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$\underline{\text { Mitrofanov I.V. }}$
On a circle, $n$ points are marked, dividing it into $n$ arcs. The circle is rotated around its center by an angle of $2 \pi k / n$ (for some natural number $k$), as a result of which the marked points move to $n$ new points, dividing the circle into $n$ new arcs.
Prove that th... | Let's assume that the radius of the circle is 1, and the rotation occurred clockwise. We can also assume that if $kk$, then the sum of the lengths of all arcs of the form $A_{i} A_{i+1} \ldots A_{i+j-1}$ is $2 \pi(j-1) \geq 2 \pi k$. On the other hand, it is less than the sum of the lengths of all arcs $A_{i} B_{i}$, w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,630 |
Petya and Vasya came up with ten polynomials of the fifth degree. Then Vasya sequentially called out natural numbers (starting from some number), and Petya substituted each called number into one of the polynomials of his choice and wrote down the obtained values on the board from left to right. It turned out that the ... | Evaluation. Since a fifth-degree equation has no more than five roots, it is similarly proven as in problem $\underline{64349}$ that the total number of numbers could not have been more than 50.
Example. Let the polynomials $P_{k}(x)=x+(x-5 k+4)(x-5 k+3) \ldots(x-5 k)$ for $k=1,2, \ldots, 10$ and
Vasya called out the... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,631 |
Frankin B.R.
Three circles touch each other externally and touch a fourth circle internally. Their centers were marked, and the circles themselves were erased. It turned out that it is impossible to determine which of the marked points is the center of the enclosing circle. Prove that the marked points form a rectangl... | Let $O$ and $R$ be the center and radius of the circumscribing circle, and $O_{1}, O_{2}, O_{3}$ and $R_{1}, R_{2}, R_{3}$ be the centers and radii of the other circles. Then
$OO_{i}=R-R_{i}(i=1,2,3), O_{i} O_{j}=R_{i}+R_{j}(i, j=1,2,3, i \neq j)$. From this, $O O_{1}-O_{2} O_{3}=O O_{2}-O_{3} O_{1}=O O_{3}-O_{1} O_{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,633 |
In a single-round-robin tournament, 10 chess players are participating. What is the minimum number of rounds after which a sole winner can be determined prematurely? (In each round, the participants are paired.
Win - 1 point, draw - 0.5 points, loss - 0).
# | Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one who has no less than 3 points. Since there are still 3 rounds ahead, the winner is stil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,634 |
[ $\left.\begin{array}{c}\text { Equations in integers } \\ \text { [Examples and counterexamples. Constructions }\end{array}\right]$
Do there exist natural numbers $a, b, c, d$ such that $a / b + c / d = 1, a / d + d / b = 2008$? | For example, $a=2009 \cdot(2008 \cdot 2009-1), b=2008 \cdot 2010 \cdot(2008 \cdot 2009-1), c=2008 \cdot 2009-1, d=2008 \cdot 2010$.
## Answer
They exist. | =2009\cdot(2008\cdot2009-1),b=2008\cdot2010\cdot(2008\cdot2009-1),=2008\cdot2009-1,=2008\cdot2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,637 |
Kozhevnikov P.
There are 2013 cards with the digit 1 written on them, and 2013 cards with the digit 2 written on them. Vasya forms a 4026-digit number from these cards. In one move, Petya can swap two cards and pay Vasya 1 ruble. The process ends when Petya gets a number that is divisible by 11. What is the maximum am... | Consider a 4026-digit number $A$, consisting of 2013 ones and 2013 twos. Let in this number, there are $k$ ones and $l=2013-k$ twos in the odd positions, then in the even positions there will be $k$ twos and $l$ ones (here $k$ can take any integer value from 0 to 2013). The difference between the sums of the digits in ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,638 |
Senderov V.A.
A hundred-digit natural number $n$ is called unusual if the decimal representation of the number $n^{3}$ ends with $n$, and the decimal representation of the number $n^{2}$ does not end with $n$. Prove that there exist at least two hundred-digit unusual numbers. | For example, such numbers are $n_{1}=10^{100}-1=99 \ldots 9$ and $n_{2}=5 \cdot 10^{99}-1$. Indeed, the numbers
$n_{1}^{3}-n_{1}=\left(n_{1}+1\right) n_{1}\left(n_{1}-1\right)=10^{100} n_{1}\left(n_{1}-1\right)$ and
$n_{2}^{3}-n_{2}=\left(n_{2}+1\right) n_{2}\left(n_{2}-1\right)=5 \cdot 10^{99} n_{2}\left(n_{2}-1\rig... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,639 |
Bogdanov I.I.
In the language of the AU tribe, there are two letters - "a" and "u". Some sequences of these letters are words, and each word has no fewer than one and no more than 13 letters. It is known that if you write down any two words in a row, the resulting sequence of letters will not be a word. Find the maxim... | If all sequences, the number of letters in which is not less than 7 and not more than 13, are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^{7}+\ldots+2^{13}=2^{14}-2^{7}$. It remains to show that this number is the largest possible.
The first method. T... | 16256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,640 |
Berrov S.L.
There are $10^{1000}$ natural numbers standing in a circle. Between every two adjacent numbers, their least common multiple is written.
Can these least common multiples form $10^{1000}$ consecutive numbers (arranged in some order)? | Let $n=10^{1000}$. Denote the initial numbers (in the order of traversal) as $a_{1}, \ldots, a_{n}$; we will assume that $a_{n+1}=$ $a_{1}$. Let $b_{i}=\operatorname{HOK}\left(a_{i}, a_{i+1}\right)$. Suppose that $b_{1}, \ldots, b_{n}$ are $n$ consecutive natural numbers.
Consider the highest power of two $2^{m}$ that... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,641 |
Marayev A.
Cubes of size $1 \times 1 \times 1$ are glued together to form a cube of size $3 \times 3 \times 3$. What is the maximum number of cubes that can be removed from it so that the remaining figure has the following two properties:
- from the side of each face of the original cube, the figure looks like a $3 \... | Example. The layers from bottom to top are shown in the figure. The 13 remaining cubes are marked in black. Each layer has cubes in all columns and rows. When the layers are overlaid, a black $3 \times 3$ square is formed. Therefore, the first condition is satisfied. The bottom and middle layers are connected and glued... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,643 |
Bogdanov I.I.
In a store, 21 white and 21 purple shirts are hanging in a row. Find the smallest $k$ such that for any initial order of the shirts, it is possible to remove $k$ white and $k$ purple shirts so that the remaining white shirts hang together and the remaining purple shirts also hang together. | First, let's show that \( k \), equal to 10, is sufficient.
First method. We will walk along the row of shirts and count the white and purple shirts separately. As soon as we count 11 of one color - let's assume, without loss of generality, purple - shirts, we will stop. Now we will remove all the white shirts that we... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,644 |
Gooovanovo A.C. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation... | By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
Then $a=\frac{m}{10^{30}-1}, b=\frac{n}{10^{30}-1}$. We also know t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,645 |
Bakayev E.B.
