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[ [ Planar graphs. Euler's formula]
Prove that for a planar graph, the inequality $2 E \geq 3 F$ holds.
# | Each piece is bounded by no fewer than three edges.
## Solution
We place a point on each edge: one in each of the adjacent pieces. In each piece, there will be no fewer than three points, so the total number of points is no less than $3 F$. On the other hand, the number of points is exactly $2 E$.
Send a comment | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,766 |
$3+$ [ Sorting in ascending (descending) order ]
$k, l, m$ - natural numbers. Prove that $2^{k+l}+2^{k+m}+2^{l+m} \leq 2^{k+l+m+1}+1$. | We can assume that $k \geq l \geq m$. Then $2^{k+l+m+1} \geq 2^{2} \cdot 2^{k+l} > 3 \cdot 2^{k+l} \geq 2^{k+l} + 2^{k+m} + 2^{l+m}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,768 |
In the expression $1 * 2 * 3 * . . * 9$, the asterisks are replaced with minus or plus.
a) Can the result be 0?
b) Can the result be 1?
c) What numbers can result? | a) An even number cannot be obtained, since the number of odd addends (which is 5) is odd.
c) To get the number 45, all asterisks must be replaced with plus signs, and to get the number -45, with minus signs.
We will prove that if we can obtain a number $n<45$, then we can change the placement of the signs so that we... | all\odd\\from\-45\to\45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,769 |
3+ [Proof by Contradiction]
In a $25 \times 25$ square, the numbers 1 and -1 are placed. The products of these numbers are calculated for each row and each column. Prove that the sum of these products is not zero.
# | The product of all 50 works is equal to the square of the product of all numbers in the table, which is 1. See problem $\underline{30299}$ for further details. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,770 |
|
In the club, each member has one friend and one enemy. Prove that
a) the number of members is even.
b) the club can be divided into two neutral clubs.
# | a) The entire circle is divided into pairs of friends.
b) Since the degree of each vertex of the corresponding graph is 2, it breaks down into cycles. In each cycle, the edges of "friendship" and "enmity" alternate, so there are an equal number of each. By placing every other member of each cycle into one circle, we w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,772 |
In a bus, there are 20 passengers, and each has many coins of 10, 15, and 20 kopecks. Each passenger needs to pay 5 kopecks.
Can they do this using (including exchanging among themselves) a) 24 coins; b) 25 coins? | a) Let it be so. Each passenger received change, which means that out of 24 coins, no more than four, or 80 kopecks, ended up in the cash register. However, the total fare is one ruble.
b) It is sufficient to prove that four passengers can pay using five coins. The first puts a 20-kopeck coin in the cash register, the... | ) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,774 |
On the plane, there are 1983 points and a unit circle.
Prove that there is a point on the circle such that the sum of the distances from this point to the given points is not less than 1983.
# | Let's denote the points as $M_{1}, M_{2}, \ldots, M_{1983}$. Consider two arbitrary diametrically opposite points $N_{1}$ and $N_{2}$ on the circle. According to the triangle inequality, given that $N_{1} N_{2}=2$, for each point $M_{i}$ we can write $M_{i} N_{1}+M_{i} N_{2} \geq 2$. Summing these inequalities, we get ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,777 |
Case enumeration $\quad]$ [ Classical inequalities (other) ]
Find all numbers that are 12 times the sum of their digits.
# | Let the desired number contain $n$ digits. Then it is not less than $10^{n-1}$ (this is the smallest $n$-digit number). On the other hand, the sum of its digits is not more than $9 n$ (since each digit is not more than 9). Therefore, the desired number does not exceed $12 \cdot 9 n=108 n$. But according to Bernoulli's ... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,779 |
$3+$ [Examples and counterexamples. Constructions]
Can 12 circles be drawn on a plane so that each touches exactly five others? | The required arrangement is shown in the figure.
Answer
It is possible. | Itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,780 |
Bakayev E.B.
A boy and a girl were sitting on a long bench. Twenty more children approached them one by one, and each of them sat between two of the already seated children. We will call a girl brave if she sat between two neighboring boys, and a boy brave if he sat between two neighboring girls. When everyone was sea... | The first method. Let's look at the number of pairs of adjacent boys and girls. Initially, it is equal to 1. Notice that if a boy sits between two boys, the number of such pairs does not change. If he sits between a boy and a girl, he "destroys" one such pair and "creates" one, so the number of such pairs does not chan... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,781 |
[ Fundamental Theorem of Arithmetic. Factorization into Prime Factors ]
Principle of Extremality (miscellaneous)
Recurrence Relations (miscellaneous).
In the sequence of triples of integers $(2,3,5),(6,15,10), \ldots$ each triple is obtained from the previous one as follows: the first number is multiplied by the seco... | 
## Solution
It is easy to verify that each triplet has the form: $\left(2^{a_{3}} 3^{5^{c}}, 2^{c} 3^{a} 5^{b}, 2^{b} 3^{c} 5^{a}\right)$, where $a, b, c$ are some integers. Indeed, the triple... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,782 |
What is the smallest number of points that need to be marked inside a convex $n$-gon so that each triangle with vertices at the vertices of this $n$-gon contains at least one marked point? | Place points "near" the vertices of an $n$-sided polygon.
## Solution
Estimate. Since an $n$-sided polygon can be divided into $n-2$ triangles using diagonals, and each triangle must contain one point, at least $n-2$ points need to be marked.
Example. Let the vertices of the $n$-sided polygon be $A_{0}, A_{1}, \ldot... | n-2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,783 |
Prove that $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2^{n}}>\frac{n}{2}$ for $n>1$. | Evaluate the sum $\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\ldots+\frac{1}{2^{k+1}}$.
## Solution
We will divide the terms in the original sum into $n$ groups:
$\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\ldots+\left(\frac{1}{2^{n-1}+1}+\frac{1}{2^{n-1}+2}+\ldots+\frac{1}{2^{n}}\right)$. It is sufficient to show th... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,784 |
[ Induction (miscellaneous). ]
$[$ Division with remainder $]$
For which $n>3$ can a set of weights with masses $1,2,3, \ldots, n$ grams be divided into three equal-mass piles? | Six weights, whose masses are six consecutive numbers, can easily be divided into three equal-mass piles.
## Solution
Let's call a number $n$ good if a set of weights with masses $1,2,3, \ldots, n$ can be divided into three equal-mass piles. For a good $n$, the total mass of the weights, which is $1 / 2 n(n+1)$, must... | n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,786 |
What is the maximum number of parts that $n$ circles can divide a plane into?
# | If $k$ circles have already been drawn, then the ( $k+1$ )-th circle is divided by them into no more than $2 k$ arcs.
