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Tomongeo A.K.
Can two exact cubes be fitted between two consecutive perfect squares?
In other words, does the inequality have a solution in integers: $n^{2}<a^{3}<b^{3}<(n+1)^{2}$? | Note that $n^{2}a^{3}+2 a^{2}+1>n^{2}+2 n+1=(n+1)^{2}$.
## Answer
No. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,524 |
Bakayev E.V.
A hundred bear cubs found berries in the forest: the youngest managed to grab 1 berry, the next older cub - 2 berries, the next one - 4 berries, and so on, with the oldest getting \(2^{99}\) berries. The fox suggested they divide the berries "fairly." She can approach two bear cubs and distribute their be... | Notice that at each step, if both participating cubs had at least one berry, then each of them will have at least one berry left. Therefore, in the end, each cub will have at least one berry.
We will prove that the fox can leave each cub exactly one berry, that is, transition from the position (1, $2, \ldots, 2^{99}$)... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,526 |
Chernyaev N.L.
In each cell of an $8 \times 8$ board, a natural number is written. It turns out that for any cutting of the board into dominoes, the sums of the numbers in all the dominoes are different. Can it be that the largest number written on the board is no more than 32? | Let's record ones in the black cells, and all numbers from 1 to 32 in the white cells. With any domino tiling, each domino will contain exactly one white and one black cell. Therefore, the sums in the dominoes will be $2,3, \ldots$, 33.
## Answer
It can. | Itcan | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,528 |
$3+$ [Examples and counterexamples. Constructions]
In the zoo, there are 10 elephants and huge pan scales. It is known that if any four elephants stand on the left pan, and any three on the right, the left pan will outweigh. Five elephants stood on the left pan and four on the right. Will the left pan definitely outwe... | Counterexample. Let five elephants weigh 7 tons each, and another five weigh 9 tons each. Then any four elephants on the left pan of the scales together weigh no less than $4 \cdot 7=28$ tons, while any three elephants on the right pan weigh no more than $3 \cdot 9=27$ tons, and the left pan will indeed outweigh.
But ... | Notnecessarily | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,529 |
[ Pythagorean Theorem (direct and inverse). ] [ Examples and counterexamples. Constructions ]
Is there a right triangle in which the lengths of two sides are integers, and the length of the third side is $\sqrt{2016}$? | Consider, for example, a right-angled triangle with legs $\sqrt{2016}$ and 3. Its hypotenuse is equal to $\sqrt{2016+9}=\sqrt{2025}=45$.
## Answer
It exists. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,530 |
On the table lies a rectangular sheet of paper. Sasha cuts it along a straight line into two pieces and puts the pieces on the table. Then he takes one of the pieces, cuts it again along a straight line into two pieces, and puts the pieces back on the table. Then he takes one piece from the table and cuts it again, and... | Notice that after one cut, the total number of vertices increases by two (if the cut passes through two vertices), by three (if the cut passes through a vertex and a point inside a side), or by four (if the cut passes through the interior points of two sides). Suppose $k$ cuts have been made, resulting in $k+1$ pieces,... | 2015 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,531 |
The code lock will open if the numbers from 1 to 16 are arranged in the cells of a $4 \times 4$ square such that the sum of the numbers in each $2 \times 2$ square is divisible by 17. Is it possible to open such a lock? | See fig. The sum of the numbers in each $2 \times 2$ square is 34.
| 1 | 16 | 2 | 15 |
| :---: | :---: | :---: | :---: |
| 4 | 13 | 3 | 14 |
| 5 | 12 | 6 | 11 |
| 8 | 9 | 7 | 10 |
Answer
It is possible.
Send a comment | Itispossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,532 |
3+ [ Examples and counterexamples. Constructions ]
Is there a natural number less than one hundred that can be represented as the sum of the squares of two different natural numbers in two different ways
# | For example, $65=1^{2}+8^{2}=4^{2}+7^{2}$ or $85=2^{2}+9^{2}=6^{2}+7^{2}$.
## Answer
It exists. | 65=1^{2}+8^{2}=4^{2}+7^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,533 |
$3+$ [ Examples and counterexamples. Constructions]
In the product of seven natural numbers, each factor was decreased by 3. Could the product have increased exactly 13 times as a result?
# | The product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 16=32$ after the specified operation turns into $(-2)^{5} \cdot(-1) \cdot 13=13 \cdot 32$.
## Answer
It could. | 13\cdot32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,535 |
Paradoxes
In the school council, a chairman is being elected. There are four candidates: A, B, C, and D. A special procedure has been proposed - each council member must write down the candidates in order of their preferences. For example, ABCD means that the council member ranks A first, does not strongly object to C... | Suppose that seven members voted in the council, and the votes were distributed as shown in the table on the left.
| Council Member | 3 | 2 | 1 | 0 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | A | B | В | Г |
| 2 | A | B | В | Г |
| 3 | A | B | В | Г |
| 4 | Г | S | A | B |
| 5 | Г | S | A | B |
| 6 | Г | В | A |... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,536 |
$\left[\begin{array}{ll}{\left[\begin{array}{l}\text { Common fractions } \\ \text { [Examples and counterexamples. Constructions ] }\end{array}\right]}\end{array}\right]$
Author: Akonn E, Kaminin D.
Can the stars in the equation $\frac{*}{*}+\frac{*}{*}+\frac{*}{*}+\frac{*}{*}=$ * be replaced with the digits from 1 ... | From the possible examples, let's give two: $7 / 4+6 / 8+5 / 1+3 / 2=9$, $5 / 4+6 / 8+9 / 3+2 / 1=7$.
## Answer
It can be done. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,537 |
Frankin B.R.
There are 100 non-zero numbers written in a circle. Between every two adjacent numbers, their product was written, and the original numbers were erased. The number of positive numbers did not change. What is the minimum number of positive numbers that could have been written initially?
# | Evaluation. Suppose there were no more than 33 positive numbers. Negative numbers in the product can only be formed if one of the factors is positive, and each positive number can participate in no more than two such products. Therefore, there can be no more than 66 negative numbers. But then the total number of number... | 34 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,538 |
[Examples and counterexamples. Constructions] [ $\quad$ Cauchy's Inequality
Author: Tominio A.K.
We took several positive numbers and constructed a sequence from them as follows: $a_{1}$ is the sum of the original numbers, $a_{2}$ is the sum of the squares of the original numbers, $a_{3}$ is the sum of the cubes of t... | a) Example 1. Take the number 2 and 1024 numbers, each equal to $1 / 2$. Then $a_{n}=2^{n}+1024 \cdot 2^{-n}=32\left(2^{n-5}+2^{5-n}\right)$. The sum of two positive reciprocal numbers is the smallest when they are closest to each other. Therefore, the constructed sequence decreases up to $n=5$, and then increases.
