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11.4. In the vertices of a heptadecagon (17-sided polygon), different integers were written (one in each vertex). Then all the numbers were simultaneously replaced with new ones: each was replaced by the difference of the two following it in a clockwise direction (subtracting the next from the adjacent one). Could the ... | Answer: could not.
Solution. Let the numbers initially recorded at the vertices of the heptadecagon be: $a_{1}$, $a_{2}, \ldots, a_{17}$ (numbering - clockwise). Then, after the specified replacement, the numbers at the vertices will be: $a_{2}-a_{3}, a_{3}-a_{4}, \ldots, a_{16}-a_{17}, a_{17}-a_{1}, a_{1}-a_{2}$.
No... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,636 |
11.5. In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$. | Answer: $30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
First method. We use the "folding" method. Reflect triangle $P Q R$ symmetrically relative to line $P R$, and triangle $T S R$ - relative to line $T R$ (see Fig. 11.5a). From the equality (*... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,637 |
11.6. A stack consists of 300 cards: 100 white, 100 black, and 100 red. For each white card, the number of black cards lying below it is counted; for each black card, the number of red cards lying below it is counted; and for each red card, the number of white cards lying below it is counted. Find the maximum possible ... | Answer: 20000.
Solution. First method. The number of different permutations of cards is finite. Therefore, their arrangement with the largest indicated sum exists (possibly not unique).
Let the cards lie in such a way that this sum is maximal. Without loss of generality, we can assume that the top card is white. Then... | 20000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,638 |
4. In a right triangle $ABC$ with angle $\angle A=60^{\circ}$, a point $N$ is marked on the hypotenuse $AB$, and the midpoint of segment $CN$ is marked as point $K$. It turns out that $AK=AC$. The medians of triangle $BCN$ intersect at point $M$. Find the angle between the lines $AM$ and $CN$. | Solution. By the property of a right triangle with an angle of $30^{\circ}$, we get that $A B: A C=2: 1$. Therefore, $A B: A K=2: 1$. By the property of medians in a triangle, $\mathrm{BM}: \mathrm{MK}=2: 1$. Then, by the converse of the angle bisector theorem, we get that $A M$ is the bisector of angle $\angle \mathrm... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,643 |
7.2. On the table lies a wooden triangle with angles $70^{\circ}, 70^{\circ}$ and $40^{\circ}$ and a pencil, and no other instruments. On a large sheet of paper, you need to draw some equilateral triangle. How can this be done? You are allowed to place the triangle on the paper several times and trace its outline with ... | Solution: Notice that $40^{\circ} \cdot 3=120^{\circ}$, and that $180^{\circ}-120^{\circ}=60^{\circ}$. Therefore, if we sequentially trace the triangle three times, placing it on the paper as shown in the left figure (positions $1-3$), we will have laid out an angle of $120^{\circ}$, and the supplementary angle will be... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,645 |
7.3. Valera, Seryozha, and Dima received three grades each for three tests, and all the grades turned out to be threes, fours, or fives. Valera said: "I have higher grades for two tests than Seryozha." Seryozha replied: "But I have higher grades for two tests than Dima." Dima countered: "And I wrote two tests better th... | Solution: If Valera had grades $5,4,3$, Seryozha - 4, 3, 5, and Dima - 3, 5, 4, then all the boys told the truth: Valera performed better than Seryozha on the 1st and 2nd tests, Seryozha performed better than Dima on the 1st and 3rd tests, and Dima performed better than Valera on the 2nd and 3rd tests.
Note: The given... | Theycould | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,646 |
7.4. Katya's hair grows twice as fast as she does, while Alyona, who grows at the speed of Katya's hair, has hair that grows one and a half times faster than Alyona herself. At the present moment, Alyona and Katya have their hair at the same height above the floor. Whose hair will reach the floor first? Justify your an... | Solution: Let Katya grow by $x$ cm. Then her hair grows by $2 x$ cm and the distance from it to the floor decreases by $2 x - x = x$ cm. During this time, Alyona grows by $2 x$ cm, her hair by $3 x$ cm, and the distance from it to the floor decreases by $3 x - 2 x = x$ cm. Conclusion: Alyona's and Katya's hair will rea... | Simultaneously | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,647 |
7.5. Point $B$ lies on segment $A C, A B=2$, $B C=1$. Indicate all points $M$ on line $A C$ for which $A M+M B=C M$, and show that there are no other points $M$ with this property. | # Solution:
Method 1. Depending on the position of point $M$ on line $A C$, there are 4 cases.
Case 1. Point $M$ lies outside segment $A C$ on the side of point $C$. Then the equality is impossible, as $A M > C M$.
Case 2. Point $M$ lies outside segment $A C$ on the side of point $A$. Then
$$
C M = A M + A B + B C ... | There\\two\such\points:\the\midpoint\of\segment\AB\\the\point\located\1\\to\the\left\of\point\A | Geometry | proof | Yes | Yes | olympiads | false | 12,648 |
1. Vitya Pereperepkin always calculates percentages incorrectly in surveys: he divides the number of people who answered in a certain way by the number of all the others. For example, in the survey "What is your name?", conducted among 7 Anyas, 9 Ols, 8 Yuls, Vitya counted $50 \%$ Yuls.
Vitya conducted a survey in his... | Answer: $a=110$ (in the $2-nd$ variant $a=104$). When Vitya calculates that some part of the students constitutes $5 \%$ of the entire class, in reality, it constitutes $5 / 100=1 / 20$ of the remaining students, $1 / 21$ of the entire class. Similarly, Vitya's $50 \%$ is one third of the class. Therefore, all those wh... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,649 |
4. We will call a number anti-triangular if it can be written in the form $\frac{2}{n(n+1)}$ for some natural number $n$. For how many numbers $k(1000 \leqslant k \leqslant 2000)$ can the number 1 be written as the sum of $k$ anti-triangular numbers (not necessarily distinct)? | Answer: all 1001 possible values of $k$ work.
Notice that $1 / 3=2 / 2 \cdot 3$ and $1 / 6=2 / 3 \cdot 4$ are anti-triangular numbers. Therefore, unity can be written as the sum of three, four, and five such numbers: $1=1 / 3+1 / 3+1 / 3=1 / 6+1 / 6+1 / 3+$ $1 / 3=1 / 6+1 / 6+1 / 6+1 / 6+1 / 3$. We will prove that the... | all1001possiblevaluesofkwork | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,650 |
1. Over the summer, the price of a one-bedroom apartment increased by $21 \%$, a two-bedroom apartment by $11 \%$, and the total cost of the apartments by $15 \%$. How many times cheaper is the one-bedroom apartment compared to the two-bedroom apartment? | Answer. One and a half times.
Solution. Let the cost of a one-bedroom apartment be $a$ rubles, and a two-bedroom apartment be $b$ rubles. Then, from the condition of the problem, it follows that $1.21 a + 1.11 b = 1.15(a + b)$, from which $1.5 a = b$.