## Solution
One circle divides the plane into two parts. Suppose $k$ circles have already been drawn. Consider the ( $k+1$ )-th circle. It intersects the previous $k$ circles in no more than $2 k$ poi... | n(n-1)+2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,787 |
$2 n$ candies are distributed among $n$ boxes. A girl and a boy take one candy each in turns, with the girl choosing first.
Prove that the boy can choose the candies in such a way that the last two candies end up being from the same box. | A boy can take candies in such a way that after he takes the $k$-th candy, at least $k$ boxes are empty.
## Solution
The boy can take his first candy in such a way that at least one box is empty after this.
Indeed, after the girl's first move, there are $2n-1$ candies left in $n$ boxes, and therefore, in some box, t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,789 |
In a rectangular table, some cells are marked: stars are drawn in them. It is known that for any marked cell, the number of stars in its column coincides with the number of stars in its row. Prove that the number of rows in the table that contain at least one star is equal to the number of columns in the table that con... | We will replace each asterisk with the number $1 / k$, where $k$ is the number of asterisks in its row (column). Then the sum of the numbers in each "non-empty" row is 1, therefore, the sum of all the numbers in the table is equal to the number of "non-empty" rows. However, similarly, it is also equal to the number of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,791 |
Prove that if in the number 12008 any number of threes are inserted between the zeros, the resulting number is divisible by 19.
# | $3 \cdot 1203 \ldots 308=3609 \ldots 924=3610 \ldots 000-76$, and the numbers 361 and 76 are divisible by 19. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,792 |
On the plane, there are several lines (more than one), no two of which are parallel.
Prove that either there is a point through which exactly two of the given lines pass, or all lines
pass through one point. | Consider some line $m$ and the intersection point of two lines, different from $m$, lying closest to the line $m$.
## Solution
Let not all lines pass through one point. Consider the set of points $M$, where at least two of the given lines intersect. Choose one of the lines - $m$. By assumption, there is a point in th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,793 |
$3+$ [ Dirichlet's Principle (finite number of points, lines, etc.).]
On the plane, there are 2004 points. We will write down all pairwise distances between them.
Prove that among the recorded numbers, there are at least thirty different ones.
# | Otherwise, all points except one are located on 29 circles.
## Solution
Suppose there is a counterexample. Choose one of the given points - A. Since the distance from point $A$ to any other point takes on fewer than 30 different values, the remaining 2003 points lie on 29 circles centered at A. Since $69 \cdot 29=200... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,794 |
Given 12 sticks of the same length. How can you cut them into smaller sticks so that you can form 13 equal triangles, with each of the smaller sticks being a side of one of these triangles?
# | We need to form triangles with sides in the ratio of $3: 4: 5$.
## Solution
Let's assume the length of each stick is 13. We will cut three sticks into segments of length $3,3,3,4$; four sticks into segments of length 3, 5, 5; and five sticks into segments of length 4, 4, 5. As a result, we will have 13 sticks of leng... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,795 |
In the TV series "The Secret of Santa Barbara", 20 characters are involved. In each episode, one of the following events occurs: a certain character learns the Secret, a certain character learns that someone knows the Secret, a certain character learns that someone does not know the Secret. What is the maximum number o... | Calculate the maximum number of times each of the three types of events specified in the condition can occur.
## Solution
Each hero can learn something in no more than 39 episodes (one episode to learn the Secret, 19 episodes to learn that each of the others does not know the Secret, and another 19 episodes to learn ... | 780 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,796 |
We are considering coverings of a chessboard with dominoes, each containing two adjacent squares.
Which coverings are more numerous - those that contain the domino a1-a2, or those that contain the domino b2-b3?
# | Match each tiling that contains the domino b2-b3 with some tiling that contains the domino a1-a2.
## Solution
Exclude from consideration all tilings that contain both the domino a1-a2 and the domino b2-b3. Consider the remaining tilings. We will call a tiling of type $A$ if it contains the domino a1-a2, and a tiling ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,797 |
Four villages are located at the vertices of a square with a side length of 1 km. To allow travel from each village to any other, two straight roads were laid along the diagonals of the square. Can the road network between the villages be laid out in a different way so that the total length of the roads is reduced, but... | Try to find a road network consisting of five straight segments.
## Solution
Let $A, B, C, D$ be the villages at the vertices of the square. On the segment connecting the midpoints of sides $A B$ and $C D$, take points $M$ and $N$, which are $3 / 8$ away from sides $A B$ and $C D$. Then $A M=B M=C N=D N=5 / 8, M N=1 ... | 2.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,798 |
Someone has arranged a 10-volume collection of works in a random order. Let's call a disorder a pair of volumes where the volume with the higher number is to the left. For this arrangement of volumes, the number $S$ of all disorders has been calculated. What values can $S$ take? | If two adjacent volumes are swapped, the number of inversions will change (decrease or increase) by 1.
## Solution
It is clear that the number $S$ of inversions is not less than 0 and not more than the number of all possible pairs of 10 volumes (the number of such pairs is $10 \cdot 9: 2=45$). We will show that any v... | Anyintegervaluefrom0to45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,799 |
Prove that triangles with side lengths $a, b, c$ and $a_{1}, b_{1}, c_{1}$ are similar if and only if
$\sqrt{a a_{1}}+\sqrt{b b_{1}}+\sqrt{c c_{1}}=\sqrt{(a+b+c)\left(a_{1}+b_{1}+c_{1}\right)}$. | By squaring both sides of the given equality, it can easily be reduced to the form
$\left(\sqrt{a b_{1}}-\sqrt{b a_{1}}\right)^{2}+\left(\sqrt{c a_{1}}-\sqrt{a c_{1}}\right)^{2}+\left(\sqrt{b c_{1}}-\sqrt{c b_{1}}\right)^{2}=0$, which means ${ }^{a} / a_{a_{1}}=b / b_{b_{1}}=c / c_{c_{1}}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,803 |
Using problem $\underline{60752}$, prove that there are infinitely many prime numbers of the form $p=4k+1$.
# | Let there be $n$ such numbers: $p_{1}, \ldots, p_{n}$. Consider the number $\left(2 p_{1} \ldots p_{n}\right)^{2}+1$. According to problem 60752, it has a prime divisor of the form $p=4 k+1$. On the other hand, this number is not divisible by any of the numbers $p_{1}, \ldots, p_{n}$. Contradiction | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,807 |
[ Dirichlet's Principle (continued). ]
[ [decimal number system]
Prove that the decimal representation of the number $2^{30}$ contains at least two identical digits, without calculating it. | Apply the divisibility rule for 3.
## Solution
$2^{30}=(1024)^{3}>10^{9}$. On the other hand, $(1024)^{3}<(2000)^{3}=8 \cdot 10^{9}$. Therefore, the number $2^{30}$ is a ten-digit number. For further details, see the solution to problem $\underline{78742}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,809 |
For what $x$ and $y$ is the number $\overline{x x y y}$ a square of a natural number? | The number $\overline{x x y y}$ is divisible by 11. If it is a perfect square, then the number $\overline{x 0 y}$ is also divisible by 11.