Ex... | )Itcould;b)itcouldnot | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 53,539 |
[ Proof by contradiction $\quad]$ [ Ordering by increasing (decreasing) ]
On the board, $n>3$ different natural numbers, less than $(n-1)!$, are written. For each pair of these numbers, Seryozha divided the larger by the smaller with a remainder and wrote down the obtained quotient (so, if he divided 100 by 7, he woul... | Suppose the opposite. Let $a_{1}, a_{2}, \ldots, a_{n}$ be the numbers on the board in ascending order, and $q_{i}$ be the quotient of the division of $a_{i+1}$ by $a_{i}(i=1,2, \ldots, n-1)$; then $a_{i+1} \geq q_{i} a_{i}$. Since all $q_{i}$ are distinct, $q_{1} q_{2} \ldots q_{n-1} \geq(n-1)$!. Thus, $a_{n / a_{1}}=... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,540 |
Two 2007-digit numbers are written on the board. It is known that from both numbers, seven digits can be erased so that the resulting numbers are the same. Prove that in the original numbers, seven digits can be inserted so that the resulting numbers are also the same.
# | Let these 2007-digit numbers be $A$ and $B$. Instead of crossing out seven digits from $A$, cover them with stickers. The remaining digits form a 2000-digit number $C$, which can also be obtained from $B$. Therefore, we can obtain $B$ from $C$ by inserting 7 digits (if this can be done in more than one way, choose any ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,541 |
Blinkov A.D:
Does there exist a convex polyhedron in which the number of edges is equal to the number of diagonals? (A diagonal of a polyhedron is a segment connecting two vertices that do not lie in the same face.) | For example, in a hexagonal prism, from each vertex of the upper base, there are three diagonals to the vertices of the lower base that do not lie in the same lateral face. Therefore, the total number of diagonals is 18, the same as the number of edges.
## Answer
It exists. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,542 |
Evochkimov M.A.
A football match was played between a team of liars (who always lie) and a team of truth-lovers (who always tell the truth). After the match, each player was asked: "How many goals did you score?" Some of the participants answered "one", Misha said "two", some answered "three", and the rest said "five"... | Suppose Misha is a truth-teller. Then nine out of ten players on the team of truth-tellers scored an odd number of goals (one, three, or five), while Misha scored an even number. But then the team of truth-tellers would have scored an odd number of goals, which contradicts the condition. Therefore, our assumption is in... | Lies | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,543 |
In a theater troupe, there are 60 actors. Any two of them have at least once played in the same play. In each play, no more than 30 actors are involved.
What is the minimum number of plays the theater could have staged? | Example. Let's divide the troupe into four groups of 15 people and hold 6 performances, in each of which some two groups are involved. The number of ways to choose two groups out of four is 4$\cdot$3:2=6.
Estimate. In total, the actors played no more than $30 \cdot 5=150$ roles, so if there are five performances, ther... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,545 |
Aladdin has several identical gold ingots, and sometimes he asks the genie to increase their number. The genie adds a thousand such ingots, but then takes exactly half of the resulting total mass of gold as a fee. Could Aladdin end up in a better position after ten such requests, if no ingot had to be cut? | Let Aladdin have $\$ 1000+$ x $\$$ ingots. After a request, it will become $\$ 1000+\mid f r a c\{x\}\{2\} \$$, and after ten requests - \$ $1000+\backslash$ frac $\{x\}\{2 \wedge\{10\}\} \$$$. Therefore, \$x \leq
## Answer
Could not. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,546 |
Authors: Dumanov I., Krotovsky R.
Discussing winter holidays in class, Sasha said: "Now, after I flew to Addis Ababa, I have celebrated New Year's in all possible hemispheres of the Earth, except one!"
What is the minimum number of places where Sasha has celebrated New Year's?
Consider the places where Sasha celebra... | Evaluation. Suppose three points were enough. They define a plane. If it passes through the center of the sphere, then Sasha has not been in two hemispheres cut by this plane. Contradiction.
If it does not pass, then we can cut off a hemisphere in which Sasha has not been with a parallel plane. In all hemispheres obta... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,547 |
Lobov A.
Let \( n \) be a natural number. We will call a sequence \( a_1, a_2, \ldots, a_n \) interesting if for each \( i = 1, 2, \ldots, n \), one of the equalities \( a_i = i \) or \( a_i = i + 1 \) holds. We will call an interesting sequence even if the sum of its terms is even, and odd otherwise. For each odd int... | Denoting the sum containing the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$ by $A_n$, and the other by $B_n$, we will prove the equality $A_n - B_n = 1$ by induction.
Base case. $A_1 - B_1 = 2 - 1 = 1$.
Inductive step. Represent the sum $A_n$ as $A' + A''$, where $A'$ contains all terms of the form
$a_1a_2...a_{n-1}(... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,550 |
[ GCD and LCM. Mutual simplicity ]
Prove that for any odd natural number $a$ there exists a natural number $b$ such that $2^{b}-1$ is divisible by $a$.
# | Consider the numbers $2^{0}-1, 2^{1}-1, \ldots, 2^{a}-1$. There are $a+1$ such numbers. Some two of them give the same remainder when divided by $a$, because there are only $a$ different such remainders. Let's say, for example, the numbers $2^{k}-1$ and $2^{m}-1$ give the same remainder when divided by $a$ and $k<m$. T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,552 |
Polynomial $p$ and number $a$ are such that for any number $x$, the equality $p(x)=p(a-x)$ holds.
Prove that $p(x)$ can be represented as a polynomial in $(x-a / 2)^{2}$.
# | Let $b=a / 2, t=x-b$ and $q(t)=p(b+t) . \quad q(-t)=p(b-t)=p(2 b-x)=p(x)=p(b+t)=q(t)$ (i.e., $q$ is an even function). Therefore,
$q(t)=1 / 2(q(t)+q(-t))$. When adding $q(t)$ and $q(-t)$, all odd powers will obviously cancel out, that is, $q(t)=$ $h\left(t^{2}\right)$. Hence,
$p(x)=q(t)=h\left(t^{2}\right)=h\left((x-... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,553 |
Solve the equation $n^{x}+n^{y}=n^{z}$ in natural numbers. | Obviously, $n \geq 2$. Considering $x \leq y \leq z$, we rewrite our equation as $n^{z-x}-n^{y-x}=1$.
If $y>x$, then 1 is divisible by $n$. But this is not true ( $n \geq 2$ ), so $y=x$. From this, $n^{z-x}=2$, which means $n=2, z-x=1$.
## Answer
$n=2, y=x, z=x+1, x$ - any natural number. | 2,x,x+1,x | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,555 |
In a chess tournament, two 7th-grade students and a certain number of 8th-grade students participated. The two 7th-graders scored 8 points, and each of the 8th-graders scored the same number of points. How many 8th-graders participated in the tournament? (Each participant plays one game with each of the others. A win e... | Let $x$ be the number of eighth graders, $y$ be the number of points scored by each eighth grader. By calculating the total points scored by all participants in the tournament in two ways, we arrive at the equation $x y + 8 = \frac{(x+2)(x+1)}{2}$, which simplifies to $2 y = \frac{(x+2)(x+1)-16}{x} = x + 3 - \frac{14}{... | 7or14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,556 |
Prove that for any integer $n$, among the numbers $n, n+1, n+2, \ldots, n+9$ there is at least one that is coprime with the other nine.
