problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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value | source stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Kovaldji A.k.
There are two houses, each with two entrances. The residents keep cats and dogs, and the share of cats (the ratio of the number of cats to the total number of cats and dogs) in the first entrance of the first house is greater than the share of cats in the first entrance of the second house, and the share... | Here is a counterexample:
| | 1 building, 1 entrance | 1 building, 2 entrance | 1 building |
| :---: | :---: | :---: | :---: |
| cats | 5 | 1 | 6 |
| dogs | 1 | 5 | 6 |
| | 2 building, 1 entrance | 2 building, 2 entrance | 2 building |
| cats | 49 | 1 | 50 |
| dogs | 10 | 6 | 16 |
## Answer
Incorrect. | Incorrect | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,100 |
Kanel-Belov A.Y.
At the ball, each young man danced a waltz with a girl who was either more beautiful than the previous dance partner, or more intelligent, and one of them danced with a girl who was simultaneously more beautiful and more intelligent. Could this have been possible? (There were an equal number of young ... | The dumbest girl should be beautiful enough, and the ugliest one should be smart enough.
## Solution
Let's assume that three girls - Anna, Vera, and Svetlana - were present at the ball, in order of increasing beauty. The order of increasing intelligence is as follows: Vera - Svetlana - Anna (Anna is the smartest). Th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,101 |
Proizvolov V.V.
There are 19 weights of $1, 2, 3, \ldots, 19$ grams: nine iron, nine bronze, and one gold. It is known that the total weight of all iron weights is 90 grams more than the total weight of the bronze weights. Find the weight of the gold weight. | Prove that the nine lightest weights are bronze, and the nine heaviest are iron.
## Solution
The difference between the total weight of the nine heaviest weights and the total weight of the nine lightest weights is $(19+18+\ldots+11)-(9+8+\ldots+1)=90$ grams. Therefore, the iron weights are the heaviest, and the bron... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,104 |
Zhendarov R.G.
In the cells of a $4 \times 4$ table, numbers are written such that the sum of the neighbors of each number is 1 (cells are considered neighbors if they share a side).
Find the sum of all the numbers in the table.
# | Let's divide all cells into 6 groups (in the figure, cells of each group are marked with their own symbol). Each group consists of all neighbors of some one cell, so the sum of the numbers in it is 1. Therefore, the sum of all numbers is 6.
.
Prove that it is impossible to transform the number 1 into the number 811 using such operations.
# | If it were possible to get 811 from 1, then by performing the operation of permutation and division of the number by 2, it would be possible to get 1 from 811. Let's try: permutation of digits results only in the numbers 811, 181, 118. Two of these numbers are odd. Dividing 118 by 2 results in the odd number 59, and pe... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,106 |
What is the maximum number of kings that can be placed on an $8 \times 8$ chessboard without any of them attacking each other?
# | In a $2 \times 2$ square, no more than one king can be placed. There are 16 non-overlapping $2 \times 2$ squares on a chessboard, so no more than 16 kings can be placed. Indeed, 16 kings can be placed if, for example, they are placed in the top-left corner of each square.
 Find this rule.
b) Find all natural numbers that transform into themselves (according to this rule).
c) Prove that the number $2^{1991}$ will become a single... | a) The law can be guessed, noticing, for example, that while the number is a single digit, it doubles, and then - apparently not. And the fact that 10 turns into 2 suggests that it is not the number itself that doubles, but the sum of its digits. Thus, the sought law has been discovered: "Double the sum of the digits."... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,109 |
6,7
Authors: Spivak A.V., Yatsenko I.V.
Can nails be hammered into the centers of 16 cells of an $8 \times 8$ chessboard so that no three nails lie on the same straight line?
# | See the figure:
## Answer
It is possible. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,111 |
Kovaldzhi A.K.
Find at least two pairs of natural numbers for which the equality $2 x^{3}=y^{4}$ holds. | Noticing that (2, 2) is a solution, try to find another one in the form $x=2^{k}, y=2^{n}$.
## Solution
Notice that $x=2, y=2$ is a solution. Let's try to find another one in the form $x=2^{k}, y=2^{n}$.
We have $2 \cdot\left(2^{k}\right)^{3}=\left(2^{n}\right)^{4}$, or $2^{3 k+1}=2^{4 n}$.
It remains to choose $k$... | (2,2),(32,16) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,112 |
Yaishchenko I.V.
In Mexico, environmentalists have succeeded in passing a law according to which each car must not be driven at least one day a week (the owner reports to the police the car's number and the "day off" for the car). In a certain family, all adults wish to drive daily (each for their own business!). How ... | b) If no more than one car "rests" each day, then there are no more than 7 cars in total.
## Solution
a) Five cars are not enough, because on the day when one of the cars is "resting," someone will have no car to ride in. Six cars, obviously, are enough.
b) If no more than one car "rests" each day, then the total nu... | 6;10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,113 |
Evdokimov M.A.
Place 32 knights on a chessboard so that each of them attacks exactly two others.
# | In a $3 \times 3$ square, it is possible to visit all cells except the central one with a knight's move and return to the starting cell.
## Solution
If knights are placed on all cells of a $3 \times 3$ square except the central one, each knight will attack exactly two others. Now, place four such "squares" on the boa... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,114 |
At the Olympiad, $m>1$ schoolchildren solved $n>1$ problems. All schoolchildren solved a different number of problems. All problems were solved by a different number of schoolchildren. Prove that one of the schoolchildren solved exactly one problem. | If a student is found who has not solved a single problem, we will not consider him. Then, if there is a problem that has not been solved by any of the students, we will not consider it. Still, all students have solved a different number of problems, and all problems have been solved by a different number of students. ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,118 |
Auto: Kovamidjei A.K.
Is the number $4^{9}+6^{10}+3^{20}$ prime? | $4^{9}+6^{10}+3^{20}=\left(2^{9}\right)^{2}+2 \cdot 2^{9} \cdot 3^{10}+\left(3^{10}\right)^{2}=\left(2^{9}+3^{10}\right)^{2}$.
This expression can be translated into English as follows:
$4^{9}+6^{10}+3^{20}=\left(2^{9}\right)^{2}+2 \cdot 2^{9} \cdot 3^{10}+\left(3^{10}\right)^{2}=\left(2^{9}+3^{10}\right)^{2}$. | (2^{9}+3^{10})^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,119 |
The numbers 1 and 2 are written on the board. Each day, the scientific consultant Vybegallo replaces the two written numbers with their arithmetic mean and harmonic mean.
a) One day, one of the written numbers (it is unknown which one) turned out to be 941664/665857. What was the other number at that moment?
b) Will ... | a) 665857/470832: the product of the numbers written on the board is always equal to 2.
b) No: the arithmetic mean decreases each time, while the harmonic mean increases. After the second step, the arithmetic mean will be equal to $17 / 12$ < 35/24, so 35/24 will never be encountered again.
