problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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[ Limitedness, monotonicity]
Given a strictly increasing function \$f \colon \mathbb{\mathrm{N}}_0 \to \mathbb{\mathrm{N}}_0\$ (where $\mathbb{\mathrm{N}}_0$ is the set of non-negative integers), which satisfies the relation $\mathrm{f}(\mathrm{n}+\mathrm{f}(\mathrm{m}))=\mathrm{f}(\mathrm{n})+\mathrm{m}+1$ for any $m... | 1) Substituting $\$ \mathrm{~m}=0 \$, \$ n=0 \$$, we get $\$ \mathrm{f}(\mathrm{f}(0))=\mathrm{f}(0)+1 \$$. If $\$ \mathrm{f}(0)=0 \$$, then we get $\$ \mathrm{f}(0)=\mathrm{f}(0)+1 \$$, which is impossible.
 $a_{100}$;
b) $a_{1983}$. | a) Immediately note that the sequence $a_{k}$ is strictly increasing. Indeed, the assumption $a_{k}=a_{k+1}=n$ immediately leads to a contradiction: $a_{n}=3 k=3(k+1)$.
Moreover, $a_{1}>1$ (otherwise $a_{a_{1}}=a_{1}=1 \neq 3$). Therefore, $a_{k}>k$ for all $k$. On the other hand, $a_{1}<a_{a_{1}}=3$. Hence, $a_{1}=2$... | 181 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Pascal's Triangle and Newton's Binomial ]
How many rational terms are contained in the expansion of
a) $(\sqrt{2}+\sqrt[4]{3})^{100}$
b) $(\sqrt{2}+\sqrt[3]{3})^{300}$? | a) For the number $(\sqrt{2})^{100-k}(\sqrt[4]{3})^{k}$ to be rational, $k$ must be a multiple of 4. There are 26 such integers: $0, 4, 8$, ..., 96, 100.
b) For the number $(\sqrt{2})^{300-k}(\sqrt[3]{3})^{k}$ to be rational, $k$ must be a multiple of 3, and $300-k$ must be even. Therefore, $k$ must be a multiple of 6... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Monotonicity, boundedness $\quad$ ]
Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0,1]$.
# | Notice that $a+b+c+d-a b-b c-c d-d a=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y, 0$ $\leq x \leq 2$ and $0 \leq y \leq 2$.
$x+y-x y=(x-1)(1-y)+1$, where $|x-1| \leq 1$ and $|1-y| \leq 1$. Therefore, $(x-1)(1-y) \leq 1$, and $x+y-x y \leq 2$.
The value 2 is achieved, for example, if $a=c=1, b=d=0$.
## Answer
2
Solv... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Does there exist a natural number $n$ such that $\sqrt[n]{17 \sqrt{5}+38}+\sqrt[n]{17 \sqrt{5}-38}=2 \sqrt{5}$? | Note that $(\sqrt{5} \pm 2)^{3}=5 \sqrt{5} \pm 30+12 \sqrt{5} \pm 8=17 \sqrt{5} \pm 38$. Therefore, the given equality is true for $n=3$.
## Answer
It exists. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Rybnikov I.g.
A store received 20 kg of cheese, and a queue formed. After selling cheese to the next customer, the saleswoman accurately calculates the average weight of the purchase for all the cheese sold and informs how many people the remaining cheese will last if everyone buys exactly this average weight. Could t... | Let $s_{k}$ be the average weight of cheese sold to the first $k$ customers. According to the condition, $20 - k s_{k} = 10 s_{k}$, hence $s_{k} = \frac{20}{k+10}$ kg, and after the $k$-th customer, there remains
200
$k+10$
message after each customer (not limited to the first ten). After the 10th customer, exactly ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4-$ [Increasing and Decreasing. Function Analysis]
Solve the equation $2 \sin \pi x / 2 - 2 \cos \pi x = x^{5} + 10 x - 54$.
# | Transferring all terms to one side, we get the equation $x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x=0$. Consider the function
$f(x)=x^{5}+10 x-54-2 \sin ^{\pi x} / 2+2 \cos \pi x$. Note that $f^{\prime}(x)=5 x^{4}+10-\pi \cos \pi x / 2-2 \sin \pi x>5 x^{4}+10-3 \pi>0$. Therefore, $f(x)$ is an increasing function, so ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Oleg drew an empty 50×50 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 irrational. Then, in each cell of the table, he wrote the sum of the numbers written next to its row and its column (... | Evaluation. Let along the left side of the table be written \( x \) irrational and
\( 50-x \) rational numbers. Then along the top side, \( 50-x \) irrational and \( x \) rational numbers are written. Since the sum of a rational and an irrational number is always irrational, in the table there are at least \( x^2 + (5... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Pooodiisskii 0. ..
Oleg drew an empty 50×50 table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers were 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 written ne... | 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 |
Ilya Muromets meets the three-headed Zmei Gorynych. And the battle begins. Every minute Ilya cuts off one of Zmei's heads. With a probability of $1 / 4$, two new heads grow in place of the severed one, with a probability of $1 / 3$ only one new head grows, and with a probability of $5 / 12$ - no heads grow. The Zmei is... | Blows by Ilya Muromets, in which the number of heads changes, are called successful.
Let's find the probability that at some point there will be a last successful blow. This means that starting from this point, there will be no more successful blows, that is, all blows will be unsuccessful. The probability of this is ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sequence $A$.
The sequence of numbers $x_{1}, x_{2}, \ldots$ is such that $x_{1}=1 / 2$ and $x_{k+1}=x_{k}^{2}+x_{k}$ for any natural number $k$.
Find the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{1}+1}+\ldots+\frac{1}{x_{100}+1}$. | Note that $\frac{1}{x_{k+1}}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, which means $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$. Therefore,
$\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\ldots+\frac{1}{x_{100}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\ldots+\frac{1}{x_{100}}-\frac{1}{x_{101}}=\frac... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Rubaev I.S.
What is the maximum finite number of roots that the equation
$$
\left|x-a_{1}\right|+. .+|x-a 50|=\left|x-b_{1}\right|+. .+|x-b 50|
$$
can have, where $a_{1}, a_{2}, \ldots, a_{50}, b_{1}, b_{2}, \ldots, b_{50}$ are distinct numbers? | Let $f(x)=\left|x-a_{1}\right|+\ldots+|x-a 50|-\left|x-b_{1}\right|-. .-|x-b 50|$ and rewrite the original equation in the form $f(x)=0$.