## Grading Criteria.
- Any complete correct solution - 7 points.
... | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,651 |
2. Find any pair of natural numbers $a$ and $b$, both greater than 1, that satisfy the equation $a^{13} \cdot b^{31}=6^{2015}$. | Solution. It is enough to provide one example. Since $2015=13 \cdot 155=31 \cdot 65$, the values $a=2^{155}, b=3^{65}$ work. Indeed,
$$
\left(2^{155}\right)^{13} \cdot\left(3^{65}\right)^{31}=2^{2015} \cdot 3^{2015}=6^{2015}
$$
Comment. Besides the obvious alternative $a=3^{155}, b=2^{65}$, there can be others. For e... | =2^{155},b=3^{65} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,652 |
3. Can the sum of 2015 consecutive natural numbers end with the same digit as the sum of the next 2019 numbers? | Answer: It cannot.
Method 1 (searching for numbers). Let the sum of numbers from $a$ to $a+2014$ end with the same digit as the sum of numbers from $a+2015$ to $a+4033$. Then the difference between these sums, which is a multiple of 10, is equal to
\[
\begin{gathered}
\frac{2015 \cdot(a+a+2014)}{2}-\frac{2019 \cdot(a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,653 |
4. Teacher Maria Ivanovna is preparing tasks for a math lesson. She wants to substitute three different natural numbers for $a$, $b$, and $c$ in the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ so that the roots of the equation are integers. Help her: choose such numbers and solve the equation. | Solution. There are many such equations, it is sufficient to provide one of them. For example, the roots of the equation
$$
\frac{1}{x+6}+\frac{1}{x+3}=\frac{1}{2}
$$
are the numbers 0 and -5.
Comment. The equation provided in the solution can be derived. Consider the equality
$$
\frac{1}{6}+\frac{1}{3}=\frac{1}{2}... | \frac{1}{x+6}+\frac{1}{x+3}=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,654 |
5. Petya marked a point $X$ on the edge $AB$ of the cube $ABCD A_{1} B_{1} C_{1} D_{1}$, dividing the edge $AB$ in the ratio $1:2$, starting from vertex $A$. Provide an example of how Petya can mark points $Y$ and $Z$ on the edges $C C_{1}$ and $A_{1} D_{1}$, respectively, so that the triangle $XYZ$ is equilateral. Jus... | Solution. Mark points $Y$ and $Z$ such that $A_{1} Z: Z D_{1}=2: 1, C_{1} Y: Y C=2: 1$. The equality of the sides of triangle $X Y Z$ follows, for example, from the equality of the broken lines $X A A_{1} Z, Z D_{1} C_{1} Y$ and $Y C B X$.
Let $a$ be the length of the edge of the given cube. Then the segments of the b... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,655 |
6. In a checkers tournament, students from 10th and 11th grades participated. Each player played against every other player once. A player received 2 points for a win, 1 point for a draw, and 0 points for a loss. There were 10 times more 11th graders than 10th graders, and together they scored 4.5 times more points tha... | Answer: 20.
Solution: 1. Let $a$ be the number of tenth graders who participated in the tournament, earning $b$ points. Then, $10a$ eleventh graders played, earning $4.5b$ points. In each match, 2 points are played for, and a total of $11a$ players play $\frac{11 a(11 a-1)}{2}$ matches. Therefore, from the condition o... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,656 |
9.1. Two given quadratic trinomials $f(x)$ and $g(x)$ each have two roots, and the equalities $f(1)=g(2)$ and $g(1)=f(2)$ hold. Find the sum of all four roots of these trinomials. | Answer: 6.
First solution. Let $f(x)=x^{2}+a x+b, g(x)=x^{2}+$ $+c x+d$. Then the conditions of the problem can be written as
$$
1+a+b=4+2 c+d \quad \text { and } \quad 4+2 a+b=1+c+d
$$
Subtracting the second equation from the first, we get $-3-a=3+c$, which means $a+c=-6$. By Vieta's theorem, $-a$ is the sum of the... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,657 |
9.2. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some integer. Then the first said: “My number is greater than 1”, the second said: “My number is greater than 2”, \ldots, the tenth said: “My number is greater than 10”. After that, all ten, sp... | Answer: 8 knights.
Solution. We will prove that none of the knights could have said either of the phrases "My number is greater than 9" or "My number is greater than 10." Indeed, if this were possible, the integer thought of by the knight would be at least 10. But then he could not have said any of the phrases "My num... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,658 |
9.3. One hundred different natural numbers are arranged in a circle. Vasya divided each of them by the next one in the clockwise direction with a remainder; it turned out that the remainders obtained by Vasya take only two different values. Petya divided each of the numbers by the next one in the counterclockwise direc... | Solution. Let us number the numbers clockwise $a_{1}, a_{2}$, $\ldots, a_{100}$ so that the number $a_{100}$ is the smallest. Then the remainder of dividing $a_{100}$ by $a_{1}$ will be $a=a_{100}$ (since $a_{1}>a_{100}$), and the remainder $b$ of dividing $a_{99}$ by $a_{100}$ will be less than $a_{100}$. Therefore, $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,659 |
9.4. The altitudes of an acute-angled triangle $ABC$ intersect at point $H$. Points $B_{1}$ and $C_{1}$ are marked on segments $BH$ and $CH$ respectively such that $B_{1}C_{1} \| BC$. It turns out that the center of the circle $\omega$, circumscribed around triangle $B_{1}HC_{1}$, lies on the line $BC$. Prove that the ... | Solution. Let $\Gamma^{\prime}$ denote the circumcircle of triangle $B H C$ (see Fig. 1). On the tangent to $\Gamma^{\prime}$ at point $H$, mark point $X$ lying inside angle $B C H$. Then $\angle B H X = \angle B C H = \angle B_{1} C_{1} H$ (the last equality follows from the fact that $B C \| B_{1} C_{1}$). Therefore,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,660 |
9.5. Each face of a cube $1000 \times$ $\times 1000 \times 1000$ is divided into $1000^{2}$ square cells with side 1. What is the maximum number of these cells that can be painted so that no two painted cells share a side?
$$
\text { (S. Dolgikh) }
$$ | Answer. $3 \cdot 1000^{2}-2000=$ $=2998000$ cells.
Solution. Consider an arbitrary coloring that satisfies the condition. Divide all the cells on the surface into "edges" as shown in Fig. 2 -... | 2998000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,661 |
# 1. CONDITION
Write the number 1997 using 11 fours and arithmetic operations. | Solution. $1997=4 \cdot 444+44 \cdot 4+44+4: 4$. | 1997=4\cdot444+44\cdot4+44+4:4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,662 |
# 2. CONDITION
Which of the numbers is greater: $2^{1997}$ or $5^{850}$? | Solution. From the inequality $2^{7}=128>125=5^{3}$, it follows that
$$
2^{1997}=\left(2^{7}\right)^{285} \cdot 2^{2}>\left(5^{3}\right)^{285}=5^{855}>5^{850}
$$
Answer: $2^{1997}>5^{850}$. | 2^{1997}>5^{850} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,663 |
9.1. On a plane, 5 points are marked. Prove that some of them can be chosen and moved in such a way that the distance between any two of the moved points does not change, and as a result, a set of 5 points remains on the plane, symmetric with respect to some line.