## Answer
## Problem | No | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,120 |
$[$ Modular arithmetic (miscellaneous). $]$
Ivan the Tsarevich has two magic swords. With the first one, he can cut off 21 heads of Zmey Gorynych. With the second one, he can cut off 4 heads, but then 2006 heads grow back on Zmey Gorynych. Can Ivan cut off all of Zmey Gorynych's heads if he initially had 100 heads? (I... | The remainder of the number of heads of the Zmey Gorynych modulo 7 does not change, and at the beginning, this remainder is equal to 2.
## Answer
It will not be able to. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,121 |
## [Coordinate method in space ]
[ equation of a plane $]$
Form the equation of the plane passing through the point $M 0(x 0 ; y 0 ; z 0)$ perpendicular to a non-zero vector $\vec{n}$ $=(a ; b ; c)$. | Point $M(x ; y ; z)$ belongs to the desired plane $\alpha$ if and only if the vector $\overrightarrow{M_{0} M}=(x-x 0 ; y-y 0 ; z-z 0)$ is perpendicular to vector $\vec{n} \cdot$ Therefore,
$$
\begin{aligned}
& M(x ; y ; z) \in \alpha \Leftrightarrow \vec{n} \perp \overrightarrow{M_{0} M} \Leftrightarrow \vec{n} \cdot... | (x-x_0)+b(y-y_0)+(z-z_0)=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,122 |
[ Coordinate method in space ] [ Parametric equations of a line ]
A line $l$ passes through the point $M 0\left(x 0 ; y_{0} ; z 0\right)$ parallel to a non-zero vector $\vec{m}=(a ; b ; c)$. Find the necessary and sufficient condition for the point $M(x ; y ; z)$ to lie on the line $l$.
# | Point $M$ lies on the line $l$ if and only if the vector $\overrightarrow{M_{0} M}$ is collinear with the vector $\vec{m}$, which means,
$$
\begin{aligned}
& M\left(x 0 ; y_{0} ; z 0\right) \in l \Leftrightarrow \overrightarrow{M_{0} M}=\vec{m} \Leftrightarrow \\
& \left\{\begin{array} { l }
{ x - x _ { 0 } = a t } \... | {\begin{pmatrix}x_{0}+\\y_{0}+\\z_{0}+,\text{or}\frac{x-x_{0}}{}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{}\end{pmatrix}.} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,123 |
## [Coordinate method in space $]$
[ equation of a plane $]$ Two planes are given by the equations $A 1 x+B 1 y+C 1 z+D 1=0$ and $A 2 x+B 2 y+C 2 z+D 2=0$. Let $\alpha-$ be the magnitude of the non-obtuse angle formed by the planes. Prove that
$$
\cos \alpha=\frac{\left|A_{1} A_{2}+B_{1} B_{2}+C_{1} C_{2}\right|}{\s... | The non-obtuse angle $\phi$ between planes is either equal to the angle between vectors $\overrightarrow{n_{1}}=(A 1 ; B 1 ; C 1)$ and $\overrightarrow{n_{2}}=(A 2 ; B 2 ; C 2)$, which are respectively perpendicular to the given planes, or complements it to $90^{\circ}$. Therefore,
$$
\cos \phi=|\cos \alpha|=\left|\fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,125 |
$\left[\begin{array}{ll}\text { Coordinate Method in Space } \\ \text { Regular Tetrahedron }\end{array}\right]$
Find the angle between the intersecting medians of two faces of a regular tetrahedron. | ## First Method.
Let's find the angle $\alpha$ between the medians $D M$ and $A K$ of the faces $A D B$ and $A B C$ of a regular tetrahedron $A B C D$ (Fig.1). For this, we draw a line through the center $Q$ of the face $A B C$ parallel to $D M$. Let this line intersect the edge $C D$ at point $S$. Then the desired an... | \arccos\frac{1}{6},\arccos\frac{2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,126 |
$\left[\begin{array}{l}\text { Substitution of variables (other) }\end{array}\right]$
$\left[\begin{array}{l}\text { Completing the square. Sums of squares }\end{array}\right]$
$\left[\begin{array}{l}\text { Polynomials (other) }\end{array}\right]$
Given the polynomial $x(x+1)(x+2)(x+3)$. Find its minimum value. | $x(x+3)(x+1)(x+2)=\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)$. Let's denote $x^{2}+3 x$ by $z$. Then $\left(x^{2}+3 x\right)\left(x^{2}+3 x+2\right)=z(z+2)=(z+1)^{2}-1$. The minimum value -1 of this function is reached when $z=-1$. The equation $x^{2}+3 x+1=0$ has solutions (the discriminant is greater than zero), ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,128 |
10,11
Prove that the sums of the squares of the distances from an arbitrary point in space to the opposite vertices of a rectangle are equal to each other.
# | Let $A B C D$ be a rectangle with sides $A B=a$ and $A D=b$. We choose a rectangular coordinate system, directing the $O X$ axis along the ray $A B$, the $O Y$ axis along the ray $A C$, and the $O Z$ axis along the ray starting at point $A$ and perpendicular to the plane of the rectangle. Let $M(x ; y ; z)$ be an arbit... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,131 |
[ Investigation of a quadratic trinomial ]
Do there exist numbers $p$ and $q$ such that the equations $x^{2}+(p-1) x+q=0$ and $x^{2}+(p+1) x+q=0$ each have two distinct roots, while the equation
$x^{2}+p x+q=0$ has no roots? | Let's take, for example, $p=0, q=0.1$. Then the first two equations have the same positive discriminant ( $D=1^{2}-4 \cdot 0.1>0$ ), while the third equation has the form: $x^{2}+0.1=0$.
## Answer
They exist. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,132 |
Tolpygo A.K.
A blindfolded magician gives the audience five cards numbered from 1 to 5. The audience hides two cards and gives the remaining three to the magician's assistant. The assistant points to two of these cards, and the audience calls out the numbers on these cards to the magician (in any order they choose). A... | Number the vertices of a regular pentagon from 1 to 5. By segments, we mean its sides and diagonals. A pair of cards hidden by the audience corresponds to one of the segments. Among the three cards the assistant has, there is a pair corresponding to a parallel segment. It is this pair that the assistant names to the ma... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,135 |
Kozhevnikov P.A.
In 10 boxes, there are pencils (there are no empty boxes). It is known that the number of pencils in different boxes is different, and in each box, all pencils are of different colors. Prove that it is possible to choose a pencil from each box so that all of them are of different colors. | Let's arrange the boxes in ascending order of the number of pencils. Note that the $n$-th box contains no less than $n$ pencils (of no less than $n$ colors). From the first box, we will take any pencil, from the second box - a pencil of a different color, from the third box - a pencil of a third color (different from t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,136 |
3 [ Coloring $\quad]$
Tom Sawyer has taken on the task of painting a very long fence, adhering to the condition: any two boards, between which there are exactly two, exactly three, or exactly five boards, must be painted in different colors. What is the minimum number of paints Tom will need for this job? | Two colors (let's say white and red) are not enough: painting board number 1 in white, Tom will be forced to paint boards with numbers 4, 5, and 7 in red. Then between the red boards numbered 4 and 7, there will be exactly two boards, which violates the condition.
