problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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In the village, there are 100 houses. What is the maximum number of closed, non-intersecting fences that can be built so that each fence encloses at least one house and no two fences enclose the same set of houses?
# | In the maximum set of fences, there is a fence that limits exactly two houses. By combining these two houses into one, we reduce the number of houses by 1 and the number of fences by 2. In this process, both conditions of the problem are maintained. Continue this process. After 99 steps, one house and one fence will re... | 199 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10,11
During an interview, ten people were offered a test consisting of several questions. It is known that any five people together answered all the questions (that is, at least one of the five gave the correct answer to each question), but any four did not. What is the minimum number of questions for which this coul... | For each question, the number of people who did not answer it is no more than four.
## Solution
According to the condition, for every group of four people, there is a question they did not answer. On the other hand, the number of people who did not answer this question is no more than four. Therefore, the number of q... | 210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10,11 |
Let $x, y, z$ be positive numbers and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$. | Make the substitutions $x=p-a, y=p-b, z=p-c$.
## Solution
Let $a=y+z, b=x+z, c=x+y, p=x+y+z$. Consider a triangle with sides $a, b, c$ (the triangle inequalities are obviously satisfied). The perimeter of this triangle is $2 p$, and let the area be denoted by $S$. By Heron's formula,
$S^{2}=p(p-a)(p-b)(p-c)=(x+y+z) ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
The Tsar summoned two sages. He gave the first 100 blank cards and ordered him to write a positive number on each (the numbers do not have to be different), without showing them to the second. Then the first can tell the second several different numbers, each of which is either written on some card or eq... | Let's prove that it is sufficient to lose 101 hairs for each sage. Let the first write the numbers $1, 2, 4, \ldots, 2^{99}$ on the cards, and inform the second of these numbers and their sum. When the second hears the number 1, he will understand that there is a card not exceeding 1. Hearing the next number $2^{k}$, h... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Yatsenko I.V.
Vanya is coming up with a number consisting of non-repeating digits without zeros - a password for his phone. The password works as follows: if, without lifting his finger from the screen, he sequentially connects the points corresponding to the digits of the password with lines, the phone will unlock. H... | The password 12769 meets Vanya's requirements. Let's see how we can connect its digits without any intersections. The digit 7 must be connected to either 2 or 6. Suppose we draw the segment 7-6. Now, 9 can only be connected to 6. Next, we inevitably need to draw the segments 7-2 and 2-1, and we get the line shown in th... | 12769 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.i.
Initially, there are 111 pieces of plasticine of the same mass on the table. In one operation, you can choose several groups (possibly one) with the same number of pieces and in each group, combine all the plasticine into one piece. What is the minimum number of operations required to get exactly 11 piec... | Let the mass of one original piece be 1. If in the first operation in each group there are $k$ pieces, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to get 11 pieces of different masses in one operation.
We will show that the required result can be achieved in two operations. For t... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Counting in two ways ] $[\quad$ Trees $]$
A travel agency ran a promotion: "Buy a trip to Egypt, bring four friends who also buy a trip, and get the cost of your trip back." During the promotion, 13 buyers came on their own, and the rest were brought by friends. Some of them brought exactly four new customers, while... | Each of the $x$ "lucky ones" brought 4 friends. Therefore, the number of "brought" customers is $4x$, and 13 more came on their own, so the total number of tourists was $13 + 4x$. On the other hand, $x$ people brought new customers, while 100 did not, meaning the total number of tourists was $x + 100$. Thus, $13 + 4x =... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Case enumeration $\quad]$ [ Classical inequalities (other) ]
Find all numbers that are 12 times the sum of their digits.
# | Let the desired number contain $n$ digits. Then it is not less than $10^{n-1}$ (this is the smallest $n$-digit number). On the other hand, the sum of its digits is not more than $9 n$ (since each digit is not more than 9). Therefore, the desired number does not exceed $12 \cdot 9 n=108 n$. But according to Bernoulli's ... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.B.
A boy and a girl were sitting on a long bench. Twenty more children approached them one by one, and each of them sat between two of the already seated children. We will call a girl brave if she sat between two neighboring boys, and a boy brave if he sat between two neighboring girls. When everyone was sea... | The first method. Let's look at the number of pairs of adjacent boys and girls. Initially, it is equal to 1. Notice that if a boy sits between two boys, the number of such pairs does not change. If he sits between a boy and a girl, he "destroys" one such pair and "creates" one, so the number of such pairs does not chan... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Given 12 sticks of the same length. How can you cut them into smaller sticks so that you can form 13 equal triangles, with each of the smaller sticks being a side of one of these triangles?
