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[ Divisibility rules for 3 and 9 ] $[$ Counting in two ways $]$
In the cells of a $5 \times 5$ table, non-zero digits are placed. In each row and in each column, ten five-digit numbers are formed from all the digits there. Can it happen that exactly one of these numbers is not divisible by 3? | Recall the divisibility rule for 3.
## Solution
If a number is divisible by 3, then the sum of its digits is divisible by 3. Let's assume, for definiteness, that the number in the top row is not divisible by 3. Then the sum of the digits in each column is divisible by 3. This means that the sum of all the digits in t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,034 |
[ Numerical inequalities. Comparisons of numbers. ] [ Examples and counterexamples. Constructions ]
Write five numbers in a row so that the sum of any two adjacent numbers is negative, but the sum of all the numbers is positive. | Note: The sum of positive numbers should be greater in modulus than the sum of negative numbers.
## Otvet
For example: $3,-4,3,-4,3$. | 3,-4,3,-4,3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 55,035 |
6,7 |
Do there exist such natural numbers $a$ and $b$ that the fractions $\frac{a}{b}$, $\frac{a+1}{b}$, $\frac{a+1}{b+1}$ are irreducible? | For example: $2 / 7,3 / 7,3 / 2$.
## Answer
They exist.
## [Equilateral (regular) triangle] Problem $\underline{98334}$ Topics: [ axial and sliding symmetries (other). ] [ Examples and counterexamples. Constructions ] Author: Evdokimov M.A. Baron Munchausen claims that he launched a ball from the edge of a billia... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,036 |
[ Arithmetic. Mental calculation, etc. ] $[\quad$ Case enumeration $\quad]$
109 apples are distributed into bags. In some bags, there are $x$ apples, and in others, there are three apples.
Find all possible values of $x$, if the total number of bags is 20. | If there were 3 apples in each package, there would be 60 apples in total. But there are 49 more apples, so the "extra" apples need to be evenly distributed among some packages. Since $49=7 \cdot 7=49 \cdot 1$ and there are 20 packages in total, either 7 packages contain 7 "extra" apples each, or one package contains 4... | 10or52 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,039 |
10,11 |
Does there exist a real $\alpha$ such that the number $\cos \alpha$ is irrational, while all the numbers $\cos 2 \alpha, \cos 3 \alpha, \cos 4 \alpha, \cos 5 \alpha$ are rational? | Assume the opposite. Then the number $A=\cos \alpha+\cos 5 \alpha$ is irrational as the sum of a rational and an irrational number. On the other hand,
$A=2 \cos 2 \alpha \cos 3 \alpha$ is rational as the product of three rational numbers. Contradiction.
## Answer
There does not exist. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,040 |
Shaparoo A.i.
Does there exist a non-convex pentagon in which no two of the five diagonals have any common points
(except vertices)? | Let five points $A, B, C, D$ and $E$, two of which $D$ and $E$ lie inside the triangle $A B C$, be connected by segments. These 10 segments can be divided into two non-self-intersecting broken lines, each consisting of 5 segments (see figure). Any one of these broken lines can be taken as a pentagon. Then the second br... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,041 |
Petya wrote natural numbers from 1 to 6 on the faces of a cube. Vasya hasn't seen the cube, but he claims that:
a) this cube has two adjacent faces with consecutive numbers;
b) there are no fewer than two such pairs of adjacent faces on the cube.
Is he right in both cases? Why? | In the set of natural numbers from 1 to 6, you can find five pairs of adjacent numbers: 1 and 2, 2 and 3, 3 and 4, 4 and 5, 5 and 6. Each such pair can be written either on two adjacent faces of a cube or on two opposite faces. However, a cube has only three pairs of opposite faces. Therefore, no more than three pairs ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,043 |
Let $\alpha, \beta, \gamma$ and $\delta$ be the degree measures of the angles of some convex quadrilateral. Can we always choose three of these four numbers such that they represent the lengths of the sides of some triangle (for example, in meters)? | Let's consider, for example, the quartet of numbers: $28,50,110,172$. A convex quadrilateral with such angles exists, since their sum is 360 and each number is less than 180.
At the same time, for each triplet of these four numbers, the triangle inequality does not hold: $28+50<110<$
$172 ; 28+110<172 ; 50+110<172$.
... | no,notalways | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,044 |
[Combinations and Permutations] [Case Enumeration]
How many ways can you choose from a full deck (52 cards)
a) 4 cards of different suits and ranks?
b) 6 cards such that all four suits are represented among them?
# | a) A card of the spades suit can be chosen in 13 ways, after which a card of the diamonds suit can be chosen in 12 ways...
b) $6=1+1+1+3=1+1+2+2$.
## Answer
a) $13 \cdot 12 \cdot 11 \cdot 10=17160$ ways; b) $4 C_{13}^{3} \cdot 13^{3}+C_{4}^{2} \cdot 13^{2} \cdot\left(C_{13}^{2}\right)^{2}=8682544$ ways. | 8682544 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,045 |
$3-$ [ Dirichlet's Principle (finite number of points, lines, etc.).]
A line is painted in two colors.
Prove that there exist three points $A, B$, and $C$, all painted in the same color, such that point $B$ is the midpoint of segment $A C$.
# | Consider on a line two arbitrary points $X$ and $Y$ of the same color, as well as point $X_{1}$, symmetric to $X$ with respect to $Y$, point $Y_{1}$, symmetric to $Y$ with respect to $X$, and the midpoint $O$ of segment $X Y$ (see figure). If at least one of these points is colored the same color as points $X$ and $Y$,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,047 |
A frog jumps from vertex to vertex of triangle $A B C$, moving each time to one of the adjacent vertices.
In how many ways can it get from $A$ to $A$ in $n$ jumps? | Let $a_{n}$ be the number of ways to return to the initial vertex in $n$ jumps, and $b_{n}$ be the number of ways to reach a neighboring vertex in $n$ jumps. It is easy to see that $a_{n+1}=2 b_{n}, b_{n+1}=a_{n}+b_{n}$. From this, it is not difficult to derive that $b_{n+2}=b_{n+1}+2 b_{n}, a_{n+2}=a_{n+1}+2 a_{n}$.
... | a_{n}=\frac{2^{n}+2\cdot(-1)^{n}}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,050 |
[ $[$ Products and factorials $]$ $[$ Case enumeration $\quad]$
Find all natural $m$ and $n$ for which $m!+12=n^{2}$. | When $m \geq 5$, $m!$ is even and a multiple of 5, meaning the decimal representation of this number ends in zero, and the number $m!+12$ ends in the digit 2. Therefore, it cannot be a perfect square.
By checking the values $m=1,2,3,4$, we get that $m=4, n=6$.