Three colors are sufficient: Tom can paint three board... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,137 |
The function $f(x)$ is defined for all $x$, except 1, and satisfies the equation: $(x-1) f\left(\frac{x+1}{x-1}\right)=x+f(x)$. Find $f(-1)$.
# | Substitute the values $x=0$ and $x=-1$ into the given equation. We get: $\left\{\begin{array}{c}-f(-1)=f(0), \\ -2 f(0)=-1+f(-1)\end{array}\right.$. Therefore, $2 f(-1)=-1+f(-1)$, which means $f(-1)=-1$.
## Answer
$-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,138 |
Binkov Yu.A.
A square and a rectangle of the same perimeter have a common angle. Prove that the point of intersection of the diagonals of the rectangle lies on the diagonal of the square.
# | The first method. Let $A B C D$ and $A B_{1} C_{1} D_{1}$ be the given square and rectangle (see Fig. a). Note that it is sufficient to prove that point $O$ is the midpoint of segment $B_{1} D_{1}$ and is equidistant from the lines $A B$ and $B C$. From the condition, it follows that $A B_{1} + A D_{1} = A B + A D$, th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,139 |
$[\quad$ Triangles (miscellaneous) [Examples and counterexamples. Constructions]
In triangles $A B C$ and $A_{1} B_{1} C_{1}: \angle A=\angle A_{1}$, the altitudes drawn from vertices $B$ and $B_{1}$ are equal, as well as the medians drawn from vertices $C$ and $C_{1}$. Are these triangles necessarily equal? | Let's provide an example of two unequal triangles for which all the equalities from the problem's condition are satisfied. On one of the sides of the acute angle $A$, we mark an arbitrary segment $A B$ (see the figure). From the midpoint $K$ of segment $A B$, we drop a perpendicular $K H$ to the other side of the angle... | Notnecessarily | Geometry | proof | Yes | Yes | olympiads | false | 54,141 |
Proizvolov V.V.
There are 40 weights with masses of 1 g, 2 g, ..., 40 g. From these, 10 weights with even masses were chosen and placed on the left pan of the scales. Then, 10 weights with odd masses were chosen and placed on the right pan of the scales. The scales turned out to be in equilibrium.
# | Let's divide the weights into pairs with a difference of 20 g: $(1,21),(2,22), \ldots,(20,40)$. If there is exactly one weight from each pair on the scales, then (regardless of the choice of weights in the pairs) the weight on the odd side is divisible by 20, while the weight on the even side is not. Contradiction. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,142 |
Baranov D.V.
The guests at a round table ate raisins from a basket containing 2011 raisins. It turned out that each person ate either twice as many or 6 fewer raisins than their right neighbor. Prove that not all the raisins were eaten.
# | The left neighbor of the one who ate the least ate twice as much, which is an even number of raisins. Then his left neighbor also ate an even number of raisins. Going around the circle, we see that everyone ate an even number of raisins. Therefore, the total number of raisins eaten is even. Since the number 2011 is odd... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,144 |
There are 100 boxes, numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all of t... | Since the order of the answers to his questions is unknown to the audience, he must make a mistake-free choice, knowing only the number of "no" answers. If $N$ notes are sent, the number of "no" answers heard can take any integer value from 0 to $N$, meaning there are $N+1$ possible outcomes. This number must determine... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,145 |
Agakhanov N.X.
At a round table, 30 people are sitting - knights and liars (knights always tell the truth, while liars always lie). It is known that each of them has exactly one friend at the same table, and a knight's friend is a liar, while a liar's friend is a knight (friendship is always mutual). When asked, "Is y... | All those sitting at the table are paired as friends, which means there are an equal number of knights and liars. Consider any pair of friends. If they are sitting next to each other, the knight will answer "Yes" to the given question, and the liar will answer "No." If they are not sitting next to each other, their ans... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,146 |
Eemenennov L.A.
The Magician lays out 36 cards in a 6×6 square (6 columns of 6 cards each) and asks the Spectator to mentally choose a card and remember the column containing it. After this, the Magician collects the cards in a certain way, lays them out again in a 6×6 square, and asks the Spectator to name the number... | Let the Magician, after the first action, not shuffle the cards but collect them without disrupting the order in the columns, and stack them into the deck one column at a time. The second time, he lays out the cards row by row, meaning the former columns become rows. After the Spectator's response, the Magician knows t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,147 |
$\underline{\text { Folklore }}$
Solve the equation: $2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}=\frac{10}{x-4}$. | Since the left, and therefore the right, part of the equation takes only positive values, then $x>4$. On the interval $(4,+\infty)$, the function
$f(x)=2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}$ is increasing, while the function $g(x)=\frac{10}{x-4}$ is decreasing, so the equation $f(x)=g(x)$ has no more than one root.
It rem... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,149 |
Bogdanov I.I.
For some 2011 natural numbers, all their 2011$\cdot$1005 pairwise sums were written on the board.
Could it be that exactly one third of the written sums are divisible by 3, and another exactly one third of them give a remainder of 1 when divided by 3? | Let's consider, for example, the numbers from 1 to 2011. Among them, $2010: 3=670$ numbers are divisible by 3, the same number give a remainder of 2 when divided by 3, and 671 numbers give a remainder of 1. Therefore, $670 \cdot 669: 2 + 670 \cdot 671 = 335 \cdot 2011 = 2011 \cdot 1005$: 3 sums are divisible by three a... | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,150 |
$\underline{\text { Berlov S.L. }}$
A five-digit number is called non-decomposable if it cannot be factored into the product of two three-digit numbers.
What is the maximum number of consecutive non-decomposable five-digit numbers? | The smallest number that can be represented as the product of two three-digit numbers is $100 \cdot 100=10000$. The next such number is:
$100 \cdot 101=10100$, so the numbers $10001,10002, \ldots, 10099$ - are non-decomposable. Thus, there are 99 consecutive non-decomposable five-digit numbers.
More than 99 non-decom... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,151 |
3 [ Examples and Counterexamples. Constructions ]
Authors: $\underline{\text { Martynov P., }}$ Martynova H.
Six positive irreducible fractions are recorded, the sum of whose numerators is equal to the sum of their denominators. Pasha converted each of the improper fractions to a mixed number. Is it necessarily true ... | For example, consider the fractions: $3 / 2=1 \frac{1}{2}, 7 / 3=2 \frac{1}{3}, 13 / 4=3 \frac{1}{4}, 21 / 5=4 \frac{1}{5}, 31 / 6=5 \frac{1}{6}$ and $1 / 56$. The sum of their numerators, as well as the sum of their denominators, is 76. At the same time, both the integer parts and the fractional parts of these numbers... | notnecessarily | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,152 |
Folklore
In a box, there are 2011 white and 2012 black balls. Two balls are randomly drawn. If they are of the same color, they are discarded and a black ball is placed back into the box. If they are of different colors, the black ball is discarded and the white ball is placed back into the box. The process continues ... | For any combination of balls drawn, the number of white balls either remains unchanged (if two black balls or balls of different colors are drawn) or decreases by 2 (if two white balls are drawn). Thus, the number of white balls remains odd. Since there is one ball left in the box, it can only be white.
## Answer
Whi... | White | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,153 |
Kosukhin O.n.