# | We need to form triangles with sides in the ratio of $3: 4: 5$.
## Solution
Let's assume the length of each stick is 13. We will cut three sticks into segments of length $3,3,3,4$; four sticks into segments of length 3, 5, 5; and five sticks into segments of length 4, 4, 5. As a result, we will have 13 sticks of leng... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the TV series "The Secret of Santa Barbara", 20 characters are involved. In each episode, one of the following events occurs: a certain character learns the Secret, a certain character learns that someone knows the Secret, a certain character learns that someone does not know the Secret. What is the maximum number o... | Calculate the maximum number of times each of the three types of events specified in the condition can occur.
## Solution
Each hero can learn something in no more than 39 episodes (one episode to learn the Secret, 19 episodes to learn that each of the others does not know the Secret, and another 19 episodes to learn ... | 780 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
For what value of $a$ does the polynomial $P(x)=x^{1000}+a x^{2}+9$ divide by $x+1$?
# | $P(-1)=1+a+9=a+10$. By the theorem of Bezout, this number should equal zero.
## Answer
For $a=-10$. | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The principle of the extreme (etc.). $\quad]$ Symmetry helps solve the task_ $\quad]$
There are thirty cards, each with a number: on ten cards - $a$, on ten others - $b$, and on the remaining ten - $c$ (the numbers $a, b, c$ are all different). It is known that for any five cards, it is possible to find another five s... | Let $a<b<c$. We will mark on the number line all possible sums of the numbers on five cards. For each of them, the opposite is also marked, so the marked points are symmetrically located relative to zero. In particular, the largest (5c) and the smallest (5a) sums are opposite, so $5 a+5 c=0$, which means $c=-a$. The su... | 0 | Combinatorics | proof | Yes | Yes | olympiads | false |
Bakayev E.V.
Cells of a $5 \times 7$ table are filled with numbers such that in each $2 \times 3$ rectangle (vertical or horizontal), the sum of the numbers is zero. By paying 100 rubles, you can choose any cell and find out what number is written in it. What is the smallest number of rubles needed to definitely deter... | Let the central cell of the table contain the number $a$. Then the sum $S$ of all the numbers in the table is $-a$, since the sum $S+a$ can be broken down into six zero sums, each with six terms, as shown in the figure.
Therefore, it is sufficient to know one number - the central one.
. Each accountant took a copy of the list and found an approximate sum of the expenses, acting as follows. First, he arbitrarily chose two numbers from... | Evaluation. Accountants each time calculated the integer part of the sum of some two numbers, which is equal to the sum of their integer parts plus, possibly, one. Let's call this addition of one an incident. It could only happen when both addends were non-integers. Each accountant ended up with the sum of the integer ... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
An abstract artist took a wooden cube $5 \times 5 \times 5$, divided each face into unit squares, and painted each of them in one of three colors - black, white, or red - such that no two adjacent squares (sharing a side) are of the same color. What is the minimum number of black squares that could resu... | Evaluation. Three squares at the vertex of a cube form a cycle of adjacent squares of length 3. Around it, another cycle of length 9 is formed from adjacent cells. And around that, a cycle of length 15. Taking three such cycles around two opposite vertices of the cube, and two small cycles around the remaining vertices... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Podlipsky o.k.
Oleg drew an empty $50 \times 50$ table and wrote a non-zero number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbe... | Evaluation. The fact that there are no more than 1250 rational numbers in the table is proven in the same way as in problem 66016 (since the product of a non-zero rational and an irrational number is irrational).
Example, when there are exactly 1250 rational numbers. Place along the left side the numbers $1,2, \ldots,... | 1250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Podlipsky O.K.
Oleg drew an empty $50 \times 50$ table and wrote a number above each column and to the left of each row. It turned out that all 100 written numbers are distinct, with 50 of them being rational and the other 50 being irrational. Then, in each cell of the table, he wrote the product of the numbers writte... | Evaluation. Suppose that among the rational numbers there is 0 and it is written at the top side of the table. Let along the left side of the table be written $x$ irrational and $50-x$ rational numbers. Then along the top side are written $50-x$ irrational and $x$ rational numbers. Note that the product of a non-zero r... | 1275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
In each cell of a $5 \times 5$ board, there is either a cross or a zero, and no three crosses stand in a row either horizontally, vertically, or diagonally. What is the maximum number of crosses that can be on the board? | Example of placing 16 crosses according to the condition, see the figure on the left.