## Answer
$m=4, n=6$. | =4,n=6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,052 |
[ Linear inequalities and systems of inequalities ]
$$
\text { [ The extremal principle (miscellaneous). ] }
$$
Nine digits: $1,2,3, \ldots, 9$ are written in some order (forming a nine-digit number). Consider all triples of consecutive digits, and find the sum of the corresponding seven three-digit numbers. What is ... | Let $\overline{a b c d e f g h i}$ be a nine-digit number. From it, we form the sum $S=\overline{a b c}+\overline{b c d}+\overline{c d e}+\overline{d e f}+\overline{f f g}+\overline{f g h}+\overline{g h i}=$ $100 a+110 b+111(c+d+e+f+g)+11 h+i$. The sum $S$ reaches its maximum when the larger digits enter it with the la... | 4648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,053 |
Provolov V.V.
A chess king has traversed the entire $8 \times 8$ board, visiting each square exactly once and returning to the starting square on the last move.
Prove that he made an even number of diagonal moves. | With each non-diagonal move, the color of the square the king stands on changes; with a diagonal move, it does not change. Since the king has toured the entire board and returned to the starting point, the number of times the color changed from white to black is the same as the number of times it changed from black to ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,054 |
Fegoov A.
Baron Munchausen claims that he managed to form a certain rectangle from several similar non-right triangles. Can we believe him? (Among the similar triangles, there can be equal ones.) | In the figure, the rectangle is divided into isosceles triangles with a vertex angle of $120^{\circ}$.
## Answer
It is possible. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,055 |
48 blacksmiths need to shoe 60 horses. What is the minimum time they will spend on the work if each blacksmith spends 5 minutes on one horseshoe
# | Notice that four blacksmiths cannot shoe one horse at the same time.
## Solution
It will take no less than $60 \cdot 4 \cdot 5: 48=25$ minutes. For this, it is necessary that none of the blacksmiths are idle. It is sufficient to show that four blacksmiths can shoe five horses in 25 minutes. For this, arrange the five... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,058 |
Spivak A.V.
A natural number had the sum of its digits subtracted from it, then from the resulting number, the sum of its (resulting number's) digits was subtracted, and so on. After eleven such subtractions, zero was obtained. From which number did they start? | The difference between a number and the sum of its digits is divisible by 9.
## Solution
The difference between a number and the sum of its digits is divisible by 9. Therefore, all numbers we obtained were divisible by 9 (except, perhaps, the initial one). Let's go from the end. Zero can be obtained from any single-d... | 100to109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,060 |
Shaovalov A.V.
What is the maximum number of white and black chips that can be placed on a chessboard so that in each row and each column there are exactly twice as many white chips as black ones? | Evaluation. The number of chips on each vertical is a multiple of 3, meaning there are no more than 6, and on the entire board - no more than 48. Example: 32 white chips are placed on white squares, and 16 black chips are placed along the main "black" diagonal and along two parallel diagonals of "length" 4.
## Answer
... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,062 |
$\underline{\text { Folklore }}$
Can the product of three three-digit numbers, for the recording of which nine different digits were used, end with four zeros? | For example, $125 \cdot 360 \cdot 748=33660000$.
## Answer
It can. | 33660000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,064 |
[ Processes and Operations ] $[$ Invariants $]$
On the board, the numbers 1, 2, 3, ... 1984, 1985 are written. It is allowed to erase any two numbers and write down the absolute value of their difference instead. In the end, one number will remain on the board. Can it be zero? | For the given operations, the parity of the sum of all numbers written on the board does not change.
## Solution
The modulus of the difference of two integers has the same parity as the sum of these numbers. Therefore, the parity of the sum of all numbers does not change with the specified replacement. The sum of int... | Itcannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,065 |
[ Number of divisors and their sum ] [ Pairing and grouping; bijections ]
Prove that a number with an odd number of divisors is a perfect square.
# | If $d$ is a divisor of the number $n$, then $n / d$ is also a divisor.
## Solution
If $d$ is a divisor of the number $n$, then $n / d$ is also a divisor. Thus, all divisors of the number can be divided into pairs. If the number has an odd number of divisors, then the divisors of some pair coincide. Then $d=n / d$, fr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,066 |
Bogdanov I.I.
In a $2005 \times 2006$ table, numbers $0,1,2$ are arranged such that the sum of the numbers in each column and each row is divisible by 3.
What is the maximum possible number of ones that can be in this table?
# | Evaluation. Let there be $n$ zeros and $d$ twos in the table. We have 2005 rows of length 2006 and 2006 columns of length 2005. For the sum in a row to be divisible by 3, there must be at least one two or two zeros. Hence, $d + n / 2 \geq 2005$. Similarly, in each column, there must be at least one zero or two twos, so... | 4019356 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,067 |
What is the maximum number of natural numbers not exceeding 2016 that can be marked so that the product of any two marked numbers is a perfect square?
# | Since $44^{2}1$, we will get a set of fewer numbers, as $1936 q>2016$.
## Answer
44 numbers.
\section*{Problem 65909 Topics:
Germann and Chekalin laid out 13 different cards on the table. E... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,068 |
[ Pairings and groupings; bijections ]
$$
\text { [ Theory of algorithms (miscellaneous). ] }
$$ Author: Shrapovalov A.B. Among 49 schoolchildren, each is acquainted with at least 25 others. Prove that they can be divided into groups of two or three people such that each person is acquainted with everyone in their ... | Let's assume that initially all 49 students are standing in the corridor, and we will gradually let them into the classroom. We will do this in such a way that at any moment, the students in the classroom are divided into the required groups. Let's say that Fyodor, a student, is standing in the corridor. If he is acqua... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,070 |
[ The ratio of the areas of triangles with a common base or common height ]
In one island tribe, there is a custom - during a ritual dance, the shaman tosses three thin straight rods of the same length, tied in a shape resembling the letter П, high into the air. Adjacent rods are connected by a short thread and can th... | A broken line formed by rods is called $A B C D$. Let $\alpha=\angle A B C$ be the angle between the first and second links, and $\beta=\angle B C D$ be the angle between the second and third (see the left figure). We can assume that $0 \leq \alpha \leq \pi$ and then $0 \leq \beta \leq 2 \pi$. The elementary outcomes a... | \frac{11}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,071 |
$4-$ $[$ Pigeonhole Principle (miscellaneous). ] Authors: Shapovalov A.V., Usov S.B. One hundred gnomes, whose weights are $1,2,3, \ldots, 100$ pounds, gathered on the left bank of a river. They cannot swim, but on the same bank, there is a rowing boat with a capacity of 100 pounds. Due to the current, it is difficul... | Let's number the gnomes by their weight. Suppose the gnomes managed to cross the river. Let's call the trips from the left bank to the right direct, and from the right bank to the left - reverse.
First method. Let there be $k$ reverse trips; then there were $k+1$ direct trips.
In $k$ reverse trips, $k$ different gnom... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,074 |
Bogognov i.i.