Answer
$88^{2}=7744$
## Problem
Translate the text above into English, preserving the original text's line breaks and format, and output the translation result directly. | 88^{2}=77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,810 |
$3+$
Prove that if $(m, 10)=1$, then there exists a repunit $E_{n}$ that is divisible by $m$. Will there be infinitely many of them? | According to Euler's theorem, $9 E_{k \varphi(m)}=10^{k \varphi(m)}-1 \equiv 0(\bmod m)$ for any $k \geq 1$. If $m$ is not divisible by 3, then $E_{k \varphi(m)} \equiv 0$ $(\bmod m)$. If, however, $m$ is divisible by 3 or 9, then $m$ divides the number $E_{9 k \varphi(m)}$.
## Answer
It will. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,811 |
[ Methods for solving problems with parameters ] [ Quadratic equations. Vieta's theorem ]
It is known that the equation $x^{2}+5 b x+c=0$ has roots $x_{1}$ and $x_{2}, x_{1} \neq x_{2}$, and some number is a root of the equation $y^{2}+2 x_{1} y+2 x_{2}=0$ and a root of the equation $z^{2}+2 x_{2} z+2 x_{1}=0$. Find $... | The common root of the equations $y^{2}+2 x_{1} y+2 x_{2}=0$ and $y^{2}+2 x_{2} y+2 x_{1}=0$ is also a root of their difference equation $2\left(x_{1}-x_{2}\right) y+2\left(x_{2}-x_{1}\right)=0$. Since $x_{1}-x_{2} \neq 0$, then $y-1=0$, that is, $y=1$. Substituting into the first equation, we get $1+2\left(x_{1}+x_{2}... | \frac{1}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,812 |
[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameters ]
For which values of the parameter $a$ are both roots of the equation $(2-a) x^{2}-3 a x+2 a=0$ greater than $1 / 2$? | When $a2$ the coefficient of $x^{2}$ and the constant term have different signs, meaning one of the roots is negative. When $a=0$ one of the roots is 0, and when $a=2$ the equation has only one root.
Consider the interval $016 / 17$. For these values of $a$
$f(1 / 2)=1 / 4(2-a)-3 / 2 a+2 a=1 / 4 a+1 / 2>0$ and the x-... | 16/17<2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,813 |
[ | [ Investigation of a quadratic trinomial ] |
| :---: | :---: |
For what values of the parameter $a$ are both roots of the equation $(1+a) x^{2}-3 a x+4 a=0$ greater than 1? | Note that $a \neq -1$.
The discriminant $9a^2 - 16a(1 + a) > 0$, therefore $7a^2 + 16a - 1 > 0$. Solving this inequality narrows the range of values for $a$ to $(-\frac{16}{7}, -1)$. For such values of $a$, the branches of the parabola are directed downwards. Therefore, the value of the quadratic polynomial at the poi... | -\frac{16}{7}<-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,814 |
[ Methods for solving problems with parameters ] [ Quadratic equations. Vieta's theorem ]
For what values of the parameter $a$ does the equation $(a-1) x^{2}-2(a+1) x+2(a+1)=0$ have only one non-negative solution
# | When $a=1$, the equation becomes linear: $4-4 x=0$, the only root of which is $x=1$.
Let $a \neq 1$. The equation can have one repeated non-negative root or two distinct roots, one of which is non-negative.
In the first case, the discriminant $(a+1)^{2}-2(a-1)(a+1)=-a^{2}+2 a+3=0$, that is, $a=-1$ or $a=3$. When $a=-... | -1\leq\leq1,=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,815 |
For what value of $a$ does the polynomial $P(x)=x^{1000}+a x^{2}+9$ divide by $x+1$?
# | $P(-1)=1+a+9=a+10$. By the theorem of Bezout, this number should equal zero.
## Answer
For $a=-10$. | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,817 |
For what $a$ and $b$ does the polynomial $P(x)=(a+b) x^{5}+a b x^{2}+1$ divide by $x^{2}-3 x+2$?
# | $x^{2}-3 x+2=(x-1)(x-2)$. Therefore, the equalities $P(1)=P(2)=0$ should hold, that is, $(a+b)+a b+1=(a+1)(b+1)=0,32(a+b)+4 a b+1=0$. From the first equality, we see that one of the numbers $a, b$ equals -1. Substituting into the second equality, we find that the second number equals ${ }^{31} / 28$.
## Otvet
$\{a, b... | {-1,\frac{31}{28}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,818 |
Among all solutions of the system
$$
\begin{aligned}
& x^{2}+y^{2}=4 \\
& z^{2}+t^{2}=9 \\
& x t+y z=6
\end{aligned}
$$
select those for which the quantity $x+z$ takes the maximum value. | Consider the vectors $\boldsymbol{a}=(x, y)$ and $\boldsymbol{b}=(t, z)$. By the condition $|\boldsymbol{a}|=2,|\boldsymbol{b}|=3,(\boldsymbol{a}, \boldsymbol{b})=6=|\boldsymbol{a}||\boldsymbol{b}|$. Therefore, the vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are directed in the same way, that is,
$\boldsymbol{a}=2(u... | \frac{4}{\sqrt{13}},\frac{6}{\sqrt{13}},\frac{9}{\sqrt{13}},=\frac{6}{\sqrt{13}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,819 |
$3+$ Prove the inequality $(a+b+c+d+1)^{2} \geq 4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$ for $a, b, c, d \in[0,1]$.
# | We can assume that $a \geq b \geq c \geq d$. Then
$(a+b+c+d+1)^{2}=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+2(a b+a c+a d+b c+b d+c d)+2(a+b+c+d)+1 \geq$ $\geq\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+2\left(b^{2}+c^{2}+d^{2}+c^{2}+d^{2}+d^{2}\right)+2\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+a^{2} \geq 4\left(a^{2}+b^{2}+c^{2}+d^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,820 |
$\underline{\text { Avilov N.I. }}$.
The teacher chose 10 consecutive natural numbers and told them to Petya and Vasya. Each boy must divide these 10 numbers into pairs, calculate the product of the numbers in each pair, and then add the five products obtained. Prove that the boys can do this in such a way that their ... | Let Petya from the first quartet of numbers $n, n+1, n+2, n+3$ form the sum $n(n+1)+(n+2)(n+3)=2 n^{2}+6 n+6$, and Vasya -
$n(n+3)+(n+1)(n+2)=2 n^{2}+6 n+2$. Then on the first quartet of numbers, Petya will score a sum 4 more than Vasya. With the second quartet of numbers, they will do the opposite, equalizing the tot... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,825 |
# | $\left(n^{\wedge} 7\right)^{\wedge}(7 \wedge 7)=n^{7 \cdot 7^{7}}=(n \wedge(7 \wedge 7))^{\wedge} 7$.
## Oтвет
Можно. | n^{7\cdot7^{7}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,828 |
The numbers 3, 9, and 15 were written on the board. It was allowed to add two of the written numbers, subtract the third from this sum, and write the result on the board instead of the number that was subtracted. After performing this operation multiple times, three numbers appeared on the board, the smallest of which ... | Note that $9-3=6$ and $15-9=6$. We will show that at any moment, one of the numbers on the board will be 6 less than the second and 6 more than the third.