# | Among the given numbers, there are five odd ones. Consider the remainders of these five numbers when divided by 3, 5, and 7. Among the remainders when divided by 3, there are no three identical ones, and among the remainders when divided by 5 and 7, there are no two identical ones. Therefore, among the five numbers, on... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,557 |
Can 10 bus routes be laid out in a city and stops be set up on them in such a way that no matter which 8 routes are taken, there will be a stop that does not lie on any of them, while any 9 routes pass through all stops. | Let's draw 10 pairwise intersecting (at different points) lines. Let the routes pass through these lines, and the stops be the points of intersection of the lines. Any nine routes pass through all the stops, since each stop lying on the remaining line is passed by one of the nine lines corresponding to these routes. An... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,559 |
If for any positive $p$ all roots of the equation $a x^{2}+b x+c+p=0$ are real and positive, then the coefficient $a$ is zero. Prove it. | If $a>0$, then for large positive $p$ the discriminant $D=b^{2}-4 a c-4 a p$ is negative, so the given equation has no real roots at all.
If $a<0$, then for large positive $p$ the product of the roots is negative ( $c+p$ and $a$ have opposite signs), so one of the roots is negative. | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,560 |
Prove that a polynomial of the form $x^{200} y^{200}+1$ cannot be represented as a product of polynomials in only $x$ and only $y$.
# | Suppose there are polynomials $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n}$ and $g(y)=b_{0} y^{m}+b_{1} y^{m-1}+\ldots+b_{m}$, for which $f(x) g(y)=x^{200} y^{200}+1$. Setting
$x=0$, we get $a_{n} g(y)=1$, that is, $g(y)=1 / a_{n}$ for all $y$. Setting $y=0$, similarly we get that $f(x)=1 / b_{m}$ for all $x$. Thus, $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,561 |
10,11 |
The numbers $1,2, \ldots, k^{2}$ are arranged in a square table
| 1 | 2 | $\ldots$ | $k$ |
| :---: | :---: | :---: | :---: |
| $k+1$ | $k+2$ | $\ldots$ | $2 k$ |
| $\ldots$ | $\ldots$ | $\ldots$ | $\ldots$ |
| $(k-1) k+1$ | $\ldots$ | $\ldots$ | $k^{2}$ |
An arbitrary number is written down, after which the ... | Let's write the given table as
| $k \cdot 0+1$ | $k \cdot 0+2$ | $\ldots$ | $k \cdot 0+k$ |
| :---: | :---: | :---: | :---: |
| $k \cdot 1+1$ | $k \cdot 1+2$ | $\ldots$, | $k \cdot 1+k$ |
| $\ldots$ | $\ldots$ | $\ldots$ | $\ldots$ |
| $(k-1) k+1$ | $(k-1) k+2$ | $\ldots$ | $(k-1) k+k$ |
Each number in the table is r... | \frac{1}{2}k(k^2+1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,564 |
Given the equation $x^{n}-a_{1} x^{n-1}-a_{2} x^{n-2}-\ldots-a_{n-1} x-a_{n}=0$, where $a_{1} \geq 0, a_{2} \geq 0, a_{n} \geq 0$.
Prove that this equation cannot have two positive roots. | Let's rewrite the given equation as $1=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\ldots+\frac{a_{n}}{x^{n}}$. For $x>0$, the function on the right-hand side is monotonically decreasing, so it cannot take the value 1 at two different positive values of $x$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,565 |
$\left.\begin{array}{l}\text { [Quadratic inequalities and systems of inequalities } \\ {[\quad \text { Pairing and grouping; bijections }]}\end{array}\right]$
Given two sets of numbers $a_{1}>a_{2}>\ldots>a_{n}$ and $b_{1}>b_{2}>\ldots>b_{n}$. Prove that $a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}>a_{1} b_{n}+$ $a_{2... | Let's group the terms in both expressions that are equidistant from the ends. As a result, we get that the difference between the expression on the left and the expression on the right is the sum of expressions of the form \(a_{k} b_{k} + a_{n+1-k} b_{n+1-k} - a_{k} b_{n+1-k} - a_{n+1-k} b_{k} = (a_{k} - a_{n+1-k})(b_{... | 26 | Inequalities | proof | Yes | Yes | olympiads | false | 53,566 |
[ Divisibility rules for 3 and 9 $]$
There is a three-digit number $\overline{a b c}$, we take $\overline{c b a}$ and subtract the smaller from the larger. We get the number $\overline{a_{1}^{1}} \overline{b_{1}} \overline{1} \bar{c}_{1}^{-}$, and do the same with it, and so on.
Prove that at some step we will get ei... | 1) $b_{1}=9$
2) $\bar{a}_{1}^{-} \bar{b}_{1}^{-} \bar{c}_{1}^{-}$ is divisible by 9.
Therefore, $\overline{a_{1}} \overline{\overline{1}} \overline{\overline{1}} \bar{c}_{1}^{-}$ is one of the numbers $99,198,297,396,495,594,693,792,891,990$. It remains to note that $594 \rightarrow 99$
$\rightarrow 891 \rightarrow 6... | 495 | Number Theory | proof | Yes | Yes | olympiads | false | 53,567 |
$3+$ [ Examples and counterexamples. Constructions ]
Prove that it is possible to arrange the numbers from 1 to $n^{2}$ in an $n \times n$ table so that the sums of the numbers in each column are equal.
# | First, let's arrange the numbers as follows: in the first row $1, 2, \ldots, n$, in the second row $n+1, n+2, \ldots, n+n$, and so on. Then in the $k$-th row, the numbers will be $(k-1) n+1$,
$(k-1) n+2, \ldots, (k-1) n+n$, and thus in the $j$-th column $-j, n+j, \ldots, (n-1) n+j$. This table still does not satisfy t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,568 |
Processes and Operations Invariants
Authors: Gorsky E.A., Dorichenko S.A.
Three natural numbers are written on the board. Petya writes down the product of any two of these numbers on a piece of paper, and decreases the third number by 1 on the board. With the new three numbers on the board, he repeats the same operat... | At each step, Petya reduces the product of the numbers on the board by the number he writes on the paper: $x y(z-1)$ $=x y z-x y$, so the product of the numbers on the board added to the sum of the numbers on the paper does not change.
Since at the end the product on the board will be equal to 0, the sum on the paper ... | xyz | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,569 |
[ Regular Polyhedra. Duality and Relationships ] [ Distance between_two points. Equation of a sphere ]
Authors: Rabzimiotskyl., Giadkih A.
Can an octahedron be inscribed in a cube such that the vertices of the octahedron lie on the edges of the cube?
# | In the figure, an octahedron is inscribed in a cube with an edge length of 4; the vertices of the octahedron divide the edges of the cube in the ratio $1: 3$. The square of the length of each edge of the octahedron is 18 (either $3^{2}+3^{2}$, or $4^{2}+1^{2}+$ $\left.1^{2}\right)$, that is, all its edges are equal.
!... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,570 |
Folklore
Is there a triangular pyramid among the six edges of which:
a) two edges are less than 1 cm in length, while the other four are more than 1 km?
b) four edges are less than 1 cm in length, while the other two are more than 1 km? | a) Let's take two isosceles triangles with sides 0.9 cm, 1001 km, 1001 km. Place them on top of each other in the same plane, and then "lift" one of them above the plane without moving its base. In other words, rotate one of the triangles around the line containing the bases of the triangles. It is clear that this can ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,572 |
Evoikinov M.A.