(S. Volchonkov) | Solution. Let's denote the given points as $A, B, C, D$ and $E$. Choose the two points that are farthest from each other, let these be $A$ and $B$. We will show that it is possible to move them in the required manner.
Draw the perpendicular bisector $\alpha$ of the segment $C D$. If $E$ lies on $\alpha$, then it is su... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,664 |
9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial
$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
has at least one integer root?
(P. Kozlov) | Answer. For $n=6$.
Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.
Suppose the numbers $a_{1}, a_{2}, \ldots, ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,665 |
9.3. Circle $\Omega$ with center at point $O$ is circumscribed around an acute-angled triangle $ABC$, where $AB < BC$; its altitudes intersect at point $H$. On the extension of segment $BO$ beyond point $O$, point $D$ is marked such that $\angle ADC = \angle ABC$. The line passing through point $H$ parallel to line $BO... | Solution. Let $P$ be the second intersection point of $BO$ with the circle $\Omega$ (see Fig. 1). Then $BP$ is the diameter of $\Omega$, and $\angle BCP = 90^\circ = \angle BAP$. Therefore, $CP \parallel AH$ and $AP \parallel CH$. Consequently, quadrilateral $AHCP$ is a parallelogram. Let $M$ be the intersection point ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,666 |
9.4. 10,000 children arrived at the camp, each of whom is friends with exactly 11 other children in the camp (friendship is mutual). Each child wears a T-shirt of one of the seven colors of the rainbow, and any two friends have different colors. The counselors demanded that some children (at least one) change the color... | Solution. Let's move to a graph where the vertices correspond to children, and the edges correspond to friendships. Recall that a vertex coloring is called proper if the colors of any two vertices connected by an edge are different. Thus, the graph is properly colored in 7 colors, 100 stable vertices are selected, and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,667 |
9.6. On the side $AC$ of the isosceles triangle $ABC$ with base $BC$, a point $D$ is taken. On the smaller arc $CD$ of the circumcircle of triangle $BCD$, a point $K$ is chosen. The ray $CK$ intersects the line parallel to $BC$ and passing through $A$ at point $T$. Let $M$ be the midpoint of segment $DT$. Prove that $\... | Solution. Extend segment $A M$ by its length beyond point $M$, obtaining point $N$ such that $A D N T$ is a parallelogram. Since $\angle A N T = \angle C A M$, to solve the problem, it is sufficient to show that $\angle A K T = \angle A N T$, or that points $A, T, N, K$ lie on the same circle (see Fig. 1).
Let $N D$ i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,669 |
9.7. Among 16 coins, there are 8 heavy ones weighing 11 g each, and 8 light ones weighing 10 g each, but it is unknown which coins are of which weight. One of the coins is a commemorative coin. How can you determine in three weighings on a two-pan balance without weights whether the commemorative coin is heavy or light... | Solution. Let's denote the commemorative coin as Ю. Set aside two non-commemorative coins $A$ and $B$, and distribute the remaining 14 coins into 7 on each side of the balance, ensuring that Ю is placed on the left. We will call a coin left or right if it was placed on the left or right side of the balance, respectivel... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,670 |
9.8. Given numbers $a, b, c$, not less than 1. Prove that
$$
\frac{a+b+c}{4} \geqslant \frac{\sqrt{a b-1}}{b+c}+\frac{\sqrt{b c-1}}{c+a}+\frac{\sqrt{c a-1}}{a+b}
$$
(K. Tyszuk) | Solution. By the AM-GM inequality, we have $b+c \geqslant 2 \sqrt{b c}$, hence
$$
4 \frac{\sqrt{a b-1}}{b+c} \leqslant 2 \sqrt{\frac{a b-1}{b c}}=2 \sqrt{\left(a-\frac{1}{b}\right) \cdot \frac{1}{c}} \leqslant\left(a-\frac{1}{b}\right)+\frac{1}{c}
$$
where in the last transition the AM-GM inequality is applied again.... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,671 |
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
 | Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,672 |
Problem 9.1. In a notebook, a triangular grid is drawn (see figure). Tanya placed integers at the nodes of the grid. We will call two numbers close if they are in adjacent nodes of the grid. It is known that
- the sum of all ten numbers is 43;
- the sum of any three numbers such that any two of them are close is 11.
... | Answer: 10.
Solution. Let's denote the numbers by variables as shown in the figure.
Then
\[
\begin{gathered}
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}=c_{1}+c_{2}+c_{3}=11 \\
\left(a_{1}+a_{2}+a... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,673 |
Problem 9.2. The least common multiple of four pairwise distinct numbers is 165. What is the maximum value that the sum of these numbers can take? | Answer: 268.
Solution. Since 165 is the least common multiple of four numbers, these numbers are divisors of 165. To maximize the sum of these numbers, it is sufficient to take the four largest divisors of 165. If one of them is the number 165 itself, then the LCM will definitely be equal to it.
Then the maximum sum ... | 268 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,674 |
Problem 9.3. The teacher wrote a fraction on the board, where the numerator and the denominator are natural numbers. Misha added 30 to the numerator of the given fraction and wrote the resulting fraction in his notebook, while Lesha subtracted 6 from the denominator of the fraction written on the board and also wrote t... | Answer: 5.
Solution. Let $\frac{a}{b}$ be the original fraction. Then Misha wrote down the fraction $\frac{a+30}{b}$ in his notebook, and Lёsha wrote down $-\frac{a}{b-6}$.
Let's write the equation
$$
\frac{a+30}{b}=\frac{a}{b-6}
$$
Transforming it, we get
$$
\begin{gathered}
(a+30)(b-6)=a b \\
a b+30 b-6 a-180=a ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,675 |
Problem 9.4. Given a cyclic quadrilateral $A B C D$. It is known that $\angle A D B=48^{\circ}, \angle B D C=$ $56^{\circ}$. Inside triangle $A B C$, a point $X$ is marked such that $\angle B C X=24^{\circ}$, and ray $A X$ is the angle bisector of $\angle B A C$. Find the angle $C B X$.
$.
The length of segment $A B$ is equal to the absolute value of the difference of the roots of the equation $x^{2}+a x+b=s$. We can express the difference of the... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,677 |
Problem 9.6. On a line, two red points and several blue points are marked. It turned out that one of the red points is contained in exactly 56 segments with blue endpoints, and the other - in 50 segments with blue endpoints. How many blue points are marked? | Answer: 15.
