problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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Agakhanov N.X.
At a round table, 30 people are sitting - knights and liars (knights always tell the truth, while liars always lie). It is known that each of them has exactly one friend at the same table, and a knight's friend is a liar, while a liar's friend is a knight (friendship is always mutual). When asked, "Is y... | All those sitting at the table are paired as friends, which means there are an equal number of knights and liars. Consider any pair of friends. If they are sitting next to each other, the knight will answer "Yes" to the given question, and the liar will answer "No." If they are not sitting next to each other, their ans... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Solve the equation: $2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}=\frac{10}{x-4}$. | Since the left, and therefore the right, part of the equation takes only positive values, then $x>4$. On the interval $(4,+\infty)$, the function
$f(x)=2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}$ is increasing, while the function $g(x)=\frac{10}{x-4}$ is decreasing, so the equation $f(x)=g(x)$ has no more than one root.
It rem... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
For some 2011 natural numbers, all their 2011$\cdot$1005 pairwise sums were written on the board.
Could it be that exactly one third of the written sums are divisible by 3, and another exactly one third of them give a remainder of 1 when divided by 3? | Let's consider, for example, the numbers from 1 to 2011. Among them, $2010: 3=670$ numbers are divisible by 3, the same number give a remainder of 2 when divided by 3, and 671 numbers give a remainder of 1. Therefore, $670 \cdot 669: 2 + 670 \cdot 671 = 335 \cdot 2011 = 2011 \cdot 1005$: 3 sums are divisible by three a... | 52 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Berlov S.L. }}$
A five-digit number is called non-decomposable if it cannot be factored into the product of two three-digit numbers.
What is the maximum number of consecutive non-decomposable five-digit numbers? | The smallest number that can be represented as the product of two three-digit numbers is $100 \cdot 100=10000$. The next such number is:
$100 \cdot 101=10100$, so the numbers $10001,10002, \ldots, 10099$ - are non-decomposable. Thus, there are 99 consecutive non-decomposable five-digit numbers.
More than 99 non-decom... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kosukhin O.n.
Sasha found that there were exactly $n$ working digit buttons left on the calculator. It turned out that any natural number from 1 to 99999999 can either be entered using only the working buttons, or obtained as the sum of two natural numbers, each of which can be entered using only the working buttons. ... | Let's show that the conditions of the problem are met if the buttons with digits $0,1,3,4,5$ remain functional.
Indeed, any digit from 0 to 9 can be represented as the sum of some two "functional" digits. Let the number from 1 to 99999999 that we want to obtain consist of digits $a_{1}, a_{2}, \ldots, a_{8}$ (some of ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Authors: Shapovalov A.V., Yatsenno I.V.
At the "Come on, creatures!" competition, 15 dragons are standing in a row. The number of heads of neighboring dragons differs by 1. If a dragon has m... | It is convenient to represent a row of dragons as a graph: instead of each dragon, we draw a point at a height corresponding to the number of heads the dragon has, and connect these points.
a) See the figure.
b) First, note that somewhere between every two cunning dragons stands a strong one. Indeed, if we walk along... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Among the actors of Karabas Barabas theater, a chess tournament was held. Each participant played exactly one game with each of the others. One solido was given for a win, half a solido for a draw, and nothing for a loss. It turned out that among any three participants, there would be ... | Example. Let's denote the participants with letters A, B, V, G, D. Suppose A won against B, B won against V, V won against G, G won against D, D won against A, and all other matches ended in a draw. The condition of the problem is satisfied.
Evaluation. From the condition, it follows that for this tournament, two stat... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Vasiliev N.B. }}$
How many maximum parts can the coordinate plane xOy be divided into by the graphs of 100 quadratic trinomials of the form
$y=a_{n} X^{2}+b_{n} x+c_{n}(n=1,2, \ldots, 100) ?$ | We will prove by induction that $n$ parabolas of the specified type can divide the plane into no more than $n^{2}+1$ parts.
Base case. One parabola divides the plane into $2=1^{2}+1$ parts.
Inductive step. The $n$-th parabola intersects each of the others at no more than two points. These intersection points, of whic... | 10001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
On a chessboard, 32 white and 32 black pawns are placed in all squares. A pawn can capture pawns of the opposite color by moving diagonally one square and taking the place of the captured pawn (white pawns can only capture to the right-up and left-up, while black pawns can only capture to the left-down a... | Note that a pawn that stood on a black square will always move only on black squares. Then after each move (on black squares) there will always be at least one pawn on black squares - the one that made the move. Similarly, at least one pawn will remain on white squares, and there will be no less than two pawns in total... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bazyan A.i.
In a $10 \times 10$ square, numbers from 1 to 100 are arranged: in the first row - from 1 to 10 from left to right, in the second row - from 11 to 20 from left to right, and so on. Andrey plans to cut the square into dominoes $1 \times 2$, calculate the product of the numbers in each domino, and sum the re... | Let's number the dominoes in the partition. Let the numbers in the $i$-th domino be $a_{i}$ and $b_{i}$. Notice that $a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2}$. Summing these equalities over all dominoes, we get that the sum of all 50 products is $\frac{a_{1}^{2}+\ldots+a_{100}^{... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
The distance between two cells on an infinite chessboard is defined as the minimum number of moves in the path of a king between these cells. On the board, three cells are marked, the pairwise distances between which are 100. How many cells exist such that the distances from them to all three marked cell... | Consider two arbitrary cells $A$ and $B$. Let the difference in the abscissas of their centers be $x \geq 0$, and the difference in the ordinates be $y \geq 0$. Then the distance $\rho(A, B)$ between these cells is $\max \{x, y\}$.
Let cells $A, B, C$ be marked. Then for each pair of cells, there exists a coordinate i... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underset{\text { Tokarev } C . \text {. }}{ }$
Numbers $a, b$ and $c$ are such that $(a+b)(b+c)(c+a)=a b c,\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)=a^{3} b^{3} c^{3}$. Prove that $a b c=0$. | First, note that $x^{2}-x y+y^{2}>|x y|$ for any distinct numbers $x$ and $y$.
Assume that $a b c \neq 0$. Then, dividing the second equality by the first, we get $\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}\right.$
$\left.-c a+a^{2}\right)=|a b| \cdot |b c| \cdot |a c|$.