Indeed, suppose this property is satisfied, and the numbers on the board are $x-6$, $x$, and $x+6$. If we add the two outer numbers and subtract the middle one, the... | 20192025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,829 |
[ $\quad \underline{\text { Invariants }} \quad]$ [Evenness and Oddness]
On the computer screen is the number 12. Every second, the number on the screen is either multiplied or divided by 2 or 3. The result of the operation appears on the screen in place of the written number. Exactly one minute later, a number appear... | Note that $12=2^{2} \cdot 3^{1}$, which means the total exponent of the factors (twos and threes) is 3. Regardless of the operation performed, each change in the number alters the total exponent of the factors by 1. A total of 60 such changes should occur. Therefore, after 60 seconds, the total exponent should have the... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,830 |
The principle of the extreme (etc.). $\quad]$ Symmetry helps solve the task_ $\quad]$
There are thirty cards, each with a number: on ten cards - $a$, on ten others - $b$, and on the remaining ten - $c$ (the numbers $a, b, c$ are all different). It is known that for any five cards, it is possible to find another five s... | Let $a<b<c$. We will mark on the number line all possible sums of the numbers on five cards. For each of them, the opposite is also marked, so the marked points are symmetrically located relative to zero. In particular, the largest (5c) and the smallest (5a) sums are opposite, so $5 a+5 c=0$, which means $c=-a$. The su... | 0 | Combinatorics | proof | Yes | Yes | olympiads | false | 52,832 |
[Sum of angles in a triangle. Theorem about the exterior angle.] [Examples and counterexamples. Constructions]
A paper triangle, one of whose angles is equal to $\alpha$, was cut into several triangles. Could it happen that all the angles of all the resulting triangles are less than $\alpha$
a) in the case where $\alp... | a) In the triangles of the partition, all angles must be greater than $180^{\circ}-70^{\circ}-70^{\circ}=40$. But the original angle of $70^{\circ}$ cannot be cut into angles greater than $40^{\circ}$.
b) The figure shows a triangle with an angle of 80º, composed of triangles with smaller angles.
, the sum of the numbers is zero. By paying 100 rubles, you can choose any cell and find out what number is written in it. What is the smallest number of rubles needed to definitely deter... | Let the central cell of the table contain the number $a$. Then the sum $S$ of all the numbers in the table is $-a$, since the sum $S+a$ can be broken down into six zero sums, each with six terms, as shown in the figure.
Therefore, it is sufficient to know one number - the central one.
. What is the minimum number of cells Petya must mark so that Vasya ... | Example. Petya can mark the cells indicated in the left figure. Then Vasya will not be able to cover more than one marked cell with one corner. But nine corners without overlaps cannot be placed on the board, since 27 $>25$.
+\ldots+1 / 2\left(3^{n+1}-1\right), \ldots=1,9,81, \ldots, 9^{n}, \ldots$
Example 2. $3,5+7, \ldots,\left(2^{n}+1\right)+\ldots+\left(2^{n+1}-1\right), \ldots=3,12, \ldots, 3 \cdot 4^{n-1}, \ldots$
## Answer
Could.
| | [Numerical tables and their p... | ForallevenN | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,836 |
Shaovalov A.V.
Given a prime number $p$. We will call a triangle allowed if all its angles are of the form $m / p \cdot 180^{\circ}$, where $m$ is an integer. We will consider allowed triangles the same if they have the same set of angles (i.e., similar). Initially, one allowed triangle is given. Every minute, one of ... | The angles of a permitted triangle are ${ }^{i} / p \cdot 180^{\circ}, j / p \cdot 180^{\circ}, k / p \cdot 180^{\circ}$. We will denote such a triangle as $T(i, j, k)$, where $i+j+k=p$. For brevity, we will measure angles in units: when we speak of an angle $m$, we mean the angle ${ }^{m} /{ }^{2} \cdot 180^{\circ}$.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,837 |
[ Orthogonal projection (other). ] [ Examples and counterexamples. Constructions ]
Is there a polyhedron whose projections onto three mutually perpendicular planes are: a triangle, a quadrilateral, and a pentagon? | From a large number of possible examples, let's take two, using a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ as the base (see figures).
On the left figure - a non-convex polyhedron $A B C O D A_{1} D M$. Its projections onto three mutually perpendicular faces are triangle $A B A_{1}$, quadrilateral $... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,838 |
Evochkimov M.A.
The company recorded its expenses in rubles under 100 budget items, resulting in a list of 100 numbers (each number has no more than two decimal places). Each accountant took a copy of the list and found an approximate sum of the expenses, acting as follows. First, he arbitrarily chose two numbers from... | Evaluation. Accountants each time calculated the integer part of the sum of some two numbers, which is equal to the sum of their integer parts plus, possibly, one. Let's call this addition of one an incident. It could only happen when both addends were non-integers. Each accountant ended up with the sum of the integer ... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,840 |
Evdokimov M.A.
An abstract artist took a wooden cube $5 \times 5 \times 5$, divided each face into unit squares, and painted each of them in one of three colors - black, white, or red - such that no two adjacent squares (sharing a side) are of the same color. What is the minimum number of black squares that could resu... | Evaluation. Three squares at the vertex of a cube form a cycle of adjacent squares of length 3. Around it, another cycle of length 9 is formed from adjacent cells. And around that, a cycle of length 15. Taking three such cycles around two opposite vertices of the cube, and two small cycles around the remaining vertices... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,841 |
Auto: Bogdanov I.i.
Pasha chose 2017 (not necessarily distinct) natural numbers $a_{1}, a_{2}, \ldots, a_{2017}$ and plays the following game with himself. Initially, he has an unlimited supply of stones and 2017 large empty boxes. In one move, Pasha adds $a_{1}$ stones to any box (of his choice), $-a_{2}$ stones to a... | $2017=43 \cdot 46+39$. Let among Pasha's numbers there be 39 twos, 46 numbers equal to 44, and the rest are ones. To achieve the required result in 43 moves, Pasha selects 39 boxes, into which he always places 2 stones; after 43 moves, there will be $43 \cdot 2=86$ stones in each of them. The remaining boxes he divides... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,846 |
Podlipsky o.k.