In some cells of a $200 \times 200$ square, there is one chip - either red or blue; the other cells are empty. One chip sees another if they are in the same row or column. It is known that each chip sees exactly five chips of the other color (and possibly some chips of its own color). Find the maximum po... | Example. Let's highlight a "border" of width 5 around a 200×200 square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place 3800 chips in these four rectangles: in the left and top ones - red, and in the right and bottom ones - blue. It is easy to see that all the requirem... | 3800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,075 |
Tolesnikov A. Several points are marked on a plane, and not all of these points lie on the same line. A circle is circumscribed around each triangle with vertices at the marked points. Can the centers of all these circles coincide with the marked points? | Consider the smallest of the circles $\omega$. Let it be circumscribed around triangle $A B C$, and $O$ - its center. If triangle $A B C$ is not equilateral, then one of its angles, for example, angle $C$, is less than $60^{\circ}$. But then $\angle C < \sin \angle C$. By the Law of Sines, the radius of the circumscrib... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,076 |
Evochkimov M.A.
a) Can a sphere of some radius intersect the faces of a regular tetrahedron in circles of radii 1, 2, 3, and 4?
b) The same question for a sphere of radius 5. | a) Let's take a regular tetrahedron such that the sphere touching its edges cuts out circles on the faces with a radius greater than 4. Then we will move one of the faces away from the center of the sphere until a circle of radius 4 is cut out in this face. The tetrahedron will remain regular, and the radii of the circ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,077 |
[^0]
There were 100 doors, each with its own key (opening only that door). The doors are numbered $1, 2, \ldots, 100$, and the keys are also numbered, but possibly with errors: the key number matches the door number or differs by 1. In one attempt, you can choose any key, any door, and check if this key fits this door... | a) Let's try if the first key fits the first door. If it does, the first key is identified. If not, it fits the second door, and the second key fits the first door. In this case, two keys are identified. We take the next unidentifiable key from the left and check it against the corresponding door, identifying at least ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,078 |
Aeto: : Let $M$ be a set. Do there exist 2018 positive irreducible fractions with distinct natural denominators such that the denominator of the difference between any two of them (after reducing to the simplest form) is less than the denominator of any of the original 2018 fractions? | Consider the fractions $\$ 1$ dfrac $\{1+q\}\{1\}, \backslash \operatorname{dfrac}\{2+q\}\{2 q\}, \ldots$, , dfrac $\{2018+q\}\{2018 q\}, \$$ where $\$ q=2018!+1 \$$. They are irreducible, since $\$(q, i)=1 \$$ for $\$ 1$ \eqslant $\mathrm{i}$ leqslant $2018 \$$.
The difference of such fractions is $\$ 1 d f r a c\{i+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,079 |
Let $M$.
On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 residents were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The residents answered in turn, and everyone could hear their answers. Knights answered... | Let's call knights and liars principled people.
Evaluation. First method. (Buchaev Abdulkadyr) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values of the balance divide the sequ... | 1009 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,080 |
Kenodarov R.g.
A board $7 \times 7$ is either empty or has an invisible ship $2 \times 2$ placed on it "by cells". It is allowed to place detectors in some cells of the board, and then turn them on simultaneously. An activated detector signals if its cell is occupied by the ship. What is the smallest number of detecto... | Evaluation. In each rectangle $2 \times 3$, there should be at least two detectors: the rectangle consists of three dominoes $1 \times 2$ (strips), and if a detector is in the edge domino, we might not know if there is a ship on the other two dominoes, and if a detector is in the middle domino, we might not know which ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,081 |
Chernyaev N.l.
On a checkered board, dominoes lie without touching even at the corners. Each domino occupies two adjacent (by side) cells of the board. The lower left and upper right cells of the board are free. Is it always possible to move from the lower left cell to the upper right cell, making moves only up and to... | On the left, the figure shows the placement of dominoes on a board $\$ 6 \mid$ times $7 \$$, which does not allow a path from the bottom-left cell to the top-right cell. Indeed, it is impossible to enter the (gray) area to the right of the lowest domino, as we must first move above the first domino, and then we are alr... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,082 |
Salimov.
The first derivative of an infinite sequence $a_1, a_2$, ... is the sequence $a'_n = a_{n+1} a_n$ (where $n = 1, 2, \ldots$), and its $k$-th derivative is the first derivative of its $(k-1)$-th derivative ($k=2, 3, \ldots$). A sequence is called good if it and all its derivatives consist of positive numbers. ... | Let \(c_n = a_nb_n\). Then \(c'_n = a_{n+1}b_{n+1} - a_nb_n = a_{n+1}(b_{n+1} - b_n) + b_n(a_{n+1} - a_n) = a_{n+1}b'_n + b_na'_n\). Since all the terms in the sum are positive, the first derivative of \(c_n\) (and of the product of any two good sequences) consists of positive numbers. Moreover, we have represented \(c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 55,083 |
Frankin B.R.
Some cells of a $100 \times 100$ board are painted black. In all rows and columns that contain black cells, the number of black cells is odd. In each row that contains black cells, we place a red chip on the middle black cell. In each column that contains black cells, we place a blue chip on the middle bl... | Remove all rows and columns from the board that do not contain black cells. Note that after this, the condition of the problem will still be satisfied. Now all black cells lie within some rectangle \(n \times m\), in each row and each column of which there is at least one black cell.
By the condition, there is exactly... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,084 |
Lukin $^{2}$.
Given the segment $[0,1]$. In one move, it is allowed to split any of the existing segments by a point into two new segments and write the product of the lengths of these two new segments on the board.
Prove that at no point will the sum of the numbers on the board exceed $1 / 2$. | We will write on the board the doubled products of the lengths and prove that their sum is less than 1. We will set up a second board, on which we will record the squares of all segment lengths. Initially, the number 1 is written on it. In the future, when a segment of length $\$ \mathrm{a} \$$ is divided into segments... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 55,085 |
Svyatov M.
On a plane, grasshopper Kolya and 2020 of his friends are sitting. Kolya is going to jump over each of the other grasshoppers (in any order) such that the starting and ending points of each jump are symmetric relative to the grasshopper being jumped over. We will call a point a finishing point if Kolya can ... | Evaluation. Method 1. Introduce a coordinate system on the plane so that Kolya is sitting at the point \$(0,0) \$. Let \$lvec \{\mathrm{a}\} \_1 \$$, \$|vec \{\mathrm{a}\} \_2 \$$, \ldots, \$ \mid vec\{\mathrm{a}\}\{2020\} \$$ be the radius vectors of the grasshoppers in the order of their jumping over Kolya. It is not... | C_{2020}^{1010} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,086 |
Ivelev B.M.
For any natural number $n$, there exists a number composed of the digits 1 and 2 that is divisible by $2^{n}$. Prove this. (For example, 2 is divisible by 2, 12 is divisible by 4, 112 is divisible by 8, 2112 is divisible by 16...) | Let's prove that different $n$-digit numbers, composed of the digits 1 and 2, give different remainders when divided by $2^{n}$. Indeed, the difference $d$ between such numbers can be represented as $a_{n} \cdot 10^{n-1} + a_{n-1} \cdot 10^{n-2} + \ldots + a_{2} \cdot 10 + a_{1}$, where each of the numbers $a_{i}$ is e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,087 |
Auto: Rozenshtein B.