All the parentheses o... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
B.R.'s Problem
The segments connecting an inner point of a convex non-equilateral $n$-gon with its vertices divide the $n$-gon into $n$ equal triangles.
For what smallest $n$ is this possible? | Let's prove that the specified situation is impossible for $n=3,4$.
The first method. For $n=3$, the angles of the triangles in the partition that meet at an internal point are equal, since the sum of any two different angles in them is less than $180^{\circ}$. But then the sides opposite to them, which are sides of t... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Berdonikov A.
Let's call a natural number good if all its digits are non-zero. A good number is called special if it has at least $k$ digits and the digits are in strictly increasing order (from left to right).
Suppose we have some good number. In one move, it is allowed to append a special number to either end or in... | Obviously, a special number does not have more than nine digits. If $k=9$, then with each operation, the number of digits changes by exactly 9, meaning the remainder of the number of its digits divided by 9 does not change, and a two-digit number cannot be made from a one-digit number.
Let $k=8$. Since all operations ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Senderov V.A.
Find all natural numbers $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first power).
# | Let $n \geq 2$, and $2=p_{1}1$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation (*) would be divisible by $q$, which is not the case. Therefore, $a>p_{k}$.
Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, th... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
Ali-Baba and the bandit are dividing a treasure consisting of 100 gold coins, arranged in 10 piles of 10 coins each. Ali-Baba chooses 4 piles, places a cup next to each, and puts some coins (at least one, but not the entire pile) into each cup. The bandit must then rearrange the cups, changing their in... | Let's show that Ali-Baba can achieve that in 7 piles there are no more than 4 coins, while the robber can achieve that there are no piles containing fewer than 4 coins. Therefore, Ali-Baba will take 100 - 7$\cdot$4 = 72 coins.
First, let's prove that the robber can act in such a way that there are no piles containing ... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
2009 non-negative integers not exceeding 100 are arranged in a circle. It is allowed to add 1 to two adjacent numbers, and this operation can be performed with any two adjacent numbers no more than $k$ times. For what least $k$ can all the numbers be guaranteed to be made equal? | Let the numbers on the circle be denoted by $a_{1}, \ldots, a_{2009}$, and set $a_{n+2009} = a_{n} = a_{n-2009}$. Let $N = 100400$.
1. Set $a_{2} = a_{4} = \ldots = a_{2008} = 100$ and $a_{1} = a_{3} = \ldots = a_{2009} = 0$. Suppose we managed to make all the numbers equal for some value of $k$. Consider the sum $S =... | 100400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In space, there are 3 planes and a sphere. In how many different ways can a second sphere be placed in space so that it touches the three given planes and the first sphere? (In this problem, we are actually talking about the tangency of spheres, i.e., it is not assumed that the spheres can only touch externally - ed. n... | Answer: from 0 to 16 (depending on the arrangement of the planes and the sphere). Let $O$ and $R$ be the center and radius of the given sphere $S$. Suppose the sphere $S_{1}$ with center $O_{1}$ touches the given sphere and three given planes. We associate the sphere $S_{1}$ with the sphere $S_{1}'$ with center $O_{1}$... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$2+$
John had a full basket of trempons. First, he met Anna and gave her half of his trempons and another half-trempon. Then he met Banna and gave her half of the remaining trempons and another half-trempon. After meeting Vanna and giving her half of the trempons and another half-trempon, the basket was empty. How man... | Notice that before meeting Vanna, John had one tremponch left, as half of this amount was half a tremponch. Before meeting Banna, he had 3 tremponchs, because half of this amount was one and a half tremponchs, that is, one and a half. Similarly, we get that initially there were 7 tremponchs.
## Answer
7 tremponchs. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $M$ be a finite set of numbers. It is known that among any three of its elements, there are two whose sum belongs to $M$.
What is the maximum number of elements that can be in $M$? | Consider either the four largest or the four smallest numbers.
## Solution
Example of a set of 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that the set $M=\left\{a_{1}, a_{1}, \ldots, a_{n}\right\}$ of $n>7$ numbers does not have the required property. We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9,10,11 |
Author: S $\underline{\text { Saghafian M. }}$.
In the plane, five points are marked. Find the maximum possible number of similar triangles with vertices at these points. | Example. Vertices and the center of a square.
Evaluation. Let's describe all configurations of four points forming four similar triangles. Let $\$ \mathrm{~A} \$$, $\$ \mathrm{~B} \$$, \$C \$ \$ \$
1. Point $\$ \mathrm{~A} \$$ lies inside triangle $\$ B C D \$$. Let $\$ B \$$ be the largest angle of triangle $\$ B C ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Chessboards and chess pieces ] [ Examples and counterexamples. Constructions ]
What is the maximum number of queens that can be placed on an $8 \times 8$ chessboard without attacking each other? | You can place no more than eight queens that do not attack each other, as each queen controls one row.
Let's show how you can place eight queens:
Send a comment | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3[ Examples and counterexamples. Constructions]
In the cells of a $5 \times 5$ square table, the numbers 1 and -1 are arranged. It is known that the number of rows with a positive sum is greater than the number of rows with a negative sum.
What is the maximum number of columns in this table that can have a negative s... | Let's consider one of the possible examples (see the table). The sum of the numbers in each of the three upper rows is positive, while the sum of the numbers in each column is negative.
| 1 | 1 | 1 | -1 | -1 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | -1 | 1 | -1 | 1 |
| -1 | 1 | -1 | 1 | 1 |
| -1 | -1 | -1 | -1... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3 [Central symmetry helps solve the task]
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 form 24. According to the condition, no more than one pawn can be placed on the fields of each pair. In addition, no more than one pawn can be placed on the fields ... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the minimum number of sportlotto cards (6 out of 49) you need to buy to ensure that at least one number is guessed correctly in at least one of them?
# | Let's fill eight cards as follows: in the first one, we will strike out numbers from 1 to 6, in the second one - from 7 to 12, and so on, in the last one - from 43 to 48. The number 49 will remain unstruck in any card. Therefore, at least five of the winning numbers will be struck out.
We will prove that seven cards m... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the class, each boy is friends with exactly two girls, and each girl is friends with exactly three boys. It is also known that there are 31 pioneers and 19 desks in the class. How many people are in this class?