A regular tetrahedron has the property that for any two of its vertices, there is a third vertex that forms a regular triangle with these two. Are there other polyhedra that have this property? | Let's fix two points, $A$ and $B$. The set of points $X$ for which triangle $A B X$ is equilateral is a circle. Take points $C, D, E$ on this circle such that triangle $C D E$ is equilateral. The polyhedron with vertices $A, B, C, D, E$ has the required property. Indeed, every pair of its vertices belongs to one of the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,646 |
Bakayev E.V.
On the table lay a pile of silver coins. With each action, either one gold coin was added and the number of silver coins was recorded on the first sheet, or one silver coin was removed and
the number of gold coins was recorded on the second sheet. In the end, only gold coins were left on the table. Prove ... | Let there be $n$ silver coins at the beginning, and $m$ gold coins at the end. We will represent the situation at each moment by a point ( $k, l$ ) on the coordinate plane ( $k$ - the number of gold coins on the table, $l$ - the number of silver coins). Each action - a shift one unit to the right or down - will be repr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,647 |
Binkov A.d:
Is it true that there exist convex polyhedra with any number of diagonals? (A diagonal is a segment connecting two vertices of the polyhedron and not lying on its surface.) | Let's construct a convex polyhedron with $n$ diagonals. For $n=0$, any pyramid will do.
Let $n>0$. Take a ( $n+2$ )-gonal pyramid $S A_{1} \ldots A_{n+2}$. Construct outside it, on the face $S A_{n+1} A_{n+2}$ as the base, a pyramid $T S A_{n+1} A_{n+2}$ (so that all $n+4$ constructed vertices are in convex position).... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,648 |
On the computer screen, a certain finite sequence of zeros and ones has been generated. The following operation can be performed with it: the digit sequence "01" can be replaced by the digit sequence "1000". Can such a process of replacements continue indefinitely or will it inevitably come to an end at some point? | The first method. Each operation can be understood as a "move" of some unit: the unit swaps places with the zero in front of it, and then two more zeros appear after the zero. Suppose one such move is made every minute, and the process continues indefinitely. Then there will be units that have "moved" an infinite numbe... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,649 |
[ $[$ Processes and Operations $]$ [ Proof by Contradiction ]
There are three liter bottles and a 100 ml measure. The first bottle is empty, the second contains 700 ml of sweet tea, and the third contains 800 ml of sweet tea. In the second bottle, 50 g of sugar is dissolved, and in the third bottle, 60 g of sugar is ... | The initial concentration of sugar in the second tank is $50 / 700 = 1 / 14$, and in the third tank it is $-60 / 800 = 3 / 40$. Since $3 / 40 > 1 / 14$, after any transfers, the concentration of sugar in each non-empty tank will not exceed $3 / 40$. Note that after any transfers, the amount of tea in each tank is a mul... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,652 |
In the tournament, 11 chess players participated: 4 from Russia and 7 from abroad. Each player played two games with each other (win - 1 point, draw - 0.5 points, loss - 0 points). At the end of the tournament, it turned out that all participants scored a different number of points, and the total points scored by the R... | The number of games played in the tournament is $10 \cdot 11=110$. The same number of points were distributed, so both the Russians and foreigners scored a total of 55 points. The best Russian chess player could not have scored less than 14.5 points (otherwise, the total points scored by the Russians would not exceed $... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,654 |
4[ Proof by contradiction ]
Find all strictly increasing sequences of natural numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$, in which $a_{2}=2$ and $a_{n m}=a_{n} a_{m}$ for any natural numbers $n$ and $m$.
# | The sequence $a_{n}=n$ obviously satisfies the condition.
Suppose there is some other sequence. Since $a_{1}n$. If $n=2 m$, then $a_{n}=a_{2} a_{m}=2 m=n$. If $n=2 m+1$ ($m \geq$ 1), then $n<a_{n}<a_{n+1}=a_{2 m+2}=a_{2} a_{m+1}=2(m+1)=n+1$, that is, the integer $a_{n}$ lies between $n$ and $n+1$. In both cases, we ha... | a_{n}=n | Number Theory | proof | Yes | Yes | olympiads | false | 52,655 |
Goovanovo $A$.
Let $n>1$ be a natural number. We write down the fractions $1 / n, 2 / n, \ldots, (n-1) / n$ and reduce each to its simplest form; the sum of the numerators of the resulting fractions is denoted by $f(n)$. For which natural numbers $n>1$ do the numbers $f(n)$ and $f(2015 n)$ have different parity? | Let $n=2^{t} \cdot m$, where $t \geq 0$, and the number $m$ is odd. We will prove that the number $f(n)$ is even in exactly two cases: either $n$ is odd and $m \equiv 1(\bmod 4)$, or $n$ is even and $m \equiv 3(\bmod 4)$.
First method. Consider an arbitrary fraction $k / n$. If $k$ is divisible by $2^{t+1}$, then the ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,657 |
10,11}
On the skin of a Rhinoceros, there are folds - vertical and horizontal. If a Rhinoceros has $a$ vertical and $b$ horizontal folds on its left side, and $c$ vertical and $d$ horizontal folds on its right side, we will say that this Rhinoceros is in the state ( $a b c d$ ) or simply a Rhinoceros ( $a b c d$ ).
I... | Let's construct a graph where the vertices will represent the states of the Rhinoceros, and the directed edges will indicate possible transitions between states and their probabilities.
It turns out there are 8 states in total. No state is terminal - the Rhinoceros can transition from any state to any other. But will ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,658 |
Vanya wrote the number 1 on the board, and then several more numbers. As soon as Vanya writes the next number, Mitya calculates the median of the existing set of numbers and writes it down in his notebook. At some point, the numbers in Mitya's notebook are: $1 ; 2 ; 3 ; 2.5 ; 3 ; 2.5 ; 2 ; 2 ; 2 ; 5$.
a) What number i... | a) The first number is obviously one. Then, for the median of the first two numbers to be 2, the second number must be 3.
For clarity, let's write down Vanya's numbers. Vanya wrote the third number $a$: $13 a_{3}$. The median is now 3, which can only be if $a \geq 3$. Add another number: $13 a \geq 3 b$.
The median o... | )2;b)2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,659 |
Kozhevnikov P.A.
What is the minimum number of cells that need to be marked on a $15 \times 15$ board so that a bishop, from any cell on the board, attacks no fewer than two marked cells? (The bishop also attacks the cell it stands on.)
# | It is more convenient to solve the problem in general: we will show that on a board of size $(2 n+1) \times(2 n+1)(n>1)$, the answer is $4 n$. (The diagram corresponds to the case $n=3$.)