In the city, there is one blue square and $n$ green ones, and each green square is connected by streets to the blue one and to two green ones, as shown in the figure. On each of the $2 n$ streets, one-way traffic has been introduced such that you can drive into and out of each square. Prove that f... | Let's prove that
a) from each green square, you can drive to a blue one;
b) from a blue square, you can drive to any green one.
We will denote the direction of movement on the streets with arrows.
a) Suppose that from the green square \( A_{1} \) it is impossible to drive to a blue square. Since it is possible to l... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,090 |
Two people $A$ and $B$ need to get from point $M$ to point $N$, which is 15 km away from $M$, as quickly as possible. On foot, they can travel at a speed of 6 km/h. In addition, they have a bicycle that can be ridden at a speed of 15 km/h. $A$ sets out on foot, while $B$ rides the bicycle until meeting pedestrian $C$, ... | To ensure that $A$ and $B$ spend the least amount of time on the road, they should arrive at $N$ simultaneously, meaning they should walk the same distances. Indeed, if $A$ arrives at $N$ before $B$, $A$ can give part of his time to $B$: $A$ can get on the bicycle a little later, allowing $B$ to ride the bicycle a litt... | 1998 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,092 |
100 numbers, including positive and negative, are written in a row. Underlined are, first, each positive number, and second, each number whose sum with the next one is positive. Can the sum of all underlined numbers be negative? Equal to zero?
# | Answer: No, it cannot. We will underline positive numbers with one line, and negative numbers, the sum of which with the next number is positive, with two lines. It is clear that after each number $a$ underlined with two lines, there is a number $b$ underlined with one line, and $a+b>0$. The underlined numbers are divi... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,094 |
A radio tube has seven contacts arranged in a circle and plugged into a socket with seven holes. Can the contacts of the tube and the holes of the socket be numbered in such a way that, when the tube is plugged in, at least one contact will always be in its correct position (i.e., in the hole with the same number)? | Number the contacts of the lamp in order, and the holes of the plug in the reverse order. Then the contact with number $k$ falls into the hole with number
$a-k$, where $a-$ is a fixed number. (More precisely, we are talking not about the numbers themselves, but about their remainders when divided by 7.) We need to cho... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,096 |
## [ Systems of nonlinear algebraic equations ]
[ Completing the square. Sums of squares ]
Find all real solutions of the system of equations $\frac{2 x_{1}^{2}}{1+x_{1}^{2}}=x_{2}, \frac{2 x_{2}^{2}}{1+x_{2}^{2}}=x_{3}, \frac{2 x_{3}^{2}}{1+x_{3}^{2}}=x_{1}$. | If one of the numbers $x_{1}, x_{2}, x_{3}$ is zero, then the other numbers are also zero. Therefore, we will further assume that $x_{1} x_{2} x_{3} \neq 0$. Then the system can be written as $\frac{2}{x_{2}}=1+\frac{1}{x_{1}^{2}}, \frac{2}{x_{3}}=1+\frac{1}{x_{2}^{2}}, \frac{2}{x_{1}}=1+\frac{1}{x_{3}^{2}}$. Adding th... | (0,0,0)(1,1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,097 |
Draw $p$ rays from point $O$ on a plane so that the sum of all pairwise angles between them is the largest. (Only angles not exceeding $180^{\circ}$ are considered.) | There are many ways to draw rays to achieve the maximum sum of angles. For example, if $n=2 k$, then half of the rays should be drawn in one direction, and the remaining half in the opposite direction. Let the sum of pairwise angles for $n$ rays be denoted by $S(n)$, then $S(2 k)=k^{2} \cdot 180^{\circ}$. If $n=2 k+1$,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,098 |
Prove that for any three infinite sequences of natural numbers
$$
\begin{gathered}
a_{1} \ldots a_{\mathrm{n}} \ldots \\
b_{1} \ldots b_{\mathrm{n}} \ldots \\
c_{1} \ldots c_{\mathrm{n}} \ldots
\end{gathered}
$$
there exist indices \( p \) and \( q \) such that
$$
a_{\mathrm{p}} \geq a_{\mathrm{q}}, b_{\mathrm{p}} \... | Let's first prove that from any sequence of natural numbers, we can select a non-decreasing subsequence, i.e., one where each term is not less than the previous one. Let \( x_{1}, x_{2}, \ldots, x_{n}, \ldots \) be an arbitrary sequence of natural numbers. Since all terms of the sequence are non-negative, there exists ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,100 |
[ Ordering in ascending (descending) order. ]
From the numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, ten pairwise sums can be formed; denote them by $a_{1}, a_{2}, \ldots, a_{10}$. Prove that knowing the numbers $a_{1}, a_{2}, \ldots, a_{10}$ (but not knowing, of course, which two numbers each of them is the sum of), o... | Let's order the data and the sought numbers in ascending order: $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}, a_{1} \leq a_{2} \leq \ldots \leq a_{10}$. Since each number appears in exactly four sums, $a_{1}+\ldots+a_{10}=4\left(x_{1}+\ldots+x_{5}\right)$, i.e., the sum $x_{1}+\ldots+x_{5}$ can be restored from t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 55,101 |
A schoolboy during the school year must solve exactly 25 problems every 7 consecutive days. The time required to solve one problem (any problem) does not change throughout the day, but changes throughout the school year according to a known law to the schoolboy and is always less than 45 minutes. The schoolboy wants to... | Let $a_{1}, a_{2}, \ldots, a_{7}, a_{8}$ be the number of problems solved by a schoolboy over some consecutive 8 days. By the condition: $a_{1}+a_{2}+\ldots+a_{7}=25, a_{2}+a_{3}+\ldots+a_{8}=25$, hence $a_{1}=a_{8}$. Therefore, the number of problems solved by the schoolboy each day must repeat every 7 days. Thus, the... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 55,102 |
$\left[\begin{array}{ll}\text { Processes and Operations } & \\ {\left[\begin{array}{l}\text { Compositions of Symmetries }\end{array}\right]} \\ {\left[\begin{array}{l}\text { Evenness and Oddness }\end{array}\right]}\end{array}\right]$
Given a rectangular billiard table of size $26 \times 1965$ (the side of length 1... | Let's replace the reflection of the ball's trajectory with a billiard reflection. Then the ball will move along the straight line $y=x$ in a plane, divided into rectangles $26 \times 1965$. Pockets correspond to points with coordinates ( $1995 m, 26 n$ ). Due to the mutual simplicity of the numbers 1995 and 26, the bal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,103 |
In each cell of a square table of $m \times m$ cells, there is either a natural number or zero. If a zero is at the intersection of a row and a column, then the sum of the numbers in the "cross" consisting of this row and this column is not less than $m$. Prove that the sum of all numbers in the table is not less than ... | Consider all rows (both rows and columns) and select the row with the smallest sum of numbers $s$. If $s \geq m / 2$, then the required inequality is obvious.