Solution. Let there be $a$ blue points to the left of the first red point, and $b$ blue points to the right; $c$ blue points to the left of the second red point, and $d$ blue points to the right. Then $a b=56, c d=50$. Additionally, $a+b=c+d$ - the number of blue points.
Notice that among the numbers $c$ ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,678 |
Problem 9.7. On the coordinate plane, points $O(0 ; 0), A(5 ; 0), B(0 ; 4)$ are marked. The line $y=k x+b$ is such that for any point $M$ on this line, the area of the quadrilateral $A O B M$ is 20. What is $k$? | Answer: $-0.8$.
Fig. 1: to the solution of problem 9.7
Solution. For any point $M$ from the condition of the problem, the following is true
$$
S_{A B M}=S_{O A M B}-S_{A B O}=20-10=10
$$
... | -0.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,679 |
Problem 9.8. Young entomologist Dima is observing two grasshoppers. He noticed that when a grasshopper starts jumping, it jumps 1 cm, then after a second, 2 cm, then another second, 3 cm, and so on.
Initially, both grasshoppers were in the same place. One of them started jumping, and after a few seconds, the second on... | Answer: $10,15,30$.
Solution. Note that the grasshoppers are one or several consecutive jumps apart from each other. Thus, the distance between them is the sum of $m$ consecutive natural numbers, where $m$ is the number of seconds by which the second grasshopper started later.
On the other hand, four consecutive jump... | 10,15,30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,680 |
1. Late at night, three wizards met at a crossroads. Each had a magic wand in their hands. A strong wind came, snatched their wands, played with them so much that none of the wizards could distinguish their own wand from another's. The wizards had no choice but to pick up the wands at random. Find the probability that ... | Solution. The wizards have six different ways to take the magic wands. In only two of the six scenarios, none of the wizards will take their own wand, since the wizard who takes the wand first can take any of the two wands belonging to the other wizards. The remaining wizards have no choice, as one of their wands is st... | \frac{1}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,681 |
2. Given an arbitrary triangle $ABC$. Prove that for this triangle the inequality $BC^{3} + AC^{3} + 3 \cdot BC \cdot AC \cdot AB > AB^{3}$ holds. | Solution. Let the sides of the triangle be denoted by $a, b, c$. Then we need to prove the inequality $a^{3}+b^{3}+3 a b c>c^{3}$. Using the triangle inequality $a+b>c$, we obtain the following chain of inequalities:
$a^{3}+b^{3}+3 a b c=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a b c>c \cdot\left(a^{2}-a b+b^{2}\right)+3 a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,682 |
3. Pantelej and Gerasim received 20 grades each in November, and Pantelej received as many fives as Gerasim received fours, as many fours as Gerasim received threes, as many threes as Gerasim received twos, and as many twos as Gerasim received fives. At the same time, their average grade for November is the same. How m... | Solution. Let's add one to each of Gerasim's grades. His total score will increase by 20. On the other hand, it will become greater than Pantelej's total score by four times the number of Pantelej's twos.
Answer: 5 twos. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,683 |
4. Given a set of identical regular pentagons, with natural numbers from 1 to 5 written at each vertex, as shown in the figure. The pentagons can be rotated and flipped. They were stacked (vertex to vertex), and it turned out that the sums of the numbers at each of the five vertices are the same. How many pentagons cou... | Solution. Stack two pentagons, flipping one of them, and align their numbers as follows: $1-5, 2-4, 3-3, 4-2$, and 5-1. Then we get a stack that meets the condition. Let's call it a "pair." Stack five pentagons, rotating them relative to each other so that the numbers align as 1-2-3-4-5, 2-3-4-5-1, and so on. Then we g... | anynaturalexcept13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,684 |
5. Each pair of numbers $A$ and $B$ is assigned a number $A * B$. Find $2021 * 1999$, if it is known that for any three numbers $A, B, C$ the identities are satisfied: $A * A=0$ and $A *(B * C)=(A * B)+C$. | Solution.
$A *(A * A)=A * 0=A *(B * B)=A * B+B$,
$A *(A * A)=(A * A)+A=0+A=A$, then $A * B+B=A$.
Therefore, $A * B=A-B$. Hence, $2021 * 1999=2021-1999=22$.
Answer: 22. | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,685 |
8.1. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total? | 8.1. Let a pen cost $r$ rubles, and the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bought is $\frac{357}{2... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,686 |
8.2. In 8th grade class "G", there are enough underachievers, but Vovochka studies the worst of all. The pedagogical council decided that either Vovochka must correct his twos by the end of the quarter, or he will be expelled. If Vovochka corrects his twos, then the class will have $24 \%$ of underachievers, and if he ... | 8.2. Let there be $n$ students in the class now. According to the condition,
$$
0.24 n = 0.25(n-1)
$$
i.e., $0.01 n = 0.25$. Therefore, $n = 25$. One person constitutes $4\%$ of 25, so there are now $24 + 4 = 28\%$ of underachievers.
Answer: $28\%$. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,687 |
8.3. In triangle $\mathrm{ABC}$, $\mathrm{AC}=1$, $\mathrm{AB}=2$, $\mathrm{O}$ is the point of intersection of the angle bisectors. A segment passing through point O and parallel to side $\mathrm{BC}$ intersects sides $\mathrm{AC}$ and $\mathrm{AB}$ at points K and M, respectively. Find the perimeter of triangle $\mat... | 8.3. $\angle \mathrm{KCO}=\angle \mathrm{BCO}=\angle \mathrm{KOC}$ (alternate interior angles). Therefore, OK = KC, and similarly BM = OM. Then
$$
A K+A M+K M=A K+K C+A M+B M=3 .
$$
Answer: 3. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,688 |
8.4. How many natural numbers less than 1000 are divisible by 4 and do not contain the digits $1,3,4,5,7,9$ in their notation? | 8.4. The desired numbers are written only with the digits $0,2,6,8$.
There is exactly one single-digit number that satisfies the condition, which is the number 8.
There are six two-digit numbers, which are $20,28,60,68,80,88$.
The desired three-digit numbers can end with the following 8 combinations of digits: $00,0... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,689 |
8.5. Prove that if the expression $x^{2}+y^{2}$ is divisible by 3, where $x$ and $y$ are integers, then $x$ and $y$ are divisible by 3. | 8.5. Let $x=3 a+r_{1}$ and $y=3 b+r_{2}$, where $r_{1}, r_{2}$ are the remainders when divided by 3, i.e., some of the numbers $0,1,2$. Then $x^{2}+y^{2}=3\left(3 a^{2}+3 b^{2}+2 a r_{1}+2 b r_{2}\right)+r_{1}^{2}+r_{2}^{2}$. Since $x^{2}+y^{2}$ is divisible by 3, then $r_{1}^{2}+r_{2}^{2}$ is divisible by 3, i.e., $r_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,690 |
9.1. At a round table, 2015 people are sitting, each of them is either a knight or a liar. Knights always tell the truth, and liars always lie. Each person was given a card with a number on it; all the numbers on the cards are different. After looking at their neighbors' cards, each person said: "My number is greater t... | Answer: 2013.