Sasha found that there were exactly $n$ working digit buttons left on the calculator. It turned out that any natural number from 1 to 99999999 can either be entered using only the working buttons, or obtained as the sum of two natural numbers, each of which can be entered using only the working buttons. ... | Let's show that the conditions of the problem are met if the buttons with digits $0,1,3,4,5$ remain functional.
Indeed, any digit from 0 to 9 can be represented as the sum of some two "functional" digits. Let the number from 1 to 99999999 that we want to obtain consist of digits $a_{1}, a_{2}, \ldots, a_{8}$ (some of ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,154 |
| [ | Hexagons |
| :---: | :---: |
| | Regular Polygons | Author: Shapovalov A.B. In a circle of radius 1, several chords are drawn, the total length of which is also 1. Prove that a regular hexagon can be inscribed in the circle such that its sides do not intersect these chords.} | We will color the smaller of the arcs stretched by the drawn chords. If we shift the colored arcs so that the corresponding chords form a broken line, the distance between the ends of this broken line will be less than 1. Since a chord of length 1 spans an arc equal to $1 / 6$ of the circle, the sum of the colored arcs... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,155 |
Frankin B.R.
a) Does there exist a triangle in which the smallest median is longer than the largest bisector? | Let in triangle $A B C$ the lengths of sides $B C, A C, A B$ be $a, b, c$ respectively, and $a \leq b \leq c$, and $A H$ is the altitude. a) Let $C M$ be the median, $A L$ be the angle bisector. If angle $C$ is obtuse or right, then $A L > A C$. Since $B C \leq$ $A C$, angle $C M A$ is obtuse or right, so $C M \leq A C... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,156 |
10,11 |
| :---: | :---: | :---: |
| | [ Proof by contradiction ] | |
| | Even and odd | |
| | Induction (other) | |
300 positive numbers are arranged in a circle. Could it happen that each of these numbers, except one, is equal to the difference of its neighbors? | Assume that the required arrangement exists. It is clear that the largest of the numbers cannot equal the difference of its neighbors; therefore, each of the other numbers equals the difference of its neighbors. In particular, the largest number occurs exactly once; denote it by $m$.
Let $d$ be one of the smallest num... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,157 |
$\underline{\text { Zhukov G. }}$.
2015 natural numbers are written in a circle such that any two adjacent numbers differ by their greatest common divisor.
Find the largest natural number $N$ that is guaranteed to divide the product of these 2015 numbers. | Evaluation. Two odd numbers cannot stand next to each other, as they do not divide their even difference. Therefore, there are no fewer than half of the numbers that are even, that is, at least 1008. Since there are more than half of the numbers that are even, some two even numbers must stand next to each other. In thi... | 3\cdot2^{1009} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,158 |
Folkpor
For each integer from $n+1$ to $2n$ inclusive ($n$ is a natural number), take the greatest odd divisor and sum all these divisors.
Prove that the result is $n^{2}$. | Induction on $n$. The base case ($n=1$) is obvious.
Induction step. Replacing the number $n+1$ in the set of numbers from $n+1$ to $2n$ with $2n+2$ and adding $2n+1$, we obtain the set of numbers from $n+2$ to $2(n+1)$. The sum of the divisors will not change, as the greatest odd divisors of the numbers $n+1$ and $2(n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,160 |
Bystriiov A.
It is known that among the members of a certain arithmetic progression $a_{1}, a_{2}, a_{3}, a_{4}, \ldots$ there are numbers $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}$.
Prove that this progression consists of integers. | $a_{3}-a_{2}=a_{2}-a_{1}=d-$ the difference of the progression. $a_{1}+a_{2}=\frac{a_{2}^{2}-a_{1}^{2}}{d}$ and $a_{2}+a_{3}=\frac{a_{3}^{2}-a_{2}^{2}}{d}$ - integers, so $a_{3}-a_{1}=2 d$ is an integer, and $d$ is an integer or a half-integer. Since $2 a_{1}+d=a_{1}+a_{2}$ is an integer, there are three cases: $a_{1}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,161 |
Authors: Shapovalov A.V., Yatsenno I.V.
At the "Come on, creatures!" competition, 15 dragons are standing in a row. The number of heads of neighboring dragons differs by 1. If a dragon has m... | It is convenient to represent a row of dragons as a graph: instead of each dragon, we draw a point at a height corresponding to the number of heads the dragon has, and connect these points.
a) See the figure.
b) First, note that somewhere between every two cunning dragons stands a strong one. Indeed, if we walk along... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,164 |
$\underline{\text { Folklore }}$
Among the actors of Karabas Barabas theater, a chess tournament was held. Each participant played exactly one game with each of the others. One solido was given for a win, half a solido for a draw, and nothing for a loss. It turned out that among any three participants, there would be ... | Example. Let's denote the participants with letters A, B, V, G, D. Suppose A won against B, B won against V, V won against G, G won against D, D won against A, and all other matches ended in a draw. The condition of the problem is satisfied.
Evaluation. From the condition, it follows that for this tournament, two stat... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,165 |
| [ Principle of the Extreme (other) $]$ |
| | [Proof by Contradiction |
## Author: Makin M.I.
Several positive numbers, each not greater than 1, are placed on a circle. Prove that the circle can be divided into three arcs such that the sums of the numbers on adjacent arcs differ by no more than 1. (If an arc contai... | The first method. Let's call the weight of an arc the sum of the numbers on it (the weight of an arc without numbers is 0), and the range - the difference between the largest and smallest weight. The number of distinct weight partitions is finite; we choose the partition with the smallest range. Let's prove that this i... | proof | Other | proof | Yes | Yes | olympiads | false | 54,166 |
$4-$ Arithmetic progression $\quad]$ [Pairings and groupings; bijections]
Authors: Dynkin E.B., Moomanov S.A., Rozental A.L., Tomingo A.K.
For which $n$ can weights of 1 g, 2 g, 3 g, ..., $n$ g be divided into three equal-mass piles? | Let's show that this can be done if and only if $n>3$ and one of the numbers $n$ or $n+1$ is divisible by 3. The necessity of these conditions is obvious, since the total weight of the weights $1+2+3+\ldots+n=1 / 2 n(n+1)$ must be divisible by 3.
To prove sufficiency, note first that the partition is possible for $n$ ... | n>3,n\equiv0,2(\bmod3) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,169 |
Friedman G.S.
Given $n$ points, $n>4$. Prove that it is possible to connect them with arrows so that from any point to any other point it is possible to get by passing either one arrow or two (any two points can be connected by an arrow in only one direction; it is possible to travel along an arrow only in the directi... | Induction by n. Base. For $n=5$ the required graph is represented in the left figure.
Induction step. Consider $n+1$ points. Suppose $n$ of them are already connected - we get a graph with $n$ vertices. We can assume that each pair of these $n$ points is connected by an arrow: otherwise, we will draw all the missing a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,170 |
Zelivinsky A.V.
