problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Perpin A.
Does there exist a quadratic trinomial $P(x)$ with integer coefficients such that for any natural number $n$, in the decimal representation of which only ones participate, the number $P(n)$ is also written with only ones? | Consider the trinomial $P(x)=x(9 x+2)$. If $n=\underbrace{1 \ldots 1}_{k}$, then $9 n+2=\underbrace{10 \ldots 01}_{k-1}$. Therefore,
## Answer
There exists. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,920 |
[ Symmetry properties and center of symmetry ]
Given a rectangular billiard table. There are pockets at its corners, and if the ball falls into one of them, it stops. The ball is released from one corner of the billiard table at an angle of $45^{\circ}$ to the side. At some point, it hits the midpoint of one of the si... | Suppose the ball was released at time $T=0$, at time $T=T_{1}$ the ball hit the midpoint of one side, and at time $T=T_{2}>T_{1}$ - the midpoint of the opposite side. The trajectories of the ball at $T$ close to $T_{1}$ and $T_{2}$ are symmetric to each other either with respect to the center of the rectangle or with r... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,922 |
[Inscribed and circumscribed circles $]$ [ Induction in geometry $\quad]$
$n$ points are located at the vertices of a convex $n$-gon. Inside this $n$-gon, $k$ points are marked. It turns out that any three of the $n+k$ points do not lie on the same line and are vertices of an isosceles triangle. What can the number $k... | Answer: $k=1$ or 0. Lemma. Let point $O$ lie inside triangle $\triangle A B C$, and points $A, B, C, O$ satisfy the condition of the problem. Then point $O$ is the center of the circumscribed circle of triangle $\triangle A B C$. Proof of the lemma. Consider triangle $\triangle A B C$, by the condition of the problem i... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,923 |
10,11 Law of Cosines An astronomical spotlight illuminates an octant (a trihedral angle with all plane angles being right angles). The spotlight is placed at the center of a cube. Can it be turned in such a way that it does not illuminate any vertex of the cube? | Answer: Yes, it can.
First, let's prove that the spotlight can be turned so that it illuminates the adjacent vertices $A$ and $B$ of the cube. If $\angle A O B > 90^\circ$, we can turn the spotlight so that it illuminates two vertices of the cube. The planes of the faces of the illuminated angle of the spotlight divid... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,924 |
$[\quad$ Inscribed and Circumscribed Circles} The bisector of angle $A$ of triangle $ABC$ is extended to intersect with the circumscribed circle around it at $D$. Prove that $AD > 1/2(AB + AC)$. | Let's introduce the notations: $A B=a, A C=b, A D=l, \angle C=\alpha, \angle B=\beta, d$ - the diameter of the circumcircle of triangle $A B C$. According to the Law of Sines, the length of a chord is equal to the product of the diameter of the circle and the sine of half the arc on which this chord subtends. Since $\s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,925 |
Tamarkin d:
Let's call the averaging of a sequence $a_{k}$ of real numbers the sequence $a_{k}^{\prime}$ with the general term
$a_{k}^{\prime}=\frac{a_{k}+\mathbf{a}_{k+1}}{2}$. Consider the sequences: $a_{k}, a_{k}^{\prime}-$ its averaging, $a^{\prime \prime}{ }_{k}-$ the averaging of the sequence $a_{k}^{\prime}$, ... | Let's call a sequence $m$-good if it and its first $m$ averages consist of integers. We will prove, using mathematical induction, that if the sequence $x_{k}$ is good, then the sequence $x_{k}^{2}-m$ is good for any non-negative integer $m$. This implies the statement of the problem. It is obvious that if the sequence ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,926 |
Perrin A.
The plane is divided by two families of parallel lines into unit squares. We will call the border of an \( n \times n \) square, consisting of the squares of the division, the union of those squares that share at least one of their sides internally with its boundary. Prove that there is exactly one way to co... | For each border, we will draw two vertical and two horizontal strips of width 1, in which its cells lie. Then the drawn strips cover our square, so either all 100 horizontal or all 100 vertical strips crossing the square are drawn (let these be vertical). Thus, all vertical strips for different borders do not coincide ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,927 |
Prove that if $\alpha, \beta$ and $\gamma$ are the angles of an acute triangle, then $\sin \alpha+\sin \beta+\sin \gamma>2$.
# | The first method ("trigonometric"). First, let's prove an auxiliary statement: If $\alpha, \beta$, and $\gamma$ are the angles of an arbitrary triangle, then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma=1$.
Indeed, since $\gamma=180^{\circ}-(\alpha+\beta)$, we have $\cos ^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,929 |
: Aekhananov $H . X$.
Given a set of $n>2$ vectors. We will call a vector in the set long if its length is not less than the length of the sum of the other vectors in the set. Prove that if every vector in the set is long, then the sum of all vectors in the set is zero. | First solution. Let the sum $\vec{\sigma}$ of all vectors be different from zero ($|\vec{\sigma}|=s>0$). Introduce a rectangular coordinate system $Oxy$, in which the axis $Ox$ is directed along $\vec{\sigma}$. Let $\vec{a}$ be the longest vector in the set, i.e., it is not shorter than $\vec{b}=\vec{\sigma}-\vec{a}$. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,930 |
What is the minimum number of weights needed to be able to weigh any number of grams from 1 to 100 on a balance scale, if the weights can be placed on either pan of the scale?
# | When solving this problem, we need the following interesting property of the ternary numeral system: any natural number can be represented as the difference of two numbers, the ternary representation of which contains only 0 and 1.
To prove this, we need to write the original number in ternary notation and construct t... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,931 |
Find the geometric locus of points, the distances from each of which to two given points are in the ratio $m$ : n.
# | Use the properties of the bisectors of the internal and external angles of a triangle.
## Solution
## First method.
Consider the case when $m \neq n$. Let $A$ and $B$ be the given points. Let point $P$ be such that $\frac{PA}{PB} = \frac{m}{n}$. If point $P$ does not lie on the line $AB$, then draw the bisector $PC$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,932 |
$\left.\begin{array}{l}\text { [Symmetry properties and center of symmetry } \\ {\left[\begin{array}{l}\text { Composition of central symmetries }\end{array}\right]} \\ {[\quad \text { Coordinate method on the plane }}\end{array}\right]$
a) Prove that a bounded figure cannot have more than one center of symmetry.