# | The number of "friendly connections" in the class is three times the number of girls and twice the number of boys. This means that the number of boys to the number of girls is in the ratio of $3: 2$, and the total number of students is divisible by 5. On the other hand, there are no fewer than 31 and no more than 38 st... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kostrikina I.A.
At a round table, pastries are placed at equal intervals. Igor walks around the table and eats every third pastry he encounters (each pastry can be encountered several times). When there were no pastries left on the table, he noticed that the last pastry he took was the first one he encountered, and he... | Since Igor passed an integer number of circles, he met the first pastry at the moment of approaching the table. In addition, the sequence of eating pastries will not change if we remove the requirement of "equal intervals". If there were four pastries, Igor would have walked exactly five circles (see the figure).
 }\end{array}\right]} \\ {\left[\begin{array}{l}\text { Pairings and Groupings; Bijections }\end{array}\right]}\end{array}$
A token is placed on a chessboard. Two players take turns moving the token to an adjacent square. It is forbidden to place the ... | Match the second move with some move of the first.
## Solution
We will divide the board into dominoes $1 \times 2$. The strategy of the first player: if before the first player's move, the token is in one of the cells belonging to some domino, then by his move, the first player moves it to the other cell of the same ... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play? | The first player plays the rarest every second game.
## Solution
According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of every two consecutive games, the first player must participate in at least one, so the number of games was no more than $2 \cdot 10 + 1 = 21$... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Evaluation + example ] [ Combinatorial geometry (other).]
What is the minimum number of shots in the game "Battleship" on a 7*7 board needed to definitely hit a four-deck ship (a four-deck ship consists of four cells arranged in a row)?
# | If on a $7 * 7$ board, n non-overlapping four-deck ships can fit, this means that (n-1) shots would not be enough.
## Solution
In the first picture, an example sequence of 12 shots is provided, with which any four-deck ship would be hit. In the second picture, an example of the placement of twelve non-overlapping fou... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
There is a strip $1 \times 99$, divided into 99 cells $1 \times 1$, which are alternately painted in black and white. It is allowed to repaint simultaneously all cells of any rectangular cell $1 \times k$. What is the minimum number of repaintings required to make the entire strip monochromatic?
# | Each repainting destroys no more than two borders between cells of different colors.
## Solution
Estimate. Let's call a bridge an segment separating two cells of different colors. Initially, we have 98 bridges. Each repainting changes the number of bridges by no more than 2. Therefore, 48 repaintings are not enough.
... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the maximum value that the expression $a e k-a f h+b f g-b d k+c d h-c e g$ can take if each of the numbers $a, b, c, d, e, f, g, h, k$ is equal to $\pm 1$.
# | Evaluation. The product of all addends equals - $(a b c d e f g h k)^{2}$, which is negative. Therefore, the number of minus ones among the addends is odd and thus not equal to zero. Consequently, the sum does not exceed 4.
Example. Let $a=c=d=e=g=h=k=1, b=f=-1$. Then the given expression equals 4.
## Answer
4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Ordering in ascending (descending) order. ] [ Classical combinatorics (other). $\quad]$
a) A traveler stopped at an inn, and the owner agreed to accept rings from a golden chain the traveler wore on his wrist as payment for lodging. However, he set a condition that the payment should be daily: each day the owner sh... | a) It is enough to cut two rings so that pieces of three and six rings are separated. On the third day, the traveler gives the piece of three rings and receives two rings as change, and on the sixth day, the piece of six rings and receives five rings as change.
b) Arrange the resulting pieces of the chain (not countin... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
Two numbers are written on the board in the laboratory. Every day, the senior researcher Petya erases both numbers on the board and writes down their arithmetic mean and harmonic mean instead. On the morning of the first day, the numbers 1 and 2 were written on the board. Find the product of the numbe... | The product of the numbers on the board does not change. Indeed, $\frac{a+b}{2} \cdot \frac{2 a b}{a+b}=a b$. Therefore, the desired product is 2.
## Answer
2.
Send a comment | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On each kilometer of the highway between the villages of Yolkiino and Palkino, there is a post with a sign. On one side of the sign, it shows how many kilometers are left to Yolkiino, and on the other side, how many kilometers are left to Palkino. Borya noticed that on each post, the sum of all the digits is 13. What i... | Let the distance from Yolkiino to Palkiino be $n$ kilometers. Clearly, $n \geq 10$. Moreover, $n \leq 49$ (otherwise, the sum of the digits on the 49th milestone would be greater than 13).
On the tenth milestone from Yolkiino, one side reads 10, and the other side reads $n-10$, which does not exceed 39. The sum of its... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Raskina I.V.
In a singing competition, a Rooster, a Crow, and a Cuckoo participated. Each member of the jury voted for one of the three performers. The Woodpecker calculated that there were 59 judges in total, and that the Rooster and the Crow received a total of 15 votes, the Crow and the Cuckoo received 18 votes, an... | The number of votes cast for the Rooster and the Raven cannot be more than $15+13=28$. Similarly, the total number of votes for the Raven and the Cuckoo cannot exceed $18+13=31$, and for the Cuckoo and the Rooster, it cannot exceed $20+13=33$ votes. By adding these three results, we estimate the upper bound of twice th... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Through the origin, lines (including the coordinate axes) are drawn that divide the coordinate plane into angles of $1^{\circ}$.
Find the sum of the abscissas of the points of intersection of these lines with the line $y=100-x$.
# | The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the origin, 180 lines are drawn, the line $y=100-x$ intersects 179 of them. For each point on the line $y=100-x$, the sum of the coordinates is 100, therefore, the total sum of the abscissa... | 8950 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The number 61 is written on the board. Every minute, the number is erased from the board and the product of its digits, increased by 13, is written in its place. After the first minute, the number 19 is written on the board ( $6 \cdot 1+13=19$). What number will be on the board after an hour? | Let's consider the numbers that will be written on the board over the first few minutes:
| After one minute | $6 \cdot 1+\mathbf{1 3}=\mathbf{1 9}$ |
| :---: | :---: |
| After two minutes | $1 \cdot 9+13=\mathbf{2 2}$ |
| After three minutes | $2 \cdot 2+13=\mathbf{1 7}$ |
| After four minutes | $1 \cdot 7+13=\mathbf{... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kuuyggin A.K.