Oleg drew an empty $50 \times 50$ table and wrote a non-zero number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbe... | Evaluation. The fact that there are no more than 1250 rational numbers in the table is proven in the same way as in problem 66016 (since the product of a non-zero rational and an irrational number is irrational).
Example, when there are exactly 1250 rational numbers. Place along the left side the numbers $1,2, \ldots,... | 1250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,847 |
Podlipsky O.K.
Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers writte... | Evaluation. Suppose that among the rational numbers there is 0 and it is written at the top side of the table. Let along the left side of the table be written $x$ irrational and $50-x$ rational numbers. Then along the top side are written $50-x$ irrational and $x$ rational numbers. Note that the product of a non-zero r... | 1275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,848 |
Evdokimov M.A.
In each cell of a $5 \times 5$ board, there is either a cross or a zero, and no three crosses stand in a row either horizontally, vertically, or diagonally. What is the maximum number of crosses that can be on the board? | Example of placing 16 crosses according to the condition, see the figure on the left.
Evaluation. Divide the board into the central cell and 8 rectangles of size $3 \times 1$ (see the figure... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,850 |
Bakayev E.v.
10 children of different heights are standing in a circle. From time to time, one of them runs to another place (between some two children). The children want to stand in order of increasing height as quickly as possible, in a clockwise direction (from the shortest to the tallest). What is the minimum num... | Number the children in ascending order of height $-1, 2, \ldots, 10$.
Estimation. Suppose they initially stood in reverse order.
First method. If there were fewer than eight sprints, then some three children remained in their places, and their order is opposite to the required one.
Second method. Let's call the comp... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,853 |
Shapovalov A.V.
The weight of each weight in the set is a non-integer number of grams. They can balance any integer weight from 1 g to 40 g (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the smallest number of weights in such a set? | Example 1. Let's take weights of 1, 1, 3, 5, 11, 21, 43 g. The first two can measure any integer weight up to 2 g. Therefore, the first three can measure up to 5 g, the first four up to 10 g, the first five up to 21 g, the first six up to 42 g, and all seven up to 85 g. If we reduce the weight of each weight by half, a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,855 |
Shapovalov A.V.
In a convex $n$-gon, several diagonals were drawn such that no three or more of them intersect at any point inside the polygon. As a result, the polygon was divided into triangles. What is the maximum possible number of triangles? | Let $n=2 k$ or $2 k+1$.
Example. We will cut off quadrilaterals from the $n$-gon by diagonals until it is no longer possible. In total, $k-1$ quadrilaterals will be obtained (and one more triangle for odd $n$). By drawing diagonals in each quadrilateral, we get $4 k-4=2 n-4$ triangles for even $n$ and $4 k-3=2 n-5$ tr... | 2n-4forevenn,2n-5foroddn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,858 |
| [Algorithm Theory (Miscellaneous). $]$ |
| :--- |
| $\left[\begin{array}{ll}\text { Semi-invariants }\end{array}\right]$ |
| $\left[\begin{array}{l}\text { Backtracking }\end{array}\right]$ |
Authors: Mednikov L.E., Shapovalov A.V.
Captain Vrungel laid out a shuffled deck of 52 cards in a circle in his cabin, leavi... | a) Here is one possible way. Fuchs over 52 turns names $52^{2}$ cards - in each turn, all cards once in the same order, after which he says "stop". Note that every time the cards move in one direction: if card $a$ is moved, the next one to be moved will be card $b$ on the other side of the free space: it will be named ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,859 |
Bogdanov I.i.
Given two tetrahedra. Neither of them has two similar faces, but each face of the first tetrahedron is similar to some face of the second.
Are these tetrahedra necessarily similar? | Let $t$ be a number sufficiently close to 1. Then there are two tetrahedra, the bases of which are equilateral triangles with side 1, and the lateral sides of one are $t, t^{2}, t^{3}$, and of the other - $1 / t, 1 / t^{2}, 1 / t^{3}$. They obviously satisfy the condition, but are not similar.
## Answer
Not necessari... | Notnecessarily | Geometry | proof | Yes | Yes | olympiads | false | 52,860 |
Gribalko A.V.
In the left bottom cell of a $100 \times 100$ board, a chip is placed. Alternating between horizontal and vertical moves to an adjacent cell (the first move being horizontal), it first reaches the left top cell, and then the right top cell. Prove that there will be two such cells $\$ \mathrm{~A} \$$ and ... | Let's color the cells in a checkerboard pattern so that the lower left cell $X$ is black. Then the upper left cell $Y$ is white. A move is made horizontally from the black cell $X$, the next move is vertically from a white cell, then again horizontally from a black cell, and so on, alternating. This means that the piec... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,861 |
Ionin Yu. I.
Let $p$ be an arbitrary real number. Find all such $x$ that the sum of the cube roots of the numbers $1-x$ and $1+x$ equals $p$. | Let $u^{3}=1-x, v^{3}=1+x$. From the equation $u+v=p$ we get $v^{3}+v^{3}+3 u v(u+v)=p^{3}$, hence $2+3 u v p=p^{3}$, that is $u^{3} v^{3}=\left(\frac{p^{3}-2}{3 p}\right)^{3}$. Now $u^{3}$ and $v^{3}$ are found as the roots of the quadratic equation $t^{2}-2 t+\left(\frac{p^{3}-2}{3 p}\right)^{3}=0$.
In the end, we g... | \\sqrt{1-(\frac{p^{3}-2}{3p})^{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,865 |
Limanov L.G.
If in each vertex of a convex polyhedron at least four edges meet, then at least one of its faces is a triangle.
Prove this.
# | Assume the opposite: there exists a polyhedron satisfying the condition without triangular faces.
Consider the plane angles of its faces and calculate their arithmetic mean in two ways. Group the angles by faces. Since each face has at least four angles, the arithmetic mean of its angles is not less than $90^{\circ}$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,866 |
Yasnovy E.A.
Investigate how many solutions the system of equations has
$x^{2}+y^{2}+x y=a$,
$x^{2}-y^{2}=b$,
where $a$ and $b$ are some given real numbers. | Let's move to new variables: $x+y=u, x-y=v$. Then the system can be written as: $3 u^{2}+v^{2}=4 a^{2}$, $u v=b$.
$(*)$
Obviously, each pair $(x, y)$ corresponds to one pair $(u, v)$, and each pair $(u, v)$ corresponds to a pair $(x, y)$, so we need to determine how many solutions the system (*) has for different val... | -2\leq\sqrt{3}b\leq2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,867 |
Given the numbers $1,2,3, \ldots, 1000$. Find the largest number $m$, such that: no matter which $m$ of these numbers are erased, among the remaining $1000-m$ numbers, there will be two such that one divides the other. | If $m \geq 500$, then, by erasing the first $m$ numbers (from 1 to $m$), we will leave the numbers from $m+1 \geq 501$ to 1000, among which none, obviously, divides another (all pairwise ratios are less than two).