b) ... | a) Suppose a bounded figure has two centers of symmetry $O_{1}$ and $O_{2}$. Introduce a coordinate system with the x-axis directed along the ray $O_{1} O_{2}$. Since $S_{\mathrm{O}_{2}} \circ S_{\mathrm{O}_{1}} = T_{2} \sigma_{1} \mathrm{O}_{2}^{\prime}$, the figure maps onto itself under a translation by the vector $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,933 |
[Convex hull and supporting lines (planes).] [ Coordinate method in space (other). ]
In eight given points in space, a spotlight is installed at each point, each of which can illuminate an octant (a trihedral angle with mutually perpendicular sides) in space. Prove that the spotlights can be rotated so that they illum... | Let's introduce a coordinate system $Oxyz$ and select 4 points with the largest $z$ coordinate from the given points. We will consider the projections of these points onto the $Oxy$ plane and illuminate this plane with right angles at the obtained points (see the solution to problem 5 for 7th grade). We will direct the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,935 |
Given the sequence of numbers $1,1,2,3,5,8,13,21,34, \ldots$, each of which, starting from the third, is equal to the sum of the two preceding ones. Prove that every natural number $n>2$ is equal to the sum of several distinct numbers from the given sequence. | 1st method. We will prove this by mathematical induction. The number 3 can be represented as $1+2$. Suppose that the number $N$ can be represented as the sum of terms of this sequence. We need to prove that this can also be done for $N+1$. If the sum representing the number $N$ does not contain the term 1, then by addi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,936 |
9,1 [Inequalities with Circumscribed, Inscribed, and Externally Tangent Circles]
A convex polygon $M$ maps onto itself under a rotation by $90^{\circ}$. Prove that there exist two circles with a ratio of radii equal to $\sqrt{2}$, one of which contains $M$, and the other is contained in $M$. | Let $O$ be the center of rotation, $R$ be the greatest distance from point $O$ to the vertices of the polygon, and $A_{1}$ be one of the vertices such that $O A_{1}=R$. If $A_{1}$ is transformed by the rotation to vertex $A_{2}$, $A_{2}$ to $A_{3}$, and $A_{3}$ to $A_{4}$, then obviously $A_{1} A_{2} A_{3} A_{4}$ is a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,937 |
10,11
$[$ Regular Polyhedra (miscellaneous). ]
Author: Zaslavskiy A.A.
Each edge of a convex polyhedron was parallel translated by some vector so that the edges formed
the framework of a new convex polyhedron. Is it necessarily equal to the original one?
# | First solution. No. Consider, for example, a regular icosahedron. Five of its faces, which share a common vertex, are the lateral faces of a regular pentagonal pyramid. The centers of these faces form a regular pentagon, the sides of which are parallel to the sides of the pyramid's base. Therefore, the edges of the ico... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,939 |
Prove that for the numbers $\left\{x_{n}\right\}$ from problem $\underline{61297}$, the continued fraction expansions can be explicitly given as: $x_{n+1}=$
$[1 ; \underbrace{2, \ldots, 2}_{2^{n}-1}]$.
Estimate the difference $\left|x_{n}-\sqrt{2}\right|$. | The fractions $P_{k} / Q_{k}$ for the continued fraction expansion $\sqrt{2}=[1 ;(2)]$ satisfy the recurrence relations $P_{-1}=1, P_{0}=1, P_{k}=2 P_{k-1}+P_{k-2}$;
$Q_{-1}=0, Q_{0}=1, Q_{k}=2 Q_{k-1}+Q_{k-2}$ (see problems $\underline{60613}, \underline{60601}$ ). From these, it is easy to derive by induction the re... | |x_{n}-\sqrt{2}|<\frac{1}{2^{2^{n}-1}} | Number Theory | proof | Yes | Yes | olympiads | false | 52,944 |
$\left[\begin{array}{c}\text { Weighings }\end{array}\right]$
$\left[\begin{array}{l}\text { Evenness and Oddness }\end{array}\right]$
$\left[\begin{array}{l}\text { Extreme Principle (other) }\end{array}\right]$
[ Systems of Linear Equations]
There are 13 weights. It is known that any 12 of them can be distributed in... | a) A set of weights satisfies a certain system of 12 linear equations. Note that all solutions to this system (sets of 13 integers) have the following properties.
1) All numbers in the set are of the same parity. Indeed, by subtracting each of the 13 numbers from the sum of all 13 numbers, we always get an even number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,945 |
$\Phi$ rank $M$.
A number 6 was written into a computer memory cell. Then the computer makes a million steps. On step number $n$, it increases the number in the cell by the greatest common divisor of this number and $n$. Prove that on each step the computer increases the number in the cell by either 1 or a prime numbe... | Let $a_{n}$ denote the number after the $n$-th step. Suppose at some step $a_{n}=3 n$ (for example, $a_{3}=9$). Let $p$, greater than 1, be the number that is first added at the $m$-th step.
Then before this, only ones were added, so $a_{m-1}=m+2 n-1$.
Therefore, $p=$ GCD $(m, m+2 n-1)=$ GCD $(m, 2 n-1)$. It is clear... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,946 |
5 -
[ Dirichlet's Principle (etc.)]
In a state with $n$ cities, and between each pair of them, an express train runs (in both directions). For each express train, the ticket prices "there" and "back" are equal, and for different express trains, these prices are different. Prove that a traveler can choose an initial ci... | Let's remove all express trains, and then start launching them one by one in ascending order of price (the first one will be the cheapest, the second one will be the cheapest of the remaining, and so on). At each moment, in each city, we will write the maximum number of express trains on which one can travel sequential... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,949 |
Gaozirin A.
In coordinate space, all planes with equations $x \pm y \pm z=n$ (for all integers $n$) were drawn. They divided the space into tetrahedra and octahedra. Let the point ( $x_{0}, y_{0}, z_{0}$ ) with rational coordinates not lie on any of the drawn planes. Prove that there exists a natural number $k$ such t... | Lemma. Let the rational numbers $a, b, c$ and $a+b+c$ be non-integers. Then there exists a natural number $k$ such that the numbers $k a, k b$, and $k c$ are non-integers, and
$12$. Thus, there exists the smallest natural number $k$ for which $f(k)>1$ (then $f(k-1) \leq 1$). We will show that this $k$ satisfies all th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,952 |
Shapovaoov A.B.
An ant is crawling along a closed path along the edges of a dodecahedron, never turning back. The path passes over each edge exactly twice.
Prove that for some edge, the ant passes it in the same direction both times. | Suppose this is not the case. Then at each intersection, the ant turns either only left or only right three times. Let's arbitrarily place left and right intersections at the vertices of a dodecahedron and consider the corresponding routes of the ant. We will obtain a set of cycles. When the "sign" of one of the inters... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,953 |
Ionin Yu.I.
a) Does there exist an infinite sequence of natural numbers with the following property: no number in the sequence divides another, but among any three numbers, one can choose two whose sum is divisible by the third?
b) If not, how many numbers can be in a set with such a property?
c) Solve the same prob... | b) To the sequence of four numbers $3,5,7,107$ given in the condition, we can add the fifth number 10693: 10693 + 5 is divisible by $3,10693+3$ is divisible by $7,10693+7$ is divisible by $5,10693+107$ is divisible by 3 and by $5,10693+7$ is divisible by 107.
We will show that it is impossible to form a sequence satis... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,955 |
White and Black play the following game. In the corners of the chessboard stand two kings: White on a1, Black on h8. The players take turns making moves. White starts. A player can place their king on any adjacent square (provided it is free), adhering to the following rules: the distance between the kings cannot be in... | Let's describe the winning strategy for white. First, the white king should move to the right along the 1st rank until the black side reduces the distance between the kings. If the black side does not do this, the distance between the kings remains 7, and the white king will reach the right edge of the board without in... | White | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,956 |
Spivak A.V.