A pirate has five bags of coins, each containing 30 coins. He knows that one bag contains gold coins, another contains silver coins, a third contains bronze coins, and each of the two remaining bags contains an equal number of gold, silver, and bronze coins. You can simultaneously take any number of coi... | Example. Let's take one coin from each bag. Among these five coins, there are coins of all three types, so there is only one coin of a certain type. If it is, for example, a gold coin, then it was taken from the bag with gold coins. Indeed, for each coin from the "mixed" bag, there is a matching one from the correspond... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,573 |
[ Prime numbers and their properties $]$ [ Examples and counterexamples. Constructions ]
Find the smallest natural number $n$, for which the following condition is satisfied: if the number $p-$ is prime and $n$ is divisible by $p-1$, then $n$ is divisible by $p$. | Let $n$ satisfy this condition. Since $n$ is divisible by $1=2-1$, it must be divisible by 2, but then it is divisible by $3=2+1$, by
$7=2 \cdot 3+1$ and by $43=2 \cdot 3 \cdot 7+1$. Therefore, $n$ must be divisible by $1806=2 \cdot 3 \cdot 7 \cdot 43$. Hence, the minimum $n$ (if it exists) is no less than 1806.
On t... | 1806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,576 |
Vasiniev N. 5.
Let $k$ and $n$ be natural numbers, $k \leq n$. Arrange the first $n^2$ natural numbers in an $n \times n$ table so that the numbers in each row are in increasing order, and at the same time, the sum of the numbers in the $k$-th column is a) the smallest; b) the largest. | a) Example. If the numbers are arranged as shown in the table - first filling the first $k$ columns, row by row, with numbers from 1 to $k n$, and then filling the remaining $n-k$ columns with the remaining numbers (in any order, as long as the condition of increasing numbers in each row is met) - then the sum of the n... | )\frac{1}{2}kn(n+1);b)\frac{1}{2}n((n-1)^{2}+k(n+1)) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,577 |
Ionin Yu. I.
The quadratic trinomial $f(x)=a x^{2}+b x+c$ is such that the equation $f(x)=x$ has no real roots.
Prove that the equation $f(f(x))=x$ also has no real roots. | The condition means that the parabola $y=f(x)$ is entirely above or entirely below the line $y=x$. In the first case, $f(f(x))>f(x)>x$, and in the second case, $-f(f(x))<f(x)<x$ for all $x$. Thus, in both cases, the equation $f(f(x))=x$ has no solutions. | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,578 |
[ Symmetric Systems. Involutory Transformations ] Symmetric Polynomials [ Methods for Solving Problems with Parameters
Solve the system of equations:
$$
\begin{aligned}
& x y=a, \\
& x^{5}+y^{5}=b^{5} .
\end{aligned}
$$
# | Let $a=0$. If $b \neq 0$, there are no solutions; if $b=0$ - an infinite set of solutions of the form $(t,-t)$.
Let $a \neq 0$. Set $x y=t$. Then $x^{5}+y^{5}=(x+y)^{5}-5(x+y)^{3} x y+5(x+y) x^{2} y^{2}=a^{5}-5 a^{3} t+5 a t^{2}$.
Solving the quadratic equation $5 a t^{2}-5 a^{3} t+a^{5}-b^{5}=0$, we find $t=\frac{1}... | \begin{aligned}& | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,579 |
10,11
In space, a closed broken line is constructed such that all segments have the same length and every three consecutive segments are pairwise perpendicular. Prove that the number of segments is divisible by 6.
# | Let $\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}$ be the vectors of three consecutive segments of the given broken line. After the vectors $\boldsymbol{e}_{2}, \boldsymbol{e}_{3}$, there should be a vector perpendicular to them, that is, the vector $\pm \boldsymbol{e}_{1}$. Continuing these consideration... | 6 | Geometry | proof | Yes | Yes | olympiads | false | 53,580 |
8,9
The side of a cell on a grid paper is 1. A rectangle with sides $m$ and $n$ is constructed along the grid lines.
Can a closed broken line be drawn within the rectangle along the grid lines that passes exactly once through each grid node located inside or on the boundary of the rectangle? If so, what is its length? | Notice that from each node of the grid located inside or on the boundary of the rectangle $n \times m$, exactly one segment of the path goes out, otherwise, this node is either not visited at all or visited more than once. The total number of nodes inside or on the boundary of the rectangle is $(n+1)(m+1)$, and thus th... | (n+1)(+1) | Other | math-word-problem | Yes | Yes | olympiads | false | 53,581 |
The first term and the common difference of an arithmetic progression are natural numbers. Prove that there exists a term in the progression whose representation includes the digit 9.
# | Suppose the first term and the difference of an arithmetic progression in absolute value are less than $\$ 10 \wedge k \$$ . Then there exists a term in the progression whose ( $k+1$ )-th digit is any given digit. In particular, this digit can be nine. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,583 |
All integers from 1 to $2 n$ are written in a row. Then, to each number, the number of the position it occupies is added.
Prove that among the resulting sums, there will be at least two that give the same remainder when divided by $2 n$.
# | Let the numbers $a_{1}, a_{2}, \ldots, a_{2 n}$ be written in a row. Suppose that all numbers $a_{1}+1, a_{2}+2, \ldots, a_{2 n}+2 n$ give different remainders when divided by $2 n$. Then these remainders are $1,2, \ldots, 2 n$. Therefore, $\left(a_{1}+1\right)+\left(a_{2}+2\right)+\ldots+\left(a_{2 n}+2 n\right)=2(1+2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,584 |
[ Sorting in ascending (descending) order ] [ Partitioning into pairs and groups; bijections $]$
Given 1993 numbers. It is known that the sum of any four numbers is positive. Is it true that the sum of all numbers is positive? | Let's denote the numbers as $a_{1}, a_{2}, \ldots, a_{1993}$. We have the following inequalities: $a_{1}+a_{2}+a_{3}+a_{4}>0, a_{2}+a_{3}+a_{4}+a_{5}>0, \ldots, a_{1990}+a_{1991}+a_{1992}+a_{1993}>0, a_{1991}+a_{1992}+a_{1993}+a_{1}>0, a_{1992}+a_{1993}+a_{1}+a_{2}>0, a_{1993}+a_{1}+a_{2}+a_{3}>0$. By summing the left-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,591 |
7,8,9 |
| $\left[\begin{array}{l}\text { Proof by contradiction }\end{array}\right]$ | | |
Auto: Boltiansky B.
Consider $4(N-1)$ boundary cells of an $N \times N$ table. It is required to fill these cells with consecutive $4(N-1)$ integers such that the sum of the numbers at the vertices of any rectangle with sides... | a), b) The corresponding arrangements are shown in Figures 1a, 1b.
| 4 | 1 | 7 |
| :--- | :--- | :--- |
| 6 | | 2 |
| 5 | 3 | 8 |
a)
| 9 | 13 | 12 | 11 | 14 |
| :---: | :---: | :---: | :---: | :---: |
| 5 | | | | 2 |
| 4 | | | | $\mathbf{1}$ |
| 3 | | | | 0 |
| 10 | 8 | 7 | 6 | 15 |
b)
Fig. 1
b) We have... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,592 |
Anjanan A.
