problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
Pe
In the country of Distant, a province is called large if more than $7 \%$ of the country's population lives in it. It is known that for each large province, there are two provinces with a smaller population such that their combined population is greater than that of this large province. What is the smallest number ... | We will order the provinces by increasing population. The first and second provinces are not large, as for each of them, there will not be two provinces with a smaller population. In the third province, less than $14\%$ of the population lives, since in both provinces with a smaller population, the total is no more tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Chessboards and chess pieces ] [ Examples and counterexamples. Constructions ]
Author: S. Preskova
Under one of the cells of an $8 \times 8$ board, a treasure is buried. Under each of the other cells, there is a sign indicating the minimum number of steps required to reach the treasure from that cell (one step all... | Let's dig up the corner cell $U$. Suppose there is a sign there. All cells at the specified distance from $U$ form a diagonal perpendicular to the main diagonal drawn from $U$. Let's dig up the corner cell $W$ on the same side as $U$. If there is also a sign there, then another diagonal perpendicular to the first one i... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Can some of the minuses in the expression $2013^{2}-2012^{2}-\ldots-2^{2}-1^{2}$ be replaced with pluses so that the value of the resulting expression equals 2013? | $(m+1)^{2}-m^{2}=2 m+1$. Therefore, $\left((n+3)^{2}-(n+2)^{2}\right)-\left((n+1)^{2}-n^{2}\right)=(2(n+2)+1)-(2 n+1)=2 n+5-2 n-1=4$. Thus, before the squares of any four consecutive natural numbers, signs can be placed in such a way that the value of the resulting expression is equal to 4.
We will divide the first 20... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A passenger left their belongings in an automated storage locker, and when they came to retrieve their items, they realized they had forgotten the number. They only remember that the numbers 23 and 37 were in the code. To open the locker, the correct five-digit number must be entered. What is the minimum number of code... | Let's consider several cases.
1) The number contains the combination 237. It can be placed in the number in three ways: **237, *237*, 237**. In each of these, each of the remaining digits can be chosen in 10 ways. In total, we get $3 \cdot 10^{2}=300$ numbers. 2) The number contains the combinations 23 and 37, with 23... | 356 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Chess coloring ] [ Examples and counterexamples. Constructions ]
a) What is the maximum number of bishops that can be placed on a 1000 by 1000 board so that they do not attack each other?
b) What is the maximum number of knights that can be placed on an $8 \times 8$ board so that they do not attack each other? | a) Since no more than one bishop can stand on one diagonal, and there are exactly 1999 diagonals running from the bottom-left to the top-right, and on the two outermost diagonals (each consisting of one cell) no more than one bishop can stand (they are located on the same perpendicular diagonal), no more than 1998 non-... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kozhevnikov P.A.
What is the minimum number of connections required to organize a wired communication network of 10 nodes, so that the communication remains possible between any two remaining nodes (even through a chain of other nodes) in the event of any two nodes failing?
# | Evaluation. To ensure that the connection is maintained even when any two nodes fail, it is necessary that at least three lines of communication enter each node (if node $A$ is connected to two nodes $B$ and $C$, then when nodes $B$ and $C$ fail, node $A$ becomes inaccessible). Thus, the total number of communication l... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a company of 10 people, 14 pairwise arguments have occurred. Prove that it is still possible to form a group of three friends.
# | The total number of ways to choose a company of three people is $C_{10}^{3}=120$. Each quarrel destroys no more than eight such companies, so the number of destroyed companies is no more than $8 \cdot 14=112$. Therefore, at least 8 friendly companies remain. | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
Laiko 0.
Around a round table, there is a company of thirty people. Each of them is either a fool or smart. Everyone sitting is asked: Is your right neighbor smart or a fool? In response, a smart person tells the truth, while a fool can say either the truth or a lie. It is known that the number of fools does not excee... | When $F=8$. If $F=0$, then one can point to any person sitting at the table. Let now $F \neq 0$. We divide all those sitting at the table into non-empty groups of consecutive smart and consecutive foolish people; the number of these groups is denoted by $2k$ ($k$ groups of smart people and $k$ groups of foolish people)... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A nut has the shape of a regular hexagonal prism. Each side face of the nut is painted in one of three colors: white, red, or blue, and adjacent faces are painted in different colors. How many differently colored nuts exist? (Not all three colors need to be used for the coloring of the nut.)
# | Faces of the same color cannot be adjacent, so there are no more than three of them. Let's consider three cases.
1) Only two colors are used - each three times. Then the colors must alternate, and the only coloring scheme is 121212. By substituting any pair of colors for 1 and 2, we get 3 types of nuts. Swapping the c... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $\left.\begin{array}{c}\text { Work Problems } \\ \text { [ Processes and Operations }\end{array}\right]$
In a turning shop, parts are machined from steel blanks, one part from one blank. The chips left after processing three blanks can be melted down to make exactly one blank. How many parts can be made from nine b... | From nine blanks, on the first stage, you can obtain nine parts, and from the remaining swarf, you can make three blanks, from which (on the second stage) you can make three parts, and from the remaining swarf, you can make one blank and then one part. In total, 13 parts.
With 14 blanks, it is easier to proceed as fol... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Combinations and Permutations] [Case Enumeration]
How many ways can you choose from a full deck (52 cards)
a) 4 cards of different suits and ranks?
b) 6 cards such that all four suits are represented among them?
# | a) A card of the spades suit can be chosen in 13 ways, after which a card of the diamonds suit can be chosen in 12 ways...
b) $6=1+1+1+3=1+1+2+2$.
## Answer
a) $13 \cdot 12 \cdot 11 \cdot 10=17160$ ways; b) $4 C_{13}^{3} \cdot 13^{3}+C_{4}^{2} \cdot 13^{2} \cdot\left(C_{13}^{2}\right)^{2}=8682544$ ways. | 8682544 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Linear inequalities and systems of inequalities ]
$$
\text { [ The extremal principle (miscellaneous). ] }
$$
Nine digits: $1,2,3, \ldots, 9$ are written in some order (forming a nine-digit number). Consider all triples of consecutive digits, and find the sum of the corresponding seven three-digit numbers. What is ... | Let $\overline{a b c d e f g h i}$ be a nine-digit number. From it, we form the sum $S=\overline{a b c}+\overline{b c d}+\overline{c d e}+\overline{d e f}+\overline{f f g}+\overline{f g h}+\overline{g h i}=$ $100 a+110 b+111(c+d+e+f+g)+11 h+i$. The sum $S$ reaches its maximum when the larger digits enter it with the la... | 4648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
48 blacksmiths need to shoe 60 horses. What is the minimum time they will spend on the work if each blacksmith spends 5 minutes on one horseshoe
# | Notice that four blacksmiths cannot shoe one horse at the same time.
## Solution
It will take no less than $60 \cdot 4 \cdot 5: 48=25$ minutes. For this, it is necessary that none of the blacksmiths are idle. It is sufficient to show that four blacksmiths can shoe five horses in 25 minutes. For this, arrange the five... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.V.