Let $c_{1}<c_{2}<. .<c 100$ be all the numbers from the set $\left\{a_{1}, . ., a 50, b_{1}, . ., b 50\right\}$, ordered in ascending order. On each of the 101 intervals $\left[-\bo... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
Consider the sequence where the $n$-th term is the first digit of the number $2^n$.
Prove that the number of different "words" of length 13 - sets of 13 consecutive digits - is 57. | The first digit of the number $2^{n}$ is determined by the number $a_{n}=\{n \lg 2\}$, specifically: if $a_{n} \in[0, \lg 2)$, then the first digit of the number $2^{n}$ is 1; if $a_{n} \in\left[\lg 2, \lg 3\right)$, then it is 2; ..., if $a_{n} \in[\lg 9,1)$, then it is 9.
Thus, the sequence of the first digits of th... | 57 | Number Theory | proof | Yes | Yes | olympiads | false |
Fomin D:
How many pairs of natural numbers ( $m, n$ ), each not exceeding 1000, exist such that
$\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} ?$ | Consider all pairs $(m, n)$ of natural numbers for which $m+n+1=s \geq 3$. We have $\frac{1}{s}2>\sqrt{2}>1 / 2>1 / s$, and since the number $\sqrt{2}$ is irrational, it falls into exactly one of the specified intervals. Thus, among the pairs $(m, n)$ with a fixed value of $s$, there will be exactly one for which the i... | 1706 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits in $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? | The number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.
If $N$ ends in 1, then when it is doubled, there is no carry from the last digit to the second last digit. Th... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
A bag of seeds was passed around the table. The first person took 1 seed, the second took 2, the third took 3, and so on: each subsequent person took one more seed than the previous one. It is known that in the second round, the total number of seeds taken was 100 more than in the first round. How many... | Let there be $n$ people sitting at the table. Then on the second round, each took $n$ more seeds than on the first, and all together took $n \cdot n=n^{2}$ more seeds than on the first. Since $n^{2}=100$, then $n=10$.
## Answer
10 people. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Given a polynomial $P(x)$ with integer coefficients. It is known that $P(1)=2013, P(2013)=1, P(k)=k$, where $k$ is some integer. Find $k$.
# | By the Bezout's theorem for integer polynomials (see the solution of problem 35562$), \( k-2013 = P(k) - P(1) \) is divisible by \( k-1 \), and \( k-1 = P(k) - P(2013) \) is divisible by \( k-2013 \). Therefore, \( |k-2013| = |k-1| \). The solution to the obtained equation is the midpoint of the segment \([1, 2013]\), ... | 1007 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Folkolo
Find the maximum value of the expression $ab + bc + ac + abc$, if $a + b + c = 12$ (where $a, b$, and $c$ are non-negative numbers). | The first method. $a b c \leq\left(\frac{a+b+c}{3}\right)^{3}=\left(\frac{12}{3}\right)^{3}=64$.
Moreover, $a b+b c+a c \leq a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+a c)$, which means $3(a b+b c+a c) \leq(a+b+c)^{2}=$ 144.
In both cases, equality is achieved if $a=b=c=4$. Therefore, the maximum value of the given exp... | 112 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3+$
$p$ and $8 p^{2}+1$ are prime numbers. Find $p$. | If $p \neq 3$, then $8 p^{2}+1$ is divisible by 3.
Answer
$p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ] [ Divisibility rules for 3 and 9 ]
For the number $2^{100}$, the sum of its digits was found, then the sum of the digits of the result, and so on. Eventually, a single-digit number was obtained. Find it. | $2^{100}=\left(2^{6}\right)^{16} \cdot 2^{4} \equiv 1^{16} \cdot 16 \equiv 7(\bmod 9)$.
## Answer
7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Ali-Baba came to a cave where there is gold, diamonds, and a chest in which he can carry them away. A full chest of gold weighs 200 kg, a full chest of diamonds - 40 kg, an empty chest weighs nothing. A kilogram of gold is worth 20 dinars on the market, a kilogram of diamonds - 60 dinars. Ali-Baba can lift and carry no... | Suppose Ali-Baba can carry $x$ kg of gold and $y$ kg of diamonds out of the cave. In this case, he can obtain $20x + 60y$ dinars. Since Ali-Baba can carry no more than 100 kg, we have $x + y \leq 100$ (*). 1 kg of gold occupies $1/200$ of a chest, and 1 kg of diamonds occupies $1/40$ of a chest. Therefore, $x/200 + y/4... | 3000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3+ [ Divisibility of numbers. General properties ]
Humpty-Dumpty walks along a straight line, either taking 37 steps to the left or 47 steps to the right in a minute.
What is the shortest time in which he can be one step to the right of the starting point? | In fact, it is required to find a solution to the equation $47 x-37 y=1 \quad(*)$ in non-negative integers with the smallest sum $x+y$.
One of the solutions to this equation is $-x=26, y=33$. Let $(x, y)$ be another integer solution to this equation. Subtracting the equation $47 \cdot 26-37 \cdot 33=1$ from (*), we ge... | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic of residues (miscellaneous). ] [ Examples and counterexamples. Constructions ]
What is the maximum number of rooks that can be placed in an $8 \times 8 \times 8$ cube so that they do not attack each other? | Obviously, in each column of eight cube cells, only one rook can stand, so it is impossible to place more than 64 rooks.
Let's show how to place 64 rooks so that they do not attack each other. We introduce a coordinate system with axes along the edges of the cube, so that each cell has coordinates ( $x, y, z$ ) of num... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest common divisor of the numbers $9 m+7 n$ and $3 m+2 n$, if the numbers $m$ and $n$ have no common divisors other than one?