Example. Let's mark the cells along the perimeter of a rectangle $(2 n-1) \times 3$, as shown in the diagram. In this case, on the ... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,660 |
$\underline{\text { Folklore }}$
Two ten-digit numbers are called adjacent if they differ in only one digit in some place (for example, 1234567890 and 1234507890 are adjacent). What is the maximum number of ten-digit numbers that can be written down so that there are no adjacent numbers among them? | There are $9 \cdot 10^{9}$ ten-digit numbers. Let's divide them into $9 \cdot 10^{8}$ tens: each ten includes numbers that differ only in the last digit (for example, the first ten includes numbers from 1000000000 to 1000000009). In each ten, all numbers are adjacent, so no more than one number can be selected from it.... | 9\cdot10^{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,661 |
Kuzneuvv A.
A square box of candies is divided into 49 equal square cells. In each cell lies a chocolate candy - either black or white. In one sitting, Sasha can eat two candies if they are of the same color and lie in adjacent cells, either by side or by corner. What is the maximum number of candies that Sasha can de... | Evaluation. For the layout shown in the left image, none of the 16 black candies can be eaten, and out of the 33 white candies, no more than 32 can be eaten due to parity.
Algorithm. We will... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,664 |
Shapovalov A.V.
In a $10 \times 10$ table, 100 different numbers are written. In one move, you can select any rectangle composed of cells and rearrange all the numbers in it symmetrically relative to its center ("rotate the rectangle by $180^{\circ}$"). Is it always possible to achieve, in 99 moves, that the numbers i... | We will paint all numbers in red. On each move, we will rotate a certain red rectangle, after which we will repaint one number in green. Thus, green numbers will no longer move. We will maintain the following properties:
1) each green number is smaller than all red ones
2) in each row from left to right and in each co... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,667 |
[ Examples and counterexamples. Constructions ]
a) any number of the form $3 k-2$, where $k$ is an integer, is the sum of one square and two cubes of integers;
b) any integer is the sum of one square and three cubes of integers. | a) It is sufficient to note that $\$ \mathrm{k} \wedge 3-(\mathrm{k}+3)^{\wedge} 3+(3 \mathrm{k}+5)^{\wedge} 2=3 \mathrm{k}-2 \$$.
b) Subtract from the number to be represented such a cube (0, 1, or -1) so that the result has the form $3 k-2$, and apply a). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,668 |
$[\underline{\text { Polyhedra and Polygons (Miscellaneous) }}]$ ] [ Rearrangement of Areas $\quad]$ Author: Saghafian $M$. On a plane, there are eight points in general position. The areas of all 56 triangles with vertices at these points are listed in a row. Prove that it is possible to place the signs “\$+\$” and ... | Note that the statement of the problem is true for any four points. Indeed, if the points form a convex quadrilateral \$ABCD\$, then \$S_\{ABC\}+S_\{ACD\}-S_\{BCD\}-S_\{ABD\}=0\$. If point \$D\$ lies inside triangle \$ABC\$, then \$S_\{ABC\}-S_\{ABD\}-S_\{ACD\}-S_\{BCD\}=0\$.
Now, let's associate the given points with... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,670 |
$4-$
a) On 44 trees, arranged in a circle, there were 44 cheerful chizhi (one chizh on each tree). From time to time, two chizhi simultaneously fly to adjacent trees in opposite directions (one - clockwise, the other - counterclockwise). Prove that the chizhi will never gather on one tree.
b) And if there are $n$ chi... | a) Let's number the trees in order, starting from some tree. For each chaffinch, consider the number of the tree it is sitting on, and sum these 44 numbers. Notice that during the process of the chaffinches flying, this sum either remains unchanged or
. Can these cards be arranged in a row so that the zeros stand next to each other, there is exactly one card between the ones, two cards between the twos, and so on up to the nines, between w... | Suppose the cards are arranged in the required manner. Let's call a segment a set of cards located between two identical cards, and the length of such a segment is the number of other cards located between them. We will calculate the sum of the lengths of all segments in two ways. On the one hand, it is equal to $0+1+2... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,677 |
$4-$ [ Geometric inequalities (miscellaneous).]
Prove that in a circle of radius 1, it is impossible to find more than 5 points, the pairwise distances between which are all greater than 1.
# | Suppose that inside a circle of radius 1, six points can be placed such that the distance between any two is greater than 1. Since the distance from the center of a circle of radius 1 to any of its points does not exceed one, none of these six points can coincide with the center of the circle. Let $A_{1}, \ldots, A_{6}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,678 |
[ Processes and operations]
Several glass marbles are arranged in three piles. A boy, who has an unlimited supply of marbles, can, in one move, take one marble from each pile or add from his supply to one of the piles as many marbles as there are already in it. Prove that in several moves the boy can achieve that ther... | It is enough to show that the boy can reduce the situation to having 2 balls in each pile. If there is 1 ball in some pile, the boy can double the number of balls in that pile. If there are no less than two balls in all piles, then he can take one ball from each pile. Then, every four moves, if he has not yet achieved ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,679 |
Nenashev S.
Natural numbers $a_{1}, a_{2}, \ldots, a_{n}$ are such that each does not exceed its number $\left(a_{k} \leq k\right)$ and the sum of all numbers is even.
Prove that one of the sums $a_{1} \pm a_{2} \pm \ldots \pm a_{n}$ is equal to zero. | We will prove this statement by induction on $n$.
Base case $(n=2)$ is obvious, as the only possible set is $a_{1}=a_{2}=1$.
Inductive step. Take a set $a_{1}, a_{2}, \ldots, a_{n}, a_{n+1}$ that satisfies the condition. If $a_{n}=a_{n+1}$, then the sum $a_{1}+\ldots +a_{n-1}$ is even; considering the inductive hypot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,680 |
Around a round table sit 13 bogatyrs (warriors) from $k$ cities, where $1<k<13$. Each bogatyr holds a golden or silver goblet, and there are also $k$ golden goblets. The prince ordered each bogatyr to pass his goblet to the neighbor on the right and to repeat this until any two bogatyrs from the same city both receive ... | Author: Bolotov A.A.
Let the statement be false, that is, at any given time, exactly one knight from each city holds a golden cup (since the number of clans equals the number of cups). Suppose that each subsequent passing of the cups occurs every minute. Then, in 13 minutes - the time for a complete cycle of the cups ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,681 |
On the island of GrayBrownMagenta, there live 13 gray, 15 brown, and 17 magenta chameleons. If two chameleons of different colors meet, they both simultaneously change their color to the third one (gray and brown both become magenta, etc.). Can it happen that after some time all chameleons will be of the same color?