What is the maximum number of white and black chips that can be placed on a chessboard so that in each row and each column there are exactly twice as many white chips as black ones? | Evaluation. The number of chips on each vertical is a multiple of 3, meaning there are no more than 6, and on the entire board - no more than 48. Example: 32 white chips are placed on white squares, and 16 black chips are placed along the main "black" diagonal and along two parallel diagonals of "length" 4.
## Answer
... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Can the product of three three-digit numbers, for the recording of which nine different digits were used, end with four zeros? | For example, $125 \cdot 360 \cdot 748=33660000$.
## Answer
It can. | 33660000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In a $2005 \times 2006$ table, numbers $0,1,2$ are arranged such that the sum of the numbers in each column and each row is divisible by 3.
What is the maximum possible number of ones that can be in this table?
# | Evaluation. Let there be $n$ zeros and $d$ twos in the table. We have 2005 rows of length 2006 and 2006 columns of length 2005. For the sum in a row to be divisible by 3, there must be at least one two or two zeros. Hence, $d + n / 2 \geq 2005$. Similarly, in each column, there must be at least one zero or two twos, so... | 4019356 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum number of natural numbers not exceeding 2016 that can be marked so that the product of any two marked numbers is a perfect square?
# | Since $44^{2}1$, we will get a set of fewer numbers, as $1936 q>2016$.
## Answer
44 numbers.
\section*{Problem 65909 Topics:
Germann and Chekalin laid out 13 different cards on the table. E... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bogognov i.i.
In some cells of a $200 \times 200$ square, there is one chip - either red or blue; the other cells are empty. One chip sees another if they are in the same row or column. It is known that each chip sees exactly five chips of the other color (and possibly some chips of its own color). Find the maximum po... | Example. Let's highlight a "border" of width 5 around a 200×200 square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place 3800 chips in these four rectangles: in the left and top ones - red, and in the right and bottom ones - blue. It is easy to see that all the requirem... | 3800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $M$.
On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 residents were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The residents answered in turn, and everyone could hear their answers. Knights answered... | Let's call knights and liars principled people.
Evaluation. First method. (Buchaev Abdulkadyr) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values of the balance divide the sequ... | 1009 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kenodarov R.g.
A board $7 \times 7$ is either empty or has an invisible ship $2 \times 2$ placed on it "by cells". It is allowed to place detectors in some cells of the board, and then turn them on simultaneously. An activated detector signals if its cell is occupied by the ship. What is the smallest number of detecto... | Evaluation. In each rectangle $2 \times 3$, there should be at least two detectors: the rectangle consists of three dominoes $1 \times 2$ (strips), and if a detector is in the edge domino, we might not know if there is a ship on the other two dominoes, and if a detector is in the middle domino, we might not know which ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Two people $A$ and $B$ need to get from point $M$ to point $N$, which is 15 km away from $M$, as quickly as possible. On foot, they can travel at a speed of 6 km/h. In addition, they have a bicycle that can be ridden at a speed of 15 km/h. $A$ sets out on foot, while $B$ rides the bicycle until meeting pedestrian $C$, ... | To ensure that $A$ and $B$ spend the least amount of time on the road, they should arrive at $N$ simultaneously, meaning they should walk the same distances. Indeed, if $A$ arrives at $N$ before $B$, $A$ can give part of his time to $B$: $A$ can get on the bicycle a little later, allowing $B$ to ride the bicycle a litt... | 1998 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Can all non-negative integers be divided into 1968 non-empty classes such that each class contains at least one number and the following condition is met: if a number $m$ is obtained from a number $n$ by erasing two adjacent digits or identical groups of digits, then both $m$ and $n$ belong to the same class (for examp... | The first method. Let the symbol $\approx$ denote belonging to the same class. $\overline{M a b N} \approx \overline{M b b a b N} \approx \overline{M b a a b a b N} \approx \overline{M b a N}$. Thus, numbers obtained by swapping adjacent digits belong to the same class. Since any permutation can be represented as a seq... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given a 29-digit number $X=\overline{a_{1}} \overline{-} \overline{-} \overline{29}\left(0 \leq a_{k} \leq 9, a_{1} \neq 0\right)$. It is known that for every $k$ the digit $a_{k}$ appears $a_{30-k}$ times in the number (for example, if $a_{10}=7$, then the digit $a_{20}$ appears seven times). Find the sum of the digit... | Let's divide all numbers into pairs $\left(a_{k}, a_{30-k}\right)$. Notice that if among these pairs there is a pair $(a, b)$, then pairs $(a, c)$ and $(c, a)$, where $c \neq b$, cannot occur, otherwise the digit $a$ would have to appear in the number $X$ $b$ times on one side and $c$ times on the other. Therefore, if ... | 201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
It is known that in a code lock, only the buttons with numbers 1, 2, and 3 are functional, and the code for this lock is three digits long and does not contain any other digits. Write the shortest sequence of digits that will definitely open this lock (the lock opens as soon as the correct three digits are pressed in t... | There are $3^{3}=27$ different three-digit numbers, in the notation of which the digits 1, 2, 3 are used. Besides the first two digits in the sequence of button presses, each of the other digits serves as the last digit of some three-digit number. Therefore, the required sequence must contain no fewer than $27+2=29$ di... | 11123222133313121223113233211 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Fomin d:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 200 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 200 can be balanced by some number of weights from the set, and in a unique way (the b... | The correct set should correspond to the factorization of the number 201 (see the solution of problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.