Solution. Let $A$ and $B$ be the people who received the cards with the largest and smallest numbers, respectively. Since both of them said the first phrase, $A$ is a knight, and $B$ is a liar. However, if they had said the second phrase, $A$ would have lied, and $B$ would have told the truth; this is im... | 2013 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,691 |
9.2. We will call a natural number interesting if the sum of its digits is a prime number. What is the maximum number of interesting numbers that can be among five consecutive natural numbers?
(
#
| # Answer: 4.
Solution. Among five consecutive natural numbers, there can be 4 interesting numbers. For example, the numbers 199, 200, 201, 202, 203 (with digit sums 19, 2, 3, 4, and 5) will work.
Now, let's prove that all 5 numbers cannot be interesting. Among our five numbers, there are three that lie within the sam... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,692 |
9.4. In a non-isosceles triangle \(ABC\), the bisectors of angle \(ABC\) and the adjacent angle were drawn. They intersect the line \(AC\) at points \(B_1\) and \(B_2\) respectively. From points \(B_1\) and \(B_2\), tangents to the incircle of triangle \(ABC\), different from the line \(AC\), were drawn. They touch thi... | Solution. Let $I$ be the center of the circle $\omega$ inscribed in triangle $ABC$. Let $D$ be the point of tangency of $\omega$ with side $AC$ (see Fig. 3). Since the line $BB_1$ passes through the center $\omega$, and points $D$ and $K_1$ are symmetric with respect to the line $BB_1$, that is, $BI$ is the bisector of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,694 |
10.1. Integers $a, x_{1}, x_{2}, \ldots, x_{13}$ are such that $a=\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{13}\right)=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{13}\right)$. Prove that $a x_{1} x_{2} \ldots x_{13}=0$.
(V. Senderov) | Solution. If any of the numbers $x_{i}$ is equal to 0, the statement of the problem is obvious. If one of the $x_{i}$ is equal to $\pm 1$, then $a=0$, so the statement is also true. Otherwise, for each $i$, the number $\left(1+x_{i}\right)\left(1-x_{i}\right)=1-x_{i}^{2}$ is negative. On the other hand, from the condit... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,695 |
10.2. On a plane, all vertices of a regular $n$-gon and its center were marked. Then the contour of this $n$-gon was drawn, and the center was connected to all vertices; as a result, the $n$-gon was divided into $n$ triangles. Vasya wrote a number in each marked point (some of the numbers may be equal). In each triangl... | Answer. For odd $n$.
Solution. Let $A_{1} A_{2} \ldots A_{n}$ be the given polygon, and $S$ be its center.
First, we show that if $n$ is even, there exist two different arrangements where the corresponding triples will be the same. Then, Petya will not be able to restore the original numbers. In the first arrangement... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,696 |
10.3. Let $A L$ be the bisector of triangle $A B C$. The perpendicular bisector of segment $A L$ intersects the circumcircle of triangle $A B C$ at points $P$ and $Q$. Prove that the circumcircle of triangle $P L Q$ is tangent to side $B C$. (S. Berlov) | Solution. Note that triangles $P L Q$ and $P A Q$ are symmetric with respect to the line $P Q$. Through point $A$, draw the tangent $X Y$ to the circle on which points $A, B, C, P, Q$ lie (see Fig. 4). To solve the problem, it is sufficient to prove that the lines $X Y$ and $B C$ are symmetric with respect to the line ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,697 |
10.4. Positive numbers $a, b, c$ satisfy the relation $a b +$
$+b c+c a=1$. Prove that
$$
\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c}) .
$$
(P. Kozlov) | Solution. Note that
$$
a+\frac{1}{a}=a+\frac{ab+ac+bc}{a}=a+b+c+\frac{bc}{a}
$$
Applying the inequality of means for the numbers $a$ and $bc/a$, we get
$$
a+\frac{1}{a} \geqslant b+c+2\sqrt{bc}=(\sqrt{b}+\sqrt{c})^{2}
$$
from which $\sqrt{a+\frac{1}{a}} \geqslant \sqrt{b}+\sqrt{c}$. Similarly, the inequalities $\sq... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,698 |
11.1. Integers $a, x_{1}, x_{2}, \ldots, x_{13}$ are such that $a=\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{13}\right)=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{13}\right)$. Prove that $a x_{1} x_{2} \ldots x_{13}=0$.
(V. Senderov) | Solution. If any of the numbers $x_{i}$ is equal to 0, the statement of the problem is obvious. If one of the $x_{i}$ is equal to $\pm 1$, then $a=0$, so the statement is also true. Otherwise, for each $i$, the number $\left(1+x_{i}\right)\left(1-x_{i}\right)=1-x_{i}^{2}$ is negative. On the other hand, from the condit... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,699 |
11.2. Several married couples came to the New Year's Eve party, each of whom had from 1 to 10 children. Santa Claus chose one child, one mother, and one father from three different families and took them for a ride in his sleigh. It turned out that he had exactly 3630 ways to choose the necessary trio of people. How ma... | Answer: 33.
Solution: Let there be $p$ married couples and $d$ children at the party (from the condition, $d \leqslant 10 p$). Then each child was part of $(p-1)(p-2)$ trios: a mother could be chosen from one of the $p-1$ married couples, and with a fixed choice of mother, a father could be chosen from one of the $p-2... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,700 |
11.3. The continuations of the medians $A A_{1}, B B_{1}$, and $C C_{1}$ of triangle $A B C$ intersect its circumscribed circle at points $A_{0}, B_{0}$, and $C_{0}$, respectively. It turns out that the areas of triangles $A B C_{0}, A B_{0} C$, and $A_{0} B C$ are equal. Prove that triangle $A B C$ is equilateral. | Lemma. Let $A, B, C$ and $D$ be distinct points on a plane, and let lines $A C$ and $B D$ intersect at point $E$. Then $\frac{S_{A B C}}{S_{A D C}}=\frac{B E}{D E}$.