Let $s < m / 2$. In each of the $k$ columns corresponding to the zeros in the original row, the sum of the numbers is at least $m-s$, and in each of the other c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,104 |
The Spanish king decided to rearrange the portraits of his predecessors in a round tower of the castle in his own way. However, he wants to swap only two adjacent portraits at a time, and these should not be portraits of two kings, one of whom reigned immediately after the other. Moreover, he is only concerned with the... | Let's number the portraits of the kings in the order of their reign: $1,2, \ldots, n$. We will show how the king can arrange the portraits from any initial order to the order $1,2, \ldots, n$ in a clockwise direction (we will call this order the standard order). First, we align the first portrait with the second (movin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,106 |
Can all non-negative integers be divided into 1968 non-empty classes such that each class contains at least one number and the following condition is met: if a number $m$ is obtained from a number $n$ by erasing two adjacent digits or identical groups of digits, then both $m$ and $n$ belong to the same class (for examp... | The first method. Let the symbol $\approx$ denote belonging to the same class. $\overline{M a b N} \approx \overline{M b b a b N} \approx \overline{M b a a b a b N} \approx \overline{M b a N}$. Thus, numbers obtained by swapping adjacent digits belong to the same class. Since any permutation can be represented as a seq... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,107 |
Given a 29-digit number $X=\overline{a_{1}} \overline{-} \overline{-} \overline{29}\left(0 \leq a_{k} \leq 9, a_{1} \neq 0\right)$. It is known that for every $k$ the digit $a_{k}$ appears $a_{30-k}$ times in the number (for example, if $a_{10}=7$, then the digit $a_{20}$ appears seven times). Find the sum of the digit... | Let's divide all numbers into pairs $\left(a_{k}, a_{30-k}\right)$. Notice that if among these pairs there is a pair $(a, b)$, then pairs $(a, c)$ and $(c, a)$, where $c \neq b$, cannot occur, otherwise the digit $a$ would have to appear in the number $X$ $b$ times on one side and $c$ times on the other. Therefore, if ... | 201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,108 |
The sum of 100 natural numbers, each of which is no more than 100, is 200.
Prove that from them, several numbers can be chosen whose sum is 100.
# | Note that among our numbers, at least two are greater than 1 (if 99 numbers are equal to 1, then the last one equals 101, which contradicts the condition). Therefore, we can divide the numbers into two groups of 50 numbers each, such that the sum of the numbers in each group is greater than 50.
According to problem 10... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,109 |
In the cells of a $10 \times 20$ table, 200 different numbers are placed. In each row, the two largest numbers are marked in red, and in each column, the two largest numbers are marked in blue. Prove that at least three numbers are marked in the table with both red and blue colors.
# | If all marked numbers are bicolored, then there are at least 40 of them.
Let there be monochromatic marked numbers. Consider the largest of them, number A. Let it be blue (the reasoning for red is analogous). Then two red numbers in the same row are greater than $A$ and, therefore, bicolored.
Consider the marked numb... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,111 |
8,9,10 |
| | Arithmetic of residues (other). | |
Prove that $1^{1983}+2^{1983}+\ldots+1983^{1983}$ is divisible by $1+\ldots+1983$. | Let's prove that for any odd $n=2m-1$, the sum $S=1^{n}+2^{n}+\ldots+n^{n}$ is divisible by $1+2+\ldots+n=n m$. Since $n$ and $m$ are coprime, it is sufficient to check that $S$ is divisible by $n$ and by $m$.
$S=\left(1^{n}+n^{n}\right)+\left(2^{n}+(n-1)^{n}\right)+\ldots+\left((m-1)^{n}+(m+1)^{n}\right)+m^{n}$. The ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,113 |
It is known that in a code lock, only the buttons with numbers 1, 2, and 3 are functional, and the code for this lock is three digits long and does not contain any other digits. Write the shortest sequence of digits that will definitely open this lock (the lock opens as soon as the correct three digits are pressed in t... | There are $3^{3}=27$ different three-digit numbers, in the notation of which the digits 1, 2, 3 are used. Besides the first two digits in the sequence of button presses, each of the other digits serves as the last digit of some three-digit number. Therefore, the required sequence must contain no fewer than $27+2=29$ di... | 11123222133313121223113233211 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,116 |
In a class, a tug-of-war tournament is being organized. In the tournament, all possible teams that can be formed from the students of this class (except for the team of the entire class) should participate exactly once. Prove that each team of students will compete against the team of all other students in the class. | Let's call two teams complementary if one of them consists of all the students who did not join the other. Each student in the class belongs to only one of the two complementary teams, which means he or she is in exactly half of all the teams. But half the number of all teams is equal to the number of all upcoming comp... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,117 |
Konyagin S. **v.**
Can the set of integers be partitioned into three subsets such that for any integer value $n$, the numbers $n, n-50, n+1987$ belong to three different subsets? | Answer: No. We will prove this by contradiction. Suppose the partition specified in the condition exists. We will write $m \sim k$ if the integers $m$ and $k$ belong to the same subset of the partition, and $m \approx k$ if they do not. We will prove that
$$
n \sim n+1937 \text{ and } \quad n \sim n-150
$$
for any in... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,118 |
[ Pairings and groupings; bijections ] [ Pascal's Triangle and the Binomial Theorem ]
How many odd natural numbers \( n < 10000 \) are there for which the number formed by the last four digits of \( n^9 \) is greater than \( n \), and how many are there for which it is less than \( n \)? | Let $N$ be the number formed by the last four digits of the number $n^{9}$, where $n$ is an odd number less than 10000. Note that the number formed by the last four digits of the number
$(10000-n)^{9}=10000^{9}-C_{9}^{1} \cdot 10000^{8} n+\ldots+C_{9}^{8} \cdot 10000 n^{8}-n^{9}$, coincides with the number $10000-N$. ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,120 |
In the center of a square pie, there is a raisin. From the pie, you can cut off a triangular piece along a line that intersects two adjacent sides at points other than the vertices; from the remaining part of the pie - the next piece (in the same way) and so on. Can you cut out the raisin?
# | In the solution to problem 79618, it is proven that a raisin can be cut out if and only if it can be cut out with one cut. Therefore, a raisin located at the center of a square cake cannot be cut out.
## Answer
cannot. | cannot | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,121 |
[ Geometry on grid paper] [ Pigeonhole Principle (miscellaneous) ]
Authors: Perermkov D.A., Skoopnnoo A.B.
A board of size $2005 \times 2005$ is divided into square cells with a side length of one unit. Some cells of the board are numbered in some order with the numbers $1,2, \ldots$ such that within a distance of le... | Consider a checkerboard square with a side length of 105, which we will call the large square. We will divide it into 25 smaller squares of size $21 \times 21$.