## Answer
a) Two weights of 67 g and 66 weights of 1 g or 66 weights of 3 g and two of 1 g.
b) 3 sets. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Fomin D:
Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 500 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 500 can be balanced by some number of weights from the set, and in a unique way (the b... | Let the largest weight of a weight in some correct set be $M$ (grams). This means that any smaller weight can be balanced by smaller weights. Let the weight of all smaller weights be $m$. Clearly, $m \geq$ $M-1$. But if $m \geq M$, then we have two ways to balance the weight $M+r$, where $r$ is the remainder of the div... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Herrer M.L. Let $S(n)$ denote the sum of the digits of the number $n$ (in decimal notation). Do there exist three distinct natural numbers $m, n$, and $p$ such that $m+S(m)=n+S(n)=p+S(p)$? | Here is one of such triples:
$m=9999999999999, S(m)=117 ; \quad n=10000000000098, S(n)=18 ; \quad p=10000000000107, S(p)=9$.
For these numbers $m+S(m)=n+S(n)=p+S(p)=10000000000116$.
## Answer
They exist. | 10000000000116 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Rubanov I.S. }}$
Microcalculator MK-97 can perform only three operations on numbers stored in memory:
1) check if two selected numbers are equal,
2) add selected numbers,
3) find the roots of the equation $x^{2}+a x+b=0$ for selected numbers $a$ and $b$, or display a message if there are no roots.... | By adding $x$ to itself, we get $2x$. We compare $x$ and $2x$. If they are equal, then $x=0$. Otherwise, we find the roots of the equation $y^{2}+2xy+x=0$. The discriminant of this equation is $4(x^{2}-x)$, so the roots are equal if and only if $x=1$. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bernicheyni i. A square sheet of paper was cut into two parts along a straight line. One of the resulting parts was again cut into two parts, and so on, many times. What is the minimum number of cuts needed so that among the resulting parts there could be exactly 100 twenty-sided polygons? | Answer: It is possible to obtain exactly 100 20-sided polygons with 1699 cuts, and it is impossible to obtain 100 20-sided polygons with fewer cuts. With each cut, the total number of paper pieces increases by 1 (since one piece is cut into two new pieces), so after $n$ cuts, there will be ($n+1$) pieces of paper. Now ... | 1699 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $\left.\begin{array}{c}\text { Induction (other) } \\ {\left[\begin{array}{l}\text { Irrational inequalities }\end{array}\right]}\end{array}\right]$
Given:
$$
a_{1}=1966, a_{\mathrm{k}}=\left[\sqrt{a_{1}+a_{2}+\cdots+a_{k-1}}\right]
$$
Find $a_{1966}$. | We will prove by induction that the initial segment of our sequence is $1966,44,44,45,45, .$. , where after the number 1966, each natural number greater than 43 appears exactly twice in a row, except for numbers of the form $74 \cdot 2^{k}$, which appear three times in a row (compare with the solution to problem 78594)... | 1024 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Two sages are playing the following game. The numbers $0,1,2, \ldots, 1024$ are written down. The first sage crosses out 512 numbers (of his choice), the second crosses out 256 of the remaining, then the first crosses out 128 numbers, and so on. On the tenth step, the second sage crosses out one number; two numbers rem... | Answer: with correct play, the difference between the remaining numbers is 32 (as they say, the "price of the game" is 32). The following considerations lead to this answer: the first player aims to maximize the difference between the remaining numbers and should always try to "thin out" the remaining set of numbers as... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V. The host has a deck of 52 cards. The audience wants to know the order in which the cards are arranged (without specifying whether it is from top to bottom or bottom to top). They are allowed to ask the host questions of the form "How many cards lie between such-and-such and such-and-such cards?" One of... | Algorithm. Let's show how to establish the order of cards in 34 questions. We will divide all the cards in the deck (numbered from bottom to top), except for the 19th, into triples:
| The cards | 1 | 2 | 3 | $\ldots$ | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | $\ldots$ | 48 | 49 | 50 | 51 | 52 |
| :---: | :---: | :---: ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $2 S$ be the total weight of some set of weights. We call a natural number $k$ average if we can choose $k$ weights from the set such that their total weight is $S$. What is the maximum number of average numbers that a set of 100 weights can have? | Note that if the number $m$ is average, then the number $100-m$ is also average. Therefore, if the number 1 is not average, then the number 99 is also not average, and the number of average numbers is no more than 97 (the number 100 is also not average). If, however, the number 1 is average, then the weight of one of t... | 97 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Golovanov A.S.
Sasha wrote a non-zero digit on the board and keeps appending a non-zero digit to the right, until he writes a million digits. Prove that a perfect square was written on the board no more than 100 times. | Let's consider separately the numbers with an odd and even number of digits. Let $x_{1}^{2}, x_{2}^{2}, . .-$ be the squares encountered on the board with an even number of digits, and their representations contain $2 n_{1}, 2 n_{2}, . .\left(n_{1}<n_{2}<\ldots\right)$ digits respectively. From the inequality $x_{k}^{2... | 100 | Number Theory | proof | Yes | Yes | olympiads | false |
Zaslavsky A.A.
In a certain state, there are 32 cities, each of which is connected by a road with one-way traffic. The Minister of Communications, a secret villain, decided to organize traffic in such a way that, after leaving any city, it would be impossible to return to it. To achieve this, he can change the directi... | Let's prove the general formula. Suppose in the state there are $\$ 2 \wedge n$ \$ cities. Then the minister can achieve the desired result in no more than $\$ 2 \wedge\{n-2\}(2 \wedge n-n-1) \$$ days.
Lemma. Suppose in the state there are $\$ 2 \mathrm{k} \$$ cities, each two of which are connected by a one-way road.... | 208 | Combinatorics | proof | Yes | Yes | olympiads | false |
Beikkreve v.P.
The sum of the tangents of angles measuring $1^{\circ}, 5^{\circ}, 9^{\circ}, 13^{\circ}, \ldots, 173^{\circ}, 177^{\circ}$ is 45. Prove this. | Author: Berkokoayko s.T.
Let's use the well-known identity $\operatorname{tg} \alpha+\operatorname{tg}\left(\alpha+60^{\circ}\right)+\operatorname{tg}\left(\alpha-60^{\circ}\right)=3 \operatorname{tg} 3 \alpha$ (which can be verified by expressing both sides in terms of $\operatorname{tg} \alpha$). From this, it follo... | 45 | Algebra | proof | Yes | Yes | olympiads | false |
In a tournament, 25 chess players are going to participate. They all play at different levels, and the stronger player always wins when they meet.
