$ (see the left figure). Fold the triangle along the line $DE$, then the point $A$ will take the position $A'$, after which we make a cut al... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,479 |
10-6-1. In the figure, there are two circles with centers $A$ and $B$. Additionally, points $A, B$, and $C$ lie on the same line, and points $D, B$, and $E$ also lie on the same line. Find the degree measure of the angle marked with a “?”.
. Angles $ABE$ and $DBC$ are equal as vertical angles. Note that triangle $DBC$ is isosceles wi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,480 |
10-8-1. There is a magical grid sheet of size $2000 \times 70$, initially all cells are gray. The painter stands on a certain cell and paints it red. Every second, the painter takes two steps: one cell to the left and one cell down, and paints the cell red where he ends up after the two steps. If the painter is in the ... | Answer: 14000.
Solution variant 1. We need to understand how many cells will be painted by the time the painter returns to the initial cell. Note that after every 2000 moves, the painter returns to the starting column, and after every 140 moves, he returns to the starting row. Therefore, he will return to the initial ... | 14000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,481 |
1. The distance between cities $A$ and $B$ is 435 km. A train departed from $A$ at a speed of 45 km/h. After 40 minutes, another train departed from city $B$ towards it at a speed of 55 km/h. How far apart will they be one hour before they meet? | Answer: The trains are approaching each other at a speed of 100 km/h. Therefore, in the last hour, they will together cover 100 km. This is the distance that will be between them one hour before the meeting.
Evaluation. 7 points for the correct solution. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,482 |
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