problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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17. On a plane, there are 1985 points of red, blue, and green colors such that no three of them lie on the same line. Some pairs of points of different colors are connected by segments, and it turns out that the same number of segments emanate from each point. Prove that there exists a red point connected by segments t... | 85.17. Suppose this is not the case, and consider all the red points (if any). Each of them is connected either only to blue points or only to green points. In the first case, we repaint it green, and in the second case, we repaint it blue. We will get a picture without red points. Denoting by $k$ the number of segment... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,181 |
18. On a warehouse, $n$ containers numbered $1, 2, 3, \ldots, n$ are stacked in two piles in a random order. An automated forklift approaches one of the piles, removes several containers from the top, and places them on the other pile. Prove that in $2 n-1$ such operations, all containers can be arranged in one pile in... | 85.18. Suppose that in one of the stacks, the first $k$ containers are already installed in the correct order - at the bottom is No. 1, then No. 2, and so on up to No. $k$. Then we will do the following: remove all the other containers from this stack and place them on another, after which we will take from this other ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,182 |
19. An arithmetic progression consisting of natural numbers contains a perfect square. Prove that it contains infinitely many perfect squares.
114 | 85.19. Suppose that in the arithmetic progression $\{a+n d\}$ there is an exact square $x=y^{2}$. Then all numbers of the form $(y+$ $+k d)^{2}$ are terms of the same arithmetic progression. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,183 |
20. Solve the system of equations:
$$
\left\{\begin{array}{l}
(x+y)^{3}=z \\
(y+z)^{3}=x \\
(z+x)^{3}=y
\end{array}\right.
$$ | 85.20. Answer. $x=y=z=0 ; \quad x=y=z=1 / \sqrt{8} ; \quad x=y=$ $=z=-1 / \sqrt{8}$. | 0;\quad1/\sqrt{8};\quad-1/\sqrt{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,184 |
21. In the convex quadrilateral $A B C D$, angle $A B D$ is $65^{\circ}$, angle $C B D$ is $35^{\circ}$, angle $A D C$ is $130^{\circ}$, and $A B=B C$. Find the angles of the quadrilateral $A B C D$. | 85.21. Answer. $\angle D A B=57^{\circ} 30^{\prime}, \quad \angle A B C=100^{\circ}, \angle B C D=$ $=72^{\circ} 30^{\prime}$. Use the fact that points $A, C$ and $D$ lie on a circle with center $B$. | \angleDAB=5730^{\},\quad\angleABC=100,\angleBCD=7230^{\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,185 |
24. A game for two players consists of a rectangular field $1 \times 25$, divided into 25 square cells, and 25 tokens. The cells are sequentially numbered with the numbers $1,2, \ldots, 25$. In one move, a player either places a new token in one of the free cells or moves a previously placed token to the nearest free c... | 85.24. Hint. The first player wins. The main idea of his strategy is as follows: on his first move, he places a chip on field 25, and then ensures that after any of his moves, there is an even number of unoccupied cells between any adjacent groups of chips, as well as between the outermost groups and the edges of the b... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,188 |
27. The lengths of the sides of a convex quadrilateral do not exceed 7. Prove that four circles with radii of 5 and centers at the vertices of the quadrilateral completely cover the quadrilateral. | 85.27. Let $O$ be any point inside the quadrilateral $ABCD$. It is clear that one of the four angles $AOB, BOC, COD$, and $DOA$ (let this be angle $AOB$ for definiteness) is not greater than $90^{\circ}$. Then $AO^2 + BO^2 \leq AB^2 \leq 7^2 = 49$, which means either $AO^2 < 25$ or $BO^2 < 25$, i.e., point $O$ lies ins... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,189 |
29*. The sequence $x_{1}, x_{2}, \ldots$ is defined by the conditions $x_{1}=$ $=0.001, x_{n+1}=x_{n}-x_{n}^{2}$. Prove that $x_{1001}<0.0005$. | 85.29. For each $i \geqslant 1$ the inequality is satisfied
$$
\frac{1}{x_{i+1}}=\frac{1}{x_{i}-x_{i}^{2}}=\frac{1}{x_{i}\left(1-x_{i}\right)}>\frac{1+x_{i}}{x_{i}}=1+\frac{1}{x_{i}}
$$
Fi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,190 |
33. The lines drawn perpendicular to the faces of a tetrahedron through the centers of the inscribed circles in these faces intersect at one point. Prove that the sums of the lengths of opposite edges of this tetrahedron are equal to each other. | 85.33. Let $O$ be the point of intersection of the lines specified in the condition, and let points $K, L, M, N$ be the feet of the perpendiculars dropped from $O$ to the edges $AB, AC, CD$, and $BD$, respectively. Let $F$ be the center of the circle inscribed in triangle $ABC$. It is clear that $FK \perp AB, FL \perp ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,191 |
34. Prove the inequality
$$
9<\int_{0}^{3} \sqrt[4]{x^{4}+1} d x+\int_{1}^{3} \sqrt[4]{x^{4}-1} d x<9.0091
$$ | 85.34. Consider the graph of the function $f(x)=\sqrt[4]{x^{4}+1}$, which divides a $3 \times 3$ square into two parts with areas $S_{1}$ and $S_{2}$ (see Fig. 60). In this case, we have
$$
\int_{0}^{3} \sqrt[4]{x^{4}+1} d x=S_{1}+S_{3}
$$
and, considering that the function $g(x)=\sqrt[4]{x^{4}-1}$ is the inverse of ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 37,192 |
37. The number 1584 has the following properties:
a) it is not a square of an integer;
b) it is different from its reverse 4851;
c) the product of the numbers 1584 and 4851 is a square of an integer.