Evaluation. Divide the board into the central cell and 8 rectangles of size $3 \times 1$ (see the figure... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.v.
10 children of different heights are standing in a circle. From time to time, one of them runs to another place (between some two children). The children want to stand in order of increasing height as quickly as possible, in a clockwise direction (from the shortest to the tallest). What is the minimum num... | Number the children in ascending order of height $-1, 2, \ldots, 10$.
Estimation. Suppose they initially stood in reverse order.
First method. If there were fewer than eight sprints, then some three children remained in their places, and their order is opposite to the required one.
Second method. Let's call the comp... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
The weight of each weight in the set is a non-integer number of grams. They can balance any integer weight from 1 g to 40 g (weights are placed on one pan of the scales, the weight to be measured - on the other). What is the smallest number of weights in such a set? | Example 1. Let's take weights of 1, 1, 3, 5, 11, 21, 43 g. The first two can measure any integer weight up to 2 g. Therefore, the first three can measure up to 5 g, the first four up to 10 g, the first five up to 21 g, the first six up to 42 g, and all seven up to 85 g. If we reduce the weight of each weight by half, a... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given the numbers $1,2,3, \ldots, 1000$. Find the largest number $m$, such that: no matter which $m$ of these numbers are erased, among the remaining $1000-m$ numbers, there will be two such that one divides the other. | If $m \geq 500$, then, by erasing the first $m$ numbers (from 1 to $m$), we will leave the numbers from $m+1 \geq 501$ to 1000, among which none, obviously, divides another (all pairwise ratios are less than two).
The number $m=499$ has the required property. Indeed, if we take 501 numbers from the set of 1 to 1000, t... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
The periods of two sequences are 7 and 13. What is the maximum length of the initial segment that can coincide? | Example. Consider the sequence with period 7 and initial terms $0,0,0,0,0,1,0$ and the sequence with period 13 and initial terms $0,0,0,0,0,1,0,0,0,0,0,0,1$. They have a common initial segment $0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0$ of length 18.
Estimate. If there is a common initial segment of length 19, then $a_{7}=a... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
The cells of a chessboard are numbered from 1 to 64 in such a way that adjacent numbers are in adjacent (by side) cells.
What is the smallest possible sum of the numbers on the diagonal?
# | Evaluation. Consider the numbers $a_{1}<a_{2}<\ldots<a_{8}$, standing on the black diagonal (of course, on the diagonal they may not be in order). Then $a_{1} \geq 1, a_{2} \geq 3, a_{3} \geq 5, \ldots, a_{7} \geq 13$. We will prove that $a_{8} \geq 39$.
Assume that the numbers were written into the cells of the board... | 88 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The hostess made a pirog (a type of Russian pie) and wants to pre-cut it into such (not necessarily equal) pieces so that the pie can be evenly divided both among five and seven people. What is the minimum number of pieces she can manage with?
# | This task is a particular case of problem $\underline{98057 .}$
## Answer
11 pieces. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
P.M.
In the country, there are 15 cities, some of which are connected by air routes belonging to three airlines. It is known that even if any one of the airlines ceases operations, it will still be possible to travel from any city to any other (possibly with layovers), using flights of the remaining two airlines. What... | Let the number of routes of the airlines be denoted by $a, b$, and $c$. If we close the last $c$ air routes, the graph will remain connected, so $a+b \geq 14$ (see problem $\underline{31098}$ g). Similarly, $b+c \geq 14, c+a \geq 14$. Adding these inequalities, we get: $2(a+b+c) \geq 42$, which means the three companie... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The value of $a$ is chosen so that the number of roots of the first equation $4^{x}-4^{-x}=2 \cos a x, 4^{x}+4^{-x}=2 \cos a x+4$ is 2007.