# | Let $c$ be the number of gray chameleons, and $b$ be the number of brown chameleons. Note that the remainder of $c-b$ divided by 3 is an invariant. Indeed, when a gray chameleon meets a brown one, the difference does not change; when a gray chameleon meets a raspberry one, the difference decreases by 3; and when a brow... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,684 |
Folklore
In a $7 \times 7$ square of cells, 16 tiles of size $1 \times 3$ and one tile of size $1 \times 1$ are placed.
Prove that the $1 \times 1$ tile either lies in the center or touches the borders of the square. | Let's color the $7 \times 7$ square as shown in the left figure. Each tile $1 \times 3$ covers exactly one shaded cell. Since there are only 16 tiles $1 \times 3$, and 17 cells are shaded, the $1 \times 1$ tile must lie on a shaded cell. Repeating the same reasoning for the coloring shown in the center figure, we get t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,685 |
Fomin D:
The numbers $1, 1/2, 1/3, \ldots, 1/100$ are written on the board. We choose two arbitrary numbers $a$ and $b$ from those written on the board, erase them, and write the number
$a + b + ab$. We perform this operation 99 times until only one number remains. What is this number? Find it and prove that it does ... | If $a_{1}, a_{2}, \ldots, a_{n}$ are numbers written on the board, then the value $\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right)$ does not change under an allowable operation. Indeed, if $a$ and $b$ are replaced by $a+b+a b$, then the factors not containing $a$ and $b$ do not change, and the produ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,686 |
Bermov S.l.
Given a $15 \times 15$ board. Some pairs of centers of adjacent cells by side are connected by segments such that a closed non-self-intersecting broken line is formed, which is symmetric with respect to one of the diagonals of the board. Prove that the length of the broken line does not exceed 200. | Clearly, the broken line intersects the diagonal. Let $A$ be one of the vertices of the broken line lying on the diagonal. We will move along the broken line until we first reach another vertex $B$ lying on the diagonal. By symmetry, if we move along the broken line from $A$ in the other direction, $B$ will also be the... | 200 | Geometry | proof | Yes | Yes | olympiads | false | 52,691 |
Bervov s.l.
Given a quadratic trinomial $f(x)=x^{2}+a x+b$. The equation $f(f(x))=0$ has four distinct real roots, the sum of two of which is -1. Prove that $b \leq-1 / 4$.
# | Let $c_{1}$ and $c_{2}$ be the roots of the equation $f(x)=0$, and $x_{1}$ and $x_{2}$ be the roots of the equation $f(f(x))=0$, the sum of which is -1. The set of roots of the latter equation coincides with the union of the sets of roots of the equations $f(x)=c_{1}$ and $f(x)=c_{2}$. Consider two cases.
1) $x_{1}$ a... | b\leq-\frac{1}{4} | Algebra | proof | Yes | Yes | olympiads | false | 52,692 |
Asahanov $H . X$.
Let $f(x)=x^{2}+a x+b \cos x$. Find all values of the parameters $a$ and $b$ for which the equations $f(x)=0$ and $f(f(x))=0$ have coinciding non-empty sets of real roots. | Let $x_{0}$ be a common root, then $f\left(x_{0}\right)=0$ and $f\left(f\left(x_{0}\right)\right)=0$. Substituting the first equality into the second, we get $f(0)=0$, i.e., $b=0$. Thus, $f(x)=x^{2}+a x$, and the equation $f(x)=0$ has roots 0 and $-a$.
The condition of the problem is satisfied if each root of the equa... | 0,0\leq<4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,693 |
Senderov V.A.
Do there exist different pairwise coprime natural numbers $a, b$, and $c$, greater than 1, such that $2^{a}+1$ is divisible by $b$, $2^{b}+1$ is divisible by $c$, and $2^{c}+1$ is divisible by $a$? | Obviously, the numbers $a, b$, and $c$ are odd. Let $a=3$. Then $b=9, c=19$. It is not difficult to verify that these numbers work: since $c$ is odd, $2^{c}+1$ is divisible by 3.
## Answer
There exist. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,694 |
Berlov S.L.
100 people came to a party. Then those who had no acquaintances among the attendees left. Then those who had exactly one acquaintance among the remaining people also left. Then, similarly, those who had exactly 2, 3, 4, ..., 99 acquaintances among the remaining people at the time of their departure also le... | It is not difficult to check that if all the attendees, except for two people $A$ and $B$, were acquainted with each other, then at the end, everyone except $A$ and $B$ should have remained, that is, 98 people.
Let's prove that 99 people could not have remained. It is clear that person $A$, who initially had the fewes... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,696 |
$\underline{\text { Agakhanov N.X. }}$.
Positive numbers $x, y, z$ are such that the absolute value of the difference of any two of them is less than 2.
Prove that $\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1}>x+y+z$. | Given the condition $|x-y|<2 \Rightarrow x^{2}-2 x y+y^{2}<4 \Rightarrow x^{2}+2 x y+y^{2}<4(1+x y) \Rightarrow x+y<2 \sqrt{x y+1}$. Similarly, $y+z<2 \sqrt{y z+1}, z+x<2 \sqrt{z x+1} \Rightarrow 2 x+2 y+2 z<2 \sqrt{x y+1}+2 \sqrt{y z+1}+2 \sqrt{z x+1}$.
Send a comment | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,697 |
Can the set of six numbers $\left(a, b, c,{ }^{a^{2}} / b, b^{2} / c, c^{2} / a\right.$, where $a, b, c$ are positive numbers, contain exactly three distinct numbers? | Let among the numbers $a, b, c$ there be different ones, and let $a$ be the largest of these numbers (or one of the largest). If $a > b$, then $a^{2} / b > a$. Otherwise, $a = b > c$, and then $b^{2} / c > b = a$. Thus, the largest of the six given numbers is greater than the largest of the numbers $a, b, c$. Similarly... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,698 |
Polansky A.
The faces of a $9 \times 9 \times 9$ cube are divided into unit cells. The cube is covered without overlaps by paper strips $2 \times 1$ (the sides of the strips follow the sides of the cells). Prove that the number of bent strips is odd. | We will paint the cells of each face of the cube in a checkerboard pattern so that the corner cells are black. In this case, each face contains 41 black and 40 white cells. Note that all bent strips will be monochromatic, while all others will not. Since the number of black cells is 6 more than the number of white cell... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,700 |
At Vasya's birthday, there were 10 kids (including Vasya). It turned out that any two of these kids have a common grandfather. Prove that seven of them have a common grandfather.
# | Consider a graph where vertices correspond to grandfathers, and edges - to common grandchildren. If there are no more than two vertices, then each grandfather has ten grandchildren.
Suppose there are exactly three vertices. They are connected by 10 edges. By the Pigeonhole Principle, some two vertices are connected by... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,701 |
[ [proof by contradiction ]
The numbers $1,2, \ldots, 100$ are arranged in a circle in some order.
Can it happen that for any two adjacent numbers, the absolute value of their difference is not less than 30 but not more than 50?