What is the minimum number of games required to determine the two strongest players? | Complexity: 5 Classes: 10,11
Example. Let's organize a tournament according to the Olympic system in five rounds: first, 12 pairs will play, then 6, then 3. Four people will remain, two pairs will play, and then the winners will play against each other. Thus, the strongest will be determined. He played no more than fi... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10,11
The game board has the shape of a rhombus with an angle of $60^{\circ}$. Each side of the rhombus is divided into nine parts. Lines are drawn through the division points, parallel to the sides and the smaller diagonal of the rhombus, dividing the board into triangular cells. If a chip is placed on a certain cell... | Let's replace the board with an equivalent $9 \times 9$ square board, where diagonals of the same direction are drawn in all cells (see Fig. 1).
Six chips are sufficient to cover all cells (see Fig. 1).
 > 10^{4} a + 100 b + c$, which means the equation has no solutions.
Therefor... | 91125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.B.
Thieves Hapok and Glazok are dividing a pile of 100 coins. Hapok grabs a handful of coins from the pile, and Glazok, looking at the handful, decides who of the two will get it. This continues until one of them gets nine handfuls, after which the other takes all the remaining coins (the division may als... | Here is Hapka's strategy, which will ensure him no less than 46 coins: take six coins each time (as long as it is possible). The pile contains 16 such full handfuls and four more coins. With this strategy, Hapka (regardless of how Glazok acts) has only two possibilities:
a) at some point, Glazok will have nine full ha... | 46 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8,9 The area of a triangle is $6 \sqrt{6}$, its perimeter is 18, and the distance from the center of the inscribed circle to one of the vertices is $\frac{2 \sqrt{42}}{3}$. Find the smallest side of the triangle. | If the inscribed circle touches the side $A C$ of triangle $A B C$ at point $M$, and $p$ is the semiperimeter of the triangle, then $A M=p-B C$. Use this equality, and then apply the Law of Cosines.
## Solution
Let $O$ be the center of the circle inscribed in the given triangle $A B C$, $r$ its radius, $S=6 \sqrt{6}$... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Fon-err-Flaass problem: }}$
A square board is divided by a grid of horizontal and vertical lines into $n^{2}$ unit squares. For what largest $n$ can one mark $n$ cells so that every rectangle of area at least $n$ with sides along the grid lines contains at least one marked cell? | Evaluation. It is clear that if $n$ cells are marked in such a way that the condition of the problem is satisfied, then in each row and each column there is exactly one marked cell. Assuming that $n \geq 3$ (it is obvious that $n=2$ is not the largest), let's take row $A$, in which the leftmost cell is marked, row $B$,... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Rubanov 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 $\left\{a_{1}, \ldots, a 50, b_{1}, . ., b 50\right\}$, ordered in ascending order. On each of the 101 intervals $\left[-... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Maiinnikova E.
At the enterprise, 50,000 people are employed. For each of them, the sum of the number of their immediate supervisors and their immediate subordinates is 7. On Monday, each employee of the enterprise issues an order and gives a copy of this order to each of their immediate subordinates (if any). Then, e... | If on an enterprise to the top executives, then each employee should eventually see at least one of the $k$ orders issued by these executives on Monday. On Monday, no more than $7 k$ employees saw them, on Tuesday - no more than $7 k \cdot 6$, on Wednesday - no more than $7 k \cdot 36$ employees. All those who saw thes... | 97 | Combinatorics | proof | Yes | Yes | olympiads | false |
Given $n$ sticks. From any three, an obtuse triangle can be formed. What is the largest possible value of $n$? | From three sticks of lengths $a \leq b \leq c$, a triangle can be formed if $a+b>c$. According to the cosine theorem, this triangle is obtuse if and only if $a^{2}+b^{2}<c^{2}$. For the given problem, we need to check if $a_{4}^{2}+a_{3}^{2} \geqslant 2 a_{3}^{2}>2 a_{2}^{2}+2 a_{1}^{2}$. On the other hand, $a_{5}<a_{1... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kondakov G.V.
Divide the segment $[-1,1]$ into black and white segments so that the integrals of any a) linear function;
b) quadratic trinomial over the white and black segments are equal.
# | Since the integral of a polynomial over an interval is equal to the increment of its antiderivative, our task is a special case of problem $\underline{98268}$.
## Answer
For example, a) intervals $[-1,-1 / 2],[1 / 2,1]$ - black, $[-1 / 2,1 / 2]-$ white.
b) intervals $[-1,-3 / 4],[-1 / 4,0],[1 / 4,3 / 4]$ - white, $[... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Berroov S.L.
Petya is coloring 2006 points, located on a circle, using 17 colors. Then Kolya draws chords with endpoints at the marked points such that the endpoints of any chord are of the same color and the chords do not have any common points (including common endpoints). Kolya wants to draw as many chords as possi... | Note that $2006=17 \cdot 118$; therefore, there will be two colors in which at least $2 \cdot 118=236$ points are painted in total.
We will prove by induction on $k$ that through $2 k-1$ points of two colors, it is always possible to draw $k-1$ non-intersecting chords with same-colored endpoints.
Base case $(k=1,2)$ ... | 117 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berlov S.L.
A convex 2011-gon is drawn on the board. Petya sequentially draws diagonals in it such that each newly drawn diagonal intersects no more than one of the previously drawn diagonals at an internal point. What is the maximum number of diagonals Petya can draw? | We will prove that in a convex $n$-gon $A_{1} A_{2} \ldots A_{n}$, the maximum number of diagonals that can be drawn in the specified manner is $2 n-6$. Petya can sequentially draw the diagonals $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \ldots, A_{n-2} A_{n}$, and then the diagonals $A_{1} A_{3}, A_{1} A_{4}, A_{1} A_{... | 4016 | Geometry | 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? | We 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 |
$\underline{\text { Frankin B.R. }}$
Given a polynomial $P(x)$ with real coefficients. An infinite sequence of distinct natural numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that
$P\left(a_{1}\right)=0, P\left(a_{2}\right)=a_{1}, P\left(a_{3}\right)=a_{2}$, and so on. What degree can $P(x)$ have? | The constant clearly does not satisfy the condition.