# | Any common divisor of the numbers $9 m+7 n$ and $3 m+2 n$ must also be a divisor of the numbers $(9 m+7 n)-3(3 m+2 n) = n$ and $7(3 m+2 n)-2(9 m+7 n) = 3 m$. Since the numbers $m$ and $n$ are coprime, any common divisor of the numbers $n$ and $3 m$ must be a divisor of the number 3, that is, it cannot be greater than 3... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9 [ Examples and counterexamples. Constructions]
Find the maximum value of the expression $|\ldots|\left|x_{1}-x_{2}\right|-x_{3}\left|-\ldots-x_{1990}\right|$, where $x_{1}, x_{2}, \ldots, x_{1990}$ are distinct natural numbers from 1 to 1990. | The modulus of the difference of two non-negative numbers is not greater than their maximum.
## Solution
Estimate. Note that the modulus of the difference of two non-negative numbers is not greater than their maximum. Therefore, $\left|x_{1}-x_{2}\right|$ $\leq \max \left\{x_{1}, x_{2}\right\}$
||$x_{1}-x_{2}\left|-... | 1989 | Other | math-word-problem | Yes | Yes | olympiads | false |
Tokoroiev $C$. $\mathbf{L}$.
Numbers $a, b$ and $c$ are such that $(a+b)(b+c)(c+a)=a b c,\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)=a^{3} b^{3} c^{3}$. Prove that $a b c=0$. | First, note that $x^{2}-x y+y^{2}>|x y|$ for any distinct numbers $x$ and $y$.
Assume that $a b c \neq 0$. Then, dividing the second equality by the first, we get $\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right)=|a b| \cdot|b c| \cdot|a c|$
All the brackets on the left and all the... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Berdnikov A.
Let's call a natural number good if all its digits are non-zero. A good number is called special if it has at least $k$ digits and the digits are in strictly increasing order (from left to right).
Suppose we have some good number. In one move, it is allowed to append a special number to either end or ins... | Obviously, a special number does not have more than nine digits. If $k=9$, then with each operation, the number of digits changes by exactly 9, meaning the remainder of the number of its digits divided by 9 does not change, and a single-digit number cannot be turned into a two-digit number.
Let $k=8$. Since all operat... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sindarov V.A.
Find all such natural $k$ that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first power).
# | Let $n \geq 2$, and $2=p_{1}1$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation (*) would be divisible by $q$, which is not the case. Therefore, $a>p_{k}$.
Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, th... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider the product $A . B$.
On an infinite tape, numbers are written in a row. The first is one, and each subsequent number is obtained by adding to the previous number its smallest non-zero digit in its decimal representation. How many digits are in the decimal representation of the number that stands in the $9 \cd... | Since each number in the sequence, starting from the second, is greater than the previous one, the $9 \cdot 1000^{1000}$-th number in the sequence is greater than $9 \cdot 1000^{1000}$, meaning it has at least 3001 digits. Let the $n$-th number in the sequence be denoted by $a_{n}$, and let $k$ be the smallest index fo... | 3001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.i.
Given a $100 \times 100$ grid, the cells of which are colored black and white. In all columns, there are an equal number of black cells, while in all rows, there are different numbers of black cells. What is the maximum possible number of pairs of adjacent (by side) cells of different colors? | Number the rows from top to bottom and the columns from left to right with numbers from 1 to 100.
In each row, there can be from 0 to 100 black cells. Since the number of black cells in all rows are different, these numbers are all from 0 to 100, except for one (say, except for \( k \)). Therefore, the total number of... | 14751 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a white $2016 \times 2016$ table, some cells are painted black. We call a natural number $k$ successful if $k$ $\leq 2016$, and in each of the grid squares with side $k$ located in the table, exactly $k$ cells are painted black. (For example, if all cells are black, then the only successful number is 1.) What is the... | Evaluation. Consider an arbitrary coloring of the table. Suppose there are at least two successful numbers, and let $a$ be the smallest of them, and $b$ be the largest.
Divide $b$ by $a$ with a remainder: $b = q a + r$, where $0 \leq r < a$. If $r \neq 0$, then $r$ is a successful number, and $a < r < b$, which contra... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zhukov G.
A teacher is planning to give the children a problem of the following type. He will inform them that he has thought of a polynomial $P(x)$ of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will tell them $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$ and separately inform t... | Evaluation. Let the teacher use some $k \leq 2016$, and consider the polynomial $P(x)$.
Consider the polynomial $P(x)=P(x)+\left(x-n_{1}\right)\left(x-n_{2}\right) \ldots\left(x-n_{k}\right)$. Notice that the degree of the polynomial $Q(x)$ is also
2017, and its leading coefficient is also 1. Moreover, $P\left(n_{1}\... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kozhevnikov P.A.
On each of two parallel lines $a$ and $b$, 50 points were marked.
What is the maximum possible number of acute-angled triangles with vertices at these points? | Obviously, the maximum is achieved when the points on both lines are sufficiently close, so that each segment with marked ends on one of the lines is seen from any point on the other line at an acute angle. Let's paint the points on one of the lines blue, and the projections of the points of the second line onto this l... | 41650 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Ya.
Given a polynomial of degree 2022 with integer coefficients and leading coefficient 1. What is the maximum number of roots it can have in the interval $(0,1)$? | Evaluation. All 2022 roots of the polynomial cannot lie on the interval $(0,1)$: otherwise, the free term of the polynomial, equal to the product of the roots by Vieta's theorem, would lie on the interval $(0,1)$ and could not be an integer. Example 1. Consider the polynomial $\$ \mathrm{P}(\mathrm{x})=\mathrm{x}^{\wed... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Numerical tables and their properties] $[$ Counting in two ways $]$
On the cells of a chessboard, grains of rice were placed. The number of grains on any two adjacent cells differed by exactly
1. At the same time, there were three grains on one of the cells of the board, and 17 grains on another. The rooster pecked ... | Let, for example, cell $A$, which contains three grains, is $k$ cells to the left and $n$ cells below cell $B$, which contains 17 grains. Consider the shortest paths leading from cell $A$ to cell $B$. Each such path consists of $k$ steps to the neighboring cell to the right and $n$ steps to the neighboring cell upwards... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Students from different cities came to the tournament. One of the organizers noticed that they could form 19 teams of 6 people each, and at the same time, less than a quarter of the teams would have a reserve player. Another suggested forming 22 teams of 5 or 6 people each, and then mo... | According to the first condition, the number of schoolchildren who arrived is no more than $19 \cdot 6+4=118$.
From the second condition, it follows that this number is no less than $22 \cdot 5+8=118$.