All possible non-empty subsets are taken from the set of numbers $1,2,3, \ldots, n$. For each subset, the reciprocal of the product of all its numbers is taken. Find the sum of all such reciprocal values. | The desired sum will result after expanding the brackets in the expression $\left(1+\frac{1}{1}\right)(1+1 / 2)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)-1$, which is equal to $2 / 1 \cdot 3 / 2 \cdot 4 / 3 \cdot \ldots \cdot{ }^{n+1} / n-1=n+1-1=n$.
## Answer
n. | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,593 |
[Area of a triangle (using height and base).] [ Examples and counterexamples. Constructions ]
Author: Gurevich G.A.
Do there exist 100 triangles, none of which can be covered by the other 99? | Consider a set of isosceles triangles such that the height of each subsequent triangle is 200 times the diameter of the previous one, and its area is 20000 times smaller than the area of the previous one. Divide one of the triangles into 200 parts using 199 segments parallel to the base and dividing the height into equ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,594 |
Folklore
Two players alternately increase a natural number so that with each increase, the difference between the new and old values of the number is greater than zero but less than the old value. The initial value of the number is 2. The player who, as a result of their move, reaches 1987 wins. Who wins with correct ... | In the sequence $1987,993,496,248,124,62,31,15,7,3$, each subsequent number is the quotient of the division of the previous number by 2 with a remainder. We will prove that this sequence consists of winning numbers (i.e., a player who names one of these numbers has a winning strategy). The number 1987 is a winning numb... | Thefirstplayer | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,595 |
Folkior
Can we find such two natural numbers $X$ and $Y$, that $Y$ is obtained from $X$ by permuting its digits, and $X+Y=$ 9... 9 (1111 nines)
# | Let's consider the last digits of numbers $X$ and $Y$. Their sum cannot be 19, so it must be 9, and no carry is transferred to the next digit. Therefore, the sum of the second-to-last digits of $X$ and $Y$ is also 9. Similarly, we conclude that the sum of the digits in each place is 9. Consequently, the number of occur... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,596 |
[Polynomial division with remainder. GCD and LCM of polynomials]
Induction (other).
Prove that for any natural $n$ there exists a non-zero polynomial $P(x)$ with coefficients equal to $0, -1, 1$, and degree not exceeding $2^n$, which is divisible by $(x-1)^n$. | Let $P_{1}(x)=x-1, P_{n+1}(x)=\left(x^{2^{n}}-1\right) P_{n}(x)$. We will prove that these polynomials have the required properties. It is easy to verify that $\operatorname{deg} P_{n}=2^{n}-1$. The coefficients of $P_{n+1}$ are two non-overlapping sets of coefficients of $P_{n}$, therefore, just like $P_{1}$, they are... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,599 |
Ruanovo i.S.
A square of $8 \times 8$ cells is painted white. It is allowed to choose any rectangle of three cells in it and repaint all of them to the opposite color (white to black, black to white). Is it possible to repaint the entire square black with several such operations? | The first method. We will fill all the cells of the board with numbers as shown in the left figure. Now it is enough to notice that initially there are more white cells with the number 1 than cells with the number 2, and each move can repaint exactly one cell with the number 1 and one with the number 2.
| 1 | 2 | 3 | ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,601 |
Table E.
The sum of $n$ numbers is zero, and the sum of their squares is one. Prove that among these numbers, there are two whose product is no greater than $-1 / n$.
# | Let $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$ be given numbers. Then $\left(x_{1}-x_{k}\right)\left(x_{n}-x_{k}\right) \leq 0$ for each $k$ from 1 to $n$. Adding all these inequalities, we get
$n x_{1} x_{n}-\left(x_{1}+x_{n}\right) \cdot 0+1 \leq 0$, that is, $x_{1} x_{n} \leq -1 / n$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,602 |
Spivak A.V.
Given a cube with an edge length of $n$ cm. We have a long piece of insulating tape 1 cm wide. It is required to cover the cube with the tape, with the tape being able to freely cross over an edge to another face, travel in a straight line parallel to the edge on a face, and not hang off the face. Into how... | $2 n$ pieces are sufficient: we cover four faces of the cube, forming the lateral surface of a rectangular prism, with $n$ rows of tape, each row 1 unit wide, and then similarly cover another set of four faces (two of which have already been covered once, and the other two are being covered for the first time).
Let's ... | 2n | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,603 |
$:$ Folkiore
The numbers $1,2,3, \ldots, 25$ are arranged in a $5 \times 5$ table such that the numbers in each row are in increasing order.
What is the greatest and the least value that the sum of the numbers in the third column can have? | Evaluation. The sum of the numbers in the first three columns is not less than the sum of the first 15 natural numbers, which is not less than 120.
If we add 2 to each number in the first column and 1 to each number in the second column, the numbers in each row will remain in non-decreasing order. Therefore, the sum o... | 4585 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,604 |
Folklore
The coefficients of the quadratic equation $x^{2}+p x+q=0$ have changed by no more than 0.001.
Can the larger root of the equation change by more than 1000? | Make the coefficient $p$ very large.
## Solution
Consider the equations $x^{2}-2 p x+p^{2}=0$ and $x^{2}-2(p+a) x+p^{2}=0$, where $p=10^{9}, a=5 \cdot 10^{-4}$. The first equation has a single root $p$, and the larger root of the second equation is
$(p+a)+\sqrt{p^{2}+2 a p+a^{2}-p^{2}}>p+\sqrt{2 a p}=p+1000$.
## An... | Yes | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,605 |
Kanel-Belov A.Y.
On a $5 \times 5$ chessboard, the maximum number of knights are placed such that they do not attack each other. Prove that such an arrangement is unique. | Example. Consider the chessboard coloring, where the central field is black. It is clear that 13 knights on black squares do not attack each other.
Evaluation. Consider the knight's tour of the board (the cells are numbered in the order of the tour). We will relax the condition: we will only require that the knights d... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,607 |
Kanel-Belov A.Y.
A cube with a side of 20 is divided into 8000 unit cubes, and a number is written in each cube. It is known that in each column of 20 cubes, parallel to the edge of the cube, the sum of the numbers is 1 (columns in all three directions are considered). In a certain cube, the number 10 is written. Thro... | Through the given cube K, one horizontal layer G and two vertical layers pass. The sum of all numbers in 361 vertical columns, not included in the last two layers, is 361. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with G (there are 361 such cub... | 333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,608 |
Shapovalovo $A . B$.
An object in the shape of a convex polyhedron is floating in the sea.
Can it happen that $90 \%$ of its volume is below the water level and at the same time more than half of its surface is above the water level? | Consider a regular pyramid floating vertex-down. Let the area of its lateral surface be S. Then the area of the base is $S \cos \alpha$, where $\alpha$ is the angle of inclination of the lateral faces to the base. By the condition, the part of the pyramid under water is similar to the original with a similarity coeffic... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,609 |
Shapovalov A.V.
Among the angles of each lateral face of a pentagonal prism, there is an angle $\varphi$. Find all possible values of $\varphi$. | In a right prism, each lateral face is a rectangle, so $\varphi=90^{\circ}$ is suitable.
Suppose $\varphi \neq 90^{\circ}$. We can assume that $\varphi<90^{\circ}$ (if the obtuse angles in the parallelograms are equal, then the acute angles are also equal). Draw in the plane of the base through one of the vertices $V$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,610 |
Senderov B.A.