The number $m=499$ has the required property. Indeed, if we take 501 numbers from the set of 1 to 1000, t... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,868 |
Two equal disks are mounted on the same axis. On the circumference of each of them, the numbers $1,2,3, \ldots, 20$ are placed in a circle at equal distances in arbitrary order. Is it always possible to rotate one disk relative to the other so that no two identical numbers stand opposite each other? | Suppose that at each turn (where a number written on one disk is opposite a number written on the other disk), there will be two identical numbers opposite each other. Then, at each turn, there is exactly one pair of identical numbers opposite each other. Indeed, we have a total of 20 turns, so if for at least one turn... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,870 |
Given a $4 \times 4$ table, some cells of which are marked with a star. Show that it is possible to place seven stars in such a way that after erasing any two rows and any two columns of this table, there will always be at least one star in the remaining cells. Prove that if there are fewer than seven stars, it is alwa... | Let there be fewer than seven asterisks placed in the table. Then there are two possible cases.
1) There are no asterisks in some row (for example, the first). Then, by the Pigeonhole Principle, there will be a row (let's say the second) in which there are no more than two asterisks. We will cross out the third and fo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,871 |
$a_{1}, a_{2}, \ldots, a_{n}$ - arbitrary natural numbers. Denote by $b_{k}$ the number of numbers from the set $a_{1}, a_{2}, \ldots, a_{n}$, satisfying the condition: $a_{i} \geq k$. Prove that $a_{1}+a_{2}+\ldots+a_{n}=b_{1}+b_{2}+\ldots$ | Let's arrange the numbers in descending order and consider the corresponding Young diagram. It is not hard to notice that $b_{j}$ is the number of squares in the $j$-th column of this diagram. Therefore, both sums from the condition are equal to the number of squares in the obtained diagram. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,872 |
Solve the equation $\underbrace{\sqrt{n+\sqrt{n+\ldots \sqrt{n}}}}_{1964 \text { times }}=m$ in integers. | Let's solve the more general equation $A_{y}(x)=z$ in integers, where $A_{y}(x)=\underbrace{\sqrt{x+\sqrt{x+\ldots \sqrt{x}}}}_{y \text { times }}$. Clearly, $x+A_{y-1}(x)=z^{2}$, or $A_{y-1}(x)=z^{2}-x$.
Thus, if the number $A_{y}(x)=z$ is an integer, then the number $A_{y-1}(x)=\left(z^{2}-x\right)$ is also an integ... | (0,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,873 |
Does there exist a number $h$ such that for no natural number $n$ the number $\left[h \cdot 1969^{n}\right]$ is divisible by $\left[h \cdot 1969^{n-1}\right]$? | Let $h=\frac{1969^{2}}{1968}$. Then $h \cdot 1969^{n-1}=\frac{1969^{n+1}}{1968}=m+\frac{1}{1968} \quad$ and $h \cdot 1969^{n}=1969 m+1+\frac{1}{1968} \quad$ (where $m-$ is a natural number). For any natural $n$, the number $m$ is greater than 1, so $1969 m+1$ does not divide by $m$.
## Answer
It exists. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,874 |
Periodicity and non-periodicity
With a natural number $K$, the following operation is performed: it is represented as a product of prime factors $K=p_{1} p_{2} \ldots p_{n}$; then the sum $p_{1}+p_{2}+\ldots+p_{n}+1$ is calculated. The same operation is performed with the resulting number, and so on.
Prove that the r... | Let's denote by $f(K)$ the sum $p_{1}+p_{2}+\ldots+p_{n}+1$. A simple calculation verifies that if $K>9$, the sequence will eventually become periodic from some point onwards.
Lemma. Suppose either $a>2, b \geq 3$, or $a \geq 2, b>3$, then $a b>a+b+1$.
Proof. Write the inequality as $(a-1)(b-1)>2$. Now it is obvious.... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,876 |
[ Dirichlet's Principle (pigeonhole principle). ] [ Partitions into pairs and groups; bijections ]
Two identical gears have 32 teeth each. They were aligned and 6 pairs of teeth were filed off simultaneously. Prove that one gear can be rotated relative to the other so that the broken teeth of one gear align with intac... | Let $n$ denote the number of pairs of teeth removed (in our case, $n=6$). Then each gear has $n^{2}-n+2$ teeth (in our case, 32). There are $n^{2}-n+1$ such rotations of the upper gear relative to the lower one, during which all teeth of both gears align. Let's call a hole the place on the gear where a tooth is missing... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,877 |
A spherical planet is surrounded by 37 point-like asteroids. Prove that at any moment there will be a point on the surface of the planet from which an astronomer will not be able to observe more than 17 asteroids.
Note. An asteroid located on the horizon line is not visible.
# | Let's draw a plane P through the center of the planet and an arbitrary pair of asteroids (we will consider it the equatorial plane). Draw the planet's axis perpendicular to plane P. It intersects the planet's surface at the "poles" A and B. Then it is easy to see that observers at points $A$ and $B$ together see only $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,878 |
[ Proof by contradiction ] [ Examples and counterexamples. Constructions ]
There are several stones, the mass of each of which does not exceed 2 kg, and the total mass is 100 kg. From them, several stones are selected, the total mass of which differs from 10 kg by the smallest possible number $d$ for this set. What is... | Example. For a set of 55 stones weighing ${ }^{20} / 11$ kg, the number $d$ is ${ }^{10} / 11$ (you need to take 5 or 6 stones).
Suppose that for some set of stones weighing $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$, the number $d$ is greater than ${ }^{10} / 11$. Choose $k$ such that $x_{1}+\ldots+x_{k-1} \leq 10 \le... | \frac{10}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,879 |
|
| | $[$ Chessboard coloring $]$ | |
At one vertex of a cube, the number 1 is written, and zeros are written at all other vertices. You can add one to the numbers at the ends of any edge.
Can you achieve that all numbers are divisible by a) 2; b) 3? | a) Initially, the sum of the numbers at the vertices of the cube is one, which is odd. Each time we add one to two vertices, the sum remains odd. Therefore, the numbers at all vertices cannot become even.
b) Let's color the vertices in a checkerboard pattern. After this, we notice that the sum of the numbers in the wh... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,880 |
Kroooov S.C.
a) In all cells of a $20 \times 20$ square, there are soldiers. Vanya names a number $d$, and Petya rearranges the soldiers so that each one moves a distance of at least $d$ (the distance is taken between the centers of the old and new cells). For which $d$ is this possible?
b) The same problem for a $21... | The maximum distance that a soldier standing in the central cell can move is $10 \sqrt{2}$. We will show that this is indeed the maximum value of $d$.
a) Divide the square into four $10 \times 10$ squares and swap the opposite squares (each cell will move by the specified distance).
b) Divide the square as shown in t... | \leq10\sqrt{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,882 |
$\underline{\text { Folklore }}$ In the set, there are weights of 1 g, 2 g, 4 g, ... (all powers of 2), and among the weights, there can be identical ones. The weights were placed on two scales such that equilibrium was achieved. It is known that all the weights on the left scale are different. Prove that there are no... | If there are two identical weights on the right pan, replace them with one weight of double the mass (thus reducing the number of weights). We will continue to do this until there are no identical weights left on the right pan. After this (due to the uniqueness of binary representation), the sets of weights on the left... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,884 |
Nazarov $\Phi$.