A cube of size $10 \times 10 \times 10$ is composed of 500 black and 500 white smaller cubes arranged in a checkerboard pattern (cubes that touch each other by a face have different colors). From this cube, 100 smaller cubes were removed such that in each of the 300 rows of size $1 \times 1 \times 10$, par... | Let's cut the original cube $A B C D E F G H$ into $2 \times 2 \times 2$ cubes. We will divide all black cubes into 4 groups $M_{\mathrm{A}}, M_{\mathrm{C}}, M_{\mathrm{F}}$ and $M_{\mathrm{H}}$ as follows: we will include in the group $M_{\mathrm{A}}$ those black cubes that are located in their $2 \times 2 \times 2$ c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,958 |
$\underline{\text { Folklore }}$
On an infinite chessboard in all directions, a certain set of cells $A$ is marked. On all cells of the board, except for the set $A$, kings are placed. All kings can, upon command, simultaneously make a move, which consists of either staying in place or moving to an adjacent cell, that... | a) Suppose the opposite: let there exist a way to fill the entire board with kings in $k<10^{n}$ moves. Consider a square $K$ of size $10^{n+5} \times 10^{n+5}$. Kings could only come from a square with side $10^{n+5} + 2k < 10^{n+5} + 2 \cdot 10^{n}$. But there are fewer kings there than
$\left(1-10^{-4}\right)\left(... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,959 |
Fomin S.V.
In the kingdom, there are 16 cities. The king wants to build a road system such that from any city, one can reach any other city by passing through no more than one intermediate city, and no more than five roads should lead out of any city.
a) Prove that this is possible.
b) Prove that if the number 5 in ... | a) As an example, it is sufficient to consider a four-dimensional cube: place the cities at its vertices, and draw roads along its edges and main diagonals. In other words, number the cities with four-digit binary numbers from 0000 to 1111 and draw roads between cities whose numbers differ in exactly one digit, and bet... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,960 |
[ $\quad$ Minimum or maximum area (volume). $]$ Convex polygons $]$ [ Auxiliary area. Area helps to solve the task $]$
Authors: Pushkar P.E. Donnikov V.L.
Contours of convex polygons $F$ and $G$ have no common points, and $G$ is located inside $F$. A chord of polygon $F$ - a segment connecting two points of the conto... | a) Consider any supporting chord and rotate it counterclockwise so that it remains a supporting chord at all times. In each position, the supporting chord divides the polygon \( F \) into two parts, one of which does not contain the polygon \( G \). The area of this part is clearly a continuous function of the angle of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,961 |
KanellyBelov A.Y.
Prove that the leading digits of numbers of the form $2^{2^{\mathrm{n}}}$ form a non-periodic sequence. | Consider a circle of length 1 as the segment [0;1] with identified endpoints. Then the fractional part $\mathrm{f}$ of the number $\mathrm{k} * \mathrm{lg} 2$ can be considered as a point on this circle. The first digit of the number $2^{\mathrm{k}}$ is determined by the position of $\mathrm{f}$ relative to the points ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,962 |
Vasiniev N. 5.
Prove that for any $k>1$ there exists such a power of two that among its last $k$ digits, at least half are nines.
(For example, $\left.2^{12}=\ldots .96,2^{53}=\ldots 992.\right)$ | Lemma. For all $k \geq 1$, the number $2^{2 \cdot 5^{k-1}}+1$ is divisible by $5^{k}$.
Proof. Induction on $k$. The base case ($k=1$) is obvious.
Inductive step. $2^{2 \cdot 5^{k-1}}+1=4^{5^{k-1}}+1$. Let $a=4^{5^{k-1}}$. Then $4^{5^{k}}+1=a^{5}+1=(a+1)\left(a^{4}-a^{3}+a^{2}-a+1\right)$
is divisible by $5^{k+1}$, si... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,963 |
Konyagin S.V.
In a family album, there are ten photographs. In each of them, three people are depicted: a man stands in the center, to the left of the man is his son, and to the right is his brother. What is the smallest number of different people that can be depicted in these photographs, given that all ten men stand... | Let's call the ten men standing in the center of the photographs the main faces. We will divide all the men in the photographs into levels. We will assign level 0 to those who do not have fathers in the photographs, and level $k+1 (k=0,1, 2, \ldots)$ to men who have fathers at level $k$. Let $r_{k}$ denote the number o... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,965 |
[Alperin.A. Do there exist 1998 distinct natural numbers, the product of any two of which is divisible by the square of their difference?
# | Let the set $N=\left\{a_{1}, \ldots, a_{n}\right\}$ consist of numbers satisfying the given condition $U$. Then the set $N_{1}=\left\{b_{1}, \ldots, b_{n}, b_{n+1}\right\}$, where $b_{1}=a_{1}, \ldots, b_{n}=a_{n}$, $b_{n+1}=0$ also satisfies $U$. By adding to each $b_{i}$ the number $c=\left(b_{2}-b_{1}\right)^{2}\lef... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,966 |
Bgerov S.L.
In some cells of a $2 n \times 2 n$ board, there are black and white chips. First, all black chips that are in the same column as some white chip are removed, and then all white chips that are in the same row as any of the remaining black chips are removed. Prove that either the black or the white chips le... | Note that at the end, no two different colored chips stand in any row or column. Indeed, if initially a black chip stood in the same column as a white one, it was removed the first time, and if after the first removal a white chip stands in the same row as a black one, it was removed the second time. Suppose at the end... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,967 |
Dopnnikov V.L. In the country, there are 2001 cities, some pairs of which are connected by roads, with at least one road leading out of each city, and no city is connected by roads to all the others. We will call a set of cities $D$ dominating if every city not in $D$ is connected by a road to one of the cities in the ... | ## Author: Dolnikov V.L.
Consider a graph where the vertices correspond to cities, and the edges correspond to roads. It is required to color the vertices of this graph in $2001-k$ colors such that no two vertices of the same color are connected by an edge (such a coloring is called proper).
Consider a vertex $A$ of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,968 |
Agakhanovo $H . X$.
Two polynomials $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$ and $Q(x)=x^{2}+p x+q$ take negative values on some interval $I$ of length greater than 2, and are non-negative outside $I$. Prove that there exists a point $x_{0}$ such that $P\left(x_{0}\right)<Q\left(x_{0}\right)$. | From the condition, it follows that $Q(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)$ and $P(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x^{2}+A x+B\right)$, where $x_{2}-x_{1}>2$, and the quadratic polynomial $x^{2}+A x+B$ either has no roots or has a double root coinciding with $x_{1}$ or $x_{2}$.
Suppose that $P... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,969 |
Find all such odd natural $n>1$, that for any coprime divisors $a$ and $b$ of $n$, the number $a+b-1$ is also a divisor of $n$. | Let $p$ be the smallest prime divisor of the number $n$. Represent $n$ in the form $p^{m} k$, where $k$ is not divisible by $p$. By the condition, the number $p+k-1$ is a divisor of $n$.
If $(p+k-1, k)>1$, then $(p-1, k)=(l, k)>1$. Thus, the number $k$ has some divisor $d, 2 \leq d \leq p-$
1. Contradiction with the ... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,970 |
Petrov F.
Prove that there are infinitely many natural numbers $n$ for which the numerator of the irreducible fraction equal to $1+1 / 2+\ldots+1 / n$ is not a power of a prime number with a natural exponent. | Let $1+1 / 2+\ldots+1 / n=S(n)=A(n) / B(n)$, where $A(n)$ and $B(n)$ are coprime. Note that $B(n)>n / 2$. Indeed, the largest power of two not exceeding $n$ is a divisor of exactly one of the numbers 1, $2, \ldots, n$ and therefore is a divisor of the denominator of the sum $S(n)$.
Suppose that for all $n \geq n_{0}$,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,971 |
Pastor A.