Thus, exactly 118 schoolchildren arrived at the tournament.
## Answer
118.
Author: Senderov V.A.
Natural numbers... | 118 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Three square paths with a common center are spaced 1 m apart from each other (see figure). Three ants start simultaneously from the lower left corners of the paths and run at the same speed: Mu and Ra counterclockwise, and Wei clockwise. When Mu reaches the lower right corner of the largest path, the o... | The lengths of the sides of two adjacent paths differ by 2 meters. Therefore, at the moment when Mu reached the corner, Ra had run 2 meters along the right side of the path and was at a distance of $2+1=3$ meters from the "lower" side of the outer path. Since Ra is halfway between Mu and Vey, Vey is at twice the distan... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Дубанов И.С.
In five pots standing in a row, Rabbit poured three kilograms of honey (not necessarily into each and not necessarily equally). Winnie-the-Pooh can take any two adjacent pots. What is the maximum amount of honey that Winnie-the-Pooh can guarantee to eat?
# | Evaluation. Let's assume Winnie-the-Pooh cannot take at least a kilogram of honey. This means that in any pair of adjacent pots, there is less than a kilogram of honey. This is true for both the two rightmost pots and the two leftmost pots. However, then in the middle pot, there must be more than a kilogram of honey (o... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Brenner D....
All natural numbers from 1 to $N, N \geq 2$ are written in some order in a circle. For any pair of adjacent numbers, there is at least one digit that appears in the decimal representation of each of them. Find the smallest possible value of $N$.
# | Since single-digit numbers do not have common digits, then $N>9$. And since the numbers adjacent to the number 9 must contain a nine in their notation, the smaller one cannot be less than 19, and the larger one cannot be less than 29. Therefore, $N \geq 29$.
Equality $N=29$ is possible, since the conditions of the pro... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Volchenkov S.G.
$N$ digits - ones and twos - are arranged in a circle. An image is called a number formed by several consecutive digits (clockwise or counterclockwise). For what smallest value of $N$ can all four-digit numbers, whose notation contains only the digits 1 and 2, be among the images? | 
Notice that the images of the numbers 1111, 2112, and 2122 cannot have common units, while the images of the numbers 2222, 1221, and 1211 cannot have common twos. Therefore, if all these numb... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[$ The Pigeonhole Principle (continued) ]
$[\quad$ Estimation + example $\quad]$
The hostess baked a pie for her guests. She may have either 10 or 11 guests. Into what minimum number of pieces should she cut the pie in advance so that it can be evenly divided among either 10 or 11 guests? | If 10 guests arrive, each should receive no less than two pieces.
## Solution
If 10 guests arrive, each should receive no less than two pieces. Indeed, otherwise one of the 10 guests would receive one piece, which is $1 / 10$ of the pie, and if 11 guests arrived, this piece would need to be further divided. Thus, the... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Xichaturyan A.V.
Bus stop B is located on a straight highway between stops A and C. At some point after leaving A, the bus found itself at a point on the highway such that the distance from it to one of the three stops is equal to the sum of the distances to the other two. After the same amount of time, the bus was ag... | At both moments in time mentioned in the problem, the sum will obviously be the distance from the bus to the farthest stop from it. This cannot be $B$, as it is closer than $C$. Therefore, these were $C$ (before the bus had traveled halfway from $A$ to $C$) and $A$ (after this moment).
. Therefore, Uncle Fyodor made 8 parts, which is 7 parts more than the number of sandwiches Sharik ate. Since these 7 parts make up 21 sandwiches, one part is three sandwiches.... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
The numbers 2, 3, 4, ... 29, 30 are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? | Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$) the even number $2n$ is already m... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evochkimov
Find the smallest natural number $\$ \mathrm{~N}>9 \$$ that is not divisible by 7, but if any of its digits is replaced by 7, the resulting number is divisible by 7. | Let the smallest suitable number be of the form $\overline{a_1a_2\ldots a_n}$. From the condition, it follows that among its digits there are no 0 and 7. If the number contains digits 8 or 9, then they can be replaced by 1 or 2, respectively, to obtain a smaller number with the same property. Thus, the desired number c... | 13264513 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
A row of new recruits stood facing the sergeant. On the command "left," some turned left, while the rest turned right. It turned out that six times more soldiers were looking at the back of their neighbor than in the face. Then, on the command "about face," everyone turned in the opposite direction. No... | Let's assume that the sergeant has lined up the soldiers between two posts. After the first command, each recruit either looks at the back of the head of the neighbor or at the face, except for the two soldiers at the ends, who can look at the posts.
If a soldier is looking at the back of the head of a neighbor, then ... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On a paper tape, bus tickets with numbers from 000000 to 999 999 are printed. Then, those tickets where the sum of the digits in the even positions equals the sum of the digits in the odd positions are marked in blue. What will be the greatest difference between the numbers of two adjacent blue tickets?
# | Let's first show that the blue tickets with numbers 908919 and 909909 are adjacent. Suppose we have found a blue ticket with the number $908919990$. First, note that the numbers $X$ and $Y$ are divisible by 11, which means their difference is also divisible by 11. Therefore, $Y-X \geq 1001$. On the other hand, all tick... | 990 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Does there exist a natural number $n$ such that the sum of the digits of $n^{2}$ is 100? | Answer: it exists.
For example, $n=10111111111$. | 10111111111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ]
Find the smallest natural number starting with the digit 4 and decreasing by four times when this digit is moved to the end of the number.
# | Answer: 410256.
Let the desired number be $\mathrm{X}=\overline{4 a b \ldots c}=4 \cdot 10^{\mathrm{n}}+A$, where $A=\overline{a b \ldots c}$ ( $A$ is an $n$-digit number). After moving the first digit to the end, we get the number $Y=\frac{}{a b \ldots c 4}=10 A+4$. According to the condition,
$$
4 \cdot 10^{\mathrm... | 410256 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Spivak $A . B$.
Over two years, the factory reduced the volume of its production by $51 \%$. Each year, the volume of production decreased by the same percentage. By how much? | If the volume of production has decreased by $51 \%$, it means that it has become $49 \%$ of the initial volume, and $49=7^{2}$.