A numbered cell, located at a distance of less than 10 from the central cell of some small square, is within that small square. Since the small squares do not... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,122 |
Folklore
Eight volleyball teams held a round-robin tournament (each team played each other once).
Prove that it is possible to select four teams $A, B, C$, and $D$ such that $A$ won against $B, C$, and $D$; $B$ won against $C$ and $D$; and $C$ won against $D$. | The total number of losses equals the total number of wins; each team played seven matches. Therefore, the average number of wins for a team is 3.5, which means at least one team won no fewer than four matches. For each of the four teams that lost to this team, the average number of wins is 1.5 (considering only the ma... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,123 |
Folklore
In the cells of a chessboard, numbers from 1 to 64 are written (the first row is numbered from left to right with numbers from 1 to 8, the second row from 9 to 16, and so on). Before some numbers, plus signs are placed, and before the others, minus signs, such that in each row and each column there are four p... | | 57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 |
| 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 |
| 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |
| 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 |
| 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |
| ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,124 |
Fomin d:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 200 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 200 can be balanced by some number of weights from the set, and in a unique way (the b... | The correct set should correspond to the factorization of the number 201 (see the solution of problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.
## Answer
a) Two weights of 67 g and 66 weights of 1 g or 66 weights of 3 g and two of 1 g.
b) 3 sets. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,125 |
Fomin D:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 500 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 500 can be balanced by some number of weights from the set, and in a unique way (the b... | Let the largest weight of a weight in some correct set be $M$ (grams). This means that any smaller weight can be balanced by smaller weights. Let the weight of all smaller weights be $m$. Clearly, $m \geq$ $M-1$. But if $m \geq M$, then we have two ways to balance the weight $M+r$, where $r$ is the remainder of the div... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,126 |
Herrer M.L. Let $S(n)$ denote the sum of the digits of the number $n$ (in decimal notation). Do there exist three distinct natural numbers $m, n$, and $p$ such that $m+S(m)=n+S(n)=p+S(p)$? | Here is one of such triples:
$m=9999999999999, S(m)=117 ; \quad n=10000000000098, S(n)=18 ; \quad p=10000000000107, S(p)=9$.
For these numbers $m+S(m)=n+S(n)=p+S(p)=10000000000116$.
## Answer
They exist. | 10000000000116 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,127 |
дубин A.
Is there a sphere on which there is exactly one rational point? (A rational point is a point where all three Cartesian coordinates are rational numbers.)
# | Consider the sphere with the equation $(x-\sqrt{2})^{2}+y^{2}+z^{2}=2$. The point $O$ with coordinates $(0,0,0)$ is a rational point lying on this sphere. Suppose there is another rational point $(x, y, z)$ on the sphere, with $x \neq 0$, since when $x=0$ there is only one solution $(0,0,0)$. Expanding the brackets:
$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,128 |
Shaovalov A.V.
On the coordinate plane, some points with integer coordinates are marked. It is known that no four of them lie on the same circle. Prove that there exists a circle of radius 1995 that does not contain any of the marked points. | Suppose that such a circle does not exist, that is, inside each circle of radius 1995, there is a marked point. Consider an infinite horizontal strip of width 4000. Divide it into squares of $4000 \times 4000$. Consider 4001 such squares in a row, and the rectangle formed by them is denoted by P. In each square, there ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,129 |
Shen A.H.
The seats for spectators along the ski track are numbered in order: $1,2,3, \ldots, 1000$. The ticket seller sold $n$ tickets for all the first 100 seats, but $n$ is greater than 100, as she sold more than one ticket for some seats (while $n<1000$). Spectators enter the track one by one. Each, upon approachi... | For each $i$ from 1 to $n$, let $k_{\mathrm{i}}$ denote the number of tickets sold for seats from the 1st to the $i$-th. Clearly, after everyone is seated, the first $n$ seats will be occupied. For each $i$ from 1 to $n$, when transitioning from the $i$-th seat to the ($i+1$)-th seat, there will be exactly $k_{i}-i$ si... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,130 |
Senderov V.A.
Prove that there are infinitely many such triples of numbers $n-1, n, n+1$ such that:
a) $n$ can be represented as the sum of two squares of natural (positive integer) numbers, while $n-1$ and $n+1$ cannot;
b) each of the three numbers can be represented as the sum of two squares of natural numbers. | We will say that a number is representable if it is the sum of the squares of two natural numbers.
a) The first method. A number of the form $4 t+3$ is not representable; a number that is a multiple of 3 but not a multiple of 9 is also not representable. The number $n=2 \cdot 100^{m}=\left(10^{m}\right)^{2}+\left(10^{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,131 |
Shapovalov A.V.
a) A square is cut into equal right-angled triangles with legs of 3 and 4 each. Prove that the number of triangles is even.
b) A rectangle is cut into equal right-angled triangles with legs of 1 and 2 each. Prove that the number of triangles is even. | a) The side of the square is composed of segments of lengths $3, 4, 5$, so its length $a$ is an integer. The area of the triangle is 6, hence the number of triangles $k = a^2 / 6$. Therefore, $a$ is even, $a^2$ is divisible by 4, and $k$ is even.
b) The perimeter of each triangle is $3 + \sqrt{5}$. Let's calculate the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,132 |
Shapovalov A.V.
In a regular 25-gon, all diagonals are drawn. Prove that there are no nine diagonals passing through one interior point of the 25-gon. | Let's denote the 25-gon as $A_{1} A_{2} \ldots A_{24} A_{25}$. We will call the index of the diagonal $A_{i} A_{j}(i<j)$ the number $\min \{j-i, 25-j+i\}$. The index of the diagonal $A_{i} A_{j}$ passing through the point where nine diagonals intersect can take values $9, 10, 11$, and 12, since between points $A_{i}$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,133 |
Pevzner and.
The vertices of a 50-gon divide the circle into 50 arcs, the lengths of which are 1, 2, 3, ..., 50 in some order. It is known that each pair of "opposite" arcs (corresponding to opposite sides of the 50-gon) differ in length by 25. Prove that the 50-gon will have two parallel sides. | Let's reduce the circle by 25 times. Let $a_{k}$ be the $k$-th side, and $L_{k}$ be the length of the corresponding arc (we number around the circle and allow $k>50$, implying $a_{k-50}$ instead of $a_{k}$, etc.). The sides $a_{k}$ and $a_{k+25}$ are parallel when the arcs between them are equal, that is, when the numb... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,137 |
Can a $2003 \times 2003$ board be tiled with $1 \times 2$ dominoes, which are allowed to be placed only horizontally, and $1 \times 3$ rectangles, which are allowed to be placed only vertically? (Two sides of the board are considered horizontal, and the other two are considered vertical.) | We will write the digit 1 in all cells of odd columns and the digit 2 in all cells of even columns. The sum of the digits in each rectangle of size $1 \times 2$ or $3 \times 1$ is divisible by 3. However, the sum of the digits on the entire board is not divisible by 3. Therefore, it is impossible to divide it into such... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,138 |
In a sports club, 100 overweight people are training, weighing from 1 to 100 kg. What is the smallest number of teams they can be divided into so that no team has two overweight people, one of whom weighs twice as much as the other?