A token is placed on a certain square of a chessboard. Two players take turns moving the token, and on each move, starting from the second, the distance it is moved must be strictly greater than on the previous move. The player who cannot make a move loses. Who wins with correct play? (The token is alw... | Let the first player always move to the field centrally symmetric to the one where the chip is located. The first player can obviously do this on the first move. We will prove by induction that he can always do this, and after each move of the second player, the distance from the chip to the center $O$ of the board inc... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,885 |
$:$ Fomin D:
On a plane, there are $N$ lines ( $N>1$ ), no three of which intersect at the same point and no two are parallel. Prove that in the parts into which these lines divide the plane, it is possible to place non-zero integers, in absolute value not exceeding $N$, such that the sums of the numbers on either sid... | We will color the plane in two colors - white and black - so that adjacent (having a common part of the border) parts are of different colors (see problem $\underline{34930}$). At each intersection point of the lines, we will place the numbers $1,-$
$1,1,-1$ in a circle so that the ones are in the white areas. Now, in ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,886 |
Fomin S. $B$.
In the kingdom, there are eight cities. The king wants to build such a road system that from each city, one can reach any other city by passing through no more than one intermediate city, and no more than $k$ roads should lead out of each city. For what values of $k$ is this possible? | If $k=2$, then from city $A$ one can "in one move" reach no more than two cities, and from them "in the next move" - no more than four. In total, from $A$ one can reach no more than six cities, but seven are needed. For $k=3$, arrange the cities at the vertices of a regular octagon, and let the roads run along all side... | k>2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,888 |
Tolongon A.K.
Given a convex octagon $A B C D E F G H$, in which all internal angles are equal to each other, and the sides are equal every other one: $A B = C D = E F = G H$,
$B C = D E = F G = H A$ (we will call such an octagon semi-regular). We draw the diagonals $A D, B E, C F$, $D G, E H, F A, G B$ and $H C$. Am... | Let $AB = a, BC = b\sqrt{2}$. Otherwise, as is easy to calculate, the sides $a_1$ and $b_1$ of the resulting octagon are equal to $a\sqrt{2} - b$ and $b\sqrt{2} - a$ respectively. In this case, $a_1 - b_1 = (\sqrt{2} + 1)(a - b) > 2(a - b)$. Similarly, $a_2 - b_2 > 2(a_1 - b_1)$, and so on. Such doubling of the differe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,889 |
Egorov A.A.
In an $n \times n$ table, it is allowed to add 1 to all numbers along any non-self-intersecting closed rook path. In the initial table, ones were placed along the diagonal, and all other cells were zeros. Is it possible to achieve, using several allowed transformations, that all numbers in the table become... | We will color the table in a checkerboard pattern. Any route of the rook that satisfies the condition contains an equal number of white and black cells, so the difference between the sum of all "black" numbers and the sum of all "white" numbers is an invariant. In the initial table, this difference is not zero, so it i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,891 |
Kanel-Belov A.Ya.
Natural numbers are sequentially written on the board. On the $n$-th step (when the numbers $a_{1}, a_{2}, \ldots, a_{n-1}$ are written), any number is written that cannot be represented as the sum $a_{1} k_{1}+a_{2} k_{2}+\ldots+a_{n-1} k_{n-1}$, where $k_{i}$ are non-negative integers (no restricti... | Suppose an infinite sequence is written out. Since the number of remainders when divided by $a_{1}$ is finite, some remainder will occur infinitely many times in this sequence, that is, there will be an infinite subsequence $b_{1}, b_{2}, \ldots$, all members of which have the same remainder when divided by $a_{1}$. By... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,892 |
At each integer point on the number line, there is a light bulb with a button. When the button is pressed, the light bulb changes its state - it turns on or off. Initially, all the light bulbs are off. A finite set of integers $S$ is given, which can be moved along the number line as a rigid shape and, when applied at ... | Let there be more than one number in $S$ (for one number everything is obvious). By the current position, we will understand the segment of bulbs from the leftmost lit one to the rightmost (the position of the segment on the line is not important for us yet). Starting from a position with one lit bulb, we will each tim... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,893 |
Kanel-Belov A.Y.
The periods of two sequences are 7 and 13. What is the maximum length of the initial segment that can coincide? | Example. Consider the sequence with period 7 and initial terms $0,0,0,0,0,1,0$ and the sequence with period 13 and initial terms $0,0,0,0,0,1,0,0,0,0,0,0,1$. They have a common initial segment $0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0$ of length 18.
Estimate. If there is a common initial segment of length 19, then $a_{7}=a... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,894 |
Fimevich V.P.
Do such numbers exist:
a) 4 different natural numbers;
b) 5 different natural numbers;
c) 5 different integers;
d) 6 different integers,
such that the sum of any three of them is a prime number? | a) An example of such a quartet of numbers: $1,3,7,9$. From these numbers, four triplets can be formed; their sums are 11, 13, 17, and 19.
Another example where the original numbers are also prime: 7, 13, 23, 53.
b) Among any five integers, either there will be three that give the same remainder when divided by 3, or... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,895 |
Kanel-Belov A.Y.
On the board, several positive integers are written: $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$. We write on another board the following numbers: $b_{0}$ - the total number of numbers on the first board, $b_{1}$ - the number of numbers greater than one, $b_{2}$ - the number of numbers greater than two, and ... | We can assume that $a_{0} \geq a_{1} \geq \ldots \geq a_{n}$.
First method. For convenience, add the number $a_{n+1}=0$ to this sequence.
The first $b_{k}$ numbers in this sequence (i.e., the numbers $a_{0}, a_{1}, \ldots, a_{b_{k}-1}$) are greater than $k$, and the rest are not greater than $k$. Thus, the numbers $b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,898 |
On a board of size $15 \times 15$ cells, 15 rooks were placed so that they do not attack each other. Then each rook was moved by a knight's move.
Prove that now some two rooks will attack each other. | Let's introduce a "coordinate system" on the board: number the horizontals and verticals with numbers from 1 to 15. If the rooks do not attack each other, then in each horizontal and each vertical there is exactly one rook, so the sum of the coordinates of all rooks is $2(1+2+\ldots+15)$, which is even. When a rook is ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 52,899 |
Shapovalov A.V.