Among any five nodes of a regular square grid, there will always be two nodes such that the midpoint of the segment between them is also a node of the grid. What is the minimum number of nodes of a regular hexagonal grid that must be taken so that among them there will always be two nodes such that the m... | Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof of the lemma. Introduce the o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Coordinate method on the plane ] [Triangle inequality (other).]
Point $M$ lies on the line $3 x-4 y+34=0$, and point $N-$ lies on the circle $x^{2}+y^{2}-8 x+2 y-8=0$. Find the minimum distance between points $M$ and $N$. | Notice that
$$
x^{2}+y^{2}-8 x+2 y-8=0 \Leftrightarrow x^{2}-8 x+16+y^{2}+2 y+1=25 \Leftrightarrow(x-4)^{2}+(y+1)^{2}=5^{2}
$$
This means the center of the circle is the point $Q(4, -1)$, and the radius is 5.
Let $d$ be the distance from the point $Q$ to the line $3 x-4 y+34=0$. Then
$$
d=\frac{|3 \cdot 4-4 \cdot(-... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\frac{[\text { Counting in two ways }]}{[\text { Trees }}\right]$
Tsar Guidon had 5 sons. Among his descendants, 100 each had exactly 3 sons, and the rest died childless.
How many descendants did Tsar Guidon have
# | When counting the descendants, don't forget about the sons of Gvidon.
## Solution
Every descendant of King Gvidon is either a son of one of his descendants or a son of Gvidon himself. According to the condition, all the descendants of Gvidon had a total of 300 sons. And Gvidon himself had 5 sons, so the total number ... | 305 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Rubanovo I.S.
There is a set of weights with the following properties:
a. It contains 5 weights, all of different weights.
b. For any two weights, there are two other weights with the same total weight.
What is the smallest number of weights that can be in this set? | Let $A$ be one of the lightest weights, and $B$ be one of the weights that follow $A$ in weight. Clearly, the pair of weights $\{A, B\}$ can only be balanced by an identical pair.
Therefore, there are at least two weights $A$ and two weights $B$. The pair $\{A, A\}$ can also only be balanced by an identical pair. Ther... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zaslavsky A.A.
Two ants each crawled along their own closed path on a $7 \times 7$ board. Each ant only crawled along the sides of the cells and visited each of the 64 vertices of the cells exactly once. What is the minimum possible number of such sides that both the first and the second ants crawled along? | An example where the number of "common" sides equals 16 is shown in the figure (one path is black, the other is red).
Estimate. Each ant visited 64 different sides. In total, there are 7$\cd... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Klepitsyn V.A.
Cells of a $5 \times 5$ board are painted in a checkerboard pattern (corner cells are black). A figure, a mini-bishop, moves along the black cells of this board, leaving a trail on each cell it visits and not returning to that cell again. The mini-bishop can move either to free (diagonally) adjacent cel... | Let's first provide an example of the mini-elephant's route that ensures it visits twelve cells (see the figure, numbers from 1 to 12 show the order of cell visits).
| 1 | | 4 | | 6 |
| :--- | :--- | :--- | :--- | :--- |
| | 2 | | 5 | |
| 3 | | 7 | | 9 |
| | 11 | | 8 | |
| 12 | | 10 | | |
We will prove t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
The dragon imprisoned the knight and gave him 100 different coins, half of which are magical (which ones exactly - only the dragon knows). Every day, the knight sorts all the coins into two piles (not necessarily equal). If the piles end up with an equal number of magical coins or an equal number of or... | b) First, we will divide the coins into two piles: the first pile with 49 coins, and the second with 51. Then, each day we will move one coin from the first pile to the second. On the 25th day, the first pile will have 25 coins, and the second pile will have 75. Therefore, on the 25th day, the first pile will have no m... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Examples and counterexamples. Constructions]
A snail woke up, crawled from the mushroom to the spring, and fell asleep. The journey took six hours. The snail moved sometimes faster, sometimes slower, and stopped. Several scientists observed the snail. It is known that:
1) At every moment of the journey, the snail wa... | Example. Let's divide the entire time of the snail's movement into 30 intervals of 12 minutes each. Suppose the snail crawls 1 m in each of the 10 intervals numbered $1, 6, 7, 12, 13, 18, 19, 24, 25$, and 30, and rests the rest of the time. Scientists observe it in intervals 1-5 (from 1st to 5th - exactly one hour), 2-... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
|
A city in the shape of a triangle is divided into 16 triangular blocks, and at the intersection of any two streets, there is a square (there are a total of 15 squares in the city). A tourist started touring the city from a certain square and ended the tour at a different square, visiting each square exactly once. Pr... | Paint the blocks in black and white in a checkerboard pattern. If a tourist has passed through two streets of one block, then he has made a turn of $120^{0}$.
## Solution
There are 15 areas in total, so the tourist has passed through 14 streets connecting pairs of adjacent areas. We will paint the blocks in black and... | 4 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Methods for solving problems with parameters ] [ Phase plane of coefficients ]
Plot on the phase plane $O p q$ the set of points $(p, q)$ for which the equation $x^{3}+p x+q=0$ has three distinct roots belonging to the interval $(-2,4)$. | See solutions of problems $\underline{61272}, \underline{61273}$.
## Answer
The set of points defined by the inequalities $4 p^{3}+27 q^{2}<0,-4 p-64<q<2 p+8$.[^0]
How many roots does the equation $8 x\left(1-2 x^{2}\right)\left(8 x^{4}-8 x^{2}+1\right)=1$ have on the interval $[0,1]$?
## Solution
Notice that $8 x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
M. V. Murraikin
The city is a rectangular grid, with a five-story building in each cell. The renovation law allows choosing two adjacent cells (by side) with buildings and demolishing the building with fewer floors (or the same number). At the same time, the number of floors from the demolished building is added to th... | a) A square $20 \times 20$ can be divided into 25 squares $4 \times 4$, in each of which one house can be left. Indeed, in a $2 \times 2$ square, it is easy to gather all the houses in one cell. We will gather them in the marked cells (left image). Similarly, all the marked cells are gathered.
 on either heads or tails. Then a coin is flipped. If the player guesses correctly how it will land, they get back their bet and an equal am... | The main observation in this problem is this: as soon as a player guesses correctly how the coin will land, they can guess incorrectly in all remaining games. Therefore, all bets after a win should be $1.