## Solution
Let the production decrease by $x \%$ per year. Taking the initial volume of production as 1, we get that after one year, the production volume is
$\alpha=1-x / 100$, and after... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Spivak A.B.
In the Rhind Papyrus (Ancient Egypt), among other information, there are decompositions of fractions into the sum of fractions with a numerator of 1, for example,
$2 / 73=1 / 60+1 / 219+1 / 292+1 / x$. One of the denominators here is replaced by the letter $x$. Find this denominator. | $1 / x=2 / 73-1 / 60-1 / 219-1 / 292=1 / 365$. Therefore, $x=365$.
## Answer
365. | 365 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$[$ [Arithmetic. Mental calculation, etc.] $]$
Authors: Gaityerri G.A., Grierenko D.:
2002 is a palindrome year, which means it reads the same backward as forward. The previous palindrome year was 11 years earlier (1991). What is the maximum number of non-palindrome years that can occur consecutively (between 1000 an... | Let the current year be a palindrome of the form $\overline{a b b a}$. When will the next such year occur? Consider two cases:
a) $b=9$ (year of the form $\overline{a 99 a}, a<9)$. Then, 11 years later, another palindromic year will occur: $\overline{(a+1) 00(a+1)}$. For example, the years 3993 and 4004.
b) $b<9$. In... | 109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Divisibility of numbers. General properties ]
The year of the current mathematics festival is divisible by its number: $2006: 17=118$.
a) Name the first number of the math festival for which this was also true.
b) Name the last number of the math festival for which this will also be true.
# | a) The first math festival was in 1990. It is clear that the year of its holding is divisible by its number, because the number is one.
b) Let $N$ be the number of the math festival. Then the year of its holding is (2006 - 17) $+N=1989+N$. Suppose the year of the festival is divisible by its number, that is, $1989+N$ ... | 1989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Work problem. $]$
The bathtub fills up in 23 minutes from the hot water tap, and in 17 minutes from the cold water tap. Masha first turned on the hot water tap. After how many minutes should she turn on the cold water tap so that by the time the bathtub is full, 1.5 times more hot water is used than cold water? | To have 1.5 times more hot water than cold water in the bathtub, the cold water tap should fill 2/5 of the bathtub, and the cold water $-3 / 5$. Then the hot water tap should be open for $3 / 5 \cdot 23=69 / 5$ minutes, and the cold water tap for $2 / 5 \cdot 17=34 / 5$ minutes. Therefore, the cold water tap should be ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Inequality problems. Case analysis]
Lёsha and Ira live in a house where each floor has 9 apartments (the house has one entrance). The floor number of Lёsha is equal to the apartment number of Ira, and the sum of their apartment numbers is 329. What is the apartment number of Lёsha? | Let $K$ be the apartment number of Ira (and the floor number of Lёsha). Then the apartment number of Lёsha is between $9K-8$ and $9K$, which means the sum of the apartment numbers of the kids is from $10K-8$ to $10K$. The system of inequalities $10K-8 \leq 329 \leq 10K$ has a unique solution: $K=33$. And Lёsha lives in... | 296 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The Tsar allocated 1000 rubles per year for the maintenance of the scribes' office (all scribes received an equal share). The Tsar was advised to reduce the number of scribes by $50 \%$, and to increase the salary of the remaining scribes by $50 \%$. By how much will the Tsar's expenses for the scribes' office change i... | Let the tsar first reduce half of the scribes. Then his expenses will be halved. After increasing the salaries, they will increase by 1.5 times. As a result, they will amount to $1.5 \cdot 0.5 = 0.75$ of the previous amount, that is, they will decrease by $25 \%$.
## Answer
They will decrease by $25 \%$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ya Chenno I.v.
A cooperative receives apple and grape juice in identical barrels and produces an apple-grape drink in identical cans. One barrel of apple juice is enough for exactly 6 cans of the drink, and one barrel of grape juice is enough for exactly 10 cans. When the recipe for the drink was changed, one barrel o... | The first method. For one can of the drink, $1 / 6$ of a barrel of apple juice and $1 / 10$ of a barrel of grape juice are used, which means the volume of the can is $1 / 6 + 1 / 10$ of the barrel. After the recipe change, $1 / 5$ of a barrel of apple juice and $4 / 15 - 1 / 5 = 1 / 15$ of a barrel of grape juice are u... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On a straight line, ten points were placed at equal intervals, and they occupied a segment of length $a$. On another straight line, one hundred points were placed at the same intervals, and they occupied a segment of length $b$. How many times is $b$ greater than $a$? | On a segment of length $a$, 9 intervals fit, and on a segment of length $b-99$. Therefore, $b=11a$.
C
11 times. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Cities $A, B$, and $C$ together with the straight roads connecting them form a triangle. It is known that the direct route from $A$ to $B$ is 200 km shorter than the detour through $C$, and the direct route from $A$ to $C$ is 300 km shorter than the detour through $B$. Find the distance between cities ... | The route $B-A-C$ (from $B$ to $C$ via $A$) is 500 km shorter than the route $B-C-A-B-C$: the length of segment $B A$ is 200 km less than the length of the route $B-C-A$, and the length of segment $A C$ is 300 km less than the length of the route $A-B-C$. The second route differs from the first by two segments $B C$. T... | 250 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Petya was traveling from Petrov to Nikolayev, while Kolya was going in the opposite direction. They met when Petya had traveled 10 km and another quarter of the remaining distance to Nikolayev, and Kolya had traveled 20 km and a third of the remaining distance to Petrov. What is the distance between Petrov an... | The first method. Let Petya have driven $10 + x$ km to the meeting point, then he had $3x$ km left to Nikolayev. Kolya drove $20 + y$ km to the meeting point, and he had $2y$ km left to Petrovo (see the figure).
 and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he ... | The additional 100 rubles spent for the second time brought the merchant an additional 20 rubles in profit. This means that the first time, to get $5 \cdot 20=100$ rubles in profit, the merchant had to pay $5 \cdot 100=500$ rubles.