How many roots does the second equation have for the same $a$? | Transform the second equation:
$$
\left.4^{x}+4^{-x}=2 \cos a x+4 \Leftrightarrow 4^{x}-2+4^{-x}=2(1+\cos a x) \Leftrightarrow\left(2^{x}-2^{-x}\right)^{2}=4 \cos ^{2 a x} / 2\right) \Leftrightarrow
$$
$$
\left[\begin{array} { l }
{ 4 ^ { x / 2 } - 4 ^ { - x / 2 } = 2 \cos \frac { a x } { 2 } , } \\
{ 4 ^ { x / 2 } ... | 4014 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the minimum number of weights needed to be able to weigh any number of grams from 1 to 100 on a balance scale, if the weights can be placed on either pan of the scale?
# | When solving this problem, we need the following interesting property of the ternary numeral system: any natural number can be represented as the difference of two numbers, the ternary representation of which contains only 0 and 1.
To prove this, we need to write the original number in ternary notation and construct t... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Ionin Yu.I.
a) Does there exist an infinite sequence of natural numbers with the following property: no number in the sequence divides another, but among any three numbers, one can choose two whose sum is divisible by the third?
b) If not, how many numbers can be in a set with such a property?
c) Solve the same prob... | b) To the sequence of four numbers $3,5,7,107$ given in the condition, we can add the fifth number 10693: 10693 + 5 is divisible by $3,10693+3$ is divisible by $7,10693+7$ is divisible by $5,10693+107$ is divisible by 3 and by $5,10693+7$ is divisible by 107.
We will show that it is impossible to form a sequence satis... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Konyagin S.V.
In a family album, there are ten photographs. In each of them, three people are depicted: a man stands in the center, to the left of the man is his son, and to the right is his brother. What is the smallest number of different people that can be depicted in these photographs, given that all ten men stand... | Let's call the ten men standing in the center of the photographs the main faces. We will divide all the men in the photographs into levels. We will assign level 0 to those who do not have fathers in the photographs, and level $k+1 (k=0,1, 2, \ldots)$ to men who have fathers at level $k$. Let $r_{k}$ denote the number o... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Hrramcovd
Find the largest natural number $N$, for which in any arrangement of different natural numbers from 1 to 400 in the cells of a $20 \times 20$ square table, there will be two numbers in the same row or column whose difference is at least $N$.
# | Example. Let's divide the table into two $20 \times 10$ rectangles vertically. In the first rectangle, we will place numbers from 1 to 200 in increasing order by rows (from 1 to 10 in the first row, from 11 to 20 in the second row, and so on). In the second rectangle, we will place numbers from 201 to 400 in the same m... | 209 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Authors: Bogdanov I.I., Chennokov G.R.
In a rectangular table with 9 rows and 2004 columns, the numbers from 1 to 2004 are arranged, each appearing 9 times. In each column, the numbers differ by no more than 3. Find the minimum possible sum of the numbers in the first row. | Evaluation. By rearranging the columns if necessary, we will henceforth assume that the numbers in the first row are in non-decreasing order. Let $a_{i}$ be the $i$-th number in the first row. Consider the sum $S=\left(a_{1}-1\right)+\left(a_{2}-1\right)+\left(a_{3}-1\right)+$ $\left(a_{4}-2\right)+\ldots+\left(a_{i}-(... | 2005004 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ CaseAnalysis ] $[$ Divisibility Rules (etc.) $]$ Author: Fomin S.B. A natural number $n$ is written in the decimal system. It is known that if a digit appears in this representation, then $n$ is divisible by this digit (0 does not appear in the representation). What is the maximum number of different digits that t... | If the digit 5 is included in the representation of a number, then the number must end in 5. Therefore, it is odd and, consequently, contains only odd digits. Thus, it cannot have more than five digits. If 5 does not appear in the decimal representation of the number, then it can include all other 8 digits. For example... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Petya liked the puzzle, he decided to glue it together and hang it on the wall. In one minute, he glued together two pieces (initial or previously glued). As a result, the entire puzzle was assembled into one complete picture in 2 hours. How long would it take to assemble the picture if Petya glued tog... | Each gluing reduces the number of pieces on the table by 1. Since after 120 gluings, one piece (the complete puzzle) was obtained, there were 121 pieces before the work began. Now, if gluing three pieces together per minute (i.e., reducing the number of pieces by 2), one piece will remain after 60 minutes.
## Oтвет
I... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In some cells of a $10 \times 10$ table, several crosses and several zeros are placed. It is known that there is no line (row or column) completely filled with the same symbols (crosses or zeros).