# | Suppose the conditions are met. Let's call numbers from 26 to 75 "middle" numbers, and the rest "extreme" numbers. Two extreme numbers cannot be adjacent (the absolute value of their difference is less than 25 or greater than 50). However, extreme numbers make up exactly half of the total number of numbers. Therefore, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,703 |
Sprungas L.
An ATM exchanges coins: doubloons for pistoles and vice versa. A pistole costs $s$ doubloons, and a doubloon costs 1 / $s$ pistoles, where $s$ is not necessarily an integer. You can insert any number of coins of one type into the ATM, after which it will give you coins of the other type in exchange, roundi... | a) Let, for example, $s=3$. Exchanging 5 doubloons, we get 2 pistoles, and exchanging the pistoles, we get 6 doubloons.
b) Let $s1$ and after the first exchange, we get $n$ pistoles. As shown above, after two exchanges, we will not get more than $n$ pistoles, therefore, the number of doubloons after the fourth exchang... | )Can;b)cannot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,704 |
G. A. Gailerenin
The police station is located on a straight road that is infinite in both directions. Someone stole an old police car, the maximum speed of which is $90 \%$ of the maximum speed of a new car. At some point, the police station realized what had happened and sent a police officer in a new police car in ... | Let's mentally send two police assistants down both sides of the road, driving at a speed greater than that of the car thief but less than that of the police officer. Let the police officer catch up with the first assistant first, then the second, then the first again, then the second again, and so on. Clearly, at some... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,705 |
A square board is divided into $n^{2}$ rectangular cells by $n-1$ horizontal and $n-1$ vertical lines. The cells are colored in a checkerboard pattern. It is known that on one diagonal, all $n$ cells are black and square. Prove that the total area of all black cells on the board is not less than the total area of the w... | Let's number the verticals and horizontals of the board, starting from the black corner. Let the distances between the lines be $x_{1}, x_{2}, \ldots, x_{n}$ respectively. The areas of the cells are $x_{i} x_{j}$, and for black cells, $i+j$ is even, while for white cells, it is odd. The difference between the black and... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,706 |
[ Plane Transformations (Miscellaneous). ] [ Tiling with Dominoes and Tiles ]
## Author: Shapovomovo A.V.
From $N$ rectangular tiles (possibly not identical), a rectangle with unequal sides is formed. Prove that each tile can be cut into two parts such that from $N$ parts a square can be formed, and from the remaini... | Let the dimensions of the rectangle be $a \times b, a < b$ and side $b$ is horizontal. Compress the rectangle uniformly horizontally so that the sides become equal. We will get a square divided into rectangles. Each part is obtained by compressing the corresponding tile horizontally by a factor of $b / a$. Therefore, i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,709 |
$\underline{\text { Berdnikov A. }}$
In a team of guards, each has a rank (a natural number). A guard of rank $N$ works for $N$ days, then sleeps for $N$ days, works again for $N$ days, sleeps for $N$ days, and so on. It is known that the ranks of any two guards differ by at least a factor of three. Can such a team pr... | Let's number the guards in descending order of their shift length. The life of the guards is divided into equal periods of sleep and duty. Each guard's periods are, at least, three times shorter than those of the previous guard; therefore, any period of the previous guard is divided, at least, into three parts by the p... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,710 |
It is known that objects $A$ and $B$ cannot be captured in the frame of a camera located at point $O$ if the angle $AOB$ is greater than $179^{\circ}$. On a plane, 1000 such cameras are placed. Each camera takes one photo simultaneously. Prove that there will be a photo in which no more than 998 cameras are captured.
... | Suppose that in each photograph, all other cameras (except the one taking the photo) are photographed. Then the points where the cameras are located are the vertices of a convex 1000-gon, since otherwise there would be a camera from which not all others are visible. The sum of all angles of the 1000-gon is $180^{\circ}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,711 |
$\underline{\text { Blokh A. }}$.
A blot was placed on the paper. For each point of the blot, the smallest and largest distance to the boundary of the blot were determined. Among all the smallest distances, the largest was chosen, and among the largest distances, the smallest was chosen, and the two numbers were compar... | Let $A$ be a point of the blot, the smallest distance from which to the boundary of the blot is $r_{0}$ - the largest of the smallest distances. Then the circle of radius $r_{0}$ centered at point $A$ is contained within the blot (if this circle were not fully contained within the blot, some point on the boundary of th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,712 |
Over the course of 92 days, the airline carried out ten flights daily. Each day, each aircraft carried out no more than one flight. It is known that for any pair of days, there is one and only one aircraft that flew on both these days. Prove that there is an aircraft that flew every day. | Consider 10 planes that flew on the first day. At least one of them must have flown for at least 10 more days (since one of these ten planes flew on each of the remaining 91 days).
Consider a plane that flew for at least eleven days. Without loss of generality, we can assume that these were the first eleven days (and ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,713 |
11 pioneers are attending five clubs in the cultural center. Prove that there exist two pioneers A and B such that all the clubs that A attends, B also attends.
# | Let's number the circles from 1 to 5 and instead of each pioneer, consider the subset of the set $\{1,2,3,4,5\}$ that consists of the circles they attend. We need to divide the 32 subsets of the specified set into 10 groups such that in each group, for any two subsets in this group, one is contained in the other. As su... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,714 |
In the village, there are 100 houses. What is the maximum number of closed, non-intersecting fences that can be built so that each fence encloses at least one house and no two fences enclose the same set of houses?
# | In the maximum set of fences, there is a fence that limits exactly two houses. By combining these two houses into one, we reduce the number of houses by 1 and the number of fences by 2. In this process, both conditions of the problem are maintained. Continue this process. After 99 steps, one house and one fence will re... | 199 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,716 |
[Power of a set. Bijective mappings] [Equivalence relation. Equivalence classes]
In the alphabet of the Mumbu-Yumbu tribe, there are only two letters A and B. Two different words denote the same concept if one can be obtained from the other using the following operations:
1) in any place in the word, the combination ... | b) Let's consider a Mumbo-Jumbo word consisting of no fewer than four letters. If the word has two identical letters in a row, they can be removed. Otherwise, the letters in the word alternate. Therefore, the beginning of the word is either ABAB or BABA. The beginning ABAB can be replaced with BABB and then further sho... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,717 |
10,11
During an interview, ten people were offered a test consisting of several questions. It is known that any five people together answered all the questions (that is, at least one of the five gave the correct answer to each question), but any four did not. What is the minimum number of questions for which this coul... | For each question, the number of people who did not answer it is no more than four.
## Solution
According to the condition, for every group of four people, there is a question they did not answer. On the other hand, the number of people who did not answer this question is no more than four. Therefore, the number of q... | 210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,718 |
|
| :---: | :---: | :---: |
| | [Permutations and Substitutions (Miscellaneous).] | |
| | Evenness and Oddness | |
| | |
| | 2n deputies were seated around a round table for a meeting. After a break, these same 2n deputies sat down around the table again, but in a different order. |
| | Prove that there will be... | Assign to each deputy the number of seats by which he has moved clockwise after the break. If the condition of the problem is not met, then all these numbers must be different.