In all cells of a $100 \times 100$ table, there are pluses. It is allowed to simultaneously change the signs in all cells of one row or all cells of one column. Is it possible, using only these operations, to obtain exactly 1970 minuses? | Let in the $i$-th row we changed the sign of $x_{i}$ times, in the $k$-th column - $y_{k}$ times. Then in the cell at the intersection of the $i$-th row and the $k$-th column, the sign will change
$x_{i}+y_{k}$ times. Therefore, in this cell, there will be a minus if and only if $x_{i}+y_{k}$ is odd. Thus, the total n... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,175 |
Auto:: KuremndrikLl:
In a $10 \times 10$ table, you need to write the digits $0,1,2,3, \ldots, 9$ in some order, so that each digit appears 10 times.
a) Is it possible to do this in such a ... | a) See fig. (numbers are replaced by colors)
b) Let in each row there are no more than three different digits, and digit $n$ appears in $a_{n}$ rows and $b_{n}$ columns. The sum $a_{0}+\ldot... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,177 |
Sidorenko A.F.
There are 21 points on a circle.
Prove that among the arcs with these points as endpoints, there will be no fewer than a hundred such that their angular measure does not exceed $120^{\circ}$. | We will prove the statement for $2 n+1$ points and $n^{2}$ arcs by induction. The base case is $n=0$.
Inductive step. Suppose we have $2 n+3$ points. Consider a "long" (more than $120^{\circ}$) arc $A B$ with endpoints among these points (if there are none, then there are more than enough "short" arcs). Let $C$ be any... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,178 |
$\underline{\text { Vasiliev N.B. }}$
How many maximum parts can the coordinate plane xOy be divided into by the graphs of 100 quadratic trinomials of the form
$y=a_{n} X^{2}+b_{n} x+c_{n}(n=1,2, \ldots, 100) ?$ | We will prove by induction that $n$ parabolas of the specified type can divide the plane into no more than $n^{2}+1$ parts.
Base case. One parabola divides the plane into $2=1^{2}+1$ parts.
Inductive step. The $n$-th parabola intersects each of the others at no more than two points. These intersection points, of whic... | 10001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,179 |
Avor: Konieviechi $M$.
The numerical sequence $\left\{x_{n}\right\}$ is such that for each $n>1$ the condition is satisfied: $x_{n+1}=\left|x_{n}\right|-x_{n-1}$.
Prove that the sequence is periodic with a period of 9. | Consider the transformation $f$ on the coordinate plane that maps a point $(x, y)$ to the point $f(x, y) = (|x| - y, x)$. Then $f\left(x_{n}, x_{n-1}\right) = \left(x_{n+1}, x_{n}\right)$. Therefore, it is sufficient to prove that applying the transformation $f$ nine times returns all points to their original positions... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,180 |
Shapovalov A.V.
A rectangle is divided into right-angled triangles, adjacent to each other only along entire sides, such that the common side of two triangles always serves as the leg of one and the hypotenuse of the other. Prove that the ratio of the longer side of the rectangle to the shorter side is at least 2. | Let the total number of triangles be $n$. Then there are $-2n$ legs and $-n$ hypotenuses. Suppose $m$ of these $n$ hypotenuses lie on the boundary of the rectangle, then inside it lie $n-m$ hypotenuses and the same number of legs, and on the sides there are $n+m$ legs. Therefore, on the boundary lie $m+(n+m)=2m+n$ vert... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,184 |
Shapovalov A.V.
a) On the board, 100 different numbers are written. Prove that among them, one can choose eight numbers such that their arithmetic mean cannot be represented as the arithmetic mean of any nine of the numbers written on the board.
b) On the board, 100 integers are written. It is known that for any eigh... | a) Consider the eight smallest numbers. By adding the ninth number to them, we increase the arithmetic mean. Thus, the arithmetic mean of the eight smallest numbers is less than the arithmetic mean of the nine smallest numbers (and, therefore, less than the arithmetic mean of any other nine numbers).
b) Consider the n... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,186 |
}
Vanya receives exactly one grade ("3", "4", or "5") for each of seven subjects every week. He considers a week successful if the number of subjects with improved grades exceeds the number of subjects with worsened grades by at least two. It turned out that $n$ weeks in a row were successful, and in the last of these... | Let's calculate the sum of Vanya's grades for each week. Let's see how the sum of grades could change on a successful week. Suppose Vanya's grades worsened in $x$ subjects. Then they improved in at least $x + 2$ subjects. Therefore, $x + (x + 2) \leq 7$, which means $x \leq 2$. The sum of grades for the subjects where ... | n\inN\backslash{1,2,4,5,8,11} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,190 |
Tokarev S.i.
a) Eight schoolchildren were solving eight problems. It turned out that each problem was solved by five schoolchildren. Prove that there will be such two schoolchildren that each problem was solved by at least one of them.
b) If each problem was solved by four students, it may happen that such a pair doe... | a) We will call a triplet consisting of two students and a problem that neither of them solved a marked triplet. Since each problem was not solved by three students, each problem corresponds to three marked triplets. Therefore, there are 24 marked triplets in total. On the other hand, there are $7 \cdot 8: 2=28$ pairs ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,191 |
Mitkin D.
The lengths of the sides of a triangle are prime numbers. Prove that its area cannot be an integer.
# | Let the lengths of the sides of the triangle be $a, b, c$. From Heron's formula, we have: $16 S^{2}=P(P-2 a)(P-2 b)(P-2 c)$, where $S-$ is the area, and $P=a+b+c$ is the perimeter of the triangle.
Suppose that $S$ is an integer. Then $P$ is even (if $P$ is odd, then $16 S^{2}$ is odd, which is not the case). Therefore... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,194 |
There are seven glasses of water: the first glass is half full, the second is one-third full, the third is one-fourth full, the fourth is $1 / 5$ full, the fifth is $1 / 8$ full, the sixth i... | a) If the capacity of a glass is considered to be 1, then the first three glasses together contain $1 \frac{1}{12}$ of water. Pour all the water from the second glass into the first, and then from the third glass into the first until the first glass is full. After this, the third glass will contain $1 / 12$.
b) We wil... | ) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,195 |
7,8,9 |
| :---: | :---: | :---: |
| | [ Pairing and grouping; bijections | |
| | [ Pigeonhole Principle (other). | |
| | Estimation + example | |
At a joint conference of the party of li... | Let's divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups sit only truth-tellers, which is impossible.
| | $\pi$ | | | | $\pi$ | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | | | $\pi$ | | | | $... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,196 |
Gooovanov A.S.
Do there exist three natural numbers, greater than 1, such that the square of each, decreased by one, is divisible by each of the others? | Let $a \geq b \geq c$ be numbers satisfying the conditions of the problem. Since $a^{2}-1$ is divisible by $b$, the numbers $a$ and $b$ are coprime. Therefore, the number $c^{2}-1$, which by condition is divisible by $a$ and $b$, must also be divisible by their product, hence $c^{2}-1 \geq a b \geq c^{2}$. Contradictio... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,197 |
$\underline{\text { Proposition } A}$.