To begin, let's show that the player will not be able to leave having lost $1 or less. Let's see what they would h... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
Does there exist a natural number that can be represented as the product of two palindromes in more than 100 ways? (A palindrome is a natural number that reads the same backward as forward.) | Exists. Consider a palindrome consisting of \$n \$ ones \$ 1 \_n=1 \ldots 1 \$.
Method 1. If \$n\$ is divisible by \$k\$, then \$1_k\$ divides \$1_n\$. For example, \$111111\$ is divisible by \$111\$ and \$11\$. It remains to choose a number \$n\$ that has more than 100 proper divisors. For example, \$ 2 \wedge{101} \... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Franklin 5.
A convex $n$-gon $P$, where $n>3$, is cut into equal triangles by diagonals that do not intersect inside it.
What are the possible values of $n$, if the $n$-gon is cyclic? | Lemma. Let a convex $n$-gon be cut into equal triangles by diagonals that do not intersect inside it. Then, for each of the triangles in the partition, at least one side is a side (not a diagonal) of the $n$-gon.
Proof. Let a triangle in the partition have angles $\alpha \leq \beta \leq \gamma$ with vertices $A, B, C$... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
On the reverse sides of 2005 cards, different numbers are written (one on each). In one question, it is allowed to point to any three cards and find out the set of numbers written on them. What is the minimum number of questions needed to find out what numbers are written on each card?
# | Let there be $N$ questions asked. It is clear that each card participates in at least one question; otherwise, we would not be able to determine the number on it.
Suppose there are $k$ cards that participate in exactly one question.
Then, two such cards cannot appear in the same question. Indeed, if two such cards we... | 1003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
On each of 2013 cards, a number is written, and all these 2013 numbers are distinct. The cards are face down. In one move, it is allowed to point to ten cards, and in response, one of the numbers written on them will be reported (it is unknown which one).
For what largest $t$ can it be guaranteed to fin... | 1) Let's first show that it is impossible to guess $1987=2013-26$ cards. We will number the cards $A_{1}, \ldots, A_{2013}$, place the numbers from 1 to 2013 on them in the same order, and indicate how to answer so that none of the numbers on the cards $A_{1}, \ldots, A_{27}$ can be determined.
For each $i=1, \ldots, ... | 1986 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Author: Shapovalov A.V., Kuukakov A.
There is a set of weights, the masses of which in grams are: $1, 2, 4, \ldots, 512$ (consecutive powers of two) - one weight of each mass. It is allowed to weigh a load using this set, placing the weights on both pans of the scales.
a) Prove that no load can be weighed using these... | Let $K_{n}(P)$ be the number of ways to weigh a weight $P$ using weights of $1,2, \ldots, 2^{n}$, and $K_{n}=\max _{P} K_{n}(P)$ (the maximum number of ways to weigh any weight using these weights). Obviously, $K_{0}=1, K_{1}=2$.
a) Our task is to prove that $K_{9} \leq 89$. We will prove that $K_{n+1} \leq K_{n}+K_{n... | 171 | Combinatorics | proof | Yes | Yes | olympiads | false |
All integers from 1 to 100 are written in a string in an unknown order. With one question about any 50 numbers, you can find out the order of these 50 numbers relative to each other. What is the minimum number of questions needed to definitely find out the order of all 100 numbers?
# | To find the desired order $a_{1}, a_{2}, \ldots, a_{100}$ of numbers in a row, it is necessary that each pair $\left(a_{i}, a_{i+1}\right), i=1,2, \ldots, 99$, appears in at least one of the sets about which questions are asked; otherwise, for two sequences $a_{1}, \ldots, a_{i}, a_{i+1}, \ldots, a_{100}$ and $a_{1}, \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ionin Yu.i.
In each cell of an infinite sheet of graph paper, a number is written such that the sum of the numbers in any square, the sides of which lie along the grid lines, does not exceed one in absolute value.
a) Prove the existence of a number $c$ such that the sum of the numbers in any rectangle, the sides of w... | Suppose in a certain rectangle with sides $a$ and $b (a < b)$. We construct four squares, each of which has three sides along some three sides of this rectangle $a \times b$; then the lines on which the fourth sides of these squares lie form a new rectangle with sides $2b - a$ and $|2a - b|$ (see Fig. 1 and 2; the case... | 3 | Combinatorics | proof | Yes | Yes | olympiads | false |
Zaslavsky A.A.
In a single-round football tournament, $n>4$ teams played. For a win, 3 points were awarded, for a draw 1, and for a loss 0. It turned out that all teams scored the same number of points.
a) Prove that there will be four teams with the same number of wins, the same number of draws, and the same number of... | a) If two teams have scored the same number of points, then the difference between the number of draws they have is a multiple of 3.
The number of draws a team has is between 0 and $n-1$. Therefore, the number of groups, each of which consists of teams with the same number of wins, draws, and losses, does not exceed $... | 10 | Combinatorics | proof | Yes | Yes | olympiads | false |
Chebotarev A.S.
On a plane, there is a circle. What is the minimum number of lines that need to be drawn so that, by symmetrically reflecting the given circle relative to these lines (in any order a finite number of times), it can cover any given point on the plane? | 1) We will prove that three lines are sufficient. Let the horizontal line $a$ and line $b$, forming a $45^{\circ}$ angle with line $a$, contain the center $O$ of the circle, and let line $c$ be parallel to $a$ and be at a distance of $0.5R$ from it, where $R$ is the radius of the given circle (see figure).
The composi... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ostrovsky $M$.
A number from 1 to 144 is guessed. You are allowed to select one subset of the set of numbers from 1 to 144 and ask whether the guessed number belongs to it. You have to pay 2 rubles for a "yes" answer and 1 ruble for a "no" answer. What is the minimum amount of money needed to surely guess the number? | Let $a_{1}=2, a_{2}=3, a_{i}=a_{i-} 1+a_{i-} 2$ for $i \geq 3$. Then $a 10=144$. We will prove by induction that among not less than $a_{i}$ numbers, the guessed number cannot be found by paying less than $i+1$ rubles.
For $i=1$ and $i=2$, this is true.
Suppose there are not less than $a_{i}$ numbers. Then either the... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tamarkin D:
In the class, there are 16 students. Every month, the teacher divides the class into two groups.