## Solution
Let's number the deputies by the numbers $1,2, \ldots, 2 n$ in the order (clockwise) in which they were sitting before the break;... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,719 |
8,9,10 |
There are 10 natural numbers written on the board. Prove that from these numbers, one can choose several numbers and place "+" and "-" signs between them so that the resulting algebraic sum is divisible by 1001. | First, consider only sums with the "+" sign. Then show that one of these sums can be subtracted from another so that the result is divisible by 1001.
## Solution
Consider all possible sums of several of the listed numbers. The number of such sums will be $2^{10}=$ 1024 (we include the empty sum). According to the Pig... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,720 |
[ Product Rule ] [ Permutations and Substitutions (Miscellaneous).]
Prove that there exist numbers representable in at least 100 ways as the sum of 2001 addends, each of which is the 2000th power of an integer. | Consider all possible sums of $20012000-th$ powers
## Solution
Consider the numbers $1,2,3, \ldots, N$. From the 2000-th powers of these numbers, we will form all possible sums, each of which contains 2001 summands. The number of such sums (without considering the order of the summands) will be no less than ${ }^{N^{... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,721 |
[Smallest or Largest Angle]
Six circles are arranged on a plane such that a certain point $O$ lies inside each of them. Prove that one of these circles contains the center of some other circle.
# | One of the angles between the six segments connecting point $O$ with the centers of the circles does not exceed $360\% / 6 = 60^{\circ}$. Let $\angle O_{1} O O_{2} \leq 60^{\circ}$, where $O_{1}$ and $O_{2}$ are the centers of circles with radii $r_{1}$ and $r_{2}$, respectively. Since $\angle O_{1} O O_{2} \leq 60^{\c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,723 |
[Smallest or largest distance (length).]
Prove that at least one of the bases of the perpendiculars dropped from an interior point of a convex polygon to its sides lies on the side itself, and not on its extension. | Let $O$ be a given point. Draw the lines containing the sides of the polygon and choose among them the one that is the least distant from point $O$. Let side $A B$ lie on this line. We will prove that the foot of the perpendicular dropped from point $O$ to side $A B$ lies on the side itself. Suppose the foot of the per... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,724 |
[Smallest or largest distance (length).]
From each vertex of a polygon, perpendiculars are dropped to the sides not containing it. Prove that for at least one vertex, one of the feet of the perpendiculars lies on the side itself, and not on its extension. | Let's take the largest side $AB$ of the given polygon and consider the strip consisting of those points whose projections onto the line $AB$ fall on the segment $AB$. This strip must be intersected by some other side $CD$ of the polygon (one of the vertices $C$ and $D$ may coincide with $A$ or $B$). The inequality $CD ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,725 |
$[$ Extreme Principle (Miscellaneous) $]$
On the plane, there is an infinite set of rectangles, the vertices of each of which are located at points with coordinates $(0,0),(0, m),(n, 0),(n, m)$, where $n$ and $m$ are positive integers (specific to each rectangle). Prove that from these rectangles, two can be chosen su... | For a rectangle with vertices at points $(0,0),(0, m),(n, 0)$, and $(n, m)$, the horizontal side is equal to $n$, and the vertical side is equal to $m$. We select from this set the rectangle with the smallest horizontal side. Let its vertical side be $m_{1}$. Consider any $m_{1}$ of the remaining rectangles.
There are... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,726 |
What is the smallest number of points that need to be marked inside a convex $n$-gon so that within any triangle with vertices at the vertices of the $n$-gon, there is at least one marked point? | Since the diagonals emanating from one vertex divide an $n$-gon into $n-2$ triangles, $n-2$ points are necessary.
From the figure, it can be understood how to manage with $n-2$ points: it is sufficient to mark one point in each shaded triangle. Indeed, inside the triangle $A_{\mathrm{p}} A_{\mathrm{q}} A_{\mathrm{r}}$... | n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,728 |
8,9
a) Can a chessboard of size $8 \times 8$, from which two opposite corner squares have been cut out, be tiled with dominoes of size $1 \times 2$?
b) Prove that if two arbitrary squares of different colors are cut out from a chessboard of size $8 \times 8$, the remaining part of the board can always be tiled with d... | a) Squares of one color, let's say black, have been cut out. Therefore, 32 white and 30 black squares remain. Since a domino tile always covers one white and one black square, it is impossible to tile an $8 \times 8$ chessboard with two opposite corner squares removed using domino tiles.
b) The diagram shows that the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,730 |
[ Auxiliary coloring (other).]
The bottom of a rectangular box is paved with tiles of size $2 \times 2$ and $1 \times 4$. The tiles were spilled out of the box and one $2 \times 2$ tile was lost. Instead of it, a $1 \times 4$ tile was taken out. Prove that it is now impossible to pave the bottom of the box with the ti... | Let's paint the bottom of the box in two colors as shown in the figure. Then each $2 \times 2$ tile covers exactly one black cell, while a $1 \times 4$ tile covers 2 or 0. Therefore, the parity of the number of black cells on the bottom of the box matches the parity of the number of $2 \times 2$ tiles. Since replacing ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,732 |
8,9
From a sheet of graph paper measuring $29 \times 29$ cells, 99 squares of size $2 \times 2$ cells have been cut out. Prove that it is possible to cut out one more such square.
# | Let's shade the $2 \times 2$ squares on this square sheet of paper as shown in the figure. In this way, 100 shaded squares will be obtained. Each cut-out square touches exactly one shaded square, so at least one shaded square remains intact and can be cut out.
, A_{n+1}$ be the intersection point of the lines $E B_{n+1}$ and $A B$. Prove that $A_{n} B=\frac{A B}{n+1}$. | Induction by p. Base: $A_{0} B=A B$.
Induction step. Let $C_{n}$ be the intersection point of the lines $E A_{n}$ and $D C, { }^{D C} /_{A B}=k, A B=a, A_{n} B=a_{n / n+1}$ and $A_{n+1} B=x$. Since
$C C_{n+1}: A_{n} A_{n+1}=D C_{n+1}: B A_{n+1}$, then $k x:\left(a_{n}-x\right)=(k a-k x): x$, that is, $x=a a_{n / a+a_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,735 |
The numerical sequence $A_{1}, A_{2}, \ldots, A_{n}, \ldots$ is defined by the equations $A_{1}=1, A_{2}=-1, A_{n}=-A_{n-1}-2 A_{n-2}$ $(n \geq 3)$.