A perfect number greater than 6 that is divisible by 3 is also divisible by 9. Prove this. | Suppose a perfect number is equal to $3 n$, where $n$ is not divisible by 3. Then all natural divisors of the number $3 n$ (including itself) can be divided into pairs $d$ and $3 d$, where $d$ is not divisible by 3. Therefore, the sum of all divisors of the number $3 n$ (which is equal to $6 n$) is divisible by 4. Henc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,198 |
Berroov S.L.
From the interval $\left(2^{2 n}, 2^{3 n}\right)$, $2^{2 n-1}+1$ odd numbers are chosen.
Prove that among the chosen numbers, there will be two such that the square of each does not divide the other. | Note that among the selected numbers, there will be numbers $a$ and $b$ that have the same remainder when divided by $2^{2 n}$. Let's prove that they are the ones we are looking for. Suppose that $a^{2}$ is divisible by $b$. Then $(a-b)^{2}=a^{2}-2 a b+b^{2}$ is also divisible by $b$. Let $a=p \cdot 2^{2 n}+r, b=q \cdo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,199 |
Bogdanov I.I.
Do there exist such pairwise distinct natural numbers $m, n, p, q$ that $m+n=p+q$ and $\sqrt{m}+\sqrt[3]{n}=\sqrt{p}+\sqrt[3]{q}>2004 ?$ | We will look for numbers in the form $m=a^{2}, n=b^{3}, p=c^{2}, q=d^{3}$, where $a, b, c, d$ are natural numbers. Then $a+b=c+d$, $a^{2}+b^{3}=c^{2}+d^{3}$, which means $a-c=d-b$,
$(a-c)(a+c)=(d-b)\left(d^{2}+b d+b^{2}\right)$. Fix such $b$ and $d$ that $b=d-1>2004$. Then the conditions are satisfied by the pair $c=1... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,200 |
Tokarev S.I.
Find all such pairs of prime numbers $p$ and $q$ that $p^{3}-q^{5}=(p+q)^{2}$. | Let neither of the numbers $p, q$ be divisible by 3. If the remainders of $p$ and $q$ when divided by 3 are the same, then the left side is divisible by 3, but the right side is not; if these remainders are different, then the right side is divisible by 3, but the left side is not.
Let $p=3$. From the equation $27-q^{... | p=7,q=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,201 |
Aghakhanov $H . X$.
Are there such $n$-digit numbers $M$ and $N$, that all digits of $M$ are even, all digits of $N$ are odd, each digit from 0 to 9 appears in the decimal representation of $M$ or $N$ at least once, and $M$ is divisible by $N$? | Let $M=\overline{a_{1}} \overline{-. . a_{n}^{-}}$ and $N=\overline{b_{1}} \overline{-\ldots b_{n}^{-}}-$ be numbers satisfying the condition. Then $M=d N$, where $d=2,4,6$ or 8 (since $M$ is even, and $N$ is odd). Let $b_{k}=9$ and
$S=\bar{b}_{\bar{k}}^{-\ldots .} \bar{b}_{n}^{-}$. Then from the inequalities $180 \ld... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,203 |
Golovanov A.S.
Prove that for any polynomial $P$ with integer coefficients and any natural number $k$, there exists a natural number $n$ such that $P(1) + P(2) + \ldots + P(n)$ is divisible by $k$.
# | Note that the numbers $P(r)$ and $P(m k+r)$ give the same remainders when divided by $k$ (see the solution to problem $\underline{35562}$). Therefore, in the sum
$P(1)+P(2)+\ldots+P\left(k^{2}\right)$ for each $r=0,1, \ldots, k-1$ there will be $k$ terms of the form $P(m k+r)$, giving the same remainders when divided ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,204 |
Bogdanov I.I.
On a chessboard, 32 white and 32 black pawns are placed in all squares. A pawn can capture pawns of the opposite color by moving diagonally one square and taking the place of the captured pawn (white pawns can only capture to the right-up and left-up, while black pawns can only capture to the left-down a... | Note that a pawn that stood on a black square will always move only on black squares. Then after each move (on black squares) there will always be at least one pawn on black squares - the one that made the move. Similarly, at least one pawn will remain on white squares, and there will be no less than two pawns in total... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,205 |
Bogdanov I.I
In the cells of a $5 \times 5$ square, zeros were initially recorded. Each minute, Vasya chose two cells with a common side and either added one to the numbers in them or subtracted one from them. After some time, it turned out that the sums of the numbers in all rows and columns were equal. Prove that th... | Let's call Vasya's move horizontal if he chose cells adjacent horizontally, and vertical otherwise. Consider the change in the sum of the numbers in the second and fourth columns. With any vertical move, the parity of this sum did not change, while with any horizontal move, it did change. Since initially this sum is ze... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,206 |
$:$ Asahanov $\boldsymbol{H - X} . \mathrm{X}$.
In the cells of a $10 \times 10$ table, natural numbers from 1 to 100 are arbitrarily placed, each exactly once. In one move, it is allowed to swap any two numbers. Prove that in 35 moves, it is possible to achieve that the sum of any two numbers in cells sharing a side ... | Let's divide the table with a vertical line $l$ into two halves. In one of the halves, for example, the right one, there will be no more than 25 even numbers. An equal number of odd numbers will be in the left half. By swapping pairs of such numbers of different parity, in no more than 25 moves, one can obtain a table ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,207 |
Bazyan A.i.
In a $10 \times 10$ square, numbers from 1 to 100 are arranged: in the first row - from 1 to 10 from left to right, in the second row - from 11 to 20 from left to right, and so on. Andrey plans to cut the square into dominoes $1 \times 2$, calculate the product of the numbers in each domino, and sum the re... | Let's number the dominoes in the partition. Let the numbers in the $i$-th domino be $a_{i}$ and $b_{i}$. Notice that $a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2}$. Summing these equalities over all dominoes, we get that the sum of all 50 products is $\frac{a_{1}^{2}+\ldots+a_{100}^{... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,208 |
Bogdanov I.I.
The distance between two cells on an infinite chessboard is defined as the minimum number of moves in the path of a king between these cells. On the board, three cells are marked, the pairwise distances between which are 100. How many cells exist such that the distances from them to all three marked cell... | Consider two arbitrary cells $A$ and $B$. Let the difference in the abscissas of their centers be $x \geq 0$, and the difference in the ordinates be $y \geq 0$. Then the distance $\rho(A, B)$ between these cells is $\max \{x, y\}$.
Let cells $A, B, C$ be marked. Then for each pair of cells, there exists a coordinate i... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,210 |
$\underset{\text { Tokarev } C . \text {. }}{ }$
Numbers $a, b$ and $c$ are such that $(a+b)(b+c)(c+a)=a b c,\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)=a^{3} b^{3} c^{3}$. Prove that $a b c=0$. | First, note that $x^{2}-x y+y^{2}>|x y|$ for any distinct numbers $x$ and $y$.
Assume that $a b c \neq 0$. Then, dividing the second equality by the first, we get $\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}\right.$
$\left.-c a+a^{2}\right)=|a b| \cdot |b c| \cdot |a c|$.
All the parentheses o... | 0 | Algebra | proof | Yes | Yes | olympiads | false | 54,211 |
Zaslavsky A.A.