Find a 20-digit number with the same properties. | 85.37. For example, the number 15841584158415841584 fits. | 15841584158415841584 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,193 |
39. 15 volleyball teams played a round-robin tournament, where each team won exactly seven matches. How many triplets of teams are there in this tournament where each team has exactly one win against the other two? | 85.39. All triples of teams can be divided into two types: those that scored one point in matches against each other, and those that scored 2, 1, and 0 points in matches against each other. If the number of triples of the first type is $-a$, and the number of triples of the second type is $-b$, then $a+b=455$. On the o... | 140 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,194 |
42. The sequence of natural numbers $a_{1}, a_{2}, \ldots$ is such that $a_{1}=1$ and $a_{k+1}-a_{k}$ is 0 or 1 for all $k$. Prove that if $a_{m}=m / 1000$ for some $m$, then there exists $n$ such that $a_{n}=n / 500$. | 85.42. Consider the difference $500 a_{i}-i$. For $i=1$ it is equal to 499, and for $i=m$ it is equal to $\left(-500 a_{m}\right)$. Obviously, when transitioning from $i$ to $i+1$, this difference either increases by 499 or decreases by 1. Let $n$ be the first number for which our difference is non-positive. Then, obvi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,196 |
43. In the sequence $f_{1}, f_{2}, \ldots$ the first two terms are 1, and each of the subsequent terms is the sum of the two preceding ones. Prove that
$$
\frac{1}{f_{1} f_{3}}+\frac{1}{f_{2} f_{4}}+\frac{1}{f_{3} f_{5}}+\ldots+\frac{1}{f_{98} f_{100}}<1
$$ | 85.43. Hint. Prove by induction that
$$
\frac{1}{f_{1} f_{3}}+\frac{1}{f_{2} f_{4}}+\ldots+\frac{1}{f_{n} f_{n+2}}=1-\frac{1}{f_{n+1} f_{n+2}}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,197 |
52*. In the class, there are an equal number of boys and girls. Each boy is friends with an even number of girls. Prove that it is possible to choose a group of several boys such that each girl is friends with an even number of boys from this group. | 85.52. Let's correspond to each group of boys a group of girls consisting of all girls who are friends with an odd number of boys from the given group. From the condition, it follows that in each such group there is an even number of girls. If there are $n$ boys and $n$ girls, then there are $2^{n}-1$ non-empty groups ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,199 |
3. Natural numbers $a$ and $b$ are such that $34 a=43 b$. Prove that the number $a+b$ is composite. | 86.3. $77 a=34 a+43 a=43 b+43 a=43(a+b)$. Thus, 43( $a+$ $+b$ ) is divisible by 77. Therefore, the number $a+b$ is also divisible by 77, which means it is composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,200 |
6. a) Find a seven-digit number, all digits of which are different and which is divisible by all these digits.
b) Does such an eight-digit number exist?
## 6th grade | 86.6. a) 7639128. b) It is clear that the record should not contain zero. Since our number is obviously even, it cannot be divisible by five - otherwise it would end in zero. There are eight digits left, the sum of which is not divisible by three, and the digit 3 must be included in the number's record - a contradictio... | 7639128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,202 |
13. Eleven pioneers visit five clubs. Prove that among them there are two, $A$ and $B$, such that all the clubs that $A$ visits, $B$ also visits.
## 7th grade | 86.13. Let's number these circles with digits from one to five. There are 32 subsets of the set of all circles. We will divide them into 10 sets:
\[
\begin{aligned}
& \emptyset,\{1\},\{1,2\},\{1,2,3\},\{1,2,3,4\},\{1,2,3,4,5\} \\
& \{2\},\{2,5\},\{1,2,5\},\{1,2,3,5\} \\
& \{3\},\{1,3\},\{1,3,4\},\{1,3,4,5\} \\
& \{4\}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,204 |
16. In an isosceles triangle \(ABC\), the angle at vertex \(B\) is \(108^\circ\). The bisector of angle \(ACB\) intersects side \(AB\) at point \(D\). A perpendicular to this bisector at point \(D\) intersects the base \(AC\) at point \(E\). Prove that \(AE = BD\). | 86.16. Reflect point $B$ with respect to the bisector $C D$ and obtain point $M$ on segment $C E$. By calculating the angles, we can see that triangles $D E M$ and $A E D$ are isosceles, and therefore we have $B D=D M=D E=A E$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,205 |
19. A pile of 25 stones is arbitrarily divided into two piles, any of the existing piles is again divided into two, and so on, until each pile consists of one stone. Each time a pile is divided into two, the product of the number of stones in the resulting two piles is recorded. Prove that the sum of all recorded numbe... | 86.19. Let's consider the quantity $S$ as the half-sum of the squares of the number of stones in the piles. Initially, $S=312.5$, and at the end, $S=12.5$. If a pile containing $x+y$ stones is split into two piles with $x$ and $y$ stones respectively, then $S$ decreases by $x y$. Therefore, the amount by which $S$ decr... | 300 | Combinatorics | proof | Yes | Yes | olympiads | false | 37,206 |
22. Find some integers $A$ and $B$ such that
$$
\frac{A}{999}+\frac{B}{1001}=\frac{1}{999999}
$$ | 86.22. Let's rewrite the equality as $1001 A + 999 B = 1$. From this, it is clear that we can take $A = 500$ and $B = -501$. | A=500,B=-501 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,208 |
23. In a convex pentagon $A B C D E$, points $H, I, K, M, O$ are the midpoints of sides $A B, B C, C D, D E$ and $E A$ respectively. Prove that the length of the closed broken line $HKOIMH$ is less than the length of the broken line $A C E B D A$. | 86.23. Note. Each segment of the broken line NKOIMN does not exceed the sum of the lengths of two segments connecting the midpoints of adjacent sides of the pentagon, and therefore does not exceed half the sum of two certain diagonals of the pentagon. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,209 |
25*. 200 football teams are holding a championship. On the first day, all teams played one game, on the second day, they all played one game again, and so on. Prove that after the sixth day, it is possible to indicate 34 teams, no two of which have played against each other.