For example, the polynomial $P(x)=x-1$ works.
Note that the leading coefficient of the polynomial $P$ is positive (otherwise, $P(x)x$ for each $x \in N$. Suppose the sequence exists. Starting from some index, the terms less than $N$ will end, that is, there will be... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zamyatin $B$.
For what least $n$ can a square $n \times n$ be cut into squares $40 \times 40$ and $49 \times 49$ so that squares of both types are present? | Note that for $n=2000=40 \cdot 49+40$ the required cutting exists (left figure).
Assume that there exists a square $n \times n$, where $n>2000$. Contradiction.
## Answer
$n=2000$ | 2000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhenodarov R.G.
Eight cells of one diagonal of a chessboard are called a fence. The rook moves around the board, not stepping on the same cell twice and not stepping on the cells of the fence (intermediate cells are not considered visited). What is the maximum number of jumps over the fence that the rook can make? | Evaluation. Note that if the rook jumps over a fence, then either the initial or the final cell of the jump is marked gray on the diagram. Since there are 24 gray cells and through each one a maximum of two jumps can be made, there can be no more than 48 jumps in total. At the same time, if there are exactly 48 jumps, ... | 47 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shirov V.
On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint o... | Let the marked points be $A_{1}, A_{2}, \ldots, A_{2 N}$ in the order of a clockwise traversal of the circle. We will prove by induction on $N$ that the number of even matchings is one more than the number of odd matchings. For $N=1$, the statement is obvious: there is only one matching, and it is even.
Inductive step... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogogn i.i.
On the shore of the round island Gdetotam, there are 20 villages, each inhabited by 20 wrestlers. A tournament was held where each wrestler faced all wrestlers from all other villages. Village A is considered stronger than village B if at least $k$ matches between wrestlers from these villages end with a v... | Evaluation. Let's show that when $k>290$, such a situation is impossible. We will order the wrestlers in each village by decreasing strength and select the tenth strongest wrestler in each village. We will show that the village where the weakest of the selected wrestlers lives cannot be stronger than the next one. Let'... | 290 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Trushin 5.
100 thimbles are arranged in a circle. Under one of them, a coin is hidden. In one move, it is allowed to flip over four thimbles and check if the coin is under one of them. After this, they are returned to their original position, and the coin is moved to one of the adjacent thimbles. What is the minimum nu... | After each of our moves and the movement of the coin, we will turn all the thimbles (along with the coin) one position clockwise. Then we can consider that after each move, the coin either stays in place or moves two positions clockwise.
Let's number all the thimbles counterclockwise in order from 0 to 99. It is clear... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
ABCDEF - a six-digit number. All digits are different and are arranged from left to right in ascending order. The number is a perfect square.
Determine what this number is.
# | The number ABCDEF is the square of a three-digit number.
## Solution
Let $\mathrm{ABCDEF}=\mathrm{XYZ}^{2}$. Clearly, $\mathrm{F}$ is the units digit of $\mathrm{Z}^{2}$. Since the digits A, B, C, D, E, F are in ascending order, $\mathrm{A} \leq 4$,
$\mathrm{E} \geq 5$, and $\mathrm{F} \geq 6$.
If $A=4$, then $\mat... | 134689 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Ribamko A.V.
In the lower left corner of a $n \times n$ chessboard, there is a knight. It is known that the minimum number of moves it takes for the knight to reach the upper right corner is equal to the minimum number of moves it takes to reach the lower right corner. Find $n$.
# | Let $n$ be even. In this case, the left lower field has the same color as the right upper one, while the right lower field has a different color. After each move, the knight lands on a field of the opposite color. Therefore, any path from the left lower corner to the right upper one consists of an even number of moves,... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sasha drew a square of size $6 \times 6$ cells and sequentially colors one cell at a time. After coloring a cell, he writes down the number of adjacent colored cells in it. After coloring the entire square, Sasha adds up the numbers written in all the cells. Prove that no matter in what order Sasha colors the cells, he... | Consider all unit segments that are common sides for two cells. There are exactly sixty such segments - 30 vertical and 30 horizontal. If a segment separates two shaded cells, we will say that it is "colored." Note that when Sasha writes a number in a cell, he indicates the number of segments that were not colored befo... | 60 | Combinatorics | proof | Yes | Yes | olympiads | false |
Berov s.L.
Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board? | Note that no two squares of natural numbers differ by 1: $x^{2}-y^{2}=(x-y)(x+y)$, and the second bracket is greater than one. Therefore, the numbers
$a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares.
The numbers $a b$ and $a(b+2)$ cannot both be squares; otherwise, their product $a^2 b(b+2)$ would also b... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Rubanov I.S.
For four different integers, all their pairwise sums and pairwise products were calculated.
The obtained sums and products were written on the board. What is the smallest number of different numbers that could end up on the board?
# | If we take the numbers $-1,0,1,2$, then, as is easy to check, each of the numbers written on the board will be equal to $-2,-1,0$, 1,2 or 3 - a total of 6 different values.
We will show that fewer than six different numbers on the board could not have occurred. Let the numbers taken be $ac+d$. If $ac+d$.
The second m... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4-$ $[\quad$ Evaluation + example $\quad]$
Car: Shapovosoov A.B.
On a strip of $1 \times N$, 25 checkers are placed on the leftmost 25 cells. A checker can move to the adjacent free cell to the right or jump over the adjacent checker to the right to the next cell (if this cell is free), movement to the left is not a... | From left to right, we number the cells and denote the checkers as $S_{1}, S_{2}, \ldots$
The checker $S_{25}$ cannot remain in place. Indeed, in this case, on the first move, $S_{24}$ will jump over it (there are no other moves) and will remain in place until the end (since at the end it must stand next to $S_{25}$).... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kuznetsov A.
In space, there are 2016 spheres, no two of which coincide. Some of the spheres are red, and the rest are green. Each point of contact between a red and a green sphere was painted blue. Find the maximum possible number of blue points. | Evaluation. Let there be $r$ red spheres and $2016-r$ green spheres. Since any two spheres touch at no more than one point, the number of blue points does not exceed $r(2016-r)=1008^{2}-(1008-r)^{2} \leq 1008^{2}$.