In a certain state, there were 2002 cities connected by roads in such a way that if travel through any of the cities was prohibited, it would still be possible to travel from any of the remaining cities to any other. Each year, the king selects a non-self-intersecting cyclic route and orders the construction... | Consider a graph where the vertices correspond to cities, and the edges correspond to roads that existed in the country before any transformations began. According to the problem, the following operation is repeatedly performed on this graph: all edges of some simple cycle are removed, and all vertices of this cycle ar... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,972 |
Hrramcovd
Find the largest natural number $N$, for which in any arrangement of different natural numbers from 1 to 400 in the cells of a $20 \times 20$ square table, there will be two numbers in the same row or column whose difference is at least $N$.
# | Example. Let's divide the table into two $20 \times 10$ rectangles vertically. In the first rectangle, we will place numbers from 1 to 200 in increasing order by rows (from 1 to 10 in the first row, from 11 to 20 in the second row, and so on). In the second rectangle, we will place numbers from 201 to 400 in the same m... | 209 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,973 |
Authors: Bogdanov I.I., Chennokov G.R.
In a rectangular table with 9 rows and 2004 columns, the numbers from 1 to 2004 are arranged, each appearing 9 times. In each column, the numbers differ by no more than 3. Find the minimum possible sum of the numbers in the first row. | Evaluation. By rearranging the columns if necessary, we will henceforth assume that the numbers in the first row are in non-decreasing order. Let $a_{i}$ be the $i$-th number in the first row. Consider the sum $S=\left(a_{1}-1\right)+\left(a_{2}-1\right)+\left(a_{3}-1\right)+$ $\left(a_{4}-2\right)+\ldots+\left(a_{i}-(... | 2005004 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,974 |
Donniov V.
In the country, there are several cities, some pairs of which are connected by two-way non-stop flights belonging to $k$ airlines. It is known that any two lines of the same airline have a common end. Prove that all cities can be divided into $k+2$ groups such that no two cities in the same group are connec... | Induction on $k$.
Base. If $k=0$, the statement is trivial: there are no airlines.
Induction step. Consider a graph where the vertices correspond to cities and the edges to flights.
Let $E_{1}, E_{2}, \ldots, E_{k}$ be the groups of edges corresponding to the first, second, $\ldots, k$-th airline.
It is not hard to... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,975 |
Does there exist a natural number $n>10^{1000}$, not divisible by 10, such that in its decimal representation, two different non-zero digits can be swapped so that the set of its prime divisors remains unchanged? | Let's provide an example of such a number. Suppose $n=13 \cdot 1 \ldots 1=14 . . .43$, the number of ones we will choose later. If we swap the one and the three, we get the number $34 \ldots 41=31 \cdot 1 \ldots 1$.
In this case, if our number consisting of only ones is divisible by $13 \cdot 31=403$, then the prime d... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,976 |
[ррбер $\underline{\text {. }}$.
On the table, there are 365 cards, with different numbers written on their reverse sides. For one ruble, Vasya can choose three cards and ask Petya to place them from left to right in ascending order of the numbers on the cards. Can Vasya, by spending 2000 rubles, guarantee to lay out ... | Lemma. Suppose Vasya managed to lay out some $N-1$ cards in the required order on the table for $x$ rubles, where $N \leq 3^{k}$. Then he can add one more card to the laid-out cards, spending no more than $k$ rubles. We will prove this by induction on $k$. The base case $(k=1)$ is obvious.
Inductive step. Suppose $N=3... | 1845 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,977 |
Poddisky o.K.
Ten pairwise distinct non-zero numbers are such that for any two of them, either the sum of these numbers or their product is a rational number.
Prove that the squares of all numbers are rational.
# | The first method. Let our set contain an irrational number $a$. Then each of the other numbers has the form either $p-a$, or ${ }^{p} / a$, where $p$ is rational. We will show that there are no more than two numbers of the form $p-a$. Let $b_{1}=$ $p_{1}-a, b_{2}=p_{2}-a, b_{3}=p_{3}-a$, then the number $b_{1}+b_{2}=\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,978 |
In the sequence of natural numbers $\left\{a_{n}\right\}, n=1,2, \ldots$, each natural number appears at least once, and for any distinct $n$ and $m$, the inequality $\frac{1}{1998}<\frac{\lef... | From the inequality in the condition, it follows that all members of the sequence are pairwise distinct.
Lemma. If $i>n$ and $a_{i}<a_{n}$, then $i-n<2000000$.
Proof. The segment $\left[1, a_{n}\right]$ contains only a finite number of terms of the sequence, so all $a_{k}$ with sufficiently large indices $k$ will be ... | |a_{n}-n|<2000000 | Number Theory | proof | Yes | Yes | olympiads | false | 52,979 |
Doponnov V.l.
Prove that from any finite set of points on a plane, one point can be removed such that the remaining set can be divided into two parts of smaller diameter. (Diameter is the maximum distance between points in the set.)
# | Let $A$ and $B$ be any two points of a given set $M$, the distance between which is equal to the diameter $d$ of this set. Then, from the definition of the diameter, it follows that if $P \in M$, then $P$ lies inside or on the boundary of the lens formed by the intersection of circles of radius $d$ centered at $A$ and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,980 |
Authors: Doppnnyov V.L. Karopov D.V.
In a country, there are 2000 cities, some pairs of which are connected by roads. It is known that through any city, no more than $N$ different non-self-intersecting cyclic routes of odd length pass. Prove that the country can be divided into $2N+2$ republics such that no two cities... | Consider a graph $G$ with vertices in cities, the edges of which correspond to roads. We will prove that the vertices of this graph can be properly colored with $2N+2$ colors (that is, so that no two vertices of the same color are connected by an edge). This is equivalent to the statement of the problem.
Choose one ed... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,981 |
$\underline{\text { Karasev } P .}$
On the plane, there is an infinite set of points $S$, and in any $1 \times 1$ square, there is a finite number of points from the set $S$. Prove that there exist two different points $A$ and $B$ from $S$ such that for any other point $X$ from $S$, the following inequalities hold:
$... | Let's prove the statement of the problem by contradiction. We can assume that for any two different points $A$ and $B$ from $S$, there exists a point $X$ from $S$ different from them such that either $|X A| < 0.999 |A B|$, or $|X B| < 0.999 |A B|$.
Reformulating the above statement: for any segment $I$ with endpoints ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,982 |
$\underline{\text { Senderov B.A. }}$
Find all pairs $(x, y)$ of natural numbers such that $x+y=a^{n}, x^{2}+y^{2}=a^{m}$ for some natural numbers $a, n$, and $m$. | Given the condition $a^{2 n}=x^{2}+y^{2}+2 x y=a^{m}+2 x y>a^{m}$, it follows that $a^{2 n}$ is divisible by $a^{m}$. Therefore, $2 x y$ is also divisible by $a^{m} = x^{2} + y^{2}$. We obtain that
$2 x y \geq x^{2} + y^{2} \geq 2 x y$. Hence, $x^{2} + y^{2} = 2 x y, x = y$. Consequently, $2 x = a^{n}, 2 x^{2} = a^{m}... | (2^{k},2^{k}),wherek\geq0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,984 |
Kozzov P.
Prove that if a natural number $N$ is represented as the sum of three squares of integers divisible by 3, then it is also represented as the sum of three squares of integers not divisible by 3. | From the condition, it follows that the number $N$ can be represented in the form $9^{n}\left(a^{2}+b^{2}+c^{2}\right)$, where one of the numbers $a, b, c$ is not divisible by 3. According to the solution of problem $\underline{110218}$, the number $9\left(a^{2}+b^{2}+c^{2}\right)$ can be represented in the form $x^{2}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,985 |
Klepczy V.A.