## Answer
500 rubles. | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
It is known that $\operatorname{tg} A+\operatorname{tg} B=2$ and $\operatorname{ctg} A+\operatorname{ctg} B=3$. Find $\operatorname{tg}(A+B)$. | $\frac{2}{\operatorname{tg} A \operatorname{tg} B}=\frac{\operatorname{tg} A+\operatorname{tg} B}{\operatorname{tg} A \operatorname{tg} B}=\operatorname{ctg} A+\operatorname{ctg} B=3$, hence $\operatorname{tg} A \operatorname{tg} B=2 / 3$. Therefore,
$\operatorname{tg}(A+B)=\frac{\operatorname{tg} A+\operatorname{tg} ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental arithmetic, etc.]
A hundred people answered the question "Will the new president be better than the previous one?". Of them, a people think it will be better, b - that it will be the same, and c - that it will be worse. Sociologists constructed two "optimism" indicators of the respondents: m=a+b/2... | The magnitude of a-c is linearly expressed through $\mathrm{a}+\mathrm{b} / 2$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}$.
## Solution
It is known that $\mathrm{a}+\mathrm{b}+\mathrm{c}=100 ; \mathrm{a}+\mathrm{b} / 2=40$. From the second equation, we get: $2 \mathrm{a}+\mathrm{b}=80$. Subtracting the first equation from... | -20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
It is known that 9 cups of tea cost less than 10 rubles, and 10 cups of tea cost more than 11 rubles. How much does a cup of tea cost
# | Write the condition in the form of inequalities and use the fact that the cost of a glass of tea is expressed as an integer number of kopecks.
## Solution
Let the cost of a glass of tea, expressed in kopecks, be \( x \). Then, according to the condition, \( 9x \leq 1100 \).
Therefore, \( 9x \) does not exceed 999, a... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Absolute value ]
Solve the equation $|x-2|+|x-1|+|x|+|x+1|+|x+2|=6$.
# | The number |x-a| is equal to the distance from the point on the number line with coordinate x to the point with coordinate a.
## Solution
Consider a point on the number line with coordinate x. The sum $|\mathrm{x}-2|+|\mathrm{x}-1|+|\mathrm{x}|+|\mathrm{x}+1|+|\mathrm{x}+2|$ is equal to the sum of the distances from ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Raskina I.V.
In a colorful family, there were an equal number of white, blue, and striped octopus children. When some blue octopus children turned striped, Dad decided to count the children. There were 10 blue and white children in total, while white and striped children together amounted to 18. How many children are ... | First solution. Note that the white octopuses were one third of the total number, and they did not change color. If we add 10 and 18, we get the total number of all children together, plus the number of white ones, which is 4/3 of the total number of all children. Thus, 4/3 of the number of children in the family is 28... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Text problems] [Case enumeration]
In a box, there are blue, red, and green pencils. In total, there are 20 pieces. The number of blue pencils is 6 times the number of green pencils, and the number of red pencils is less than the number of blue pencils. How many red pencils are in the box? | Think about how many blue pencils there can be.
## Solution
Since there are 20 pencils in total, and blue and green pencils together make up 7 parts. This means there can be 6 or 12 blue pencils, and green and red pencils would then be 1 and 13 or 2 and 6, respectively. Since there are fewer red pencils than blue one... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
When Buratino went to the VMSh classes, Papa Carlo promised to pay him one kopeck for the first correctly solved problem, two kopecks for the second, four kopecks for the third, and so on. In a month, Buratino received 655 rubles 35 kopecks. How many problems did he solve?
# | For simplicity in calculations, let's assume that before attending the VMSh classes, Buratino had one kopeck. Then, after solving the first problem, Buratino will have 2 kopecks. Solving the second problem, he will receive two more, and he will have $2 * 2=4$ kopecks. Solving the third problem, he will receive four mor... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Fedorov P.M.
It is known that $a+\frac{b^{2}}{a}=b+\frac{a^{2}}{b}$. Is it true that $a=b$? | Let's bring the left side to a common denominator: $a+\frac{b^{2}}{a}=\frac{a^{2}+b^{2}}{a}$. Similarly, we will do the same for the right side.
We obtain the equality:
$$
\frac{a^{2}+b^{2}}{a}=\frac{a^{2}+b^{2}}{b}
$$
From the condition, it is clear that $a \neq 0, b \neq 0$. Therefore, $a^{2}+b^{2}>0$ (see comment... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ] $[\quad$ Case enumeration $\quad]$
## Author: Raskina I.V.
The year 2009 has the property that by rearranging the digits of the number 2009, it is impossible to obtain a smaller four-digit number (numbers do not start with zero). In which year will this property reappear for the first time? | Author: Raschina I.V.
In the years $2010, 2011, \ldots, 2019$ and in 2021, the number of the year contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be reduced,
## Answer
in the year 2022. | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
Solve the equation $x^{3}+x-2=0$ by trial and error and using Cardano's formula. | The equation has an obvious root $x=1$. Dividing by $x-1$, we obtain the quadratic equation $x^{2}+x+2=0$, which has no real roots.
Cardano's formula gives $x=\sqrt[3]{-1+\sqrt{\frac{28}{27}}}+\sqrt[3]{-1-\sqrt{\frac{28}{27}}}$.
From this, it follows that $\sqrt[3]{-1+\sqrt{\frac{28}{27}}}+\sqrt[3]{-1-\sqrt{\frac{28}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3 [ Text problems (other) $\quad]$
Four photographs can be arranged to form three different rectangles (see fig.). The perimeter of one of them is 56 cm. Find the perimeters of the other two rectangles, if the perimeter of a photograph is 20 cm.
(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the desired value, since $\frac{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
\begin{aligned} & [\text { Recurrence relations (other) }] \\ & {[\text { Periodicity and non-periodicity }]}\end{aligned}
The first term of the sequence is 934. Each subsequent term is equal to the sum of the digits of the previous term, multiplied by 13.
Find the 2013th term of the sequence. | Let's compute the first few terms of the sequence: $a_{2}=16 \cdot 13=208, a_{3}=10 \cdot 13=130, a_{4}=4 \cdot 13=52, a_{5}$ $=7 \cdot 13=91, a_{6}=10 \cdot 13=130=a_{3}$. Since each subsequent number is calculated using only the previous number, the terms of the sequence will repeat with a period of 3. The number 201... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { [Rectangles and squares. Properties and characteristics } \\ \text { Word problems (miscellaneous). }\end{array}\right]$
A frame for three square photographs has the same width everywhere (see figure). The perimeter of one opening is 60 cm, and the perimeter of the entire frame is 180 cm... | From the condition, it follows that the length of the side of the opening is 15 cm. Let $d$ cm be the width of the frame, then the perimeter of the rectangle is
$8 \cdot 15 + 12 d = 120 + 12 d = 180$. Therefore, $d=5$.