## 9th grade | 86.25. We will prove that for any $n=1,2, \ldots, 33$ it is possible to construct two non-intersecting groups of teams - the first with $n$ teams, and the second with $5 n+1$ teams - such that all teams that played against teams in the first group are in the second group (this means, in particular, that the teams in th... | 34 | Combinatorics | proof | Yes | Yes | olympiads | false | 37,211 |
31. Find some integers $A, B$ and $C$ such that
$$
\frac{A}{999}+\frac{B}{1000}+\frac{C}{1001}=\frac{1}{999 \cdot 1000 \cdot 1001}
$$
## 10th grade | 86.31. Answer. $A=500, B=-1, C=-500$. | A=500,B=-1,C=-500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,213 |
35. Find all positive solutions of the equation
$$
x^{1986} + 1986^{1985} = x^{1985} + 1986^{1988}
$$ | 86.35. Let's rewrite the equation as $x^{1985}(x-1)=$ $=1986^{1985}(1986-1)$. Since $x=1986$ is a solution to the equation, and the left side is an increasing function for $x>1$, there are no other solutions. | 1986 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,215 |
38. Let $a_{1}=2$ and $a_{n+1}=a_{1} a_{2} \ldots a_{n}+1$ for $n=1,2, \ldots$. Prove that
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}<1
$$ | 86.38. Hint. Use the equality
$$
\frac{1}{a_{n}}=\frac{1}{a_{1} a_{2} \ldots a_{n-1}}-\frac{1}{a_{1} a_{2} \ldots a_{n}}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,217 |
40. A Martian is born at midnight and lives exactly 100 days. It is known that throughout the history of the now-extinct Martian civilization, an odd number of Martians were born. Prove that there were at least 100 days when the number of inhabitants on Mars was odd. | 86.40. Let's conditionally accept the duration of a Martian year as one hundred days. In the life of each Martian, there was exactly one first day of the year. Since the total number of Martians was odd, the population of Mars was odd on the first day of some year. Similarly, on the second day of some year, the populat... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,218 |
41. On the side $AB$ of the square $ABCD$, a point $K$ is chosen, on the side $CD$ - a point $H$, and on the segment $KH$ - a point $M$. Prove that the second (distinct from $M$) point of intersection of the circumcircles of triangles $AKM$ and $MHC$ lies on the diagonal $AC$. | 86.41. Instruction. Prove that $\angle N C H=\angle K A N$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,219 |
42*. In Shvambania, one non-stop air route has been closed. It is known that after this, from any Shvambanian airport to any other one can fly, perhaps with layovers. Before the route was closed, this could be done with no more than $n$ layovers. Prove that now one can fly from any airport to any other with no more tha... | 86.42. Suppose that the airline $A B$ is closed, and assume that the shortest route from city $X$ to city $Y$ now has a length of $2 n+1$ stops. Let's number the cities in this flight from $X_{0}=X$ to $X_{2 n+1}=Y$. Previously, there was the shortest flight from $X$ to $X_{i+1}$, requiring no more than $n$ stops, and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,220 |
50. The set $A$ consists of positive numbers. It is known that the sum of any two of its elements is also its element and any interval $[a ; b](0<a<b)$ contains an interval that is entirely composed of elements of the set $A$. Prove that the set $A$ contains all positive real numbers. | 86.50. Consider any number $c>0$ and prove that $c \in A$. It is clear that there exist numbers $a, b$ such that $a<b<c$, and the interval $[a ; b] \subset A$. Similarly, the interval $[c-a ; c-b]$ contains an interval $[x ; y] \subset A$. In particular, $x \in A$. Since $c-x \in A$, then $c=(c-x)+x$ belongs to the set... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,223 |
53*. The King has traversed a $9 \times 9$ board, visiting each of its squares exactly once. The King's route is not closed and may be self-intersecting. What is the maximum possible length of such a route, if the length of a diagonal move is $\sqrt{2}$, and the length of a vertical or horizontal move is 1?
## 10th gr... | 86.53. Answer. $16+64 \sqrt{2}$. | 16+64\sqrt{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,225 |
55. The function $F: R \rightarrow R$ is continuous. For any real $x$, there exists a natural number $n$ such that $F(F(\ldots F(x)) \ldots) = 1$ (the symbol $F$ is written $n$ times). Prove that $F(1)=1$. | 86.55. It follows from the condition that the equation $F(x)=x$ can have a solution only at $x=1$. Suppose, however, that $F(1)>1$ (the other case is analogous). Then, from the continuity of $F(x)$, it follows that for all $x$ we have $F(x)>x$. But then $F(F(\ldots(1) \ldots)>1$, no matter how many times we apply the f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,226 |
58. Compute the integral
$$
\int_{-1}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}
$$ | 86.58. The main idea of the solution is to use the "non-evenness" of the function \( 1 /\left(1+x^{3}+\sqrt{1+x^{6}}\right) \), specifically:
\[
\begin{aligned}
& \text { of the function } 1 /\left(1+x^{3}+\sqrt{1+x^{6}}\right) \text {, specifically: } \\
& \int_{-1}^{1} \frac{d x}{1+x^{3}+\sqrt{1+x^{6}}}=\int_{-1}^{0... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 37,228 |
59*. Prove that on a plane, several circles that do not intersect at internal points can be arranged so that each circle touches exactly five others. | 86.59. Consider a sphere inscribed in the framework of a regular dodecahedron, and the circles cut out on it by the faces of the polyhedron. Under stereographic projection onto a plane, these circles will become circles, and the required property follows from the fact that each face of the dodecahedron is surrounded by... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,229 |
60. Given:
$$
\begin{array}{ll}
u_{1}=a x+b y+c z, & v_{1}=a x+b z+c y \\
u_{2}=a y+b z+c x, & v_{2}=a z+b y+c x \\
u_{3}=a z+b x+c y, & v_{3}=a y+b x+c z
\end{array}
$$
where $a, b, c, x, y, z$ are real numbers. It is known that $u_{1} u_{2} u_{3} = v_{1} v_{2} v_{3}$. Prove that by permuting the numbers in the trip... | 86.60. It is easy to check that $u_{1}+u_{2}+u_{3}=v_{1}+v_{2}+v_{3}$, $u_{1} u_{2}+u_{2} u_{3}+u_{3} u_{1}=v_{1} v_{2}+v_{2} v_{3}+v^{\prime} v_{1}$. Since $u_{1} u_{2} u_{3}=$ $=v_{1} v_{2} v_{3}$, the polynomials $\left(x-u_{1}\right)\left(x-u_{2}\right)\left(x-u_{3}\right)$ and $\left(x-v_{1}\right) \times$ $X\left... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,230 |
6. Two players take turns placing crosses and noughts in the cells of a $9 \times 9$ square (the first player places crosses, and the opponent places noughts). At the end, the number of rows and columns where there are more crosses than noughts is counted - these are the points scored by the first player. The number of... | 87.6. Hint. The first one needs to use central symmetry. The maximum guaranteed number of his points is 10, i.e., he can guarantee to win two points over the second one. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,234 |
10. In the countries of Dillia and Dallia, the monetary units are dillers and dallers, respectively, with 1 diller in Dillia being exchangeable for 10 dallers, and 1 daller in Dallia being exchangeable for 10 dillers. A novice financier has 1 diller and can freely travel between the two countries and exchange their mon... | 87.10. It is not hard to see that the difference between the number of dallers and the number of dillers always gives a remainder of 1 when divided by 11. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,235 |
16. The vertices of a closed non-self-intersecting broken line with eight segments are the vertices of some cube. Prove that one of the segments of this broken line coincides with an edge of the cube. | 87.16. Suppose this is not the case, and consider coloring the vertices of the cube in black and white such that any two adjacent vertices have different colors. Then, any edge of the broken line either connects vertices of the same color or is a major diagonal of the cube. It is clear that in our broken line, there mu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,237 |
17. The repair and construction company "Horns and Hooves" has taken on the task of building a 100 km road from Arbatov to Chernomorsk. The construction plan is as follows: 1 km of the road will be built in the first month, and thereafter, if $A$ km has already been completed by the start of any given month, then an ad... | 87.17. If we assume that the length of the road is always less than 100 km, then each month at least $1 / 100^{10}$ km is built. Thus, after $100^{11}$ months, the construction will be completed. | 100^{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,238 |
18. There is a tool for geometric constructions on a plane (set square), allowing the following:
a) if two points are given, then a line passing through them can be drawn;
b) if a line and a point on it are given, then a perpendicular to this line can be erected at the given point.