The test consists of 30 questions, each with two answer options (one correct, the other incorrect). In one attempt, Vitya answers all the questions, after which he is told how many questions he answered correctly. Can Vitya act in such a way as to guarantee that he will know all the correct answers no lat... | We can consider that the answer to each question is either YES or NO.
a) Let's say that in the $k$-th attempt $(k=1,2, \ldots, 29)$, Vitya answers YES to the $k$-th question and NO to all the others. Any two attempts differ in exactly two places, so the results either match or differ by 2. If the result of the $k$-th a... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,986 |
Muirhead's $\mathbf{M . B}$.
For positive numbers $x_{1}, x_{2}, \ldots, x_{n}$, prove the inequality
$\left(1+x_{1}\right)\left(1+x_{1}+x_{2}\right) \ldots\left(1+x_{1}+x_{2}+\ldots+x_{n}\right) \geq \sqrt{(n+1)^{n+1}} \cdot \sqrt{x_{1} x_{2} \ldots x_{n}}$. | Raising to the square, we rewrite the inequality as
$S=\frac{x_{1} x_{2} \ldots x_{n}}{\left(1+x_{1}\right)^{2}\left(1+x_{1}+x_{2}\right)^{2} \ldots\left(1+x_{1}+x_{2}+\ldots+x_{n}\right)^{2}} \leq \frac{1}{(n+1)^{n+1}}$.
Let $y_{1}=\frac{x_{1}}{1+x_{1}}, \quad y_{2}=\frac{x_{2}}{\left(1+x_{1}\right)\left(1+x_{1}+x_{2... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,987 |
[ $\left.\quad \begin{array}{rl}\text { Vieta's Theorem }\end{array}\right]$
Authors: Asahanov N.X., Bosgonov I.I.
Do there exist such non-zero numbers $a, b, c$, that for any $n>3$ one can find a polynomial of the form $P_{n}(x)=x^{n}+\ldots$ $+a x^{2}+b x+c$, having exactly $n$ (not necessarily distinct) integer ro... | Suppose such $a, b, c$ exist. Let $k$ be the maximum number of factors (greater than 1 in modulus) into which the number $c$ can be decomposed. Then each polynomial $P_{n}(x)$ has no more than $k$ roots different from $\pm 1$.
Let $x_{1}, \ldots, x_{n}$ be the roots of this polynomial. Consider the sum $S=\frac{1}{x_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,988 |
## [ Prime numbers and their properties Divisibility of numbers. General properties Examples and counterexamples. Constructions [The Fundamental Theorem of Arithmetic. Factorization into prime factors]
Author: Senorerov B.A.
Given a finite set of prime numbers $P$. Prove that there exists a natural number $x$ such t... | Lemma. Let $p$ be a prime number. Then the number $2^{n}$ can be represented as $a^{p}+b^{p}$ if and only if $n-1$ is divisible by $p$.
Proof. If $n-1=k p$, then $2^{n}=\left(2^{k}\right)^{p}+\left(2^{k}\right)^{p}$.
Suppose that $2^{n}=a^{p}+b^{p}$. Let $a=2^{s} k, b=2^{t} m$, where $k, m$ are odd. If, say, $s>t$, t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,989 |
Shapovalov A.V.
Given an integer $n>1$. Two players take turns marking points on a circle: the first one in red, the second one in blue (marking the same point twice is not allowed). When $n$ points of each color have been marked, the game ends. After that, each player finds the longest arc on the circle with endpoint... | Let's outline the winning strategy for the Second player. We can assume that the circumference length is $n$ and the First player initially marks a vertex of the inscribed regular $n$-gon. The Second player marks the vertices of this polygon as long as possible. Let's call the arcs connecting adjacent vertices of the $... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,990 |
Doonnikov V.l.
Let's call a company $k$-unbreakable if in any partition of it into $k$ groups, there will be two acquaintances in one of the groups. Given a 3-unbreakable company in which there are no four people who are all mutually acquainted, prove that it can be divided into two companies, one of which is 2-unbrea... | Each company corresponds to a graph, where vertices correspond to people, and edges - to acquaintances. A company is $k$-partitionable if and only if the vertices of this graph can be properly colored with $k$ colors (so that adjacent vertices have different colors).
Lemma. Suppose the graph has no cycles of odd lengt... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,991 |
Prove that for $n \geq 6$, a regular $(n-1)$-gon cannot be inscribed in a regular $n$-gon such that exactly one vertex of the $(n-1)$-gon lies on each side of the $n$-gon, except for one side. | Let the correct $(n-1)$-gon $B_{1} \ldots B_{n-1}$ be inscribed in the correct $n$-gon $A_{1} \ldots A_{n}$. We can assume that $A_{1}$ and $B_{1}$ are the closest vertices of these polygons, and the points $B_{2}, B_{3}, B_{4}$, and $B_{5}$ lie on the sides $A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{5}$, and $A_{5} A_{6}$. L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,993 |
Suppose that the continued fractions
$$
\alpha=p-\frac{q}{p-\frac{q}{p-\frac{q}{\ldots}}} \quad \text { and } \quad \beta=\frac{q}{p-\frac{q}{p-\frac{q}{\ldots}}}
$$
converge. According to
problem 61330, they will converge to the roots of the polynomial $x^{2}-p x+q=0$. On the other hand, the same roots will also be... | The solution is analogous to the solution of problem $\underline{61316}$. Consider the case of the fraction $\alpha$. The corresponding convergents ${ }^{P_{k /}} Q_{k}$ satisfy the recurrence relations $P_{-1}=1, P_{0}=1, P_{k}=p P_{k-1}-q P_{k-2} ; Q_{-1}=0, Q_{0}=1, Q_{k}=$ $p Q_{k-1}-q Q_{k-2}$. It is easy to verif... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,996 |
10,11 | |
Authors: Bogdanov I.i., Podlipsky O.K.
Given natural numbers $a$ and $b$, where $a<b<2a$. On a grid plane, some cells are marked such that in every grid rectangle $a \times b$ or $b \times a$ there is at least one marked cell. For what largest $\alpha$ can we assert that for any natural number $N$, there e... | Let's introduce a coordinate system on the plane such that the centers of the cells, and only they, have integer coordinates. We will say that a cell has the same coordinates as its center. We will call a rectangle $a \times b$ vertical or horizontal if its side of length $b$ is vertical or horizontal, respectively.
1... | \alpha=\frac{1}{2^{2}-2+b^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,997 |
Can a regular triangle be divided into a million polygons so that no straight line intersects more than forty of these polygons?
We say that a line intersects a polygon if it has at least one common point with it. | Solution. In the figure, a convex 12-sided polygon is cut by diagonals into 5 parts: an 8-sided polygon and four triangles. Each line intersects no more than two of these triangles. Indeed, a line intersecting a triangle must intersect at least one of its sides, common with the 12-sided polygon; on the other hand, a li... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,999 |
Aekseev B.5.