In a non-isosceles triangle, two medians are equal to two altitudes. Find the ratio of the third median to the third altitude. | Let in triangle $ABC$ the inequality $1=AB<AC<BC$ holds. Then the median $AA'$ is equal to the height dropped from vertex $B$, and the median $BB'$ is equal to the height dropped from vertex $C$. Therefore, the distance from point $A'$ to the line $AC$ is half of $AA'$, which means $\angle A'AC=30^{\circ}$. Similarly, ... | 7:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,212 |
B.R.'s Problem
The segments connecting an inner point of a convex non-equilateral $n$-gon with its vertices divide the $n$-gon into $n$ equal triangles.
For what smallest $n$ is this possible? | Let's prove that the specified situation is impossible for $n=3,4$.
The first method. For $n=3$, the angles of the triangles in the partition that meet at an internal point are equal, since the sum of any two different angles in them is less than $180^{\circ}$. But then the sides opposite to them, which are sides of t... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,213 |
Berdonikov A.
Let's call a natural number good if all its digits are non-zero. A good number is called special if it has at least $k$ digits and the digits are in strictly increasing order (from left to right).
Suppose we have some good number. In one move, it is allowed to append a special number to either end or in... | Obviously, a special number does not have more than nine digits. If $k=9$, then with each operation, the number of digits changes by exactly 9, meaning the remainder of the number of its digits divided by 9 does not change, and a two-digit number cannot be made from a one-digit number.
Let $k=8$. Since all operations ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,214 |
10,11 [ $\quad$ Proof by contradiction
Auto: Shapovosoov A.
Inside a circle, 100 points are marked, no three of which lie on the same line.
Prove that they can be paired and a line can be drawn through each pair so that all intersection points of the lines are inside the circle.
# | Let's pair the points so that the sum of the lengths of the corresponding segments is maximized. Suppose, for the pairs of points $(A, B)$ and $(C, D)$, the lines $AB$ and $CD$ intersect outside the circle (see figure).
. ] | |
Prove that a connected graph with no more than two odd vertices can be drawn without lifting the pencil from the paper and tracing each edge exactly once. | Let's consider the case when a graph has no odd vertices. We will prove by induction on the number of edges in the graph that it can be traversed by a cycle. The base case (a graph with no edges) is obvious.
Inductive step. Consider an arbitrary connected graph where all vertices have even degrees. Since this graph ha... | proof | Other | proof | Yes | Yes | olympiads | false | 54,216 |
On the plane, there are $n$ points ( $n>3$ ), no three of which lie on the same line. Prove that among the triangles with vertices at these points, acute triangles do not exceed three quarters. | Among four points in general position, there will be three that form an obtuse or right-angled triangle.
## Solution
We will consider all possible sets of four points and for each set, determine the number of acute and non-acute triangles. By summing the number of acute triangles across all sets of four points, we ob... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,220 |
On a plane, there are $n$ figures. Let $S_{i_{1} \ldots i_{k}}$... | a) Let $W_{m}$ denote the area of the part of the plane covered by exactly $m$ figures. This part consists of pieces, each of which is covered by some specific $m$ figures. The area of each such piece is counted $C_{m}^{k}$ times when computing $M_{k}$. Therefore,
$$
M_{k}=C_{k}^{k} W_{k}+C_{k+1}^{k} W_{k+1}+\ldots+C_... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,224 |
On the board, there are $n$ natural numbers. In one operation, instead of two numbers that do not divide each other, you can write their greatest common divisor and their least common multip... | a) Let's call a pair of numbers correct if one of them is divisible by the other. Note that with each operation, the number of correct pairs increases. When all pairs become correct, the process will stop.
b) Let $a_{1}, \ldots, a_{n}$ be the initial numbers. It is not hard to verify that with the specified operation, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,225 |
[ Geometric Interpretations in Algebra ]
Let $a$ and $b$ be natural coprime numbers. Consider points on the plane with integer coordinates $(x, y)$ lying in the strip $0 \leq x \leq b-1$. Assign to each such point an integer $N(x, y)=a x+b y$.
a) Prove that for each natural number $c$ there is exactly one point $(x, ... | a) From problem $\underline{60489}$, it follows that the equation $a x+b y=c$
has a solution ( $x_{0}, y_{0}$ ) in integers. Then all solutions of equation (*) have the form ( $x_{0}+k b, y_{0}-k a$ ) (see problem 60514). The "distance" between consecutive numbers of the form $x_{0}+k b$ is $b$, so exactly one of them... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,226 |
[ Methods for solving problems with parameters ] [ Theorem of Bezout. Factorization ] [ Geometric interpretations in algebra ]
Investigate the systems of equations:
a) $\left\{\begin{aligned} 2 x+3 y & =5 \\ x-y & =2, \\ x+4 y & =a ;\end{aligned}\right.$
b) $\left\{\begin{aligned} x+a y & =1, \\ 2 x+4 y & =2, \\ b x... | a) The left side of the first equation is the sum of the left sides of the other two equations.
b) Compare the last two equations.
## Solution
g) Consider the given polynomial $f(t)=t^{3}-z t^{2}-y t-x$. Let $a, b, c$ be its roots.
If $a, b, c$ are distinct numbers, then $f(t)=(t-a)(t-b)(t-c)$, which means $x=a b c... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,227 |
Prove that the 13th of the month is more likely to fall on a Friday than on any other day of the week. It is assumed that we live according to the Gregorian calendar.
# | 
## Solution
Recall that a year is not a leap year if its number is not divisible by 4 and if its number is divisible by 100 but not by 400. A non-leap year consists of 52 weeks and one day, wh... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,228 |
Senderov V.A.
Find all natural numbers $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first power).
# | Let $n \geq 2$, and $2=p_{1}1$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation (*) would be divisible by $q$, which is not the case. Therefore, $a>p_{k}$.
Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, th... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,230 |
9,10,11 |
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| | [ Lines, rays, segments, and angles (miscellaneous). | |
| | [ Examples and counterexamples. Constructions $]$ | |
On a plane, $n>2$ lines in general position are drawn (that is, no two lines are parallel and no three lines intersect at the same point). These lines have divi... | a) Consider the polygon $T$, which is the union of all bounded parts. It is clear that all angles are perpendicular to the angles of $T$ that are less than $180^{\circ}$. From the formula for the sum of exterior angles, it immediately follows that there are no fewer than three such angles.
An example with three angles... | )3;b)nforoddn,n-1forevenn | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,234 |
Shapovalov A.V.
a) In a $2 \times n$ table (where $n>2$), numbers are written. The sums in all columns are distinct. Prove that the numbers can be rearranged in the table so that the sums in the columns are distinct and the sums in the rows are distinct. b) In a $10 \times 10$ table, numbers are written. The sums in a... | a) Suppose the sums in the rows are equal. If there is a column where the numbers are different, we will swap them, and the sums of the numbers in the rows will become different. If in each column both numbers are equal, then all numbers in the row are different. Take the three smallest numbers $a<b<c$ in the top row a... | notalways | Combinatorics | proof | Yes | Yes | olympiads | false | 54,235 |
Kanel-Belov A.Y.