## Answer
5 cm. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the nearest integer to the number $x$, if $x=\frac{1}{\sqrt[4]{\frac{5}{4}+1}-\sqrt[4]{\frac{5}{4}-1}}$. | 
If the inequalities are verified by squaring, then \(1.9 < \frac{\sqrt{6} + \sqrt{2}}{2} < 2\).
## Answer
2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find all natural $n$ and $k$, satisfying the equation $k^{5}+5 n^{4}=81 k$.
# | $5 n^{4}=k\left(9+k^{2}\right)(3+k)(3-k)$. The left side of this equation is positive for any natural value of $n$, so the right side must also be positive. Therefore, it is sufficient to check two natural values of $k: k=1$ and $k=2$.
1) If $k=1$, then $5 n^{4}=80$, which means $n=2$.
2) If $k=2$, then $5 n^{4}=130$.... | 2025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Gukhanov $H . X$.
Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. From each trinomial $f_{i}(x)$, one root was chosen and denoted as $x_{i}$. What values can the sum $f_{... | Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$, since $f_{1}\left(x_{1}\right)=$ 0. Similarly, we obtain the equalities
$f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_{100}\left(x_{99}\right)=c... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems (miscellaneous). ] [ Systems of linear equations ]
Avor: Blinkov A.d:
Sasha and Vanya were born on March 19. Each of them celebrates their birthday with a cake with candles equal to the number of years they have turned. In the year they met, Sasha had as many candles on his cake as Vanya has today. It... | If we transfer the "extra" candles from today's Sasha's cake to the old Vanya's cake, we will get four cakes with the same number of candles. This means that today Vanya turned $216: 4=54$ years old.
## Answer
54 years old. | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Find the largest natural number, all digits in the decimal representation of which are different and which
is reduced by 5 times if the first digit is erased. | Given $\overline{a A}=5 A$ (where $A$ is the number composed of all digits except the first, and $a$ is the first digit). Let $n$ be the number of digits in $\overline{a A}$. Then
$4 A=a \cdot 10^{n-1}$, which means $A=25 a \cdot 10^{n-3}$.
If $n>4$, then the number $A$, and thus the desired number, has two identical... | 3750 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kalinin D.A.
In the room, there are 20 chairs of two colors: blue and red. On each chair sits either a knight or a liar. Knights always tell the truth, and liars always lie. Each of the seated individuals claimed that they were sitting on a blue chair. Then they somehow rearranged themselves, after which half of those... | Initially, all knights sit on blue chairs, and all liars on red ones. Therefore, the number of knights who moved to red chairs is equal to the number of liars who moved to blue chairs. Both groups claimed they were sitting on red chairs. In total, 10 people said they were sitting on red chairs. Therefore, the number of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Evokimov
A natural number is written on the board. If you erase the last digit (in the units place), the remaining non-zero number will be divisible by 20, and if you erase the first digit, it will be divisible by 21. What is the smallest number that can be written on the board if its second digit is not equal to 0?
... | The second last digit of the number is 0, as the number without the last digit is divisible by 20. Therefore, the number must be at least four digits long. Note that the number remaining after erasing the last digit cannot be 100 according to the condition. Also, this number cannot be 120 or 140, as numbers of the form... | 1609 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Several natural numbers are written in a row with a sum of 20. No number and no sum of several consecutive numbers equals 3. Could there be more than 10 numbers written? | Example with 11 numbers: $1,1,4,1,1,4,1,1,4,1,1$.
## Answer
It could. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shnop D.
In the park, there were lindens and maples. Maples made up $60 \%$ of them. In the spring, lindens were planted in the park, after which maples accounted for $20 \%$. In the fall, maples were planted, and maples again made up $60 \%$. By what factor did the number of trees in the park increase over the year? | At first, there were 1.5 times fewer lindens than maples, and in the summer, there became 4 times more. At the same time, the number of maples did not change. Therefore, the number of lindens became $1.5 \cdot 4=6$ times more.
By the end of the year, the ratio of the number of lindens to the total number of trees beca... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Tournaments and tournament tables]
A school table tennis championship was held according to the Olympic system. The winner won six matches. How many participants in the tournament won more games than they lost? (In a tournament according to the Olympic system, participants are divided into pairs. Those who lost a gam... | Since in each round every player found a partner and in each pair one of the players was eliminated, the total number of players decreased by half after each round. The winner participated in every round and won, so there were a total of six rounds. Therefore, the total number of participants was
$2^{6}=64$. Those who... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$3-$ $[$ Algorithm Theory (miscellaneous). $\quad]$
Author: Berdikov A.
Sasha writes a sequence of natural numbers on the board. The first number $N>1$ is written in advance. New natural numbers he gets by subtracting from or adding to the last written number any of its divisors greater than 1. Can Sasha, for any nat... | The first method. By adding $N$ each time, we get 2011N. By subtracting 2011 each time, we get 2011.
The second method. If $N$ is odd, add $N$ and get an even number. By adding or subtracting twos from it, we get 4022. By subtracting 2011, we get 2011.
## Answer
In any case.
Author: $\underline{\text { Folklore }}$... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Raskina I.V.