Proof. Let $B H_{B}$ and $D H_{D}$ be the altitudes in triangles $A B C$ and $A D C$ respectively (see Fig. 5). Then $B H_{B} \| D H_{D}$, so triangles $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,701 |
11.4. Positive numbers $a, b, c$ satisfy the relation $a b + b c + c a = 1$. Prove that
$$
\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})
$$
(P. Kozlov) | Solution. Note that
$$
a+\frac{1}{a}=a+\frac{a b+a c+b c}{a}=a+b+c+\frac{b c}{a} .
$$
Applying the inequality of means for the numbers $a$ and $b c / a$, we get
$$
a+\frac{1}{a} \geqslant b+c+2 \sqrt{b c}=(\sqrt{b}+\sqrt{c})^{2}
$$
from which $\sqrt{a+\frac{1}{a}} \geqslant \sqrt{b}+\sqrt{c}$. Similarly, the inequa... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,702 |
2. At the New Year's celebration, schoolchildren organized an exchange game: if they were given five tangerines, they would exchange them for three crackers and a candy, and if they were given two crackers, they would exchange them for three tangerines and a candy. Father Frost played this game with them several times ... | Solution. Ded Moroz conducted 50 exchanges, as he was given 50 candies. At the end of the game, he had no crackers left, meaning he exchanged all of them back. For every two exchanges of tangerines for crackers (ten given - six received), he conducted three exchanges of crackers for tangerines (six given - nine receive... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,704 |
3. For a natural even number $n$, it is known that if it is divisible by a prime number $p$, then the number $n-1$ is divisible by the number $p-1$. Prove that $n$ can only be a power of two. | Solution. Let the number $n$ be divisible by an odd prime $p$. Then the number $p-1$ is even, which means that the odd number $n-1$ is divisible by the even number $p-1$. This is a contradiction. Therefore, the number $n$ does not have any odd divisors, and in its prime factorization, only the number 2 is present, i.e.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,705 |
4. A tourist walked all the paths in the park, passing each one exactly twice. The gardener says that the paths in the park cannot be walked, passing each one exactly once. Can such a park exist? | Solution. Such a park exists. Let's provide an example. In the figure, the paths of the park form a star. This park can be walked around by going back and forth along the paths, starting from the central point $A(A-B-A-C-A-D-A)$. You can also start the walk from any terminal point. This park cannot be walked around by ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,706 |
5. There are 111 pebbles in a pile. Petya and Vasya are playing a game: they take turns taking any number of pebbles from the pile, but no more than 9. Skipping a turn is not allowed. The player who takes the last pebble wins. Petya goes first. Who will win with correct play, and how should he play? | Solution. Petya will win. He needs to take one pebble, leaving Vasya with 110. Then, in response to any move by Vasya, he should take enough pebbles to leave a number of pebbles that is a multiple of 10 before Vasya's next move. This is always possible because Vasya takes between 1 and 9 pebbles, and therefore cannot r... | Petyawillwin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,707 |
6. How many six-digit numbers exist in which four consecutive digits form the number $2021?$ | Solution. Consider three types of six-digit numbers: $\overline{a b 2021}, \overline{a 2021 b}, \overline{2021 a b}$ (the bar denotes the decimal representation of the number, where $a, b$ - are digits).
In the first case, $\overline{a b 2021}$, $a$ can be any digit from 1 to 9, and $b$ can be any digit from 0 to 9. T... | 280 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,708 |
11.5. This figure contains $9^{2}-4 \cdot 3=69$ cells. Figure 2 shows how a gardener can plant 60 apple trees. We will prove that it is impossible to plant more than 60 apple trees.
Figure 2
... | Answer: the maximum number of apple trees is 60. | 60 | Combinatorics | proof | Yes | Yes | olympiads | false | 12,710 |
7.1. Are there such numbers $a$ and $b$ for which all four numbers $a+b, a-b, a \cdot b, a: b$ are different and each of these numbers equals one of the numbers $0,9,3,6, \quad 1,6,3,9$? | Answer: Yes, $a=2.4$ and $b=1.5$.
Comments. The suitable pair is unique. If it is indicated correctly - 7 points. | =2.4,b=1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,711 |
7.3. Prove that for any positive real numbers $a, b, c, x, y$ at least one of the numbers $x /(a b+a c)$ and $y /(a c+b c)$ will be less than $(x+y) /(a b+b c)$. | Solution. If $x /(a b+a c) \leq y /(a c+b c)$, then $x /(a b+a c) \leq(x+y) /(a b+b c+2 a c)<(x+y) /(a b+b c)$. If $y /(a c+b c) \leq x /(a b+a c)$, then $y /(a c+b c) \leq(x+y) /(a b+b c+2 a c)<(x+y) /(a b+b c)$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,713 |
7.4. Prove that there are infinitely many natural numbers $n$ for which the numbers $2 n-3$ and $3 n-2$ have a common divisor not equal to 1. | Solution. It is sufficient to note that for any positive integer $k$, when $n=5 k-1$, the numbers $2 n-3=10 k-5$ and $3 n-2=15 k-5$ are divisible by 5. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,714 |
7.5. Prove that from any set of 15 positive integers, each of which does not exceed 2020, it is always possible to select two non-intersecting subsets with the same sums of the numbers in these subsets. | Solution. The sum of 15 numbers is no more than $15 \cdot 2020=30300$. From 15 numbers, we can obtain $2^{15}$ different sets. Since $2^{15}=32768>30300$, some two sums will be the same. If these two sets (subsets) have the same elements, then subtract them in each set. In this case, at least one number will remain in ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,715 |
10-1. Piglet has balloons of five colors. He managed to arrange them in a row in such a way that for any two different colors in the row, there will always be two adjacent balloons of these colors. What is the minimum number of balloons Piglet could have? | Answer: 11 balls.
Solution. Consider the balls of color $a$. Their neighbors must be balls of all 4 other colors. But one ball can have no more than two neighbors, so there must be at least 2 balls of color $a$. This is true for each of the 5 colors, so there must be at least 10 balls in total.
Notice that some ball ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,716 |
10-2. Real numbers $a, b$ and $c$ are such that $|a| \geq|b+c|,|b| \geq|c+a|$ and $|c| \geq|a+b|$. Prove that $a+b+c=0$. | Solution. Let's square both sides of each inequality:
$$
\left\{\begin{array}{l}
a^{2} \geq(b+c)^{2} \\
b^{2} \geq(c+a)^{2} \\
c^{2} \geq(a+b)^{2}
\end{array}\right.
$$
Add all three inequalities and after rearranging the terms, we get $(a+b+c)^{2} \leq 0$. From this, it is clear that $a+b+c=0$.
Criteria. Only the a... | +b+=0 | Algebra | proof | Yes | Yes | olympiads | false | 12,717 |
10-3. Solve the system $\left\{\begin{array}{c}2 y=|2 x+3|-|2 x-3| \\ 4 x=|y+2|-|y-2|\end{array}\right.$ | Solution. Let's draw the graphs of the lines described by each equation. Each of them represents a broken line consisting of 3 segments.
First equation: $y=\frac{|2 x+3|-|2 x-3|}{2}$ $x-1.5 ; y=3$, when $-1.5 \leq x \leq 1.5 ; y=2 x$.
Second equation: $x=\frac{|y+2|-|y-2|}{4}$
$$
y2 ; y=1, \text { when }-2 \leq x \l... | -1\leqx\leq1,2x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,718 |
10-4. In an acute-angled triangle $A B C$, the angle bisectors of angles $A, B$, and $C$ intersect the circumcircle of the triangle at points $A_{1}, B_{1}$, and $C_{l}$ respectively. The lines $A B$ and $B_{l} C_{l}$ intersect at point $M$, and the lines $B C$ and $A_{l} B_{1}$ intersect at point $N$. Is it true that ... | Answer: Yes, correct.