Ali-Baba and the bandit are dividing a treasure consisting of 100 gold coins, arranged in 10 piles of 10 coins each. Ali-Baba chooses 4 piles, places a cup next to each, and puts some coins (at least one, but not the entire pile) into each cup. The bandit must then rearrange the cups, changing their in... | Let's show that Ali-Baba can achieve that in 7 piles there are no more than 4 coins, while the robber can achieve that there are no piles containing fewer than 4 coins. Therefore, Ali-Baba will take 100 - 7$\cdot$4 = 72 coins.
First, let's prove that the robber can act in such a way that there are no piles containing ... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,238 |
[ Trigonometric inequalities ] [ Monotonicity and boundedness ] [ Monotonicity, boundedness ]
Authors: Senderov V.A., Yaino I.V.
Solve the equation $\cos (\cos (\cos (\cos x)))=\sin (\sin (\sin (\sin x)))$.
# | There are no roots. We will show that for all $x \in \mathrm{R}$, the inequality
$$
\cos (\cos (\cos (\cos x)))>\sin (\sin (\sin (\sin x)))
$$
holds. It is sufficient to prove this for $x \in [0, 2\pi]$. If $x \in [\pi, 2\pi]$, the statement is obvious: for such $x$, $\cos (\cos (\cos (\cos x))) > 0$, while $\sin (\s... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,241 |
Shapovalov A.V.
Does there exist a sequence of natural numbers in which each natural number occurs exactly once and for any $k=1,2,3, \ldots$ the sum of the first $k$ terms of the sequence is divisible by $k$? | Let's specify a way to construct such a sequence. The first term $a_{1}$ can be taken as the number 1. Suppose we have managed to select the first $n$ terms $a_{1}, a_{2}, a_{n}$, and let $m$ be the smallest number not included in them, and $M$ be the largest of those included. Denote by $S_{k}$ the sum of the first $k... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,242 |
[Symmetry and Involutory Transformations] $[\quad$ Counting in Two Ways $\quad]$
## Authors: Galochkin A.I., Lyako O.
Given non-constant polynomials $P(x)$ and $Q(x)$, whose leading coefficients are equal to 1.
Prove that the sum of the squares of the coefficients of the polynomial $P(x) Q(x)$ is not less than the s... | The norm of a polynomial $R(x)=c_{n} x^{n}+c_{n-1} x^{n-1}+\ldots+c_{1} x+c_{0}$ is defined as the number $|R|=\sqrt{c_{n}^{2}+c_{n-1}^{2}+\ldots+c_{0}^{2}}$ and we set
$R^{*}(x)=c_{0} x^{n}+c_{1} x^{n-1}+\ldots+c_{n-1} x+c_{n}$.
Lemma. For any polynomials $P$ and $Q$, the equality $|P Q|=\left|P Q^{*}\right|$ holds.... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,243 |
Authors: Urrarovskiy V.A., Shaovalov A.V.
Let's call a maze a chessboard $8 \times 8$, on which some fields are separated by partitions. On the command RIGHT, the rook moves one field to the right or, if there is a board edge or partition to the right, remains in place; similarly, the commands LEFT, UP, and DOWN are e... | Number all possible initial positions, that is, pairs (maze, rook's position) - their number is finite. Compile a program $\Pi_{1}$ to traverse all fields for the first initial position. Now suppose the initial position was №2. Apply the program $\Pi_{1}$ and, if the rook does not traverse all fields, append a few comm... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 54,246 |
Find all natural numbers $n$ such that for any two coprime divisors $a$ and $b$ of $n$, the number $a + b - 1$ is also a divisor of $n$. | It is easy to check that numbers of the form $p^{k}$, where $p$ is a prime, and the number 12 satisfy the condition of the problem.
We will show that no other numbers satisfy the condition.
The case of odd $n$ is considered in 109752. Let $n$ be even and not a power of two; represent it in the form $n=2^{m} k$, where... | nispoweroforn=12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,247 |
Akooyan A.V.
Can 12 rectangular parallelepipeds $P_{1}, P_{2}, \ldots, P_{12}$ be placed in space, with their edges parallel to the coordinate axes $O x, O y, O z$, such that $P_{2}$ intersects (i.e., has at least one common point) with each of the remaining, except $P_{1}$ and $P_{3}$, $P_{3}$ intersects with each of... | Suppose this is possible. Note that two of the considered parallelepipeds intersect if and only if their projections on all three coordinate axes intersect.
Consider 4 pairs of parallelepipeds: $P_{1}$ and $P_{2}, P_{4}$ and $P_{5}, P_{7}$ and $P_{8}, P_{10}$ and $P_{11}$.
If we take parallelepipeds from different pa... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,249 |
Lenonyanov L.A.
Each vertex of a convex quadrilateral of area $S$ is reflected symmetrically with respect to the diagonal that does not contain this vertex. Denote the area of the resulting quadrilateral by $S^{\prime}$. Prove that $\frac{S^{\prime}}{S}<3$.
# | Given the specified reflection, the lengths of the diagonals of the quadrilateral are preserved. Let the acute angle between the diagonals be $\alpha$, then after the reflection, one of the angles between the diagonals becomes either $3 \alpha$ or $3 \alpha - \pi$, and therefore the ratio of the areas is $\left.\frac{\... | \frac{S^{\}}{S}<3 | Geometry | proof | Yes | Yes | olympiads | false | 54,250 |
Astashov B.
Some participants of the olympiad are friends, and the friendship is mutual. Let's call a group of participants a clique if all of them are friends with each other. Their number is called the size of the clique. It is known that the maximum size of a clique is even. Prove that the participants can be seate... | Author: Kanel-Belov A.Ya.
Let's distribute people into rooms $R_{1}$ and $R_{2}$: place the maximum clique in $R_{1}$, and the remaining people in $R_{2}$. We will start moving people from $R_{1}$ to $R_{2}$ one by one. With each move, the clique in $R_{1}$ decreases by no more than 1, and the clique in $R_{2}$ increa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,252 |
[Ribamko A.V.
In a blitz tournament, $2 n+3$ chess players participated. Each player played exactly one game with every other player. The tournament schedule was such that games were played one after another, and each player rested for at least $n$ games after playing a game. Prove that one of the players who played i... | Let's call a chess player's step the number of games between two of their consecutive games (including the second one). Then all steps are at least $n+1$.
Consider any $n+3$ consecutive games $g_{1}, \ldots, g_{n+3}$; in them, there are $2 n+6$ participants. Note that only three chess players could have participated i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,254 |
Greetings $M$. Around a round table, $N$ knights are seated. Every morning, the wizard Merlin seats them in a different order. Starting from the second day, Merlin allows the knights to make as many swaps as they like during the day: two adjacent knights can swap places if and only if they were not neighbors on the fi... | Number the knights sitting on the first day clockwise from 1 to $N$. We will describe the order at the table by a string of these numbers, listing the knights clockwise. Let's call the selected orders of the form $k, k-1, \ldots, 2,1, k+1, k+2, \ldots, N$ for $k=1,2,3, \ldots, N-1$ (the $N$-th selected order coincides ... | N | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,255 |
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