Natural numbers $a, b, c, d$ are such that the least common multiple of these numbers is equal to $a+b+c+d$.
Prove that $a b c d$ is divisible by 3 or by 5 (or by both). | Let $M=\operatorname{HOK}(a, b, c, d)=a+b+c+d$. It is sufficient to prove that $M$ is divisible by 3 or 5. We can assume that $a \geq b \geq c \geq d$. If all four numbers are equal, then $M=a$, which contradicts the condition. Therefore, the number $b+c+d$ is less than $3a$ and is divisible by $a$.
If $b+c+d=2a$, the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,611 |
Galperin G.A.
There exist 1000 consecutive natural numbers, none of which is a prime number (for example, $1001!+2, 1001!+3, \ldots, \quad 1001!+1001$).
Do there exist 1000 consecutive natural numbers, among which there are exactly five prime numbers? | Let $P_{n}$ denote the number of prime numbers among a thousand consecutive numbers from $n$ to $n+999$. Note that $P_{n+1}$ differs from $P_{n}$ by no more than one. However, $P_{1}>5$, and $P_{1001!+2}=0<5$, so when $n$ changes from 1 to $1001!+2$, the value of $P_{n}$ will be equal to 5 for some $n$.
## Answer
The... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,612 |
[ Quadratic inequalities and systems of inequalities ] [ Examples and counterexamples. Constructions ]
Find all prime numbers $p$, for each of which there exists a natural number $m$ such that
$\sqrt{m}+\sqrt{m+p}$ is also a natural number. | Let $p$ be odd, that is, $p=2 n+1$, where $n$ is a natural number. Then the condition of the problem is satisfied for $m=n^{2}$. It remains to consider the case $p=2$. Let $\sqrt{m}+\sqrt{m+2}$ be a natural number. Then the number $(\sqrt{m}+\sqrt{m+2})^{2}=2 m+2+2 \sqrt{m^{2}+2 m}$, and therefore, $2 \sqrt{m^{2}+2 m}$... | p- | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,617 |
Authors: Anioov S.S., Proizvodov V.V.
Let \( a, b, c \) be non-negative integers such that \( 28a + 30b + 31c = 365 \). Prove that \( a + b + c = 12 \).
# | Let $a+b+c \leq 11$. Then $28 a+30 b+31 c \leq 31(a+b+c) \leq 11 \cdot 31=341$. And this cannot be!
Let $a+b+c=13$. The variant $a=13, b=c=0$ does not satisfy the condition: $28 \cdot 13+30 \cdot 0+31 \cdot 0=364 \neq 365$. The only remaining variant is $a+b+c=13, a>0$ and $28 a+30 b+31 c=28(a+b+c)+2 b+3 c \geq$ $28 \... | 12 | Number Theory | proof | Yes | Yes | olympiads | false | 53,620 |
8,9,1 |
| | Cutting into parallelograms | ] |
| | Rectangles and squares. Properties and criteria | ] |
Auto: [proizvooooov B.B.
A square with a side of 1 was cut into rectangles, and one side was marked on each of them. Prove that the sum of the lengths of all the marked sides cannot be less than 1. | Increase each rectangle's side, perpendicular to the marked one, to 1. In this case, its area will not decrease and will become (numerically) equal to the length of the chosen side. Thus, the sum of the lengths of the chosen sides is equal to the sum of the areas of the elongated rectangles, which, in turn, is not less... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,621 |
Blinkov A.d:
In a football championship, 16 teams participated. Each team played against each of the others once, with 3 points awarded for a win, 1 point for a draw, and 0 for a loss. We will call a team successful if it scored at least half of the maximum possible number of points. What is the maximum number of succ... | Each team played 15 games and therefore could have earned a maximum of $15 \cdot 3=45$ points. Thus, a team is successful if it has at least 23 points.
But one of the teams scored no more than the average possible number of points. Even if all matches were successful, the average is
$15 \cdot 1.5=22.5$.
We will show... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,623 |
Shapovalov A.B.
Can 1995 different natural numbers be arranged in a circle so that for any two adjacent numbers, the ratio of the larger to the smaller is a prime number? | Suppose that numbers $a_{1}, a_{2}, \ldots, a_{1995}$ have been found that can be arranged in the required manner. Let the number $a_{k}$ ( $k$ $=1,2, \ldots, 1995$ ) be represented as the product of $n_{k}$ prime factors (not necessarily distinct). Since any two adjacent numbers differ by one prime factor, for each $k... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,626 |
Iterations [ Continuity Considerations
Let $P(x)$ be a polynomial of odd degree. Prove that the equation $P(P(x))=0$ has no fewer distinct real roots than the equation $P(x)=0$.
# | Let $x_{1}, \ldots, x_{n}$ be all the distinct roots of the equation $P(x)=0$. We need to prove that the equation $P(P(x))=0$ has at least $n$ distinct roots.
Consider $n$ equations: $P(x)=x_{1}, P(x)=x_{2}, \ldots, P(x)=x_{n}$. Each of them has a solution, since $P(x)$ is a polynomial of odd degree. Let $a_{1}$ be th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,627 |
Polepsky o.K.
In the cells of a $100 \times 100$ table, non-zero digits are written. It turned out that all 100 hundred-digit numbers written horizontally are divisible by 11. Could it be that exactly 99 hundred-digit numbers written vertically are also divisible by 11? | Suppose that the required arrangement of digits is possible.
Consider the chessboard coloring of the cells of our table. According to the divisibility rule for 11, in each row, the sum of the digits on the black cells has the same remainder when divided by 11 as the sum of the digits on the white cells. Therefore, in ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,629 |
Agakhanov N.K.
The cosines of the angles of one triangle are respectively equal to the sines of the angles of another triangle.
Find the largest of the six angles of these triangles. | From the condition, it follows that the angles $\alpha_{1}, \alpha_{2}, \alpha_{3}$ of the first triangle are acute $\left(\cos \alpha_{i}=\sin \beta_{i}>0\right.$, where $\beta_{1}, \beta_{2}, \beta_{3}$ are the angles of the second triangle). Therefore, $\beta_{i}=90^{\circ} \pm \alpha_{i}$,
$i=1,2$, 3. From the equ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,630 |
Three regular polygons are inscribed in a circle, the number of sides of each subsequent one being twice that of the previous one. The areas of the first two are $S_{1}$ and $S_{2}$. Find the area of the third. | Let $S$ be the area of the third polygon, $R$ the radius of the circle, $n$ the number of sides of the first polygon, $\alpha=\pi / 2 n$. Then $S_{1}=1 / 2 n R^{2} \sin 4 \alpha$, $S_{2}=n R^{2} \sin 2 \alpha, S=n R^{2} \sin \alpha=S_{2 /} \cos \alpha$. From the first two equations, $S_{1 / S_{2}}=\cos 2 \alpha$ so, $\... | \sqrt{\frac{2S_{2}^{3}}{S_{1}+S_{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,633 |
$[. A$. Given three distinct natural numbers, one of which is equal to the half-sum of the other two. Can the product of these three numbers be a perfect 2008th power of a natural number?} | For example, $6^{669}, 2 \cdot 6^{669}, 3 \cdot 6^{669}$. Their product is $6 \cdot 6^{3 \cdot 669}=6^{2008}$.