$P$ and $Q$ are subsets of the set of expressions of the form ( $a_{1}, a_{2}, \ldots, a_{n}$ ), where $a_{i}$ are natural numbers not exceeding a given natural number $k$ (there are a total of $k^{n}$ such expressions). For each element ( $p_{1}, \ldots, p_{n}$ ) of the set $P$ and each element $\left(q_... | a) We call a set $\bar{d}$ the opposite of set $d$ if $\bar{d}$ is obtained from $d$ by replacing all ones with twos and all twos with ones. Denote by $\bar{M}$ the set of all sets that are opposite to the sets in $M$. Sets $d$ and $\bar{d}$ do not coincide at any position. Therefore, if $d$ is in $P$, then $\bar{d}$ c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,000 |
A triangular table is constructed according to the following rule: in the top row, there is a single natural number $a>1$, and below each number $k$ on the left, we write the number $k^{2}$, and on the right - the number $k+1$. Prove that in each row of the table, all numbers are different.
For example, for $a=2$, the... | Suppose that in some rows of the table, the same numbers appear. Let $n$ be the number of the topmost of these rows, and $p$ and $q$ be the equal numbers in row $n$. Since there are no equal numbers in the previous row, $p$ and $q$ are obtained from numbers in the previous row by different operations: one by squaring, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,001 |
Sheffer $\underline{\text { P }}$.
For any natural number $n$ the sum $C_{n}^{1}+1973 C_{n}^{3}+1973^{2} C_{n}^{5}+\ldots$ is divisible by $2^{n-1}$. Prove this. | Let's denote our sum by $s_{n}$. We will use induction on $n$.
Base case. $s_{1}=1$ is divisible by $2^{0}, s_{2}=2$ is divisible by $2^{1}$.
Inductive step. Let $a=1+\sqrt{1973}, \quad b=1-\sqrt{1973}$ and note that $s_{n}=\frac{a^{n}-b^{n}}{2 \sqrt{1973}}$.
Since $a^{n}-b^{n}=\left(a^{n-1}-b^{n-1}\right)(a+b)-a b\... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,003 |
A peasant needs to transport a wolf, a goat, and cabbage across a river. The boat can only hold one person, and with him either the wolf, the goat, or the cabbage. If left unsupervised, the wolf will eat the goat. If left unsupervised, the goat will eat the cabbage. How can the peasant transport his cargo across the ri... | Think about whom the peasant can leave unattended.
## Solution
The peasant cannot leave the wolf with the goat or the goat with the cabbage, but he can leave the cabbage with the wolf. Let's show on a diagram how the peasant should proceed:
$$
\begin{aligned}
& \text { 1. Peasant with goat } \rightarrow . \quad \tex... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,011 |
Uncle Chernomor wrote the number 20 on a piece of paper. 33 bogatyrs pass the paper to each other, and each either adds one to the number or subtracts one from it. Can the result be the number 10?
# | After the sheet has been in the hands of the hero, the number written on it will change its parity. Therefore, after 33 changes, the number will become odd.
## Answer
It cannot.
## Problem | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,012 |
$\left.\begin{array}{l}{[\text { Order relation }} \\ {[\text { Pigeonhole Principle (other). }]}\end{array}\right]$
Four friends came to the ice rink, each with her brother. They paired up and started skating.
It turned out that in each pair the "escort" was taller than the "lady" and no one was skating with their s... | With whom does Lusya Egorova, the tallest among the girls, skate?
## Solution
The tallest girl is Lusya Egorova, and according to the condition, she skates with a boy who is taller than her. There are two such boys, but one of them is her brother. Therefore, Lusya Egorova skates with Yura Vorobyov. Reasoning similarl... | LusyaEgorovaYuraVorobyov,OlyaPetrovaAndreyEgorov,InnaKrymovaSeryozhaPetrov,AnyaVorobyovaDimaKrymov | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,013 |
2+ [ Examples and counterexamples. Constructions ]
Place real numbers at the vertices of a pentagon so that the sum of the numbers at the ends of one side is equal to 1, at the ends of another side is equal to 2, ..., and at the ends of the last side is equal to 5.
# | 
For any value of $x$, the sums of the numbers on the four sides will be 1, 2, 3, 4, respectively. To make the sum of the numbers on the 5th side equal to 5, we need to find $x$ from the conditi... | \frac{3}{2},-\frac{1}{2},\frac{5}{2},\frac{1}{2},\frac{7}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,014 |
On the table lie three red sticks of different lengths, the sum of which is 30 cm, and five blue sticks of different lengths, the sum of which is also 30 cm. Can the red and blue sticks be cut so that they can then be paired, with the sticks in each pair being the same length but of different colors?
# | Try to add blue and red sticks, each 30 cm long, and compare them with each other.
## Solution
Let's check if there are pairs of sticks of the same length but different colors. If there are, set these pairs aside; if not, choose the shortest stick. Now, take any stick of a different color and cut a piece of the same ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,015 |
In a $4 \times 4$ square, 15 points are drawn. Prove that from it, a $1 \times 1$ square can be cut out that does not contain any points inside it. | Try to cut the original square into $1 \times 1$ squares.
## Solution
Let's cut our square into 16 smaller squares $1 \times 1$. Among these small squares, there will definitely be at least one that does not contain a point (15 points cannot fit into 16 squares).
Submit a comment | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,016 |
19 lily-of-the-valley bushes are planted in a circle.
a) Prove that there will always be two adjacent bushes, the total number of bells on which is even.
b) Is it always possible to find two adjacent bushes, the total number of bells on which is divisible by 3? | a) First method. Suppose that no such pair of adjacent lilies of the valley can be found. This means that lilies of the valley with an even and odd number of bells alternate. But an odd number (19) of lilies of the valley cannot alternate.
Second method. Consider all 19 "adjacent" sums. Their total sum is twice the nu... | notalways | Number Theory | proof | Yes | Yes | olympiads | false | 53,017 |
[ CaseAnalysis ] $[$ Divisibility Rules (etc.) $]$ Author: Fomin S.B. A natural number $n$ is written in the decimal system. It is known that if a digit appears in this representation, then $n$ is divisible by this digit (0 does not appear in the representation). What is the maximum number of different digits that t... | If the digit 5 is included in the representation of a number, then the number must end in 5. Therefore, it is odd and, consequently, contains only odd digits. Thus, it cannot have more than five digits. If 5 does not appear in the decimal representation of the number, then it can include all other 8 digits. For example... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,018 |
$\underline{\text { Folkkior }}$
On the computer screen, a number is displayed that increases by 102 every minute. The initial value of the number is 123. Programmer Fedia has the ability to change the order of the digits of the number on the screen at any moment. Can he ensure that the number never becomes a four-dig... | One way: as soon as the last digit becomes 1 or 2, change the order of the digits so that it becomes the first digit. After this, Fedya can calmly skip up to seven increments, and the last digit will become 1 or 2 no later than after five increments. This means that Fedya can always reduce the number in time.
## Answe... | Can | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,019 |
Gussarov $M$.
There are three piles of stones. It is allowed to add to any of them as many stones as there are in the other two piles, or to throw out of any pile as many stones as there are in the other two piles. For example: $(12,3,5) \rightarrow(12$, 20, 5) (or (4, 3, 5)). Is it possible, starting with piles of 19... | In each of the initial piles, there is an odd number of stones. It is not hard to verify that the allowed operations preserve this property, and the number of stones in each pile remains odd. Therefore, none of the piles can become empty.
## Answer
It is not possible.
Send a comment | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,020 |
Spivak A.V.