The cells of a chessboard are numbered from 1 to 64 in such a way that adjacent numbers are in adjacent (by side) cells.
What is the smallest possible sum of the numbers on the diagonal?
# | Evaluation. Consider the numbers $a_{1}<a_{2}<\ldots<a_{8}$, standing on the black diagonal (of course, on the diagonal they may not be in order). Then $a_{1} \geq 1, a_{2} \geq 3, a_{3} \geq 5, \ldots, a_{7} \geq 13$. We will prove that $a_{8} \geq 39$.
Assume that the numbers were written into the cells of the board... | 88 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,902 |
Shapovalov A.V.
a) There are 128 coins of two different weights, with an equal number of coins of each weight. How can you use a balance scale without weights to guarantee finding two coins of different weights in no more than seven weighings?
b) There are eight coins of two different weights, with an equal number of... | a) Let there be two groups of $2 k$ coins each, and it is known that the first group is heavier than the second. Then, with one weighing, we can obtain two groups of $k$ coins with the same property. Indeed, place $k$ coins from each group on the scales. If the scales are in balance (otherwise, the desired groups are a... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,903 |
Eemel'yanov L.A. Does there exist a correct triangular prism that can be covered (without overlaps) with various equilateral triangles? (It is allowed to bend the triangles over the edges of the prism.) | The figure shows a net of a regular triangular prism divided into three equilateral triangles, where the lateral edge is $\sqrt{3}$ times the length of the base edge.
## Answer
It exists. | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,904 |
Shen A.H.
a) A test was given in class. It is known that at least $2 / 3$ of the problems on this test were difficult: each such problem was not solved by at least $2 / 3$ of the students. It is also known that at least $2 / 3$ of the students in the class did well on the test: each such student solved at least $2 / 3... | a) Example. Let there be three problems on the test, a third of the students solved the first and third, a third solved the second and third, and the rest solved nothing.
b) Let's draw a table on the unit square, where rows correspond to students and columns to problems; in this case, we will assign the upper rows to s... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,905 |
Yashchenko I.V.
On a $4 \times 6$ board, there are two black chips (Vanya's) and two white chips (Seryozha's, see the diagram). Vanya and Seryozha take turns moving any of their chips one cell forward (vertically). Vanya starts. If after any move, a black chip ends up between two white chips horizontally or diagonally... | First, move the middle chip.
## Solution
We will move Vanya's middle chip forward, ignoring Seryozha's moves.
Let's call the number of the horizontal line on which the chip stands (counting from the bottom) its height. If Vanya's chip ends up between Seryozha's chips, its height $h$ will be equal to half the sum of ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 52,906 |
The hostess made a pirog (a type of Russian pie) and wants to pre-cut it into such (not necessarily equal) pieces so that the pie can be evenly divided both among five and seven people. What is the minimum number of pieces she can manage with?
# | This task is a particular case of problem $\underline{98057 .}$
## Answer
11 pieces. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 52,907 |
Prove that there does not exist a polynomial of degree not less than two with non-negative integer coefficients, the value of which for any prime $p$ is a prime number.
# | Suppose such a polynomial $Q(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$ exists.
If $a_{0}=0$, then $Q(x)=x\left(a_{n} x^{n-1}+a_{n-1} x^{n-2}+\ldots+a_{1}\right)$, hence for a prime $p$, the number $Q(p)$ is divisible by $p$ and greater than $p$ (since the degree of $Q$ is greater than 1), so $Q(p)$ is a com... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 52,909 |
P.M.
In the country, there are 15 cities, some of which are connected by air routes belonging to three airlines. It is known that even if any one of the airlines ceases operations, it will still be possible to travel from any city to any other (possibly with layovers), using flights of the remaining two airlines. What... | Let the number of routes of the airlines be denoted by $a, b$, and $c$. If we close the last $c$ air routes, the graph will remain connected, so $a+b \geq 14$ (see problem $\underline{31098}$ g). Similarly, $b+c \geq 14, c+a \geq 14$. Adding these inequalities, we get: $2(a+b+c) \geq 42$, which means the three companie... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,912 |
$\underline{\text { Kryzhanovsky O.F. }}$
Does there exist a polynomial $P(x)$ that has a negative coefficient, while all coefficients of any of its powers $(P(x))^{n}, n>1$, are positive? | It is sufficient to find a polynomial such that the coefficients of its square and cube are positive: any other power can be represented as a product of squares and cubes.
Let's call a polynomial positive if all its coefficients are positive.
Consider the polynomial $f(x)=x^{4}+x^{3}+x+1$. It is easy to see that the ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,913 |
[ Pairings and groupings; bijections ] [ Examples and counterexamples. Constructions ]
Author: Filovich V.P.
Integers from 1 to $n$ are written in a row. Under them, the same numbers are written in a different order. Can it happen that the sum of each number and the one written under it is a perfect square a) for $n... | a) For example:
123456789
826543917
b) It is not hard to see that under the number 11, only the number 5 can be written, but under the number 4, only the number 5 can also be written. This is a contradiction.
c) Write the number $2025-k$ under the number $k$ for all $k=29,30, \ldots, 1996$. Then the sum of the numb... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 52,914 |
The value of $a$ is chosen so that the number of roots of the first equation $4^{x}-4^{-x}=2 \cos a x, 4^{x}+4^{-x}=2 \cos a x+4$ is 2007.
How many roots does the second equation have for the same $a$? | Transform the second equation:
$$
\left.4^{x}+4^{-x}=2 \cos a x+4 \Leftrightarrow 4^{x}-2+4^{-x}=2(1+\cos a x) \Leftrightarrow\left(2^{x}-2^{-x}\right)^{2}=4 \cos ^{2 a x} / 2\right) \Leftrightarrow
$$
$$
\left[\begin{array} { l }
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = 2 \cos \frac { a x } { 2 } , } \\
{ 4 ^ { x / 2 } ... | 4014 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,917 |
$\underline{\text { Perrin A. }}$.
A quadratic trinomial $f(x)$ can be replaced by one of the trinomials $x^{2} f\left(\frac{1}{x}+1\right)$ or $(x-1)^{2} f\left(\frac{1}{x-1}\right)$. Can we obtain the trinomial $x^{2}+10 x+9$ from the quadratic trinomial $x^{2}+4 x+3$ using such operations? | Let $f(x)=a x^{2}+b x+c$. Note that under permissible operations, the discriminant of the trinomial does not change: under the first operation, $f(x)$ changes to the trinomial $(a+b+c) x^{2}+(b+2 a) x+a$ with the discriminant $(b+2 a)^{2}-4 a(a+b+c)=b^{2}+4 a b+4 a^{2}-4 a^{2}-4 a b-4 a c=b^{2}-4 a c$, and under the se... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,918 |
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