# | Consider all pairs of fat people who should not end up on the same team. Additionally, it can be assumed that all 100 fat people have different weights.
## Solution
Note that the presence of fat people with the same weight only simplifies the problem. Indeed, with each fat person, you can place in the team all fat pe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,139 |
Sharievn I.f.
There are nine wrestlers of different strengths. In any match between any two of them, the stronger one always wins. Is it possible to divide them into three teams of three wrestlers each so that in the matches between the teams on a round-robin system, the first team wins more matches than the second, t... | Let's assign each of the fighters a "rating" from 9 to $1: 9$ - the strongest, and so on. Try to form teams so that the sums of the ratings of the fighters in the teams are equal.
## Solution
We will order our fighters by strength and assign each a rating from 9 to 1. Then the sum of the ratings of the fighters is 45... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,140 |
Kanel-Belov A.Y.
On a line, there are two chips, a red one on the left and a blue one on the right. It is allowed to perform any of two operations: inserting two chips of the same color consecutively at any point on the line, and removing any two adjacent chips of the same color. Is it possible to leave exactly two ch... | Let's replace the red chips with zeros, the blue chips in even positions (counting from the left) with ones, and those in odd positions with minus ones. It is easy to see that the allowed operations do not change the sum of these numbers. In the initial position, this sum was 1, so it will always be 1. Therefore, it is... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,142 |
Gaikov S.B.
The grandfather of Baron K.F.I. von Munchausen built a square castle, divided it into 9 square halls, and placed an arsenal in the central hall. The father of the baron divided each of the eight remaining halls into 9 equal square foyers and arranged winter gardens in all the central foyers. The baron hims... | First, let's walk around one hall, for example, clockwise (left figure). Consider the adjacent hall and walk around it similarly. To walk around both halls, it is sufficient to change the direction of movement in two pairs of cells along the boundary of the halls (center figure). Consider the next hall that borders the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,143 |
$\underline{\text { Proposition } A}$.
A perfect number greater than 28, divisible by 7, is also divisible by 49. | Suppose a perfect number is equal to $7 n$, where $n$ is not divisible by 7. Then all natural divisors of the number $7 n$ (including itself) can be divided into pairs $d$ and $7 d$, where $d$ is not divisible by 7. Consequently, the sum of all divisors of the number $7 n$ (which is equal to 14n) is divisible by 8. Hen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,149 |
Bveroo s.l.
In a company of $2 n+1$ people, for any $n$ people, there exists a person different from them who is acquainted with each of them.
Prove that in this company, there is a person who knows everyone. | Obviously, there are two acquaintances, and if there are $k$ pairwise acquaintances (where $k \leq n$), then by the condition, there will be a person different from them, who is acquainted with all these $k$ people. From this, it follows that there will be $n+1$ pairwise acquaintances: $A_{1}, \ldots, A_{n+1}$. Conside... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,150 |
Doinnikov V.L.
On a line, there are $2 k-1$ white and $2 k-1$ black segments. It is known that any white segment intersects with at least $k$ black segments, and any black segment intersects with at least $k$ white segments. Prove that there exists a black segment that intersects with all white segments, and a white s... | It is sufficient to prove the following statement: if any white segment intersects with at least $k$ black segments, then there exists a black segment that intersects with all white segments.
Assume the opposite. Choose for each black segment a white segment that does not intersect with it. Such a white segment lies e... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,152 |
Auto: Dolnikov V.L.
In each cell of a square table of size $n \times n$ cells ($n \geq 3$), a number 1 or -1 is written. If you take any two rows, multiply the numbers standing above each other, and sum the $n$ resulting products, the sum will be equal to 0. Prove that the number $n$ is divisible by 4. | Note that if we change the sign of all numbers in one column, the property of the table is preserved. The same is true for swapping two columns. Therefore, we can achieve that in the first row, only ones are present, and, by the property of the table for the first and second rows, $n=2 m$. By swapping columns, we can m... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,153 |
$\underline{\text { Rubanov I.S. }}$
Microcalculator MK-97 can perform only three operations on numbers stored in memory:
1) check if two selected numbers are equal,
2) add selected numbers,
3) find the roots of the equation $x^{2}+a x+b=0$ for selected numbers $a$ and $b$, or display a message if there are no roots.... | By adding $x$ to itself, we get $2x$. We compare $x$ and $2x$. If they are equal, then $x=0$. Otherwise, we find the roots of the equation $y^{2}+2xy+x=0$. The discriminant of this equation is $4(x^{2}-x)$, so the roots are equal if and only if $x=1$. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,154 |
Eemenyakov L.A.
A whole number is written on the board. Its last digit is remembered, then erased, and, multiplied by 5, added to the number that remains on the board after erasing. Initially, the number $7^{1998}$ was written. Can the number $1998^{7}$ result after applying several such operations? | Let's see how the remainder of the number when divided by 7 changes under this operation. Let $b$ be the last digit of the number. Then the number has the form $10a + b$, and after applying the operation, it becomes $a + 5b$. Since $5(10a + b) - (a + 5b) = 49a$, the remainder is multiplied by 5. Since the original numb... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,155 |
Point Systems
On a plane, there are 25 points; it is known that among any three points, two can be chosen such that the distance between them is less than 1. Prove that among the given points, there are 13 that lie within a circle of radius 1.
# | Let $x$ be any of the given 25 points. If all points are within a distance of less than 1 from $x$, there is nothing to prove. Otherwise, let $y$ be any point whose distance from $x$ is greater than or equal to 1. Let $z$ be any of the remaining 23 points. In the triangle $xyz$, by the condition, there is a side that i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,157 |
Bernicheyni i. A square sheet of paper was cut into two parts along a straight line. One of the resulting parts was again cut into two parts, and so on, many times. What is the minimum number of cuts needed so that among the resulting parts there could be exactly 100 twenty-sided polygons? | Answer: It is possible to obtain exactly 100 20-sided polygons with 1699 cuts, and it is impossible to obtain 100 20-sided polygons with fewer cuts. With each cut, the total number of paper pieces increases by 1 (since one piece is cut into two new pieces), so after $n$ cuts, there will be ($n+1$) pieces of paper. Now ... | 1699 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,158 |
Can four lead balls and a point light source be placed in space so that every ray of light emanating from the source intersects at least one of the balls? | Answer: Yes, it is possible. Let the light source be located at the center $O$ of a regular tetrahedron $ABCD$. Consider the trihedral angle formed by the rays $OA$, $OB$, and $OC$. Construct a sphere that intersects the rays $OA$, $OB$, and $OC$ and does not contain the point $O$. Such a sphere clearly exists: for exa... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,159 |
Do there exist three non-zero digits that can be used to form an infinite number of decimal representations of squares of different integers?