The inhabitants of the Island of Misfortune, like us, divide the day into several hours, an hour into several minutes, and a minute into several seconds. But they have 77 minutes in a day and 91 seconds in an hour. How many seconds are there in a day on the Island of Misfortune? | If you divide 77 by the number of minutes in an hour, you get the number of hours in a day. If you divide 91 by the number of minutes in an hour, you get the number of seconds in a minute. This means that both 77 and 91 are divisible by the number of minutes in an hour. GCD(77, 91) = 7. Since there is more than one min... | 1001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folkopr }}$
Three companies $A, B$, and $C$ decided to jointly build a road 16 km long, agreeing to finance this project equally. In the end, $A$ built 6 km of the road, $B$ built 10 km, and $C$ contributed her share in money - 16 million rubles. How should companies $A$ and $B$ divide this money b... | Each company was supposed to build $51 / 3$ km of road. Instead of company $C$, company $A$ built $6-5 \frac{1}{3}=2 / 3$ km of road, and company $B$ built
$10-5 \frac{1}{3}=14 / 3$ km. Therefore, 16 million rubles should be divided between them in the ratio $2: 14$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Vanya went to the shooting range with his dad. The deal was this: Vanya would be given 10 bullets, and for each hit on the target, he would receive three more bullets. Vanya made 14 shots, and he hit the target exactly half of the time. How many bullets did Vanya have left? | Vanya hit the target 7 times, so he received 21 additional cartridges. Since he initially had 10 cartridges, he ended up with a total of 31 cartridges. Vanya made 14 shots, so 17 cartridges remained.
## Answer
17 cartridges. | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Folklore
In the class, there are 17 people. It is known that among any ten, there is at least one girl, and there are more boys than girls. How many girls are there in this class? | Since there are more boys than girls in the class, there are no fewer than nine boys. If there are more than nine boys, the condition that among any ten people there is at least one girl will not be met. Therefore, there are exactly 9 boys and 8 girls.
## Answer
8. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the smallest number, written with only ones, that is divisible by 333...33 (with 100 threes in the sequence). | The recording of this number consists of 300 ones. | 300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Decimal numeral system]
Find all natural numbers that increase by 9 times when a zero is inserted between the units digit and the tens digit.
# | Let's write our number in the form $10 a+b$, where $b$ is the units digit. We get the equation $100 a+b=9(10 a+b)$. From this, $10 a=8 b$, i.e., $5 a=4 b$. Thus, $b$ is divisible by 5. Considering the two cases $b=0, b=5$, we get the unique answer: 45. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and operations [ Examples and counterexamples. Constructions ]
Several boxes together weigh 10 tons, and each of them weighs no more than one ton. How many three-ton trucks are definitely enough to haul this cargo?
# | We will load the trucks with boxes in any order, only ensuring that the "overloading" of the truck does not occur. As a result, more than 2 tons will be loaded onto each truck. Therefore, five trucks will definitely be enough. Four trucks may not be enough. For example, let's say we have 13 boxes, each weighing ${ }^{1... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3 [ Central symmetry helps solve the problem]
What is the maximum number of pawns that can be placed on a chessboard (no more than one pawn per square), if:
1) a pawn cannot be placed on the e4 square;
2) no two pawns can stand on squares that are symmetric relative to the e4 square?
# | Evaluation. All fields of the board except for the a-file, the 8th rank, and the e4 square can be divided into pairs symmetrical relative to e4. Such pairs amount to 24. According to the condition, no more than one pawn can be placed on the fields of each pair. Additionally, no more than one pawn can be placed on the f... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Covering with percentages and ratios ] [ Inclusion-exclusion principle ]
Three crazy painters started painting the floor each in their own color. One managed to paint $75 \%$ of the floor red, another $70 \%$ green, and the third $65 \%$ blue. What part of the floor is definitely painted with all three colors? | Evaluation. 25% of the floor is not painted red, 30% of the floor is not painted green, and 35% of the floor is not painted blue.
25+30+35=90. From this, it follows that at least 10% of the floor is painted with all three colors.
An example where exactly 10% of the floor is painted with all three colors is clear from... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7
An engineer arrives at the train station at 8 o'clock in the morning every day. At exactly 8 o'clock, a car arrives at the station and takes the engineer to the factory. One day, the engineer arrived at the station at 7 o'clock and started walking towards the car. Meeting the car, he got in and arrived at the factor... | In 20 minutes, the car would have traveled the distance the engineer walked twice. Therefore, the car met the engineer at $7:50$.
## Answer
50 minutes. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Pairing and grouping; bijections ] [ Proof by contradiction ]
There is a set of natural numbers (it is known that there are at least seven numbers), and the sum of any seven of them is less than 15, while the sum of all numbers in the set is 100. What is the smallest number of numbers that can be in the set? | A set of 50 numbers, each equal to 2, satisfies the condition.
If the set contains no more than 49 numbers, then they can be divided into 7 groups such that each group contains 7 or fewer numbers. According to the condition, the sum of the numbers in each group is less than 15, that is, no more than 14. Therefore, the... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Prove that $M . \mathbf{B}$.
The sequence of numbers $a_{1}, a_{2}, \ldots$ is defined by the conditions $a_{1}=1, a_{2}=143$ and $a_{n+1}=5 \cdot \frac{a_{1}+a_{2}+\ldots+a_{n}}{n}$ for all $n$ $\geq 2$.
Prove that all members of the sequence are integers. | The number $a_{3}=5 \cdot 72=360$ is an integer. For $n \geq 4$
$$
a_{n}=5 \cdot \frac{a_{1}+\ldots+a_{n-1}}{n-1} \text { and } a_{n-1}=5 \cdot \frac{a_{1}+\ldots+a_{n-2}}{n-2} \text {, hence } a_{n}=\frac{5}{n-1}\left(\frac{n-2}{5} a_{n-1}+a_{n-1}\right)=\frac{n+3}{n-1} a_{n-1} \text {. Therefore, if } n \geq
$$
$$
... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
$\underline{\text { Martynova } H}$.
To play the game of Hat, Nadya wants to cut a sheet of paper into 48 identical rectangles. What is the minimum number of cuts she will have to make if she can rearrange any pieces of paper but cannot fold them, and Nadya can cut as many layers of paper at once as she likes? (Each c... | With each cut, the number of paper pieces can increase by no more than twice. Therefore, after five cuts, you can get no more than 32 pieces, which is not enough.
Six cuts will suffice. For example, you can cut the sheet in half, align the two rectangular pieces and cut them in half again (resulting in four equal rect... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Senderov V.A.
At the vertices of a cube, numbers from 1 to 8 were written, and on each edge - the absolute difference of the numbers at its ends. What is the smallest number of different numbers that can be written on the edges?
# | Three numbers must be necessarily present: for this, it is sufficient to consider the three edges of the cube coming out of the vertex where the number 1 (or 8) is written. Let's prove that there will be an arrangement of numbers for which exactly three numbers will be required. Consider 2 squares. In the vertices of t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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