Let $I$ be the center of the inscribed circle of triangle $ABC$ (point $I$ is the intersection of the angle bisectors $AA_1, BB_1$, and $CC_1$), and let line $B_1C_1$ intersect side $AC$ and bisector $AA_1$ at points $P$ and $Q$ respectively (see figure). Then $\angle AQC_1 = \frac{1}{2} \left( \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,719 |
10-5. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest fig... | Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leq 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14$ - 43 $=23$ parts, and each will have between 7 and 12 parts. One figurine is clearly... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,720 |
1.1 A courtyard table tennis tournament among 15 players is held according to certain rules. In each round, two players are randomly selected to compete against each other. After the round, the loser receives a black card. The player who receives two black cards is eliminated from the competition. The last remaining pl... | Answer: 29
Solution. In each match, there is always exactly one loser. Since 14 players were eliminated, there were a total of $14 \cdot 2+1=29$ losses. | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,721 |
2.1 Two squares are arranged as shown in the figure. If the part of the smaller square that intersects with the larger one is cut off, 52% of its area will remain, and for the larger square, without their common part, 73% of its area will remain. Find the ratio of the side of the smaller square to the side of the large... | Answer: 0.75
Solution. Note that $27\%$ of the area of the larger square is equal to $48\%$ of the area of the smaller square, that is, the ratio of the areas is $\frac{27}{48}=\frac{9}{16}$. But the area of each square is equal to the square of the ratio of the sides, so the sides are in the ratio $3 \div 4=0.75$. | 0.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,722 |
3.1 Two ants are running along a circle towards each other at constant speeds. While one of them runs 9 laps, the other runs 6. At the point where the ants meet, a red dot appears. How many red dots will there be on the circle?
. Find the smallest natural number that hides the numbers 2021, 2120, 1220, and 1202. | # Answer: 1201201
Solution. Notice that the number contains at least two twos and one zero. If there are exactly two twos, then the zero must stand both between them and after them, but then there must be at least two zeros. Therefore, only on twos and zeros, we need 4 digits (either two twos and two zeros, or three t... | 1201201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,726 |
7.1 In the example of addition and subtraction, the student replaced the digits with letters according to the rule: identical letters are replaced by identical digits, different letters are replaced by different digits. From how many different examples could the record $0<\overline{\overline{Б A}}+\overline{\text { БА ... | # Answer: 31
Solution. The sum of two two-digit numbers is no more than 199, so $\overline{\text { YAG }}$ is a three-digit number starting with 1, Y $=1$. Let's look at the last digit in each number, A. It is added twice and subtracted once, so the value of the expression is $\mathrm{A}$, and $\mathrm{A} \neq 0$. $\m... | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,727 |
8.1 On the Island of Truth, only knights, who always tell the truth, and liars, who always lie, live. A tourist met four residents of the island A, B, C, D. Resident A told the tourist: "Exactly one of us four is a liar." B said: "All of us are liars." Then the tourist asked C: "Is it true that A is a liar?". When the ... | Answer: definitely A and B, and possibly D, but possibly not
Solution. Let's consider the first two questions. B is definitely a liar. If A is a knight, then the only option is KLLK. If A is a liar, then the remaining options are LLLK, LLLL, and LLKL (the option where all are liars is impossible due to B's answer). In... | definitelyAB,possiblyD,butpossiblynot | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 12,728 |
1. Prove that the product of $n$ positive numbers with a fixed sum is maximal when they are all equal to each other. | Solution. Suppose the opposite. There is a set of numbers $x_{1}, x_{2}, \ldots, x_{n}$ with sum $S$, such that their product is maximal and not all numbers are equal. Consider two unequal numbers $a$ and $b, a>b$ from this set and replace them with equal numbers $a_{1}=a-\frac{a-b}{2}, b_{1}=b+\frac{a-b}{2}$ with the ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,729 |
3. In three-dimensional space, a Euclidean coordinate system $O x y z$ is given. The points of the curve have coordinates $(\sin \varphi, \cos \varphi, \sin \varphi)$, where $\varphi$ runs through the interval $[0 ; 2 \pi]$. Prove that all points of the curve lie in one plane. | Solution. Any plane can be defined by an equation of the form $A x+B y+C z+D=0$, where $A, B, C, D$ are real constants, and the first three are not all zero. To solve the problem, it is sufficient to specify such values of these constants that the equation of the plane remains valid for all values of $\varphi$. Obvious... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,731 |
4. In a set consisting of $n$ elements, $2^{n-1}$ subsets are selected, any three of which intersect (have at least one element belonging to these three sets). Prove that they all have a common element. | Solution. There are $2^{n}$ subsets in total, which means exactly half of the subsets are selected. Consider some subset and its complement. Since they do not intersect, only one of them should be selected. Therefore, with any two selected subsets, their intersection must also be selected, as the complement of the inte... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,732 |
5. In a circle with center $O$, a diameter $M N$ is drawn, and points $K$ - the midpoint of arc $M N$, point $E$ - the midpoint of chord $M K$, and point $B$ - the midpoint of arc $K N$ are marked. Chord $A B$ is drawn through point $E$. On segment $A B$, as a side, a rectangle $A B C D$ is constructed such that its ve... | Solution. Draw radii $O K$ and $O B$, then triangle $M O K$ is a right and isosceles triangle. Segment $E O$ is its median, altitude, and bisector, so the angles $\angle M O E=\angle E O K=\angle K O B=\angle B O N=45^{\circ}$. Angle $\angle B O E=\angle E O K+\angle K O B=45^{\circ}+45^{\circ}=90^{\circ}$, which means... | \frac{^{2}\sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,733 |
7.1. Palm oil has increased in price by $10 \%$. Due to this, the cheese of one of the manufacturers has increased in price by $3 \%$. What is the percentage of palm oil in the cheese of this manufacturer? | Answer: $30 \%$.
Solution. Let's assume that the cheese cost 100 conditional rubles per kilogram. Then it increased by 3 rubles. Since this happened due to the increase in the price of palm oil, 3 rubles is $10 \%$ of the cost of palm oil in the cheese, meaning the cost of palm oil in a kilogram of cheese is 30 rubles... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,734 |
7.2. Three cyclists leave the city. The speed of the first one is 12 km/h, the second one is 16 km/h, and the third one is 24 km/h. It is known that the first one was cycling exactly when the second and third were resting (standing still), and that at no time did two cyclists cycle simultaneously. It is also known that... | Answer: 16 km
Solution: From the condition, it follows that each of the cyclists was riding at the time when the other two were standing, and, moreover, at any given moment, one of the cyclists was riding (the first one rode when the second and third were standing). Since they all traveled the same distance, and the r... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,735 |
7.3. The number $B$ is obtained from the number $A$ according to the following rule: all digits of the number $A$ are changed simultaneously. If a digit is greater than 2, then 2 can be subtracted from it, and if a digit is less than 8, then 2 can be added to it (for example, the digit 4 can be replaced with 2 or 6, wh... | Answer: It cannot.