What is the minimum number of months that must pass so that every two students end up in different groups at some point? | Example. The figure shows how to divide a class into two groups so that any two students are in different groups in at least one of the four months. Each student corresponds to a column in the table, and each month corresponds to a row. A zero in a cell of the table means that the student is in the first group, and a o... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
a) On each of the fields of the top and bottom horizontal lines of an $8 \times 8$ chessboard, there is a chip: white ones at the bottom, black ones at the top. In one move, it is allowed to move any chip to an adjacent free cell vertically or horizontally. What is the minimum number of moves required ... | a) Evaluation. To reach the opposite side of the board, a piece needs to make seven vertical moves. However, at least one of the two pieces standing on the same vertical must make a horizontal move (otherwise, they would not be able to pass each other). Therefore, together these pieces will make no fewer than 15 moves.... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10,11
[Application of trigonometric formulas (geometry) $]$
[ Identical transformations (trigonometry). ]
In a convex quadrilateral $A B C D: A C \perp B D, \angle B C A=10^{\circ}, \angle B D ... | Let $K$ and $M$ be the points of intersection of the line $CB$ with the line $AD$ and the circumcircle of triangle $ACD$, respectively (see figure). Then
$\angle MDA = \angle MCA = 10^{\circ}$, so $DM$ is the bisector of angle $KDB$. Also note that $\angle ABD = 50^{\circ}$, $\angle CBD = 80^{\circ}$, hence $\angle KB... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Tokaeva I.
Let $F_{1}, F_{2}, F_{3}, \ldots$ be a sequence of convex quadrilaterals, where $F_{k+1}$ (for $k=1,2,3, \ldots$) is obtained by cutting $F_{k}$ along a diagonal, flipping one of the parts, and gluing it back along the cut line to the other part. What is the maximum number of different quadrilaterals that t... | Let $ABCD$ be the original quadrilateral $F_{1}$. We can assume that each time the half of the quadrilateral containing side $CD$ is flipped, while side $AB$ remains stationary. In this process, the sum of angle $A$ and the opposite angle does not change. Additionally, the set of side lengths does not change. However, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
A set of five-digit numbers $\left\{N_{1}, N_{k}\right\}$ is such that any five-digit number, all digits of which are in non-decreasing order, coincides in at least one digit with at least one of the numbers $N_{1}, N_{k}$. Find the smallest possible value of $k$. | A set with the specified properties cannot consist of a single number. Indeed, for each $N=\overline{a b c d e}$, there is a number $G=\overline{\boldsymbol{g g 9 g g}}$ that differs from $N$ in all digits, where $g$ is a non-zero digit different from $a, b, c, d, e$. We will show that the numbers $N_{1}=13579$ and $N_... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Problem } A .}$
A natural number is written on the board. If the number $x$ is written on the board, then you can add the number $2 x+1$ or ${ }^{x} / x+2$. At some point, it was found that the number 2008 is on the board. Prove that it was there from the beginning. | We can assume that no "extra" numbers were written on the board, meaning all numbers "participated" in obtaining the number 2008. Note that all the numbers written are positive. Suppose at some point the board has a rational number, in its irreducible form, written as
$x = p / q$. Then we can write the number $2x + 1 ... | 2008 | Number Theory | proof | Yes | Yes | olympiads | false |
Shapovalov A.V.
55 boxers participated in a tournament with a "loser leaves" system. The fights proceeded sequentially. It is known that in each match, the number of previous victories of the participants differed by no more than 1. What is the maximum number of fights the tournament winner could have conducted? | We will prove by induction that
a) if the winner has conducted no less than $n$ fights, then the number of participants is no less than $u_{n+2}$;
b) there exists a tournament with $u_{n+2}$ participants, the winner of which has conducted $n$ fights ( $u_{k}-$ Fibonacci numbers).
Base case $\left(n=1, u_{3}=2\right)... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A.K.
On the plane, there is an open non-intersecting broken line with 31 segments (adjacent segments do not lie on the same straight line). Through each segment, a line containing this segment was drawn. As a result, 31 lines were obtained, some of which may have coincided. What is the smallest number of different lin... | Evaluation. Except for the ends, the broken line has 30 vertices, and each is the intersection of two lines. If there are no more than eight lines, then there are no more than $7 \cdot 8: 2=28$ intersection points - a contradiction.
Example - in the figure.
 initially all the cement is located at one warehouse $S$, and it needs to be distributed evenly among all warehouses. Then, to each warehouse, except for $S$, cement must be delivered in some trip, so there should be at least 2010 trips in total.
We will prove by induction on $n$ tha... | 2010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Having defeated Koschei, Ivan demanded gold to ransom Vasilisa from the bandits. Koschei led him to a cave and said: "In the chest lie gold ingots. But you cannot simply take them: they are enchanted. Put one or several into your bag. Then I will put one or several from the bag back into the chest, but... | Ivan will act in such a way that each time Kashchey's move will be the only one: all other numbers either have already appeared in previous moves or are too large - Ivan does not have such a number of ingots at that moment. We will record the moves of the game as follows:
number of ingots moved (number of ingots Ivan ... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Gladkova E.B.
Computers $1,2,3, \ldots, 100$ are connected in a ring (the first with the second, the second with the third, ..., the hundredth with the first). Hackers have prepared 100 viruses, numbered them, and at different times, in a random order, they launch each virus on the computer with the same number. If a ... | Let's prove several statements.
1) Every computer is infected by viruses at least twice.