However, if any symbol is placed in any empty cell, this condition will be violated. What is the minimum nu... | Let us assume that we have filled the table according to the condition, and cell $A$ is free. Since placing any symbol in it should result in a line of identical symbols, there must be a line containing it where all other cells are filled with crosses, and the same is true for noughts (these lines must, of course, be a... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Theory of algorithms (other) }) \\ {[\underline{\text { Estimation + example }}]}\end{array}\right.$
You need to find out a five-digit phone number by asking questions that can be answered with "yes" or "no." What is the minimum number of questions required to guarantee finding the numbe... | We need to ask questions in such a way that each subsequent question approximately halves the number of remaining possible options.
## Solution
Initially, when we know nothing about the phone number, there are $10^{5}=100000$ possible options for the phone number. We ask the question: "Is the number greater than 5000... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems ] $[$ Case enumeration ]
In a box, there are blue, red, and green pencils. In total, there are 20. The number of blue pencils is 6 times the number of green pencils, and the number of red pencils is less than the number of blue pencils. How many red pencils are in the box? | Think about how many blue pencils there can be.
## Solution
Since there are 20 pencils in total, and blue and green pencils together make up 7 parts. This means there can be 6 or 12 blue pencils, and green and red pencils would then be 1 and 13 or 2 and 6, respectively. Since there are fewer red pencils than blue one... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
On a circular road, there are four gas stations: $A, B, C$, and $D$. The distance between $A$ and $B$ is 50 km, between $A$ and $C$ is 40 km, between $C$ and $D$ is 25 km, and between $D$ and $A$ is 35 km (all distances are measured along the circular road in the shortest direction).
a) Provide an exa... | First, determine the arrangement of gas stations $A, C$, and $D$.
## Solution
The problem provides all three distances between $A, C$, and $D$. First, let's determine the arrangement of these three gas stations. Gas stations $A$ and $C$ divide the circular road into two arcs. If gas station $D$ were on the shorter ar... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5,6,7,8 | |
A four-digit number starts with the digit 6. This digit was moved to the end of the number. The resulting number turned out to be 1152 less than the original. Find the original number. | Let's write the condition of the problem in the form of a rebus, for example:
## $-6 A B C$ $\frac{1.52}{1152}$
Then C $=8$ and the example takes the form:
Therefore, $B=3$ and the example has the following form:
## $-6 \mathrm{~A}^{438}$ $-{ }_{A 38}^{10}$
Since a carry-over has occurred, then $A=5$ :
$-6538$
... | 6538 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
\left[\begin{array}{l}{[\text { Decimal numeral system }]} \\ {[\text { Case enumeration }}\end{array}\right]
Author: Raskina I.V.
The year 2009 has the following property: by rearranging the digits of the number 2009, it is impossible to obtain a smaller four-digit number (numbers do not start with zero). In which y... | Author: Raschina I.V.
In the years $2010, 2011, \ldots, 2019$ and in 2021, the number of the year contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be reduced by rearranging the digits.
## Answer
in 2... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
There are 6 locked suitcases and 6 keys to them. However, it is unknown which key fits which suitcase. What is the minimum number of attempts needed to definitely open all the suitcases? And how many attempts would be needed if there were not 6, but 10 suitcases and keys?
# | Try to determine which of the 6 suitcases the first key fits in five attempts.
## Solution
Standard incorrect solution: "Each of the six suitcases is tried with each of the six keys, totaling $66=36$ attempts." However, it is possible to find the correspondence between the keys and the suitcases with fewer attempts. ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Cube
Can the numbers from 1 to 12 be placed on the edges of a cube so that the sums of the numbers on all faces are the same?
# | On the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, we will place numbers as follows:
$A B-10, B C-5, C D-7, A D-4$,
$A_{1} B_{1}-3, B_{1} C_{1}-9, C_{1} D_{1}-6, A_{1} D_{1}-8$
$A A_{1}-2, B B_{1}-11, C C_{1}-1, D D_{1}-12$
On each face of the cube, the sum of the numbers is 26.
## Answer
It is possible. | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Pairing and grouping; bijections $]$ [ Decimal number system ]
Find the last digit of the number $1^{2}+2^{2}+\ldots+99^{2}$. | $1^{2}+2^{2}+\ldots+99^{2} \equiv 10\left(1^{2}+2^{2}+\ldots+9^{2}\right)=0(\bmod 10)$.