## Answer
It can.
## $\begin{array}{ll}\text { Problem } 111651 & \text { Topics: } \\ \text { [Fundamental Theorem of Arithmetic. Factorization into prime factors }]\end{array}$ Difficulty: 3+ $[\quad$ Ex... | 6^{2008} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,635 |
[Ribamko A.V. On a $100 \times 100$ chessboard, 100 non-attacking queens are placed. Prove that in each corner $50 \times 50$ square, there is at least one queen.
# | Let there be no queens in the top-left $50 \times 50$ square. Then all 50 queens on the 50 top rows are in the top-right $50 \times 50$ square, and all 50 queens on the 50 left columns are in the bottom-left $50 \times 50$ square. But both of these squares are covered by 99 diagonals, so there are no more than 99 queen... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,636 |
Akopyan A.V.
Through a terminal, money can be transferred to a mobile phone, with a commission of a natural number of percent. Fedya put an integer amount of rubles on his mobile phone, and his account was credited with 847 rubles. How much money did Fedya put on the account, given that the commission is less than $30... | Let Fedia put $n$ rubles, and the commission is $k \%$. Then $(1-k / 100) n=847$, that is, $84700=$ $(100-k) n .84700=2 \cdot 2 \cdot 5 \cdot 5 \cdot 7 \cdot 11 \cdot 11$. According to the condition $70<100-k<100$, so it is necessary to find all numbers,
dividing 84700, in this range. A small enumeration shows that the... | 1100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,637 |
주мㅆㅇㅡ D: In a row of 2009 weights, the weight of each weight is an integer number of grams and does not exceed 1 kg. The weights of any two adjacent weights differ by exactly 1 g, and the total weight of all the weights in grams is an even number. Prove that the weights can be divided into two piles such that the sums... | Clearly, the weights of all weights placed at odd positions have the same parity, while the weights of all other weights have a different parity. Since the total weight is even, the 1005 weights at odd positions have even weights. Place the first weight on the left pan of the scales (let its weight be $2a \leq 1000$ g)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,638 |
Stakhanov $H . X$.
Given quadratic trinomials $x^{2}+2 a_{1} x+b_{1}, x^{2}+2 a_{2} x+b_{2}, x^{2}+2 a_{3} x+b_{3}$. It is known that $a_{1} a_{2} a_{3}=b_{1} b_{2} b_{3}>1$. Prove that at least one of these trinomials has two roots. | Assume the opposite; then the discriminants of all trinomials are non-positive, that is, $a_{k}^{2} \leq b_{k}(k=1,2,3)$. The left (and therefore the right) parts of these inequalities are non-negative, so they can be multiplied: $\left(a_{1} a_{2} a_{3}\right)^{2} \leq b_{1} b_{2} b_{3}=a_{1} a_{2} a_{3}$. But this co... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,639 |
|
| | $[$ Chess Coloring $]$ | |
Author: Raskina I.V.
At the edge of a round rotating table, 30 cups of tea were placed at equal intervals. The March Hare and Alice sat down at the table and started drinking tea from two of the cups (not necessarily adjacent). When they finished their tea, the Hare turned the table... | We will paint every other cup blue and red. Let the March Hare drink from a red cup at first. We will prove that Sonya drank from a blue cup at first. Indeed, if she drank from a red one, then after any rotation of the table, two cups of the same color would be emptied. Since there are 15 of each color, and they are em... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false | 53,640 |
Khachaturyan M.A.
The length of an adult worm is 1 meter. If a worm is adult, it can be cut into two parts in any ratio of lengths. As a result, two new worms are obtained, which immediately start growing at a rate of 1 meter per hour each. When the length of a worm reaches a meter, it becomes an adult and stops growi... | At the moment $\frac{1}{2^{10}}$ (hours) from the start of the procedure, cut off $\frac{1}{2^{10}}$ meters from the worm. At the moment $\frac{1}{2^{9}}$ from the grown part, cut off $\frac{1}{2^{9}}$ meters. At the moment $\frac{1}{2^{8}}$, cut off $\frac{1}{2^{8}}$ meters, ..., at the moment $\frac{1}{2}$, cut the a... | 1-\frac{1}{2^{10}} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,642 |
Shevyakov V.
Given a convex quadrilateral. If any diagonal is drawn in it, it will be divided into two isosceles triangles. And if both diagonals are drawn in it at the same time, it will be divided into four isosceles triangles. Is this quadrilateral necessarily a square? | All conditions are satisfied by an isosceles trapezoid with the smaller base equal to the lateral side, and angles of $72^{\circ}$ at the larger base.
## Answer
Not necessarily. | notfound | Geometry | proof | Yes | Yes | olympiads | false | 53,643 |
Folklore
Among the participants of the Olympiad, each is acquainted with at least three others. Prove that it is possible to select a group of an even number of participants (more than two people) and seat them at a round table so that each is acquainted with both neighbors.
# | We will arrange the participants in a row so that acquaintances stand next to each other. Eventually, this process will end. This means that all three acquaintances of the last person in the row (denote him as $A$) are already arranged. Let $B$ and $C$ be those of them who do not stand next to $A$. If there is an even ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,644 |
Kanel-Belov A.Ya.
The sum of the digits of a natural number $n$ is 100. Can the sum of the digits of the number $n^{3}$ be 1000000? | It is enough to take a number with ones in the 1st, $10-\mathrm{th}$, $100-\mathrm{th}$, ..., $10^{99}$-th place (counting from the right).
## Answer
It can. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,645 |
Folklore
In a certain state, the airline system is arranged in such a way that each city is connected by air routes to no more than three other cities, and from any city, you can reach any other city with no more than one layover. What is the maximum number of cities that can be in this state? | Evaluation. From a fixed city $A$, one can directly reach no more than three cities, and with one transfer - no more than $3 \cdot 2=6$ additional cities. Thus, the total number of cities can be no more than ten. An example of a network of 10 cities is shown in the figure.
. Between them lies $k_{i}-i-1$ cards, so the desired sum $S=\left(k_{... | 1260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,651 |
[Evenness and Oddness] $[$ Invariants $]$ Author: Shapovedaov A.B. Several peasants have 128 sheep. If any one of them has no less than half of all the sheep, the others conspire and dekulakize him: each takes as many sheep as they already have. If two of them have 64 sheep each, they dekulakize one of them. Seven de... | After the first expropriation, the number of sheep that everyone except the expropriated has is even, and the total number of sheep is also even, which means the remainder of the expropriated is also even. Similarly, after the second expropriation, the number of sheep each has is divisible by 4, and after the seventh e... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 53,652 |
Misha solved the equation $x^{2} + a x + b = 0$ and told Dima a set of four numbers - the two roots and the two coefficients of this equation (but did not say which of them were the roots and which were the coefficients). Will Dima be able to find out which equation Misha solved, if all the numbers in the set turned ou... | Suppose there are two different equations $x^{2}+a x+b=0$ with roots $c, d$ and $x^{2}+a_{1} x+b_{1}=0$ with roots $c_{1}, d_{1}$, for which the sets specified in the condition coincide. Then $a+b+c+d=a_{1}+b_{1}+c_{1}+d_{1}$. By Vieta's theorem, $c+d=-a$ and $c_{1}+d_{1}=-a_{1}$, so $b=b_{1}$. Since the equations cons... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,653 |
Chekanov Yu.V.