First solution. Note that the numbers $A$ and $B$ can consist of only 6 or 7 digits. The sum of the numbers $A$ and $B$ is equal to the sum of the numbers $C$ and $D$, where the number $C$ has the smaller of the corresponding digits of $A$ and $B$ at each position, and the number $D$ has the larger ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,736 |
7.5. 16 travelers, each of whom is either a liar or a knight (liars always lie, knights always tell the truth), settled into 3 rooms of a hotel. When everyone gathered in their rooms, Basil, who was staying in the first room, said: "There are more liars than knights in this room right now. Although no - there are more ... | Answer: 9 knights.
Solution: Since Vasily's statements contradict each other, Vasily is a liar. Therefore, both of Vasily's statements (about each room) are false, and in each room (when he was there) the number of liars and knights was equal. This means that in each room, without Vasily, there was one more knight tha... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,738 |
10.4 In triangle $A B C$, the median $A M$ is drawn. Can the radius of the circle inscribed in triangle $A B M$ be exactly twice the radius of the circle inscribed in triangle $A C M$? | See the figure.
\[
\begin{aligned}
& AB=c, \\
& BC=2a, \\
& CA=b, \\
& AM=m, \\
& S_{ABM}=S_{ACM}=S
\end{aligned}
\]
We will use the known formula for the radius \( r \) of the inscribed ci... | 2c | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,739 |
1. Students Leshа and Pasha received the same scholarship for winning the olympiad and decided to spend it on repairing their rooms. Pasha bought 4 rolls of wallpaper and 4 cans of paint with all his money. Leshа bought 7 rolls of the same wallpaper and 2 cans of the same paint in the same store. At the same time, Lesh... | Answer: a can of paint.
Solution. Let's subtract from the purchases of Leshа and Pasha 4 rolls of wallpaper (r.ob.) and 2 cans of paint (b.kr.) each. Then it turns out that in terms of cost 2 b.kr. = 3 r.ob. + ticket, that is, 2 b.kr. cost more than 2 r.ob. Therefore, b.kr. is more expensive.
Criteria. Only answer - ... | \can\of\paint | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,740 |
4. It is known that the numbers EGGPLANT and FROG are divisible by 3. What is the remainder when the number CLAN is divided by 3? (Letters represent digits, the same letters represent the same digits, different letters represent different digits).
Answer: 0 | Solution. By the divisibility rule for 3, the sums B+A+K+L+A+Z+A+N and Z+A+B+A are divisible by 3, and therefore the difference of these sums $K+N+A+H$ is also divisible by 3, which by the rule means that the number KLAN is divisible by 3, hence the remainder is 0.
Criteria. Only the answer - 0 points | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,741 |
8.1. Usually, we write the date in the format of day, month, and year (for example, 17.12.2021). In the USA, however, it is customary to write the month number, day number, and year in sequence (for example, 12.17.2021). How many days in a year cannot be determined unequivocally by its writing? | Answer: 132.
Solution. Obviously, these are the days where the date can be the number of the month, that is, it takes values from 1 to 12. There are such days $12 \times 12=144$. But the days where the number matches the month number are unambiguous. There are 12 such days. Therefore, the number of days sought is $144... | 132 | Other | math-word-problem | Yes | Yes | olympiads | false | 12,742 |
8.2. Which of the numbers is greater: $2^{2021}$ or $5^{864}$? | Answer. $2^{2021}>5^{864}$
Solution. From the inequality $2^{7}=128>125=5^{3}$, it follows that $2^{2021}=\left(2^{7}\right)^{288} \cdot 2^{5}>\left(5^{3}\right)^{288}=$ $5^{864}$ | 2^{2021}>5^{864} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,743 |
8.3. Find the value of the expression $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}$, if $a+b+c=0$. | Answer: 3.
Solution. $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+(a+b+c)(3 a b+3 b c+3 c a)-3 a b c$. Using the condition $a+b+c=0$, we get: $a^{3}+b^{3}+c^{3}=3 a b c$. Then $\frac{a^{2}}{b c}+\frac{b^{2}}{a c}+\frac{c^{2}}{a b}=\frac{a^{3}+b^{3}+c^{3}}{a b c}=\frac{3 a b c}{a b c}=3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,744 |
8.4. The shorter leg $AC$ of the right triangle $ABC$ has a length of $b$. A point $D$ is chosen on the hypotenuse $AB$ such that $BD = BC$. A point $E$ is taken on the leg $BC$ such that $DE = BE = m$. Find the perimeter of the quadrilateral $ADEC$.
---
Translation:
8.4. The shorter leg $AC$ of the right triangle $A... | Answer: $2 m+b$.
Solution. Let $M$ be the midpoint of the hypotenuse $AB$ of the given right triangle $ABC$. Then, $AM=BM=CM$ and $\angle BDE=\angle MBC=\angle MCB$ (angles at the base of two isosceles triangles). Since $BD=BC$, $\triangle BDE=\triangle BCM$ (by side and two adjacent angles), hence $CM=DE=m$ and $CE=B... | 2m+b | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,745 |
8.5. Two players take turns placing one checker on a 51×51 grid. In doing so, there should be no more than two checkers on each row and each column. A player loses if they cannot make their move. Who wins with correct play? | Answer. The second player wins.
Solution. Note that the rearrangement of the board's rows does not change anything. In other words, if a certain position is losing for the first player, and we cut the board into rows and rearrange them, the new position will also be losing for the first player (and if it was winning f... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,746 |
1. Does there exist a pair of unequal integers $a, b$, for which the equality
$$
\frac{a}{2015}+\frac{b}{2016}=\frac{2015+2016}{2015 \cdot 2016}
$$
holds? If such a pair does not exist, justify it. If such a pair does exist, provide an example. | # Solution.
Multiply the equation by the number $2015 \cdot 2016$. We get the following equation
$$
2016 a + 2015 b = 2015 + 2016
$$
Rewrite the equation in the following form
$$
2016(a-1) = 2015(1-b)
$$
Since the numbers 2016 and 2015 are coprime, then $(a-1)$ is divisible by 2015, i.e., there exists an integer $... | 4031 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,747 |
2. In the parliament of a certain state, there are 2016 deputies, who are divided into 3 factions: "blues," "reds," and "greens." Each deputy either always tells the truth or always lies. Each deputy was asked the following three questions:
1) Are you a member of the "blues" faction?
2) Are you a member of the "reds" f... | # Solution.
Let the number of deputies telling the truth in the "blue," "red," and "green" factions be $r_{1}, r_{2},$ and $r_{3}$ respectively, and the number of deputies lying in the "blue," "red," and "green" factions be $l_{1}, l_{2},$ and $l_{3}$ respectively.
According to the problem:
$\left\{\begin{array}{l}r... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,748 |
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