We will provide an example with this number of blue points. Let $l$ be some line, $\alpha$ a plane perpe... | 1016064 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
A chess piece can move 8 or 9 squares horizontally or vertically. It is forbidden to move to the same square twice.
What is the maximum number of squares this piece can visit on a $15 \times 15$ board? (The tour can start from any square.) | The central cell cannot be reached from any other cell. Cells in the central vertical (or horizontal) can only be reached from cells of this vertical. Conversely, starting from such a cell, the figure will not leave the central vertical, meaning it will cover no more than 14 cells.
The remaining $225-29=196$ cells can... | 196 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov $A$.
On each square of the chessboard, there is a rook at the beginning. In each move, you can remove a rook that attacks an odd number of rooks. What is the maximum number of rooks that can be removed? (Rooks attack each other if they stand on the same vertical or horizontal line and there are no other roo... | None of the rooks standing in the corner cells can be taken. Indeed, at the moment when the first of them is taken, it is attacked by exactly two rooks. Suppose it was possible to leave only the four corner rooks. Then the last rook taken was attacked by either two rooks (if it stood on the edge of the board) or none. ... | 59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Combinatorics (miscellaneous).]
$[$ Estimation + example $]$
In a pond, 30 pikes were released, which gradually eat each other. A pike is considered full if it has eaten at least three pikes (full or hungry). What is the maximum number of pikes that can become full? | The number of pikes eaten is not less than three times the number of satiated ones.
## Solution
Let $s$ be the number of satiated pikes. Then they together have eaten no less than $3 s$ pikes. Since each pike can only be eaten once, and at least one pike remains at the end, $3 s<30$. Therefore, $s \leq 9$.
We will p... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The number 123456789 is written on the board. From the written number, two adjacent digits are chosen, provided that neither of them is 0, 1 is subtracted from each digit, and the chosen digits are swapped (for example, from 123456789, one can get 123436789 in one operation). What is the smallest number that can be obt... | When performing each operation, the parity of each digit does not change.
## Solution
Notice that when performing each operation, the parity of the digit in each position does not change. Indeed, initially, we had the number 123456789, which is a number of the form OEOEOEOEO (O stands for an odd digit, and E stands f... | 101010101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For what value of the parameter $m$ is the sum of the squares of the roots of the equation $x^{2}-(m+1) x+m-1=0$ the smallest?
# | The sum of the squares of the roots $\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(m+1)^{2}-2(m-1)=m^{2}+3$ is minimal when $m=0$. Note that in this case, the equation
$x^{2}-x-1=0$ has roots.
## Answer
When $m=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the sum $+1+3+9+27+81+243+729$, any addends can be crossed out and the signs before some of the remaining numbers can be changed from "+" to "-". Masha wants to use this method to first get an expression whose value is 1, then, starting over, get an expression whose value is 2, then (starting over again) get 3, and ... | Every integer can be written as a sum of powers of three with coefficients 0, 1, and -1 instead of the usual 0, 1, and 2 (see problem 30840). Therefore, Masha will be able to get all integers from 1 to $1+3+9+27+81$ + $243+729=1093$
## Answer
Up to the number 1093 (inclusive). | 1093 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
Cheburashka has a set of 36 stones with masses of 1 g, 2 g, ..., 36 g, and Shapoklyak has super glue, one drop of which can glue two stones together (thus, three stones can be glued with two drops, and so on). Shapoklyak wants to glue the stones in such a way that Cheburashka cannot select one or several ... | Example. By gluing stones with masses 1 and 18, 2 and 17, ..., 9 and 10, respectively, Cheburashka will get a set where each stone weighs from 19 to 36 grams, so one stone will be too little, and two will be too many.
Another way. By gluing all the stones with odd masses in pairs, Cheburashka will get a set where all ... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.B.
There are 99 sticks with lengths $1,2,3, \ldots, 99$. Can a contour of some rectangle be formed from them? | For example, from sticks of lengths 1, 2, 3, we can form two sides of length 3, and the remaining matches can be divided into 48 pairs with a sum of lengths 103:
$\{4,99\},\{5,98\}, \ldots,\{51,52\}$ and form two sides of length $24 \cdot 103=2472$.
## Answer
It is possible. | 2472 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5,6 |
| :---: | :---: | :---: |
| | Classical combinatorics (miscellaneous). | |
| | Examples and counterexamples. Constructions | |
| | Estimation + example | |
Each of thirty sixth-graders has one pen, one pencil, and one ruler. After their participation in the Olympiad, it turned out that 26 students lost the... | From the condition, it follows that four sixth-graders have a pen, seven have a ruler, and nine have a pencil. Thus, at least one item can be owned by no more than $4+7+9=20$ people. Therefore, no fewer than $30-20=10$ people have lost all three items.
All three items will be lost by exactly 10 people if each of the o... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Does there exist such an $N$ and $N-1$ infinite arithmetic progressions with differences $2,3,4, \ldots, N$, such that every natural number belongs to at least one of these progressions? | Let $N=12$. Each natural number can be written in the form $12 k+r$, where $r$ is one of the numbers $0,1, \ldots, 11$. All numbers for which $r$ is even belong to the progression $2,4,6, \ldots$; all numbers for which $r$ is a multiple of $3$ belong to the progression $3,6,9, \ldots$ The remaining numbers have $r=1,5,... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Tolony A.K.
In a certain kingdom, there were 32 knights. Some of them were vassals of others (a vassal could have only one suzerain, and a suzerain was always richer than his vassal). A knight who had at least four vassals bore the title of baron. What is the maximum number of barons that could be under these conditio... | Evaluation. 8 barons should have 32 vassals, and the richest knight cannot be anyone's vassal. Example. Let 24 knights be vassals of six barons, and all these barons be vassals of the richest Baron. In total, 7 barons.
## Answer
7 barons. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Vassiyev H.E.
In each cell of an $8 \times 8$ square, one of the diagonals is drawn. Consider the union of these 64 diagonals. It consists of several connected components (a component includes points that can be reached from each other by traveling along one or more diagonals). Can the number of these components be
a... | b) In the figure, an example is given where the number of parts is 21. One of them is a multiply connected "window frame," in which single diagonals are placed in the cells.