How can a perpendicular be dropped ... | 87.18. Suppose we have a point $A$ and a line $L$. Draw an arbitrary line through point $A$,
Fig. 62 intersecting $L$, and draw a perpendicular to it at point $A$. The two perpendicular lin... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,239 |
19. On a $10 \times 10$ board, 50 checkers are placed: 25 in the lower left quarter and 25 in the upper right quarter. In one move, any checker can jump over an adjacent checker, horizontally, vertically, or diagonally, to the next square if it is free. Can all the checkers end up on the left half of the board after se... | 87.19. Answer. No, they cannot. Hint. Follow the color of the squares on which the checkers are placed. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,240 |
20. There is an unlimited supply of coins with denominations of 1, 2, 5, 10, 20, 50 kopecks, and 1 ruble. It is known that A kopecks can be exchanged using B coins. Prove that then B rubles can be exchanged using A coins. | 87.20. Suppose that $A=x+2 y+5 z+10 t+20 u+50 v+$ $+100 w$ and $x+y+z+t+u+v+w=B$. Then, multiplying the second equation by 100, we have $100 B=100 x+50(2 y)+20(5 z)+$ $+10(10 t)+5(20 u)+2(50 v)+1(100 w)$. Combining this equality with the first equation, we obtain the required change. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,241 |
27. At points $A$ and $B$ of intersection of two circles, the tangents to these circles are mutually perpendicular. Let $M$ be an arbitrary point on one of the circles, lying inside the other circle. Extend the segments $A M$ and $B M$ beyond point $M$ to intersect the circle containing $M$ inside itself at points $X$ ... | 87.27. By the property of inscribed and central angles, we have $\angle M A B + \angle M B A = \angle A O^{\prime} B / 2$; since $O A \perp O^{\prime} A$ (here $O$ and $O^{\prime}$ are the centers of the circles), $O B \perp O^{\prime} B$, then $\angle A O^{\prime} B = 180^{\circ} - \angle A O B$, i.e., $\angle X A B +... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,244 |
30. An astronomer, observing 50 stars in the sky, found that the sum of all pairwise distances between them is $S$. A cloud that rushed by obscured 25 stars. Prove that the sum of pairwise distances between the visible stars is less than $S / 2$.
## 10th grade | 87.30. Let $A_{1}, A_{2}, \ldots, A_{25}$ be the stars obscured by the cloud, and $B_{1}, B_{2}, \ldots, B_{25}$ be the other stars. We will sum all possible triangle inequalities $B_{i} B_{j} \leqslant B_{i} A_{k}+B_{j} A_{k}$ for all $1 \leqslant i \neq j, k \leqslant 25$. The left side of the resulting inequality is... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,246 |
38. Given positive numbers $a, b, c, d$. Prove that if $cd=1$, then on the interval with endpoints $ab$ and $(a+c)(b+d)$, there is at least one square of an integer. | 87.38. Hint. Prove that $\sqrt{(a+c)(b+d)} \geqslant \sqrt{a b}+1$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,248 |
41. In the hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$, there is a point $O$ from which all sides are visible at an angle of $60^{\circ}$. Prove that if $O A_{1}>O A_{3}>O A_{5} \quad$ and $\quad O A_{2}>O A_{4}>O A_{6}, \quad$ then $A_{1} A_{2}+A_{3} A_{4}+A_{5} A_{6}<A_{2} A_{3}+A_{4} A_{5}+A_{6} A_{1}$. | 87.41. Consider an angle equal to $60^{\circ}$, and mark points $B_{i}$ on its sides at a distance $O A_{i}$ from the vertex of the angle, with even points on one side of the angle and odd points on the other (see Fig. 63). We need to prove that
$$
B_{1} B_{2}+B_{3} B_{4}+B_{5} B_{6}<B_{2} B_{3}+B_{4} B_{5}+B_{6} B_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,251 |
42. Factorize the number $989 \cdot 1001 \cdot 1007 + 320$ into prime factors. | 87.42. Let's denote 991 by $x$ and rewrite the given expression
$$
\begin{gathered}
(x-2)(x+10)(x+16)+320=x^{3}+24 x^{2}-108 x= \\
=x(x+6)(x+18)=991 \cdot 997 \cdot 1009
\end{gathered}
$$
These three numbers are prime, and the factorization is complete. | 991\cdot997\cdot1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,252 |
44*. The guard catches a thief who has broken into the palace of the Sultan of Baghdad. To catch the thief, the guard must be in the same room as the thief. The palace consists of 1000 rooms connected by doors. The layout of the palace is such that one cannot move from one room to another except through the (always sin... | 87.44. Hint. a) In the palace, there will always be a room that, when closed, will divide the palace into several parts, each of which contains no more than half of the total number of rooms in the palace. Place a guard in this room, after which the remaining guards search each of the parts. Acting in this way, it is n... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,253 |
47. Eight non-negative real numbers, the sum of which is equal to one, are placed at the vertices of a cube. For each edge, the numbers at its ends are multiplied. Prove that the sum of all such products does not exceed $1 / 4$. | 87.47. Let's color the vertices of a cube in black and white such that no two vertices of the same color are connected by an edge. Then the sum of interest does not exceed the product $x y$, where $x$ is the sum of the numbers in the black vertices, and $y=1-x$ is the sum of the numbers in the white vertices. Since $x(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,254 |
51. Given a stack of $2 n+1$ cards, the following two operations are allowed:
a) a part of the cards is taken from the top and placed at the bottom while maintaining the order;
b) the top $n$ cards are inserted into $n$ gaps between the lower $n+1$ cards, maintaining the order.