## Answer
0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
There were seven boxes. In some of them, seven more boxes (not nested within each other) were placed, and so on. In the end, there were 10 non-empty boxes.
How many boxes are there in total? | At each operation, one empty box is filled. Since there are now 10 non-empty boxes, 10 operations have been performed. With each operation, seven boxes were added. Therefore, in the end, there are $7 + 10 \cdot 7 = 77$ boxes.
The Russian national football team won against the Tunisian national team with a score of $9:... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Investigation of a quadratic trinomial ]
A quadratic trinomial $y=a x^{2}+b x+c$ has no roots and $a+b+c>0$. Determine the sign of the coefficient $c$.
# | 
The quadratic polynomial has no roots, which means its graph does not intersect the x-axis. Since \( y(1) = a + b + c > 0 \), the graph is located in the upper half-plane (see the figure), th... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Nazarov } \Phi}$.
Positive numbers $a, b, c, d$ are such that $a \leq b \leq c \leq d$ and $a+b+c+d \geq 1$. Prove that $a^{2}+3 b^{2}+5 c^{2}+7 d^{2} \geq 1$.
# | $a^{2}+3 b^{2}+5 c^{2}+7 d^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2\left(b^{2}+2 c^{2}+3 d^{2}\right) \geq a^{2}+b^{2}+c^{2}+d^{2}+2(a b+(a+b) c+(a+b+c) d)=(a+b+c+d)^{2} \geq 1$
$$
\begin{aligned} &\({\left[\begin{array}{l}\text { Decimal number system } \\ \text { Problem } \underline{98333} \text { Topics: }\end{array}\right] ... | 399 | Inequalities | proof | Yes | Yes | olympiads | false |
7,8,9 |
What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other
# | If in a square of four cells there are two kings, then they attack each other.
## Solution
We will divide the board into 16 squares $2 \times 2$.
Estimate. In each of these 16 squares, there can be no more than one king.
Example. Place a king in the lower left corner of each of the 16 squares.
## Answer
16 kings. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.v.
There are 100 numbers. When each of them was increased by 1, the sum of their squares did not change. Each number was increased by 1 again.
Will the sum of the squares change this time, and if so, by how much? | $(a+2)^{2}-(a+1)^{2}=(a+1)^{2}-a^{2}+2$. Therefore, the second change is 2$\cdot$100 greater than the first.
## Answer
It will increase by 200.
Problem | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
In a circle, there are boys and girls (both are present), a total of 20 children. It is known that for each boy, the neighbor in the clockwise direction is a child in a blue T-shirt, and for each girl, the neighbor in the counterclockwise direction is a child in a red T-shirt. Can the number of boys in th... | The MHD (clockwise) arrangement cannot be due to the color of child X's T-shirt. Therefore, one boy should stand clockwise from the boy, one boy from him, and so on. This means that there are no fewer than half of all the children in the circle who are boys. By similar considerations, there are no fewer than half of th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V. A sequence of several natural numbers is written, with a sum of 20. No number and no sum of several consecutive numbers equals 3. Could there be more than 10 numbers written? | Example with 11 numbers: $1,1,4,1,1,4,1,1,4,1,1$.
## Answer
It could.
Send a comment | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.B.
In each cell of a strip of length 100, there is a chip. For 1 ruble, you can swap any two adjacent chips, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order? | Evaluation. Each chip must change the parity of its number. A free operation does not change parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Algorithm. We will number the chips in order from 0 to 99. We will color the cells in four colors: $a b c d a b c ... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.B.
What is the maximum number of different integers that can be written in a row so that the sum of any 11 consecutive numbers is 100 or 101? | Evaluation. Suppose we have managed to list such distinct numbers $x_{1}, ..., x_{23}$, that the sum of any 11 consecutive numbers equals $A$ or $B$ (otherwise $x_k = x_{k+11}$). Thus, $S_k = S_{k+2}$. Since $x_{1} + S_{2} + S_{13} = S_{1} + S_{12} + x_{23}$, then $x_{1}= x_{23}$. Contradiction.
Example. Let's choose ... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Authors: Gooovanov A.S., Sokolov A.