Consider, for example, computer 1. If the first virus to infect it came from computer 100, then it will also be infected by virus 1. If the first virus to infect computer 1 is virus 1, then the virus that first infected computer ... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On the Island of Misfortune with a population of 96 people, the government decided to carry out five reforms. Each reform is opposed by exactly half of all citizens. A citizen will go to a rally if they are dissatisfied with more than half of all the reforms. What is the maximum number of people the government can expe... | Let $x$ be the number of people who came out to the rally. Consider the total number of "dissatisfactions." On one hand, each reform is opposed by exactly 48 residents, so the total number of dissatisfactions is $48 \cdot 5 = 240$. On the other hand, each person who came out to the rally is dissatisfied with at least t... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Pairing and grouping; bijections] Proof by contradiction
What is the smallest number of weights in a set that can be divided into 3, 4, and 5 equal-mass piles? | Answer: 9. First, let's prove that the set cannot contain fewer than nine weights. Suppose this is not the case, that is, there are no more than eight. Let $60 m$ be the total mass of all the weights in the set. First, note that the set cannot contain weights with a mass greater than 12 t (since the set can be divided ... | 9 | Number Theory | proof | Yes | Yes | olympiads | false |
Geometry on graph paper
In each cell of a chessboard, there are two cockroaches. At a certain moment, each cockroach crawls to an adjacent (by side) cell, and the cockroaches that were in the same cell crawl to different cells. What is the maximum number of cells on the board that can remain free after this? | Evaluation. We will color 20 cells of the board as shown in the left figure. Each colored cell has the following property: no matter which two adjacent cells the cockroaches crawl into from it, the cockroaches will not be able to reach these cells from any other colored cell. Therefore, after all the cockroaches have m... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
M. Murraikin
On an island, there live 100 knights and 100 liars, each of whom has at least one friend. Knights always tell the truth, while liars always lie. One morning, each resident either said the phrase "All my friends are knights" or the phrase "All my friends are liars," and exactly 100 people said each phrase.... | Let's show that it is possible to find no less than 50 pairs of friends, one of whom is a knight, and the other is a liar. If the phrase "All my friends are liars" was said by no less than 50 knights, then each of them knows at least one liar, and the 50 required pairs are found. Otherwise, the phrase "All my friends a... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Gopovanov A.S. }}$The sum of the digits in the decimal representation of a natural number $n$ is 100, and the sum of the digits of the number $44 n$ is 800. What is the sum of the digits of the number $3 n$? | Note that $44 n$ is the sum of 4 instances of the number $n$ and 4 instances of the number $10 n$.
If we add these numbers digit by digit, then in each digit place, there will be the sum of the quadrupled digit from the same place in the number $n$ and the quadrupled digit from the next place. If no carries occur in t... | 300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Theory of algorithms (other). }} \\ {[\quad \text { Estimation + example }}\end{array}\right]$
There are 2004 boxes on the table, each containing one ball. It is known that some of the balls are white, and their number is even. You are allowed to point to any two boxes and ask if there... | We will number the boxes (and thus the balls in them) from 1 to 2004 and denote a question by a pair of box numbers. We will call non-white balls black.
We will show that a white ball can be found in 2003 questions.
We will ask the questions $(1,2),(1,3),(1,2004)$. If all answers are positive, then the first ball is ... | 2003 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
|
Rabbits are sawing a log. They made 10 cuts. How many chunks did they get?
# | Into how many parts is a log divided by the first cut? How does the number of pieces change after each subsequent cut?
## Solution
The number of chunks is always one more than the number of cuts, since the first cut divides the log into two parts, and each subsequent cut adds one more chunk. Answer: 11 chunks.
## An... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Rabbits are sawing a log again, but now both ends of the log are secured. Ten middle chunks have fallen, while the two at the ends remain secured. How many cuts did the rabbits make?
# | How many logs did the rabbits get?
## Solution
The rabbits got 12 logs - 10 fallen and 2 secured. Therefore, there were 11 cuts.
## Answer
11 cuts.
## Problem | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Cutting (other).]
What is the maximum number of pieces into which a round pancake can be divided using three straight cuts?
# | Recall problem 11.
## Solution
If from three lines each pair intersects inside the pancake, it will result in 7 pieces (see Fig. 11.4). If, however, any two of these lines are parallel or intersect outside the pancake, there will be fewer pieces.
## Answer
7 pieces. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Write in a line the first 10 prime numbers. How to cross out 6 digits so that the largest possible number is obtained
# | Notice that the first 10 prime numbers form a 16-digit number.
## Solution
Here are the first ten prime numbers written in sequence: 2357111317192329
## Answer
The resulting number is 7317192329. | 7317192329 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10 guests came to visit and each left a pair of galoshes in the hallway. All pairs of galoshes are of different sizes. The guests began to leave one by one, putting on any pair of galoshes that they could fit into (i.e., each guest could put on a pair of galoshes not smaller than their own). At some point, it was disco... | If there are more than five guests left, then the guest with the smallest size of galoshes among the remaining guests can put on the largest of the remaining galoshes.
## Solution
Let's number the guests and their pairs of galoshes from 1 to 10 in ascending order of the size of the galoshes. Suppose there are 6 guest... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left.\quad \begin{array}{ll}{\left[\begin{array}{l}\text { Principles of divisibility by 3 and 9 } \\ \text { [Examples and counterexamples. Constructions] }\end{array}\right]}\end{array}\right]$
Write the number 2013 several times in a row so that the resulting number is divisible by 9. | The number 201320132013 is divisible by 9, since the sum of its digits is equal to $(2+0+1+3) \cdot 3=18$.
## Answer
Given a sheet of graph paper. Each node of the grid is marked with some letter. What is the smallest number of different letters needed to mark these nodes so that on any segment (going along the sides... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Grandpa is twice as strong as Grandma, Grandma is three times stronger than Granddaughter, Granddaughter is four times stronger than Bug, Bug is five times stronger than Cat, and Cat is six times stronger than Mouse. Grandpa, Grandma, Granddaughter, Bug, and Cat together with Mouse can pull out the Turnip, but without ... | How many Mice does a Cat replace? And a Granddaughter?
## Solution
A Cat replaces 6 Mice. A Bug replaces $5 \bullet 6$ Mice. A Granddaughter replaces 4$\cdot$5$\cdot$6 Mice. A Grandma replaces $3 \cdot 4 \cdot 5 \cdot 6$ Mice. A Grandpa replaces $2 \cdot 3 \cdot 4 \cdot 5 \cdot 6$ Mice. In total, it requires:
$(2 \c... | 1237 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Chuk and Gek were decorating the Christmas tree with their mother. To prevent them from fighting, their mother gave each of them the same number of branches and the same number of ornaments. Chuk tried to hang one ornament on each branch, but he was short of one branch. Gek tried to hang two ornaments on each branch, b... | Try to do as Chuk did - hang one toy on each branch.
## Solution
Let's try to do as Chuk did — hang one toy on each branch, then one toy will be left over. Now, let's take two toys — one that is left over, and another one from one of the branches. If we now hang these toys as the second ones on the branches that stil... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many two-digit numbers are there where the digit in the tens place is greater than the digit in the units place?