Prove that using the specified operati... | 87.51. Let's number the cards in order with numbers from 0 to \(2n\). Then the condition that for any three consecutive cards with numbers \(a\), \(b\), and \(c\), \(a - 2b + c\) is divisible by \(2n + 1\), is preserved under the described operations. From this, it follows that the numbers of the first and second cards... | 2n(2n+1) | Combinatorics | proof | Yes | Yes | olympiads | false | 37,255 |
54. Continuous functions \( f, g: [0; 1] \rightarrow [0; 1] \) have the property that \( f(g(x)) = g(f(x)) \) for any \( x \) from \([0; 1]\). Prove that if \( f \) is increasing, then there exists \( a \in [0; 1] \) such that \( f(a) = g(a) = a \). | 87.54. Consider a point $x_{0}$ such that $g\left(x_{0}\right)=x_{0}$. Let $x_{1}=f\left(x_{0}\right), x_{2}=f\left(x_{1}\right)$, and so on. Then, firstly, $g\left(x_{n}\right)=x_{n}$ for any $n$, and, secondly, the sequence $\left\{x_{n}\right\}$ is monotonic, which follows from the increasing nature of the function ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 37,256 |
60*. In an $M$-element set, $S$ subsets containing $a_{1}, a_{2}, \ldots, a_{s}$ elements respectively have been selected. It is known that none of these subsets is contained in another. Prove that
$$
\frac{1}{\binom{M}{a_{1}}}+\frac{1}{\binom{M}{a_{2}}}+\ldots+\frac{1}{\binom{M}{a_{s}}} \leqslant 1
$$
where $\binom{... | 87.60. Let's denote the subsets by $A_{1}, A_{2}, \ldots, A_{s}$. For each $k$ from 1 to $M$, consider the set $C_{k}$ of chains of subsets of the form
$$
B_{1} \subset B_{2} \subset \ldots \subset B_{M}=\{1,2, \ldots, M\}
$$
where each $B_{i}$ contains exactly $i$ elements, and the set $B_{a_{k}}$ coincides with $A_... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,258 |
4. Do there exist such non-zero integers $A$ and $B$ that one of them is divisible by their sum, and the other by their difference
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be included in the translated text. Here is the final version:
4. Do there e... | 88.4. Answer. No, they do not exist. For any two non-zero numbers, either their sum or their difference has an absolute value greater than that of both given numbers. | proof | Logic and Puzzles | other | Yes | Yes | olympiads | false | 37,261 |
5. There is a pile of 1001 stones on the table. A move consists of taking a stone from any pile that contains more than one stone, and then dividing one of the piles into two. Is it possible to leave only piles consisting of three stones on the table after several moves? | 88.5. Answer. No, it cannot. Hint. Consider the quantity $S$, equal to the sum of the number of piles and the number of remaining stones.
Fig. 64 | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,262 |
9. Each of the natural numbers $a, b, c$ and $d$ is divisible by the natural number $ab - cd$. Prove that $ab - cd = 1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 88.9. If $a b - c d = p$, then $a, b, c, d$ are divisible by $p$ and therefore $p = a b - c d$ is divisible by $p^2$. Hence, $p = 1$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,263 |
18*. On a grid sheet of size $21 \times 21$, the vertices of the cells are colored red and blue. All vertices located on the top edge of the sheet and all vertices located on the right edge of the sheet, except for the bottommost one, are painted red. All other vertices on the edges are painted blue. Prove that there i... | 88.18. Let us consider all possible types of coloring the vertices of one cell (see Fig. 64). Now let us count the number of segments with red endpoints for all cells. Each segment inside the sheet is counted twice, and on the edge of the sheet there are 41 such segments, i.e., the total sum is odd. On the other hand, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,266 |
19. $a b c=1, a+b+c=1 / a+1 / b+1 / c$. Prove that one of the numbers $a, b, c$ is equal to 1. | 88.19. Instruction. Check that $(a-1)(b-1)(c-1)=0$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,267 |
21. Find a 100-digit number without zero digits that is divisible by the sum of its digits. | 88.21. Answer. For example, the number $111 \ldots 11995125$ (the record starts with 94 ones). | 111\ldots11995125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,268 |
23. In a 120-apartment building, 119 people live. An apartment is considered overcrowded if at least 15 people live in it. Every day, the residents of one of the overcrowded apartments have an argument and move to different apartments. Is it true that the moves will eventually stop? | 88.23. Answer. Yes, this is correct. If every morning any two people living in the same apartment shake hands, then the total number of handshakes decreases when people move. If on some day $n \geqslant 15$ people move to apartments where $a_{1}, a_{2}, \ldots, a_{n}$ residents already live respectively, then the total... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,269 |
24. a) \(x, y, z \geqslant 0 ; x+y+z=1 / 2\). Prove that
$$
\frac{1-x}{1+x} \frac{1-y}{1+y} \frac{1-z}{1+z} \geqslant \frac{1}{3}
$$
b) \(x_{1}, x_{2}, \ldots, x_{n} \geqslant 0 ; x_{1}+\ldots+x_{n}=1 / 2\). Prove that \(\frac{1-x_{1}}{1+x_{1}} \frac{1-x_{2}}{1+x_{2}} \times \ldots \times \frac{1-x_{n}}{1+x_{n}} \geq... | 88.24. Hint. Use the inequality
$$
\frac{1-x}{1+x} \frac{1-y}{1+y} \geqslant \frac{1-(x+y)}{1+(x+y)}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,270 |
27. It is known that $a<b<c$. Prove that the equation
$$
\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}=0
$$
has two roots $x_{1}$ and $x_{2}$, such that $a<x_{1}<b<x_{2}<c$. | 88.27. Let's rewrite the equation as
$$
f(x)=(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0
$$
By the condition $f(a)=(a-b)(a-c)>0, f(b)=(b-a)(b-c)<0$. Therefore, there is at least one root of the equation on the intervals $[a ; b]$ and $[b ; c]$. On the other hand, a quadratic equation cannot have more than two roots, and thus ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,271 |
29. $a, b$ and $c$ are natural numbers such that $a^{3}$ is divisible by $b, b^{3}$ is divisible by $c$, and $c^{3}$ is divisible by $a$. Prove that $(a+b+$ $+c)^{13}$ is divisible by $a b c$. | 88.29. Instruction. Prove that all products of the form $a^{i} b^{i} c^{k}$, where $i, j, k$ are non-negative integers, the sum of which is equal to 13, are divisible by $a b c$.
| proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,272 |
35. Given 100 consecutive natural numbers. Can they be arranged in a circle so that the product of any two adjacent numbers is a perfect square? | 88.35. Answer. No, it cannot. Indeed, if the product of two numbers is a perfect square, then the number two enters the prime factorization of these numbers with exponents of the same parity. At the same time, among 100 consecutive natural numbers, there are obviously numbers of the form $2k+1$ and of the form $4k+2$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,274 |
36. In a regular hexagonal pyramid, the center of the circumscribed sphere lies on the surface of the inscribed sphere. Find the ratio of the radius of the circumscribed sphere to the radius of the inscribed sphere
## FINAL ROUND
## 8th grade | 88.36. Answer. $1+\sqrt{7 / 3}$. | 1+\sqrt{7/3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,275 |
39. Find two coprime four-digit natural numbers $A$ and $B$ such that for any natural numbers $m$ and $n$, the numbers $A^{m}$ and $B^{n}$ differ by at least 4000. | 88.39. For example, the numbers $A=4001$ and $B=8001$ are suitable. The difference of any two of their powers is divisible by 4000 and is not equal to zero. | A=4001,B=8001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,276 |
40. $N$ cities are connected by $2N-1$ one-way roads. At the same time, from any city, you can travel to any other city without breaking the rules. Prove that there is a road, after closing which this property will be preserved. | 88.40. We will prove this by induction on the number of cities $N$. The base case is obvious. For $N>2$, we start visiting the cities, starting from any one. It is clear that sooner or later we will return to a city we have already visited, i.e., we will get a cycle of $k>2$ cities. We will declare these $k$ cities alo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,277 |
41. In trapezoid $ABCD$ (with bases $BC$ and $AD$), points $K$ and $L$ are taken on sides $AB$ and $CD$. Prove that if angles $BAL$ and $CDK$ are equal, then angles $BLA$ and $CKD$ are also equal. | 88.41. It follows from the condition that the quadrilateral $A K L D-$ is inscribed. Therefore, $\angle A B L + \angle A K L = 180^{\circ}$. Then, obviously, $\angle B K L + \angle B C L = 180^{\circ}$, and thus, the quadrilateral $B C L K$ is also inscribed. Hence, $\angle A B L = \angle L C K$ and, therefore, $\angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,278 |
43. A word is defined as an arbitrary finite sequence of zeros and ones. The tripling of a word $A$ is its threefold repetition $A A A$. For example, if $A=101$, then its tripling is the word 101101101. Two operations are allowed on words:
1) insert the tripling of any word at any position (including at the beginning o... | 88.43. We associate with each word $A=a_{1} a_{2} \ldots a_{n}$ the number $T(A)=a_{1}+2 a_{2}+\ldots+n a_{n}$. Check independently that the remainder of $T(A)$ when divided by three does not change under the specified operations. However, since $T(01)=2$ and $T(10)=1$, one cannot be obtained from the other. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,280 |
44*. In Baron Münchhausen's forest, there are fir trees and birches, and exactly 10 birches grow at a distance of exactly 1 km from each fir tree. The Baron claims that there are more fir trees than birches in his forest. Could this be possible?
## 9th grade | 88.44. Answer. Strangely enough, the baron can be believed. Such a forest does exist. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,281 |
46. Several chips are placed on a chessboard. In one move, one of the chips is moved to a free adjacent (vertically or horizontally) square. After several moves, it turned out that each chip had visited all the squares exactly once and returned to its original square. Prove that there was a moment when none of the chip... | 88.46. Suppose chip $A$ was the first to return to its original square. Then at the previous moment, none of the chips were on their original squares. Indeed, they all had to leave their places - otherwise, $A$ could not have circled the entire board. On the other hand, none of them had returned to their place yet. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,282 |
53. A snail is crawling on a plane, turning $90^{\circ}$ after each meter of its path. What is the maximum distance from the starting point it could be after crawling 300 m, having made a total of 99 left and 200 right turns? | 88.53. Answer. $100 \sqrt{2}$ meters. Since the snail made 99 turns, we can divide its path into 100 segments, each consisting of an integer number of meters (some may have zero length), moving along which the snail turned only to the right or did not turn at all. It is easy to see that the distance between the beginni... | 100\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,285 |
54. The function $f: R \rightarrow R$ is continuous. For each real $x$, the equality $f(x) \cdot f(f(x))=1$ holds. It is known that $f(1000)=999$. Find $f(500)$. | 88.54. Answer. $f(500)=1 / 500$. Indeed, $f(999)=$ $=f(f(1000))=1 / f(1000)=1 / 999$. Therefore, the function takes the values 999 and $1 / 999$, which means there exists a number $a$ such that $f(a)=500$. But then $f(500)=f(f(a))=1 / f(a)=1 / 500$. | \frac{1}{500} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,286 |
60*. On a plane, a convex $n$-gon is given. Let $a_{k}$ be the length of its $k$-th side, and $d_{k}$ be the length of its projection onto the line containing this side $(k=1,2,3, \ldots, n)$. Prove that
$$
2<\frac{a_{1}}{d_{1}}+\frac{a_{2}}{d_{2}}+\ldots+\frac{a_{n}}{d_{n}} \leqslant 4
$$
Comment. As additional prob... | 88.60. Let \( A_{1} A_{2} \ldots A_{n} \) be our polygon \( M \); the \( i \)-th side is the segment \( A_{i} A_{i+1} \left( A_{n+1} \equiv A_{1} \right) \). Consider the vectors \( \vec{v}_{i} = \)
. The road can only be assembled so that the direction of all arrows coincides with the direction of the locomotive. A correct closed path can be formed using all the rails from the set. Prove that ... | 89.3. Instruction. Prove that the difference between the number of segments of the first type and the number of segments of the second type must be divisible by four (and in fact, it should be equal to four). | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,291 |
6. On a $10 \times 10$ board, two players take turns placing crosses and noughts (each can place either a cross or a nought). The player who, after their move, has three crosses (or noughts) in a row—horizontally, vertically, or diagonally, with no gaps—wins. Can either player always ensure a win, and if so, who: the p... | 89.6. Answer. The second player wins. To do this, he needs to repeat the first player's moves symmetrically relative to the center, swapping crosses for noughts and noughts for crosses, until he gets a chance to win. The fact that such an opportunity will inevitably arise is established by a small enumeration within th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,293 |
11. There are 99 copies of a regular 101-gon, the vertices of each of which are numbered in order from 1 to 101 (see Fig. 20). Can these 99 polygons be stacked (they can also be flipped) so that the sums of the numbers along the edges of the stack are the same? | 89.11. Answer. No, it cannot. Consider a face of the stack on which none of the 101-gons lies as a side (1 - 101) - there are at least two such faces. Then the sums of the numbers on the two vertical edges of this face have different parity and cannot be equal. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,294 |
12. Find the smallest natural number greater than 1 that is at least 600 times greater than each of its prime divisors. | 89.12. Answer. This number is $1944=2^{3} \cdot 3^{5}$. | 1944 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,295 |
13. Given several (at least two) non-zero numbers. It is allowed to erase any two numbers $A$ and $B$ and write down the numbers $A+B / 2$ and $B-A / 2$ instead. Prove that after several such operations, it is impossible to obtain the original set of numbers again.