Prove that for any natural number $n$, the number $2^{n+2}-7 A_{n}^{2}$ is a perfect square. | Consider another sequence $B_{1}, \ldots, B_{n}, \ldots$, defined by the same recurrence relation but with different initial conditions:
$B_{1}=1, B_{2}=3$. We will prove by induction the relations: $7 A_{n}^{2}+B_{n}^{2}=2^{n+2}, 7 A_{n} A_{n-1}+B_{n} B_{n-1}=-2^{n}$.
Base case. $7 A_{1}^{2}+B_{1}^{2}=7+1=2^{3}, 7 A... | 2^{n+2}-7A_{n}^{2}=B_{n}^{2} | Algebra | proof | Yes | Yes | olympiads | false | 52,736 |
[Trigonometric Identities.]
$$
\text { [ Vectors help solve the task, ] }
$$
Prove the equalities:
a) $\cos \frac{\pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}$;
b) $\frac{1}{\sin \frac{\pi}{7}}=\frac{1}{\sin \frac{2 \pi}{7}}+\frac{1}{\sin \frac{3 \pi}{7}}$;
c) $\sin 9^{\circ}+\sin 49^{\circ}+\sin 89^{\circ}+\ldots+\si... | a) Use the result of problem $\underline{8.1}$ for a regular pentagon inscribed in a unit circle.
b) Consider a regular heptagon $A_{1} A_{2}, \ldots, A_{7}$. Let $M$ be the point of intersection of the diagonals $A_{1} A_{4}$ and $A_{2} A_{5}$.
The equality of the problem follows from the similarity of triangles $A_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,737 |
10,11 |
Let $x, y, z$ be positive numbers and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$. | Make the substitutions $x=p-a, y=p-b, z=p-c$.
## Solution
Let $a=y+z, b=x+z, c=x+y, p=x+y+z$. Consider a triangle with sides $a, b, c$ (the triangle inequalities are obviously satisfied). The perimeter of this triangle is $2 p$, and let the area be denoted by $S$. By Heron's formula,
$S^{2}=p(p-a)(p-b)(p-c)=(x+y+z) ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,738 |
Bogdanov I.I.
The Tsar summoned two sages. He gave the first 100 blank cards and ordered him to write a positive number on each (the numbers do not have to be different), without showing them to the second. Then the first can tell the second several different numbers, each of which is either written on some card or eq... | Let's prove that it is sufficient to lose 101 hairs for each sage. Let the first write the numbers $1, 2, 4, \ldots, 2^{99}$ on the cards, and inform the second of these numbers and their sum. When the second hears the number 1, he will understand that there is a card not exceeding 1. Hearing the next number $2^{k}$, h... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,740 |
Bakayev E.B.
In the country, there are 100 cities, and non-stop flights operate between any two cities. All flights are paid and cost a positive (possibly non-integer) number of tugriks. For any pair of cities A and B, the flight from A to B costs the same as the flight from B to A. The average cost of a flight is 1 t... | a) Let all 99 flights from the home city cost 49.6 tugriks each. This is possible since the total cost of all flights (regardless of direction) is $99 \cdot 50$ tugriks. Then, to fly out of the home city and return to it, one would already need to spend more than 99 tugriks.
b) Consider all 99! possible circular routes... | ) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,741 |
Shapovalov A.V. An even number of nuts is divided into three piles. In one operation, you can transfer half of the nuts from a pile with an even number of nuts to any other pile. Prove that, no matter how the nuts are initially distributed, it is possible to gather exactly half of all the nuts in one pile using such o... | Note that if we achieve pile sizes $x, 2x$, and $y$, we can get piles of sizes $x, x$, and $x+y$, where $x+y$ is even, and then - sizes $x$,
$1/2(3x+y)$ and $1/2(x+y)$, where the middle pile has exactly half the nuts.
Let's describe an algorithm to achieve the desired pile sizes. Choose a pair of piles such that at l... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,742 |
Kuzneuov A.
On the board, several reduced polynomials of the 37th degree are written, all coefficients of which are non-negative. It is allowed to choose any two written polynomials $f$ and $g$ and replace them with two reduced polynomials of the 37th degree $f_{1}$ and $g_{1}$ such that $f+g=f_{1}+g_{1}$ or $f g=f_{1... | Note that when replacing polynomials $f$ and $g$ with $f_{1}$ and $g_{1}$, the sum of the coefficients of the $36-x$ polynomials does not change. For the first type of replacement (where $f+g=f_{1}+g_{1}$), this is obvious, and for the second type of replacement, it follows from the equality
$\left(x^{37}+a x^{36}+\ld... | proof | Algebra | proof | Yes | Yes | olympiads | false | 52,743 |
Yatsenko I.V.
Vanya is coming up with a number consisting of non-repeating digits without zeros - a password for his phone. The password works as follows: if, without lifting his finger from the screen, he sequentially connects the points corresponding to the digits of the password with lines, the phone will unlock. H... | The password 12769 meets Vanya's requirements. Let's see how we can connect its digits without any intersections. The digit 7 must be connected to either 2 or 6. Suppose we draw the segment 7-6. Now, 9 can only be connected to 6. Next, we inevitably need to draw the segments 7-2 and 2-1, and we get the line shown in th... | 12769 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,744 |
Gribalko A.V.
On the board, there are \$1000\$ consecutive integers. In one move, you can partition the written numbers into pairs in any way and replace each pair of numbers with their sum and difference (it is not necessary to subtract the smaller from the larger; all replacements occur simultaneously). Prove that \... | Since $\$(x+y)^{\wedge} 2+(x-y)^{\wedge} 2=2\left(x^{\wedge} 2+y^{\wedge} 2\right) \$$, the sum of the squares of all numbers on the board doubles with
$(\mathrm{n}+4)^{\wedge} 2=8 \mathrm{n}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,745 |
Bogdanov I.I.
101 sages stand in a circle. Each of them either believes that the Earth revolves around Jupiter, or believes that Jupiter revolves around the Earth. Once a minute, all sages simultaneously announce their opinions. Immediately after this, each sage whose both neighbors think differently from him changes ... | Let's associate each sage with a certain opinion marked by a "+", and with the opposite opinion by a "-". Then the arrangement of sages corresponds to the arrangement of 101 signs in a circle.
Suppose at some moment two identical signs stand next to each other. Then in the next minute they will not change, and therefo... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 52,746 |
Bogdanov I.i.
Initially, there are 111 pieces of plasticine of the same mass on the table. In one operation, you can choose several groups (possibly one) with the same number of pieces and in each group, combine all the plasticine into one piece. What is the minimum number of operations required to get exactly 11 piec... | Let the mass of one original piece be 1. If in the first operation in each group there are $k$ pieces, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to get 11 pieces of different masses in one operation.
We will show that the required result can be achieved in two operations. For t... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,747 |
Folklore
The length of a rectangular plot is 4 meters, and the width is 1 meter.