## Answer
It c... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Petya wants to make an unusual die, which, as usual, should have the shape of a cube, with dots drawn on its faces (different numbers of dots on different faces), but at the same time, on any two adjacent faces, the number of dots must differ by at least two (at the same time, it is al... | Let's arrange six numbers (according to the number of dots on the faces of a cube) in ascending order. We will prove that among them, there are no three consecutive numbers. Suppose such three numbers exist: \(a\), \(a+1\), and \(a+2\). Then the numbers \(a\) and \(a+1\) must be on opposite faces of the cube. No matter... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
For what values of the parameter $a$ is the sum of the squares of the roots of the equation $x^{2}+2 a x+2 a^{2}+4 a+3=0$ the greatest? What is this sum? (The roots are considered with their multiplicity.)
# | The equation has roots when $D / 4=a^{2}-\left(2 a^{2}+4 a+3\right)=-\left(a^{2}+4 a+3\right) \geq 0$, that is, when $-3 \leq a \leq-1$. The sum of the squares of the roots is
$\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(2 a)^{2}-2\left(2 a^{2}+4 a+3\right)=-8 a-6$. This expression takes its maximum value at the smal... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Zamyatin $B$.
Vladimir wants to make a set of cubes of the same size and write one digit on each face of each cube so that he can use these cubes to form any 30-digit number. What is the smallest number of cubes he will need? (The digits 6 and 9 do not turn into each other when flipped.)
# | No less than 30 units, twos, ..., nines, and no less than 29 zeros. In total, no less than 50 cubes. It is not difficult to arrange no less than 30 instances of each digit on 50 cubes so that the digits on each cube do not repeat.
## Answer
50 cubes. | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3+ [ Examples and counterexamples. Constructions]
In a square table of size $100 \times 100$, some cells are shaded. Each shaded cell is the only shaded cell either in its column or in its row. What is the maximum number of cells that can be shaded
# | Example. We will color all the cells of one row and all the cells of one column, except for their common cell. In this case, the condition of the problem is satisfied and exactly 198 cells are colored.
Evaluation. For each colored cell, we will highlight the line (row or column) in which it is the only colored one. In... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berlov S.L.
A natural number $n$ is called good if every its natural divisor, increased by 1, is a divisor of the number $n+1$.
Find all good natural numbers. | Clearly, $n=1$ satisfies the condition. Also, all odd prime numbers satisfy it: the divisors of such a number $p$, increased by 1, are 2 and
$p+1$; both of them divide $p+1$.
On the other hand, any number $n$ that satisfies the condition has a divisor 1; hence, $n+1$ is divisible by 1 +1, which means $n$ is odd.
Sup... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evokimov M.A.
By the river, there lives a tribe of Mumbo-Jumbo. One day, with urgent news, a young warrior Mumbo and a wise shaman Yumbo set out simultaneously for the neighboring tribe. Mumbo ran at a speed of 11 km/h to the nearest raft storage and then sailed on a raft to the neighboring tribe. Meanwhile, Yumbo, wit... | Let the habitat of the Mumbo-Jumbo tribe be denoted by $O$, the storage that Mumbo ran to by $M$, and the storage that Jumbo went to by $U$. Clearly, $M$ is located upstream from $O$, and $U$ is downstream. Let the distances from $O$ to $M$ and $U$ be $x$ and $y$ km, respectively
$(x<y)$, and the speed of the river be... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6,7 $[\quad$ Examples and Counterexamples. Constructions
Authors: Shaovalov A.V., Raskina I.V.
From an equal number of squares with sides 1, 2, and 3, form a square of the smallest possible size.
# | Evaluation. Let the desired square be composed of $n$ squares of each type. Then its area is $n \cdot\left(1^{2}+2^{2}+\right.$ $\left.3^{2}\right)=14 n=2 \cdot 7 \cdot n$. Since the side length of the desired square must be an integer, the obtained number must be a perfect square. Therefore, the number $n$ must contai... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kalinin D.,.
What is the minimum number of $1 \times 1$ squares that need to be drawn to create an image of a $25 \times 25$ square divided into 625 $1 \times 1$ squares? | To draw the border of a square, it will be necessary to draw all 96 border squares. Let's divide the inner square $23 \times 23$ into 264 dominoes $1 \times 2$ and one square $1 \times 1$. One of the squares of each domino needs to be drawn (otherwise, the segment of the grid inside the domino will not be shown). In to... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the plane, $n$ lines are drawn such that every two intersect, but no four pass through the same point. There are a total of 16 intersection points, and through 6 of them, three lines pass. Find $n$.
# | "Move" the given construction in such a way that any two lines still intersect, but no three lines pass through the same point. Then, if some three lines intersected at a certain point $O$, now instead of one point $O$, there will be three points of pairwise intersection of these lines. Therefore, as a result of "movin... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
On the first horizontal line of the chessboard, there are 8 black queens, and on the last - 8 white queens. In what minimum number of moves can the white queens exchange places with the black ones? The white and black queens take turns, moving one queen per move. | The queen on the b1 or b8 square that moved first could not have moved to the opposite rank with that move, meaning these two queens together made at least three moves. Similarly, the pairs of queens on the c, ..., g files each made at least three moves. The first of the four corner queens to move did not land on a cor... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In a $10 \times 10$ square, all cells of the left upper $5 \times 5$ square are painted black, and the rest of the cells are white. Into what maximum number of polygons can this square be cut (along the cell boundaries) so that in each polygon, the number of black cells is three times less than the number... | In each polygon of the partition, there must be cells of both colors. This means that there must be a black cell adjacent to a white one. However, there are only 9 such cells. See the example of cutting into 9 polygons in the figure.
![](https://cdn.mathpix.com/cropped/2024_05_06_c07b2c4be0d3dcbcc4e5g-27.jpg?height=42... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Frankin B.R. }}$ 5.
A billiard table has the shape of a rectangle $2 \times 1$, with pockets located at the corners and at the midpoints of the longer sides. What is the smallest number of balls that need to be placed inside the rectangle so that each pocket lies on a line with some two balls? | Let there be only three balls. A line passing through a pair of balls inside a rectangle intersects the boundary of the rectangle exactly at two points. We have six pockets, so we need at least three lines. Three balls will give three lines only if these lines form a triangle. However, there are only seven lines passin... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Chernyaev N.L. }}$
Ten children were given 100 spaghetti noodles each in their plates. The children did not want to eat and started playing. In one move, one of the children transfers one noodle from their plate to each of the other children. After what minimum number of moves can all the children ... | Evaluation. Consider the difference between the number of macaroni pieces Petya has and the number of macaroni pieces Vasya has. Petya's action will reduce this difference by 10, Vasya's action will increase it by 10, and the actions of other children will not change it. Therefore, this difference will not be zero if a... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On the board, two-digit numbers are written. Each number is composite, but any two numbers are coprime.