## 7th grade | 89.13. Since the sum of the squares of the numbers in the set decreases with the specified operations, the original set cannot repeat. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,296 |
16. Prove that if $x+y+z \geqslant x y z$, then $x^{2}+y^{2}+z^{2} \geqslant$ $\geqslant x y z$. | 89.16. If the numbers $x, y, z$ are not less than 1 in absolute value, then $x^{2}+y^{2}+z^{2} \geqslant x+y+z$ and the inequality is obvious. If, however, one of them (for example, $x$) has an absolute value less than 1, then $x^{2}+y^{2}+z^{2} \geqslant y^{2}+z^{2} \geqslant|y z| \geqslant|x y z|$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,298 |
18. A librarian approaches a shelf with an eight-volume collection every minute and swaps two adjacent volumes. Can he do this in such a way that after some time, all possible arrangements of the volumes have been realized, each exactly once? | 89.18. Answer. Yes, the librarian can act this way. Instruction. Prove this by induction on the number of volumes. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,299 |
19. Two players play the following game: a chip is placed on the central node of a $10 \times 10$ grid. On each turn, each player has the right to move it to any other node of the grid, but the length of their move (i.e., the distance by which they moved the chip) must be greater than the length of the previous move ma... | 89.19. The second player wins. He must move the chip to the node symmetric to the one it is on, relative to the center of the square. | The\\player\wins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,300 |
20. Does there exist a set of one hundred different natural numbers such that the product of any five of them is divisible by the sum of these same five numbers?
## 8th grade | 89.20. Answer. Yes, it exists. To construct it, it is sufficient to consider any hundred natural numbers, after which multiply them all by a number equal to the product of the sums of all possible quintets of numbers from the original set. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,301 |
22. $A$ is a natural number greater than 1, and $B$ is a natural divisor of the number $A^{2}+1$. Prove that if $B-A>0$, then $B-A>\sqrt{A}$. | 89.22. Let $B=A+k$. Then $A^{2}+1=(A+k)(A-k)+k^{2}+1$ and, consequently, $k^{2}+1$ is divisible by $A+k=B$. But if $k \leqslant \sqrt{A}$, then $k^{2} \leqslant A$ and $k^{2}+1 \leqslant A+1<B\left(B \neq A+1\right.$, since $A^{2}+1$ is not divisible by $A+1$ ) - a contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,302 |
24. In the cells of an infinite chessboard, $m$ chips are placed. For each chip, the product of the number of chips in its row and the number of chips in its column is calculated. Prove that the number of chips for which this number is not less than $10 m$ does not exceed $m / 10$. | 89.24. Suppose all chips are located in $k$ columns: $a_{1}, a_{2}, \ldots, a_{k}$ chips respectively. We compute the product for each chip in the first column - their sum does not exceed $a_{1} m$ - and similarly for the other columns. Therefore, the sum of all products does not exceed $a_{1} m + a_{2} m + \ldots + a_... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,303 |
25. At the end of a chess tournament that lasted for $k$ rounds, it turned out that the number of points scored by the participants formed a geometric progression with a natural denominator greater than 1. How many participants could there have been:
a) when $k=1989$
b) when $k=1988$? | 89.25. Answer. If \( k=1989 \), there were two participants in the tournament, and if \( k=1988 \), there could have been two or three chess players.
Suppose there were \( n \) participants: the first scored \( A \) points, the second \( A p \) points, ..., and the winner \( A p^{n-1} \) points. The winner cannot scor... | n=2whenk=1989,n=2orn=3whenk=1988 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,304 |
26. On a plane, $N$ lines are drawn, no two of which are parallel and no three intersect at the same point. For which $N$ is it always possible to place one of the numbers $1,2, \ldots, N-1$ at each intersection point of the lines so that on any line all these numbers appear exactly once?
## 9th grade | 89.26. Answer. This is only possible for even $N$. Consider all points marked with a one. Two lines pass through each of them, and therefore $N$ must be divisible by two.
Now let $N$ be even. Number the lines from 1 to $N$ and for any $i$ and $j$ less than $N$, at the intersection of the $i$-th and $j$-th lines, write... | N | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,305 |
37. Determine whether the operation *, which associates with any two natural numbers $X$ and $Y$ a natural number $X * Y$, can simultaneously possess the following three properties:
a) $A * B=|A-B| *(A+B)$ when $A \neq B$;
b) $(A C) *(B C)=(A * B)(C * C)$;
c) $(2 k+1) *(2 k+1)=2 k+1$.
## 10th grade | 89.37. Consider the operation that associates with any two natural numbers their greatest common odd divisor. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,308 |
44. Given an operation *, which associates with any two integers X and Y an integer X * Y. It is known that every integer is equal to X * Y for some integers X and Y. Prove that such an operation cannot simultaneously have the following two properties:
a) $A * B=-(B * A)$
b) $(A * B) * C=A *(B * C)$. | 89.44. Using alternately properties $a$ and b, we get $(A * B) * C=-(C *(A * B))=-((C * A) * B)=B *(C * A)=$ $=-(A *(B * C))=-((A * B) * C)$, from which $(A * B) * C=0$. But by the condition, $A * B$ can take any integer value $X$, as well as $X \% C$, which leads to a contradiction. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,311 |
47. Prove that if the equation $A x^{2}+(C-B) x+(E-$ $-D)=0$ has a real root greater than 1, then the equation $A x^{4}+B x^{3}+C x^{2}+D x+E=0$ has at least one real root.