# | The problem can be solved by simple enumeration.
1) If the units digit is 0, then the tens digit can take values from 1 to 9.
2) If the units digit is 1, then the tens digit can take values from 2 to 9.
3) If the units digit is 2, then the tens digit can take values from 3 to 9.
4) If the units digit is 8, then the te... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The scent from a blooming lily-of-the-valley bush spreads in a radius of 20 m around it. How many blooming lily-of-the-valley bushes need to be planted along a straight 400-meter alley so that every point along it smells of lily-of-the-valley
# | Notice that to meet the conditions of the problem, the distance between adjacent lilies of the valley should not exceed $40 \mathrm{M}$.
## Solution
Let's mentally move the leftmost 20 m of the alley to the right end. Then we will have segments of the alley that are 40 m long, and to the right of each segment, a lily... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kalinin D.A.
On a line, several points were marked. After that, between each pair of adjacent points, one more point was marked. This "densification" was repeated two more times (a total of three times). As a result, 113 points were marked on the line. How many points were initially marked? | Find how many points there were before the last compaction, i.e., solve the problem "from the end".
## Solution
If there were $n$ points marked (before compaction), then after compaction, there will be $2 n-1$ points (of which $n$ are old and $n-1$ are new). If after compaction, there are $k$ points, then $2 n-1=k$ o... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The digits of a three-digit number A were written in reverse order to obtain number B. Can the number equal to the sum of A and B consist only of odd digits?
# | Let, for example, $A=219$. Then $B=912, A+B=1131$.
## Answer
Yes, it can. | 1131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For a two-digit number, the first digit is twice the second. If you add the square of its first digit to this number, you get the square of some integer. Find the original two-digit number.
# | The first digit is twice the second only in the following two-digit numbers: 21, 42, 63, and 84. By checking, we find that only the number 21 satisfies the condition of the problem.
## Answer
21.00 | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Semi-invariants $]$ [ $\left[\begin{array}{l}\text { Processes and operations }\end{array}\right]$
A chocolate bar has a size of $4 \times 10$ squares. In one move, it is allowed to break one of the existing pieces into two along a straight line. What is the minimum number of moves required to break the entire choco... | After each break, the number of pieces increases by 1.
## Solution
Notice that after each break, the number of pieces increases by 1. Initially, we have a whole chocolate bar, i.e., one piece, and in the end, we need to get 40 pieces of one tile each. Therefore, we need to perform exactly 39 breaks.
## Answer
In 39... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Combinatorics (miscellaneous). ] [Principle of the extreme (miscellaneous) ]
How many ways are there to rearrange the numbers from 1 to 100 such that adjacent numbers differ by no more than 1?
# | Where can the number 1 be placed?
## Solution
Next to the number 1, only the number 2 can stand, so 1 must be at the edge. Suppose 1 is at the beginning. Then the next number is 2, the next is 3 (no other numbers can be next to 2), the next is 4, and so on. We get the arrangement $1, 2, \ldots, 99, 100$.
If 1 is at ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Weighing $\quad]$ [ Examples and Counterexamples. Constructions ]
What weights can three weights have so that with their help, it is possible to weigh any whole number of kilograms from 1 to 10 on balance scales (weights can be placed on both pans)? Provide an example. | Let's take, for example, weights of 3, 4, and 9 kg: $1=4-3, 2=9-3-4, 5=9-4, 6=9-3, 7=3+4, 8=3+9-4, 10=4+9-3$.
## Answer
For example, weights of 3, 4, and 9 kg.
In a corridor 100 m long, 20 r... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tomonvo A.K.
Petya's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300
dollars or adding 198 dollars.
What is the maximum amount Petya can withdraw from his account if he has no other money? | Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars. The maximum number that is a multiple of 6 and does not exceed 500 is 498.
Let's show how to withdraw 498 dollars. Perform the following operations: $500-300=200, 200+198=398, 398-300=98, 98+198=296, 29... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Case Analysis ] [ Proof by Contradiction ]
In the cells of a $3 \times 3$ table, numbers are arranged such that the sum of the numbers in each column and each row is zero. What is the smallest number of non-zero numbers that can be in this table, given that this number is odd?
# | ## Example.
| 0 | -1 | 1 |
| :---: | :---: | :---: |
| -1 | 2 | -1 |
| 1 | -1 | 0 |
Evaluation. We will prove that it is impossible to use fewer non-zero numbers.
If the table contains exactly one non-zero number, then the sum of the numbers in the row containing this number is not zero. Suppose the table contains e... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ [ CaseAnalysis $\quad$]
Find all odd natural numbers greater than 500 but less than 1000, for each of which the sum of the last digits of all divisors (including 1 and the number itself) is 33.
# | For an odd number, all its divisors are odd. Since the sum of their last digits is odd, there must be an odd number of divisors. If a number has an odd number of divisors, then it is a square of a natural number (see problem $\underline{30365}$). Let's consider all the squares of odd numbers within the given range: $23... | 729 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic progression $]$ [ Extreme principle (other) . ] Author: Shapovalov $A . B$. In a $29 \times 29$ table, the numbers $1,2,3, \ldots, 29$ were written, each 29 times. It turned out that the sum of the numbers above the main diagonal is three times the sum of the numbers below this diagonal. Find the number... | Above (below) the diagonal there are $29 \cdot 14=406$ numbers. It is not difficult to check that the sum of the 406 largest numbers in the table (16, 17, ..., 29, taken 29 times each) is exactly three times the sum of the 406 smallest numbers (1, 2, ..., 14, taken 29 times each). Therefore, all the numbers on the diag... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
On the table, 28 coins of the same size are arranged in a triangular shape (see figure). It is known that the total mass of any three coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle.
First solution. Let's take a rhombus of 4 coins. As can be seen from the figure, the masses of two coins in it are equal. Cons... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
On a chessboard, $n$ white and $n$ black rooks are placed such that rooks of different colors do not attack each other. Find the maximum possible value of $n$.
# | Let's prove that for $n>16$, the specified arrangement is impossible. Note that on each row and each column, rooks can only be of one color (or the row/column can be free of rooks). We will denote a row (column) by the same color as the rooks standing on it.
Since there are more than 16 rooks, there are at least three... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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