What is the maximum number of numbers that can be written? | Evaluation. Since any two written numbers are coprime, each of the prime numbers 2, 3, 5, and 7 can appear in the factorization of no more than one of them. If there are five or more numbers on the board, then all prime factors in the factorization of one of them must be at least 11. But this is a composite number, so ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the expression $10: 9: 8: 7: 6: 5: 4: 3: 2: 1$, parentheses were placed such that the value of the expression is an integer.
What is the smallest number that could have been obtained?
# | In order for the value of the expression to be an integer, after placing the parentheses and writing the resulting expression as a common fraction, the number 7 must end up in the numerator. Therefore, the value of this expression is no less than 7. This can be achieved, for example, as follows:
$10: 9:(8: 7:(6:(5: 4:... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zaslavsky A.A.
In a volleyball tournament, each team met every other team once. Each match consisted of several sets - until one of the teams won three sets. If a match ended with a score of $3: 0$ or $3: 1$, the winning team received 3 points, and the losing team received 0. If the set score was
$3: 2$, the winner r... | Evaluation. If there are no more than three teams, then the "Simpletons" won all the matches, which means they have the most points. Contradiction.
If there are four or five teams, then each team will play three or four matches. This means that the "Cunning" won no more than one match and scored a maximum of $5=3+1+1$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In each cell of a $1000 \times 1000$ square, a number is inscribed such that in any rectangle of area $s$ that does not extend beyond the square and whose sides lie along the cell boundaries, the sum of the numbers is the same. For which $s$ will the numbers in all cells necessarily be the same? | It is clear that when $s=1$, the numbers in all cells are the same.
Let $s>1$ and $p$ be a prime divisor of $s$. In cells where the sum of the coordinates is divisible by $p$, we write ones, and in the other cells, we write zeros. In any rectangle $T$ of area $s$, one of the sides is divisible by $p$, so it is divided... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
M . B.
On some cells of a $10 \times 10$ board, there is a flea. Once a minute, the fleas jump simultaneously, and each one jumps to an adjacent cell (along a side). A flea jumps strictly in one of the four directions parallel to the sides of the board, maintains its direction as long as possible, and changes it to th... | Evaluation. On one vertical line, there can be no more than two fleas jumping along the vertical (otherwise, fleas located in cells of the same color will meet). The same is true for horizontals. In total, on 20 horizontals and verticals, there are no more than 40 fleas.
Example. It is clear that fleas from cells of d... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
|
Does there exist a quadrilateral $A B C D$ of area 1 such that for any point $O$ inside it, the area of at least one of the triangles $O A B, O B C, O C D, D O A$ is irrational. | For example, any trapezoid of area 1 with bases $BC=1$ and $AD=\sqrt[3]{2}$ will do. Indeed, the height of such a trapezoid is $\frac{2}{1+\sqrt[3]{2}}$. Suppose that for some point $O$ the areas of triangles $AOD$ and $BOC$ are rational. Then the heights of these triangles, dropped from point $O$, are $\alpha$ and $\f... | 16 | Geometry | proof | Yes | Yes | olympiads | false |
In a square with side 100, $N$ circles of radius 1 are placed, and any segment of length 10, entirely located inside the square, intersects at least one circle. Prove that $N \geq 400$.
Note from Problems.Ru: Open circles are considered, that is, circles without their bounding circumference. | Consider the figure F, consisting of a set of points that are no more than 1 unit away from a segment of length 10. We will divide the entire $100 \times 100$ square into 50 vertical strips, each 2 units wide, and place 8 non-overlapping figures, each equal to F, in each of these strips. It then follows from the condit... | 400 | Geometry | proof | Yes | Yes | olympiads | false |
[Dirichlet's Principle (finite number of points, lines, etc.)]
In a park, there are 10,000 trees planted in a square grid (100 rows of 100 trees each). What is the maximum number of trees that can be cut down so that the following condition is met: if you stand on any stump, you will not see any other stump? (The tree... | Let's divide the trees into 2500 quadruples, as shown in the figure. In each such quadruple, no more than one tree can be cut down. On the other hand, all the trees growing in the upper left corners of the squares formed by our quadruples of trees can be cut down. Therefore, the maximum number of trees that can be cut ... | 2500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Andjans A.
$N$ friends simultaneously learned $N$ pieces of news, with each person learning one piece of news. They started calling each other and exchanging news.
Each call lasts 1 hour. Any number of news items can be shared in one call.
What is the minimum number of hours required for everyone to learn all the news?... | a) A piece of news known to one of the friends will be known to no more than two (including the first) after 1 hour, no more than four after the second hour, ..., and no more than 32 after the fifth hour. Therefore, it will take no less than 6 hours.
We will show that 6 hours are sufficient. The conversations can proc... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a set of 100 numbers. If one number is removed, the median of the remaining numbers will be 78. If another number is removed, the median of the remaining numbers will be 66. Find the median of the entire set.
# | Let's arrange the numbers in ascending order. If we remove a number from the first half of the sequence (with a number up to 50), the median of the remaining numbers will be the 51st number in the sequence. If we remove a number from the second half, the median of the remaining numbers will be the 50th number, but it i... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow Metro from "Taganskaya" station to "Kievskaya" station, and in the evening - back (see diagram).
Upon e... | Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is $11 p + 17(1-p) = 17 - 6p$.
On the return journey from "Kievskaya" to "Taganskaya," the expected travel time is $17 p + 11(1-p) = 11 + 6p$.
According to the condition, $(11 +... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the class, there are fewer than 30 people. The probability that a randomly chosen girl is an excellent student is $3 / 13$, and the probability that a randomly chosen boy is an excellent student is $4 / 11$. How many excellent students are there in the class? | The probability that a randomly chosen girl is an excellent student is the ratio of the number of excellent girl students to the total number of girls in the class. Therefore, the number of girls is divisible by 13, meaning it is either 13 or 26.
Similarly, the number of boys is either 11 or 22. Considering that there... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.