An alien from the star Tau Ceti, upon arriving on Earth on Monday, exclaimed: "A!". On Tuesday, he exclaimed: "AU!", on Wednesday — "AUUA!", on Thursday — "AUUAUAAU!". What will he exclaim on Saturday? | Divide the "following" statement in half and compare it with the "previous".
## Solution
By dividing the "following" statement into two equal parts, we see that the first half matches the "previous", while the second half is a "mirror reflection" of the previous, meaning the letters A and U are swapped.
## Answer
"... | AUUAAUAAUUAUAAUUAUAAUUAUAAUUAU! | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,021 |
Proizvolov V.V.
A rectangle was cut by six vertical and six horizontal cuts into 49 rectangles (see figure). It turned out that the perimeter of each of the resulting rectangles is an integer number of meters. Is it necessarily true that the perimeter of the original rectangle is an integer number of meters?
.
## Solution
Consider the shaded rectangles on the diagram ("diagonal"). The horizontal side of the original rectangle is composed of their horizontal sides. The same applies to the vertical side.
Therefore, the perimeter of the original rectan... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,022 |
| $\left.\begin{array}{c}\text { Processes and operations } \\ \text { Examples and counterexamples. Constructions }\end{array}\right]$ |
| :---: | :---: |
On the table, there are five coins arranged in a row: the middle one is heads up, and the rest are tails up. In one operation, it is allowed to simultaneously flip... | First, let's flip the first 3 coins, then the first two will lie heads up, and the next three will lie tails up. Now let's flip the last three coins, as a result, all five coins will lie heads up.
## Answer
yes, it can. | yes,itcan | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,024 |
[ $\quad \underset{\text { Counting in two ways }}{\text { even and odd }}]$
In a $4 \times 4$ table, natural numbers were written. Could it be that the sum of the numbers in each subsequent row is 2 more than in the previous one, and the sum of the numbers in each subsequent column is 3 more than in the previous one?... | Suppose this is possible. Let the sum of the numbers in the first row be $a$, and the sum of the numbers in the first column be $b$. Then the sum of all the numbers in the table is $a+(a+2)+(a+4)+(a+6)=4a+12$, which is divisible by 4. On the other hand, it is also equal to $b+(b+3)+(b+6)+(b+9)=4b+18$, which is not divi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,026 |
Shapovalov A.V.
Petya liked the puzzle, he decided to glue it together and hang it on the wall. In one minute, he glued together two pieces (initial or previously glued). As a result, the entire puzzle was assembled into one complete picture in 2 hours. How long would it take to assemble the picture if Petya glued tog... | Each gluing reduces the number of pieces on the table by 1. Since after 120 gluings, one piece (the complete puzzle) was obtained, there were 121 pieces before the work began. Now, if gluing three pieces together per minute (i.e., reducing the number of pieces by 2), one piece will remain after 60 minutes.
## Oтвет
I... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,029 |
2+ [Examples and counterexamples. Constructions]
On some cells of a $4 \times 4$ square board, gold coins are placed in a stack, and on the remaining cells, silver coins. Is it possible to place the coins so that in each $3 \times 3$ square there are more silver coins than gold, and on the entire board there are more ... | Let's place a stack of nine silver coins on one of the cells of the central $2 \times 2$ square, and one gold coin on each of the other cells of the board. Then, in each $3 \times 3$ square, there will be 9 silver coins and 8 gold coins, and on the entire board, there will be 15 gold coins and 9 silver coins.
## Answe... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,030 |
$\underline{\text { Folklore }}$
Can three vectors be placed on a plane such that the magnitude of the sum of any two of them is equal to 1, and the sum of all three is equal to the zero vector?
# | Note that if the sum of three unit vectors is zero, then the sum of any two of them is equal to the third with the opposite sign, meaning its modulus is 1. Three such unit vectors can be arranged in different ways.
The first method. Consider an equilateral triangle \(ABC\) with sides of unit length (see figure). Then ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,031 |
$2+$ [Coordinate Method in Space (Miscellaneous)] In space, a parallelogram $A B C D$ and a plane $M$ are given. The distances from points $A, B$, and $C$ to the plane $M$ are $a, b$, and $c$, respectively. Find the distance $d$ from vertex $D$ to the plane $M$. | What is the distance from the center of the parallelogram to the plane $M$?
## Solution
We can assume that the plane $M$ is perpendicular to the x-axis. The sums of the x-coordinates of opposite vertices of the parallelogram are equal (this is twice the x-coordinate of the center $O$ of the parallelogram). The x-coor... | |\\b| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,035 |
[ $\left[\begin{array}{l}\text { Arithmetic. Mental calculation, etc. } \\ {[\text { Counting in two ways }}\end{array}\right]$
A very cunning kiosk owner received several packs of envelopes for sale, with 100 envelopes in each pack. He counts 10 envelopes in 10 seconds. How many seconds will it take him to count 60 e... | Note that $100-60<60$.
## Solution
The kiosk owner realized that, having a pack of 100 envelopes, one can count out 60 envelopes by simply counting out 40
- then 60 envelopes will remain in the pack. The same applies to 90: it is enough to remove 10 envelopes from the pack.
Send a comment | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,036 | |
Bogdanov I.I.
In some cells of a $10 \times 10$ table, several crosses and several zeros are placed. It is known that there is no line (row or column) completely filled with the same symbols (crosses or zeros).
However, if any symbol is placed in any empty cell, this condition will be violated. What is the minimum nu... | Let us assume that we have filled the table according to the condition, and cell $A$ is free. Since placing any symbol in it should result in a line of identical symbols, there must be a line containing it where all other cells are filled with crosses, and the same is true for noughts (these lines must, of course, be a... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,037 |
$\left[\begin{array}{l}\text { Theory of algorithms (other) }) \\ {[\underline{\text { Estimation + example }}]}\end{array}\right.$
You need to find out a five-digit phone number by asking questions that can be answered with "yes" or "no." What is the minimum number of questions required to guarantee finding the numbe... | We need to ask questions in such a way that each subsequent question approximately halves the number of remaining possible options.
## Solution
Initially, when we know nothing about the phone number, there are $10^{5}=100000$ possible options for the phone number. We ask the question: "Is the number greater than 5000... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,039 |
Does there exist a quadrilateral pyramid in which two opposite lateral faces are perpendicular to the base? | Take a quadrilateral base, draw two perpendicular planes to it through two opposite sides. These will be the planes of the lateral faces. Take the vertex of the pyramid on the line of intersection of these planes.
## Solution
Consider the tetrahedron OABS, where the dihedral angles SOA and SOB are right angles. Next,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,040 |
Let $a$ and $b$ be integers. Write the number $b$ to the right of the number $a$. If the number $a$ is even, then divide it by 2; if it is odd, then subtract one from it and then divide it by 2. The resulting number $a_{1}$ write under the number $a$. To the right of the number $a_{1}$, write the number $2 b$. Perform ... | Consider the binary representation of the number $a=\overline{\alpha_{n} \ldots \alpha_{0}}$. Let $i_{1}, \ldots, i_{\mathrm{s}}$ be the positions where 1s are located, then $a=\sum_{i=0}^{n}$ $\alpha_{i} 2^{\mathrm{i}}=\sum_{t=1}^{s} \alpha_{i_{t}} 2^{\mathrm{i}_{\mathrm{t}}}=\sum_{t=1}^{s} 2^{\mathrm{i}^{\mathrm{t}}}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,042 |
For the encryption of telegraph messages, it is required to divide all possible ten-digit "words" - sets of ten dots and dashes - into two groups such that any two words in the same group differ in at least three positions. Indicate a method for such a partition or prove that it does not exist.