Can three trees be planted on it so that the distance between any two trees is at least 2.5 meters? | Let there be three trees growing on the plot $A B C D$. We will divide it into two rectangles of size $2 \times 1$ (see figure).
In one of these rectangles, at least two trees grow. The dis... | No | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,748 |
Shapovalov A.V.
Some students in the class watch football, some watch cartoons, but there are no students who do not watch either. The average math grade for cartoon lovers is less than 4, and for football lovers - also less than 4. Can the average math grade for the entire class be greater than 4? | Let's say there are 7 people in the class, and two of them have only fives in math and watch only cartoons, three have only threes in math and watch both, and the other two have fives in math but watch only football. Then the average score of each of the two groups is less than 4 (in each group, there are fewer "excell... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,749 | |
$\underline{\text { Zhukov } \Gamma}$.
Let $C(n)$ be the number of distinct prime divisors of the number $n$. (For example, $C(10)=2, C(11)=1, C(12)=2$.)
Is the number of such pairs of natural numbers $(a, b)$, where $a \neq b$ and $C(a+b)=C(a)+C(b)$, finite or infinite? | For example, all pairs of the form $\left(2^{n}, 2^{n+1}\right)$ work. Here $C(a)=C(b)=1, C(a+b)=C\left(3 \cdot 2^{n}\right)=2$.
## Answer
Infinite. | Infinite | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,750 |
[ Counting in two ways ] $[\quad$ Trees $]$
A travel agency ran a promotion: "Buy a trip to Egypt, bring four friends who also buy a trip, and get the cost of your trip back." During the promotion, 13 buyers came on their own, and the rest were brought by friends. Some of them brought exactly four new customers, while... | Each of the $x$ "lucky ones" brought 4 friends. Therefore, the number of "brought" customers is $4x$, and 13 more came on their own, so the total number of tourists was $13 + 4x$. On the other hand, $x$ people brought new customers, while 100 did not, meaning the total number of tourists was $x + 100$. Thus, $13 + 4x =... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,752 |
Podlipsky O.K.
Do there exist 2013 different natural numbers such that the sum of any 2012 of them is not less than the square of the remaining one?
# | Let $a$ be the largest of the given 2013 different natural numbers. Then $a \geq 2013$, so $a^{2} \geq 2013 a$. But the sum of all the other numbers does not exceed $2012 a$.
## Answer
## Problem | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,754 |
From a pie having the shape of a convex polygon, it is allowed to cut off a triangular piece ABC, where A is some vertex, and B and C are points lying strictly inside the sides having vertex A. Initially, the pie has the shape of a square. At the center of this square is a raisin. Prove that at no step can a piece cont... | A piece of cake always has sides belonging to the sides of the original square cake.
## Solution
The cutting operation is arranged in such a way that after cutting, two sides of the polygon are shortened, and a new side also appears. Thus, no side is completely cut off. Therefore, after any number of cuts, a piece of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,755 |
$[$ Theory of algorithms (miscellaneous) $]$ $[$ Extreme principle (miscellaneous) ]
Vasya solved problems for 10 days - at least one problem each day. Each day (except the first), if the wea... | Consider any two consecutive days. Each day at least one problem is solved, but solving exactly one problem on both days is not allowed, so at least three problems are solved over these two days. Thus, over the first 8 days, Vasya solved at least \(4 \cdot 3 = 12\) problems. If he solved at least two problems on the ni... | Cloudy | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,756 |
8,9 On the coordinate plane ( $x ; y$ ), a circle of radius 4 with its center at the origin is drawn. A line given by the equation $y=4-(2-\sqrt{3}) x$ intersects it at points $A$ and $B$. Find the sum of the lengths of the segment $A B$ and the smaller arc $A B$. # | Solving the system of equations, find the coordinates of the points of intersection of the line and the circle. Then, applying the cosine theorem, find the angle between the radii of the circle drawn to the found points of intersection.
## Solution
Solving the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{... | \frac{2\pi}{3}+4\sqrt{2-\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,758 |
[Central Angle. Arc Length and Circumference]
On the coordinate plane ($x ; y$), a circle with radius 4 and center at the origin is drawn. The line given by the equation $y=\sqrt{3} x-4$ intersects it at points $A$ and $B$. Find the sum of the lengths of segment $A B$ and the larger arc $A B$. | Solve the system of equations to find the coordinates of the points of intersection between the line and the circle. Then, using the cosine theorem, find the angle between the radii of the circle drawn to the found points of intersection.
## Answer
$\frac{16 \pi}{3}+4 \sqrt{3}$ | \frac{16\pi}{3}+4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,759 |
$\underline{\text { Folklore }}$
Prove that there are infinitely many prime numbers.
# | Assume the opposite. Let $p_{1}, p_{2}, \ldots, p_{n}$ be all the prime numbers. Consider the number $p_{1} p_{2} \ldots p_{n}+1$. This number is not divisible by any of the numbers $p_{1}, p_{2}, \ldots, p_{n}$ and, therefore, cannot be factored into a product of primes. Contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,760 |
[ Pairing and Grouping; Bijections $]$
$$
\text { [ Factorization ] }
$$
Prove that $1^{n}+2^{n}+\ldots+(n-1)^{n}$ is divisible by $n$ for odd $n$.
# | Consider the sum of symmetric terms.
## Solution
Let's break down the terms into pairs. $k^{n}+(n-k)^{n}$ is divisible by $k+(n-k)=n-k$.
Send a comment | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,761 |
## [ Symmetry and Involution Transformations ]
Let's call a natural number $n$ convenient if $n^{2}+1$ is divisible by 1000001. Prove that among the numbers $1,2, \ldots$, 1000000, there is an even number of convenient numbers. | If $n$ is a convenient number, then $1000001-n$ is also convenient.
## Solution
$(1000001-n)^{2}+1=1000001(1000001-2 n)+n^{2}+1$. Therefore, if $n$ is a convenient number, then $1000001-n$ is also convenient. It remains to check that the number 500000 is inconvenient. Indeed, $\left(5 \cdot 10^{5}\right)^{2}+1=25 \cd... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,762 |
Solve the equation $x^{2}+y^{2}=x+y+2$ in integers.
# | Let's write the equation as $(2 x-1)^{2}+(2 y-1)^{2}=10$.
From this, it is clear that one term equals 9, and the other equals 1.
## Answer
$\{2,0\},\{2,1\},\{-1,0\},\{-1,1\}$. | {2,0},{2,1},{-1,0},{-1,1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,763 |
$\left.\begin{array}{l}\text { 3 } \quad \text { Fermat's Little Theorem }\end{array}\right]$
Let $p$ be a prime number, and $a$ not divisible by $p$. Prove that there exists a natural number $b$, for which $a b \equiv 1$ $(\bmod p)$. | $a \cdot a^{p-2}=a^{p-1} \equiv 1(\bmod p)$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,764 |
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