# | Answer: Yes, they do exist. For example, the digits 1, 5, 6 are suitable. Indeed, let's prove that $\underbrace{3 \ldots 3}_{n \text { threes }} 4^{2}=$ $\underbrace{1 \ldots 1}_{n+1} \underbrace{5 \ldots 5}_{n} 6$. For this, note that $\underbrace{3 \ldots 3}_{n \text { threes }} 5 \cdot 3=\underbrace{0 \ldots 0}_{n} ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,160 |
9,10 A square field is divided into 100 equal plots, 9 of which are overgrown with weeds. It is known that weeds spread to those and only those plots that have at least two adjacent plots already affected by weeds (plots are adjacent if they share a common side). Prove that the entire field will not be completely overg... | It is easy to check that the length of the boundary of the entire overgrown area (or several areas) does not increase. At the initial moment, it does not exceed $9 \cdot 4=36$, therefore, at the final moment, it cannot be equal to 40. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,161 |
What is the smallest number of weights in a set that can be divided into 4, 5, and 6 equal-mass piles? | Let's first prove that the set cannot contain fewer than eleven weights. Suppose this is not the case, that is, there are no more than ten. Let \(60 \mathrm{~m}\) be the total mass of all weights in the set. First, note that the set cannot contain weights with a mass greater than \(10 \mathrm{~m}\) (since the set can b... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,163 |
[ Completing the square. Sums of squares ]
[ Substitution of variables (other) $\quad]$
Find all real solutions of the equation
$$
36 \sqrt{x-2}+4 \sqrt{y-1}=28-4 \sqrt{x-2}-\sqrt{y-1}
$$ | From the condition, it is clear that $x > 2, y > 1$. Let's introduce the notations: $\sqrt{x-2}=a^{2}, \sqrt{y-1}=b^{2}$. Our equation can be rewritten as: $36 / a^{2}+4 / b^{2}+4 a^{2}+b^{2}-28=0$. We can break the constant term into two addends and regroup the equations:
$$
\left(36 / a^{2}-24+4 a^{2}\right)+\left(4... | 11,5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 55,165 |
фзestiev I.V.
There are three piles of stones. Sisyphus carries one stone from one pile to another. For each transfer, he receives from Zeus an amount of coins equal to the difference in the number of stones in the pile where he places the stone and the number of stones in the pile from which he takes the stone (the s... | We will call stones from the same pile acquaintances, and from different piles - strangers. Then Sisyphus's income for one relocation is equal to the change in the number of pairs of acquainted stones. Since in the final moment all stones ended up in their original piles, the total change in the number of acquaintances... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,166 |
A labyrinth is called a 10*10 grid, some pairs of adjacent nodes in which are connected by a segment - a "wall" in such a way that by moving from one cell to an adjacent one along a side and not passing through walls, one can visit all cells of the square. The boundary of the square will also be considered to be surrou... | Let's enumerate all initial states (determined by the maze and the initial position of the robot). For each state, we will write a program that is a continuation of the program for the previous state.
## Solution
Consider all initial states. Each state is defined by the maze and the initial position of the robot in t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,170 |
If the difference between the largest and smallest of $n$ given real numbers is $d$, and the sum of the absolute values of all $n(n-1) / 2$ pairwise differences of these numbers is $s$, then
$$
(n-1) d \leq s \leq n^{2} d / 4
$$
Prove this. | Solution. Let's place the points $a_{1}, a_{2}, \ldots a_{n}$ on the number line. Then $d$ is the distance between the extreme points, the leftmost and the rightmost, and $s=\sum_{i}\left|a_{i} a_{j}\right|$ is the sum of all pairwise distances between these points. Clearly, we can assume that the points are labeled as... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 55,171 |
Vasilev N.B.
In an infinite chain of nerve cells, each can be in one of two states: "rest" and "excitement." If a cell is excited at a given moment, it sends a signal that reaches both neighboring cells after one unit of time (say, one millisecond). A cell becomes excited if and only if a signal comes from one of its ... | Solution. It is sufficient to trace how the excitation spreads from a single cell for the first ten to fifteen cycles (see fig.) to notice the following patterns. 1. At time $t=2^{k}$, where $k=0$, $1,2, \ldots$ only two cells are excited: $x=-2^{k}$ and $x=2^{k}$. 2. At time $t=2^{k}-1$ $2^{k}$ cells from $x=-2^{k}+1$... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,172 | |
[ $\left.\begin{array}{c}\text { Induction (other) } \\ {\left[\begin{array}{l}\text { Irrational inequalities }\end{array}\right]}\end{array}\right]$
Given:
$$
a_{1}=1966, a_{\mathrm{k}}=\left[\sqrt{a_{1}+a_{2}+\cdots+a_{k-1}}\right]
$$
Find $a_{1966}$. | We will prove by induction that the initial segment of our sequence is $1966,44,44,45,45, .$. , where after the number 1966, each natural number greater than 43 appears exactly twice in a row, except for numbers of the form $74 \cdot 2^{k}$, which appear three times in a row (compare with the solution to problem 78594)... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,175 |
Two sages are playing the following game. The numbers $0,1,2, \ldots, 1024$ are written down. The first sage crosses out 512 numbers (of his choice), the second crosses out 256 of the remaining, then the first crosses out 128 numbers, and so on. On the tenth step, the second sage crosses out one number; two numbers rem... | Answer: with correct play, the difference between the remaining numbers is 32 (as they say, the "price of the game" is 32). The following considerations lead to this answer: the first player aims to maximize the difference between the remaining numbers and should always try to "thin out" the remaining set of numbers as... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,177 |
Konizin S.V.
Prove that in a convex equilateral (but not necessarily regular) pentagon, a regular triangle can be placed such that one of its sides coincides with a side of the pentagon, and the entire triangle lies inside this pentagon.
# | Suppose that for some equilateral convex pentagon $A B C D E$ with a side equal to one, the statement of the problem is false. We can assume that the largest diagonal is $A D$, that points $A$ and $D$ lie on a horizontal line ($D$ to the right of $A$), points $B$ and $C$ are in its upper half-plane, and $C$ is no close... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,178 |
[ Induction in Geometry $]$ $[$ Integer Lattices (miscellaneous). ]
## Authors: Buhretov A.I., Kanet-Bepov A.Ya.
On a meadow in the shape of a square, there is a circular pit. A grasshopper is jumping around the meadow. Before each jump, it chooses a vertex and jumps in the direction of this vertex. The length of th... | It is sufficient to prove the following statement. Let each side of the square have a length of 1 and be divided into $2^{\text {n }}$ equal parts ( $\mathrm{n} \geq 0$ ), and through the division points, draw lines parallel to the sides. Then the grasshopper can
reach any of the $4^{\mathrm{n}}$ resulting cells.
For ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,179 |
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