# | Consider all words starting with the same sequence of eight dots and dashes. There are four such words, and they all need to be in different groups. But there are only two groups. Therefore, the required partition is impossible.
Send a comment | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,043 |
Among 4 people, there are no three with the same first name, middle name, or last name, but any two people have either the same first name, middle name, or last name. Is this possible?
# | ## Solution
Yes, it could be like this, for example: Ivan Ivanovich Ivanov, Ivan Petrovich Petrov, Petr Ivanovich Petrov, Petr Petrovich Ivanov.
Send comment | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,044 |
Plot $\boldsymbol{m} \times \boldsymbol{n}$. A rectangular plot of size $m \times n$ is divided into squares $1 \times 1$. Each square is a separate plot, connected by gates to adjacent plots. For what plot sizes can you walk through all the square plots, visiting each one exactly once, and return to the starting point... | We will color the squares in a checkerboard pattern. At each transition, the color of the cell changes, so if such a route is possible, it can only be with an even number of cells (returning to the same cell), i.e., either $m$ or $n$ must be even numbers. It remains to check that for any dimensions, a traversal is poss... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,045 |
Perfect squares. Prove that all numbers of the form 16; 1156; 11156 and so on (the number 15 is inserted into the middle of the previous number) are perfect squares. | At first, we guess a few numbers, such as $16=4 \times 4, 1156=34 \times 34, 111556=334 \times 334$. We will prove that the numbers satisfying the condition are of the form where all digits are threes, and the last digit is a four.
$$
\begin{aligned}
& \underbrace{33 \ldots .34^{2}}_{n-1}=(3 \cdot \underbrace{11 \ldot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,046 |
[ Text problems ] $[$ Case enumeration ]
In a box, there are blue, red, and green pencils. In total, there are 20. The number of blue pencils is 6 times the number of green pencils, and the number of red pencils is less than the number of blue pencils. How many red pencils are in the box? | Think about how many blue pencils there can be.
## Solution
Since there are 20 pencils in total, and blue and green pencils together make up 7 parts. This means there can be 6 or 12 blue pencils, and green and red pencils would then be 1 and 13 or 2 and 6, respectively. Since there are fewer red pencils than blue one... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,047 |
Yashchenko I.V.
On a circular road, there are four gas stations: $A, B, C$, and $D$. The distance between $A$ and $B$ is 50 km, between $A$ and $C$ is 40 km, between $C$ and $D$ is 25 km, and between $D$ and $A$ is 35 km (all distances are measured along the circular road in the shortest direction).
a) Provide an exa... | First, determine the arrangement of gas stations $A, C$, and $D$.
## Solution
The problem provides all three distances between $A, C$, and $D$. First, let's determine the arrangement of these three gas stations. Gas stations $A$ and $C$ divide the circular road into two arcs. If gas station $D$ were on the shorter ar... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,048 |
[ Triangle inequality (miscellaneous). ] [ Minimum or maximum distance (length).]
Petya bought a "Constructor" set, which contained 100 sticks of different lengths. The instructions for the "Constructor" state that any three sticks from the set can form a triangle. Petya decided to test this statement by forming trian... | Pete only needs to check if it is possible to form a triangle from the two shortest sticks and the longest one. If the triangle cannot be formed, the statement in the instruction is refuted. If, however, the triangle can be formed, then the sum of the lengths of the two shortest sticks is greater than the length of the... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,050 |
Is it possible to arrange 12 identical coins along the walls of a large square box so that along each wall there are exactly
a) 2 coins;
b) 3 coins;
c) 4 coins;
d) 5 coins;
e) 6 coins;
f) 7 c... | a) Since, by condition, all coins must be placed along the walls, and each wall is touched by exactly two coins, the total number of coins is no more than 8.
b)-d) Examples of the required arrangements are shown in the figure
,e)no;b),),)yes | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,051 |
5,6,7,8 | |
A four-digit number starts with the digit 6. This digit was moved to the end of the number. The resulting number turned out to be 1152 less than the original. Find the original number. | Let's write the condition of the problem in the form of a rebus, for example:
## $-6 A B C$ $\frac{1.52}{1152}$
Then C $=8$ and the example takes the form:
Therefore, $B=3$ and the example has the following form:
## $-6 \mathrm{~A}^{438}$ $-{ }_{A 38}^{10}$
Since a carry-over has occurred, then $A=5$ :
$-6538$
... | 6538 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,052 |
\left[\begin{array}{l}{[\text { Decimal numeral system }]} \\ {[\text { Case enumeration }}\end{array}\right]
Author: Raskina I.V.
The year 2009 has the following property: by rearranging the digits of the number 2009, it is impossible to obtain a smaller four-digit number (numbers do not start with zero). In which y... | Author: Raschina I.V.
In the years $2010, 2011, \ldots, 2019$ and in 2021, the number of the year contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be reduced by rearranging the digits.
## Answer
in 2... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,053 |
Does there exist a convex polyhedron such that any section by a plane, not passing through a vertex, is a polygon with an odd number of sides?
# | Shift the section parallel and observe how the parity of the number of intersected edges changes when the section passes through a vertex.
## Solution
Suppose such a polyhedron exists. If it had a vertex A, where an even number $\mathrm{m}$ of edges meet, then the polyhedron would have an $\mathrm{m}$-sided section s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,055 |
[ Triangle inequality (miscellaneous). ] [ Pairings and groupings; bijections ]
In Old Kalitva, 50 schoolchildren live, and in Middle Boltay - 100 schoolchildren. Where should a school be built to minimize the total distance traveled by all schoolchildren? | Let's take 50 schoolchildren from Old Kalitva and 50 from Middle Boltayev and divide them into 50 pairs, with one schoolchild from each city in each pair. For each such pair, the distance they travel to school does not depend on the location of the school (of course, it must be built somewhere on the road between the c... | InMiddleBoltayev | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,057 |
$\underline{35714}$ Topics [Algorithm Theory (other).] $[\quad \underline{\text { Induction (other) }} \quad]$
There are eight glasses of water on the table. It is allowed to take any two glasses and equalize the amount of water in them by pouring water from one glass to another. Prove that using such operations, it i... | 8 is a power of two. Solve this problem first for four glasses, and then for eight.
## Solution
Divide the glasses into pairs A - A', B - B', C - C', D - D', and equalize the amount of water in each pair of glasses. Now we have two identical sets of four: A, B, C, D and A', B', C', D'. Equalize the amount of water in... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 53,060 |
[ Methods for solving problems with parameters ] [
Specify all points $(x, y)$ in the plane through which at least one curve of the family $y=p^{2}+(2 p-1) x+$ $2 x^{2}$ passes.
# | The point $(x, y)$ lies on one of the given curves if and only if the equation $p^{2}+2 p x+\left(2 x^{2}-x-y\right)=$ 0 has a solution (with respect to $p$), which means that $D / 4=x^{2}-\left(2 x^{2}-x-y\right)=y+x-x^{2} \geq 0$.
## Answer
All points above the parabola $y=x^{2}-x$, including the parabola itself. | x^{2}-x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,061 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.