## FINAL ROUND
## 8th grade | 89.47. Let $t>1$ be the square root of a number that is the solution to the mentioned quadratic equation. Then $A t^{4} + (C-B) t^{2} + (E-D) = 0$, i.e., $A t^{4} + C t^{2} + E = B t^{2} + D$. Denote the left and right sides of this equality by $F$ and $G (F=G)$, and let $f(x) = A x^{4} + B x^{3} + C x^{2} + D x + E$. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,312 |
50. $k$ - a natural number greater than 1. Prove that in the cells of a $k \times k$ table, it is impossible to arrange the numbers $1,2,3, \ldots$ ..., $k^{2}$ such that all sums of the numbers in the rows and columns are powers of two. | 89.50. Suppose such an arrangement exists. Let $2^{a}$ be the smallest sum of numbers in a row. Then $2^{a} \geqslant 1+2+$ $+\ldots+k=k(k+1) / 2$. On the other hand, the sum of all numbers in the table should be divisible by $2^{a}$, and we get that $k^{2}\left(k^{2}+1\right) / 2$ is divisible by $2^{a}$. If $k$ is od... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,314 |
51. On the margins of a $10 \times 10$ board, there are 91 white checkers. The painter takes one of them, repaints it black, and places it on any free cell of the board. Then he takes another white checker, repaints it black, and so on until all the checkers are black. Prove that at some point, there will be checkers o... | 89.51. Suppose this is not the case. Then at any moment in time, there is a row and a column entirely filled with checkers, and the entire cross must consist of checkers of the same color. It remains to note that initially the cross was white, and in one recoloring, a cross of one color cannot disappear and a cross of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,315 |
54*. In the language of the Tru-la-la tribe, a word is any sequence of 10 digits 0 and 1. It is known that two words are synonyms if and only if one can be obtained from the other by a series of operations of the following kind: from the word, several consecutive digits, the sum of which is even, are erased, after whic... | 89.54. Answer. 56 words. Instruction. Consider all words of the Tru-lya-ya tribe language of the form $11 \ldots 100 \ldots 0100 \ldots 0$. Prove that any word can be brought to such a form. To prove that these words are distinct, use the following quantity defined for any word in the Tru-lya-ya language: the number of... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,318 |
58. $x, y, z$ - real numbers from the interval $[0 ; 1]$. Prove that
$$
2\left(x^{3}+y^{3}+z^{3}\right)-\left(x^{2} y+y^{2} z+z^{2} x\right) \leqslant 3
$$ | 89.58. Hint. Use the obvious inequality
$$
\begin{aligned}
& \left(1-x^{2}\right)(1-y)+\left(1-y^{2}\right)(1- \\
& \quad-z)+\left(1-z^{2}\right)(1-x) \geqslant 0
\end{aligned}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,320 |
61. The sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that for any natural number $k$ the equality
$$
a_{k+1}=\frac{k a_{k}+1}{k-a_{k}}
$$
holds. Prove that this sequence contains infinitely many positive and negative numbers. | 89.61. Consider the sequence $\left(b_{k}\right)$ such that $b_{1}=\operatorname{arctg} a_{1}, b_{k+1}=b_{k}+$ $+\operatorname{arctg}(1 / k)$. Then, using the formula for the tangent of
Fi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,322 |
62*. In triangle $ABC$, point $M$ lies on side $AB$, point $N$ lies on side $BC$, and $O$ is the intersection point of segments $CM$ and $AN$. It is known that $AM + AN = CM + CN$. Prove that $AO + AB = CO + CB$. | 89.62. Note. Both the equality $A M + A N = C M + C N$ and the equality $A O + A B = C O + C B$ are equivalent to the fact that a circle can be inscribed in the "quadrilateral" $X A O C Y$ (see Fig. 65), i.e., the quadrilateral $O M B N$ (or $A B C O$) is "exscribed". | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,323 |
71. The sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that for any $m$ and $n$, the inequality
$$
\left|a_{m}+a_{n}-a_{m+n}\right| \leqslant \frac{1}{m+n}
$$
holds. Prove that this sequence is an arithmetic progression. | 89.71. Let's fix a natural number $k$ and consider an arbitrary natural number $n$. Then
$$
\begin{aligned}
& \left|a_{n+1}+a_{k}-a_{n+k+1}\right|<\frac{1}{n+k+1} \\
\text { and } \quad & \left|a_{n}+a_{k+1}-a_{n+k+1}\right|<\frac{1}{n+k+1} .
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
& \qquad\left|a_{n+1}+a_{k}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,327 |
72*. Two players play the following game. There is a board on which the number 1000 is written, and a pile of 1000 matches. On each turn, each player (they take turns) can either take from the pile,
or add no more than five matches to it (initially, neither player has any matches), and then the number of matches in the... | 89.72. Answer. The second player wins. He mentally divides all numbers from 0 to 999 into 125 sequences of eight consecutive numbers, and then acts as follows: after the first player writes the first (not necessarily the largest) number in any of these sequences on the board, the second player makes a countermove by wr... | The\\player\wins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,328 |
1. Petya bought a general notebook with a volume of 96 pages and numbered all its pages in order with numbers from 1 to 192. Vasya tore out some 25 sheets from this notebook and added up all 50 numbers written on them. Prove that he could not have gotten a sum of 1990. | 90.1. Use the fact that the sum of the page numbers on one sheet is always odd. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,329 |
4. The Wolf and the Hare are playing the following game: a number is written on the board, and a move consists of subtracting any non-zero digit of the number and writing the resulting number in place of the old one. They take turns. The one who gets zero wins. Initially, the number 1234 is written on the board. The Wo... | 90.4. The Wolf wins. For this, it is enough for him to subtract the last digit of the number on each of his moves, making it divisible by 10. | TheWolfwins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,331 |
9. On the computer screen is the number 123. The computer adds 102 to the number on the screen every minute. Programmer Fedia can, at any moment, change the number on the screen by rearranging its digits arbitrarily. Can Fedia act in such a way that a three-digit number always remains on the screen? | 90.9. Yes, it can. Indeed, denoting the computer's actions by the arrow $\rightarrow$, and Fedia's actions by the arrow $\Rightarrow$, we can write
$$
122 \rightarrow 225 \rightarrow 327 \rightarrow 429 \rightarrow 531 \Rightarrow 135 \rightarrow 247 \Rightarrow 327 \rightarrow \ldots \ldots
$$
and then the repetitio... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,335 |
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