A square board is divided by seven lines parallel to one side of the board and seven lines parallel to the other side of the board into 64 rectangular cells, which are painted in white and black in a checkerboard pattern. The distances between adjacent lines are not necessarily the same, so the cells ca... | Let the top-left cell be white. All odd columns are shifted to the left, and then all odd rows are shifted up. Now all the white cells are gathered in two rectangles in the top-left and bottom-right corners, and the ratio of the area of each white rectangle to the area of each black one is no more than 2. If the width ... | \frac{5}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,656 |
Class $\mathbf{M}$.
In the class, there are 15 boys and 15 girls. On March 8, some boys called some girls to congratulate them on the holiday (no boy called the same girl twice). It turned out that the children could be uniquely divided into 15 pairs so that in each pair there was a boy who called the girl. What is th... | Let's denote the boys as $M_{1}, M_{2}, \ldots, M_{15}$, and the girls as $-D_{1}, D_{2}, \ldots, D_{15}$, such that $M_{1}-D_{1}, M_{2}-D_{2}, \ldots, M_{15}-D_{15}$ is the only pairing that satisfies the condition of the problem. Suppose each boy called at least two girls. We will draw an arrow from each girl $D_{i}$... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,657 |
Chebotarev A.S.
In each cell of a $101 \times 101$ square, except the central one, there is one of two signs: "turn" or "straight." A machine enters from outside into any cell on the border of the square, after which it drives parallel to the sides of the cells, adhering to two rules:
1) in a cell with the "straight"... | Note that if the car can drive from cell $A$ to cell $B$, then it can also drive from cell $B$ to cell $A$ - by driving the same route in reverse. Therefore, it is sufficient to prove that, when leaving the house, the car can exit the boundaries of the square.
Suppose the signs are somehow placed. "Release" the car fr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,659 |
Perepechko A.yu.
Given an increasing infinite sequence of natural numbers $a_{1}, \ldots, a_{n}, \ldots$, where each term is either the arithmetic mean or the geometric mean of its two neighbors. Is it necessarily true that from some point onward, this sequence becomes either an arithmetic or a geometric progression? | Consider the sequence where $a_{2 n-1}=n(n+1), a_{2 n}=(n+1)^{2}: 2,4,6,9,12,16,20,25 \ldots$ Each even term is the arithmetic mean, and each odd term is the geometric mean of its neighbors.
It remains to note that, since the arithmetic mean of two different numbers is not equal to the geometric mean, our sequence doe... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,660 |
Yasinsky V.
On a plane, $n$ points are given, which are the vertices of a convex $n$-gon, $n>3$. It is known that there are exactly $k$ equilateral triangles with side 1, whose vertices are the given points.
a) Prove that $k \leq 0.666 n$.
# | a) For each of the given points, there exists a line passing through it such that all other given points lie on one side of this line. This allows us to select two triangles among all unit triangles with a vertex at the considered point - the "farthest left" triangle and the "farthest right" triangle (it is not exclude... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,662 |
A.Y. Evnin
2000 people registered on a new website. Each of them invited 1000 people to be their friends. Two people are considered friends if and only if each of them invited the other to be a friend. What is the minimum number of pairs of friends that could have formed? | Evaluation. A total of 2,000,000 invitations were sent, while the number of pairs on the site is $1000 \cdot 1999 = 1999000$. There are 1000 more invitations than pairs, so within at least 1000 pairs, two invitations were sent. Therefore, at least 1000 pairs were formed.
Example: Place everyone at the vertices of a re... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,663 |
4-
Can we, by applying the functions sin, cos, tan, cot, arcsin, arccos, arctan, arccot to the number 1 in some order, obtain the number 2010? (Each function can be used any number of times.) | Let $f(x)=\operatorname{ctg}(\operatorname{arctg} x)=1 / x, g(x)=\sin (\operatorname{arctg} x)=\frac{x}{\sqrt{1+x^{2}}}$. By induction, we check that $g^{n}(x)=g\left(g^{n-1}(x)\right)=$ $\frac{x}{\sqrt{1+n x^{2}}}$
Taking $n=2010^{2}-1$, we get that $g^{n}(1)=1 / 2010$, from which $f\left(g^{n}(1)\right)=2010$.
## A... | 2010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,665 |
Bogosiov I.I.
2011 numbers are written on the board. It turns out that the sum of any three of the written numbers is also a written number.
What is the smallest number of zeros that can be among these numbers?
# | An example from 2009 zeros and numbers $1, -1$ satisfies the condition.
Assume the number of zeros is no more than 2008. Then on the board, there will be either three non-negative numbers, among which at least two are strictly positive, or three non-positive numbers, among which at least two are strictly negative. Sup... | 2.8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,667 |
Highness and.
The Rhinoceros has vertical and horizontal folds on its skin. In total, there are 17 folds. If the Rhinoceros scratches itself against a tree, either two horizontal or two vertical folds on that side disappear, and on the other side, one horizontal and one vertical fold are added. (If there are not two f... | Let it succeed. The number of folds on the left side decreases by 2 when the left side is scratched and increases by 2 when the right side is scratched. Since this number did not change, the left and right sides were scratched the same number of times. With each scratch, the total number of vertical folds changes by 1,... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,668 |
Aeakhanov $H . X$.
Let $a_{1}, \ldots, a_{10}$ be distinct natural numbers, each not less than 3, with their sum equal to 678. Can the sum of the remainders when some natural number $n$ is divided by the 20 numbers $a_{1}, a_{2}, \ldots, a_{10}, 2 a_{1}, 2 a_{2}, \ldots, 2 a_{10}$ equal $2012$? | Suppose such a number $n$ exists. The maximum possible remainder when dividing by a natural number $m$ is $m-1$. Therefore, the sum of the remainders when dividing an arbitrary number by $a_{1}, \ldots, a_{10}$ is no more than $678-10=668$, and the sum of the remainders when dividing it by $2 a_{1}, 2 a_{2}, \ldots, 2 ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,670 |
5,6 |
Kolya claims that it is possible to determine whether the sum of all four-digit numbers, which do not contain the digit 0 or 9, is divisible by 101 without calculating the sum itself. Is Kolya right? | A number paired with a number composed of the digits $a, b, c, d$ is a number composed of the digits $9-a, 9-b, 9-c, 9-$ $d$, placed in the same order. Thus, all four-digit numbers satisfying the given conditions are divided into pairs, the sum of the numbers in which is 9999. Therefore, the sum of all such numbers wil... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,672 |
Shapovalov A.V.
Prove that it is possible to write a non-negative number on each edge of an arbitrary tetrahedron such that the sum of the numbers on the edges of each face is numerically equal to its area. | Let the area of the smallest face be $s$, the largest face be $S$, and the two remaining faces be $a$ and $b$. Write the number $s$ on the edge common to the smallest and largest faces, and write zero on the other two edges of the smallest face. On the two other edges of the largest face, write the numbers $1 / 2(S-b+a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,673 |
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