Sasha writes down the numbers 1, 2, 3, 4, 5 in some order, places arithmetic operation signs "\$\$\$", "\$-\$", "\$|times\$", and parentheses, and looks at the result of the obtained expression. For example, he can get the number 8 using the expression $\$(4-3) \backslash t i m e s(... | For example, $\$ 3 \backslash$ times $(2 \backslash$ times 4 \times $5+1)=123 . \$$
## Answer
Yes, it can. | 123 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9,10
In a chess tournament, 12 people participated. After the tournament, each participant made 12 lists. The first list includes only themselves, the second list includes them and those they defeated, the third list includes everyone from the second list and those they defeated, and so on. The 12th list includes ever... | Answer: 54. If the $(k+1)$-th list is the same as the $k$-th, then the lists numbered $k+2, \ldots, 11, 12$ will also be exactly the same. But according to the condition, the 11-th list and the 12-th are different. Therefore, each participant's $k$-th list contains exactly $k$ people. In particular, the 2-nd list conta... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10,11
The mass of each of the 19 weights does not exceed 70 grams and is an integer number of grams. Prove that it is impossible to form more than 1230 different mass sets from these weights.
# | Let's divide all sets into two parts: those with no more than seventeen weights and all the others. The mass of each set from the first part does not exceed $17 \cdot 70=1190$, so among them, there are no more than 1190 different by mass. On the other hand, the number of sets in the second part is $19+1=20$. Therefore,... | 1210 | Combinatorics | proof | Yes | Yes | olympiads | false |
Electrician I.V. was called to repair a string of four lamps connected in series, one of which had burned out. It takes 10 seconds to unscrew any lamp from the string, and 10 seconds to screw it back in. The time spent on other actions is negligible. What is the minimum time the electrician can definitely find the burn... | If after replacing one bulb the garland does not light up, then we replaced a working bulb.
## Solution
Suppose we did not replace some two bulbs. Then, if one of them is burnt out, we will not be able to determine which one. Therefore, to definitively identify the burnt-out bulb, we need to unscrew at least three of... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Fedorov R.M.
In the Banana Republic, parliamentary elections were held in which all residents participated. All those who voted for the "Mandarin" party love mandarins. Among those who voted for other parties, 90% do not like mandarins. What percentage of the votes did the "Mandarin" party receive in the elections, if... | Find in two ways how many people love tangerines.
## Solution
Let the entire population of the republic be $-N$ people, of which $M$ people voted for "Tangerine." Then, on one hand, tangerines are loved by $0.46 N$ people, and on the other hand, this number is equal to the number of people who voted for "Tangerine" p... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ [Reverse Process ]
Lёnya thought of a number. He added 5 to it, then divided the sum by 3, multiplied the result by 4, subtracted 6, divided by 7, and got 2. What number did Lёnya think of?
Try to form an equation to determine the unknown number.
# | Let's denote the number thought of by Lёna as $x$. Then we can form the equation
$$
\{[((x+5): 3)-4]-6\}: 7=2
$$
By sequentially moving all numbers from the left side to the right, we get a new equation
$$
x=\{[((27)+6): 4] 3\}-5
$$
from which it is easy to determine that $x=10$. From this, it is also clear that to... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the lake, lotuses grow. Over the course of a day, each lotus splits in half, and instead of one lotus, two appear. After another day, each of the resulting lotuses splits in half, and so on. After 30 days, the lake is completely covered with lotuses. How long did it take for the lake to be half full?
# | Note that the number of lotus flowers doubles in one day.
## Solution
If you read the condition of the problem carefully, you will understand that the lake was half full after 29 days. The day before the lake is completely filled, it will be exactly half full.
## Answer
After 29 days | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bootin D.A.
Find in the sequence $2,6,12,20,30, \ldots$ the number standing a) at the 6th; b) at the 1994th position. Explain your answer. | $2=1 \cdot 2 ; 6=2 \cdot 3$
## Solution
We can notice that $2=1 \cdot 2, 6=2 \cdot 3, 12=3 \cdot 4$, and assume that the $n$-th term of the sequence is $n \cdot (n+1)$. Checking on the 4th ($20=4 \cdot 5$) and 5th ($30=5 \cdot 6$) terms of the sequence shows that we are correct. Therefore, the number at the 6th posit... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Reverse Process]
A peasant, buying goods, paid the first merchant half of his money and another 1 ruble; then he paid the second merchant half of the remaining money and another 2 rubles, and finally, he paid the third merchant half of what was left and another 1 ruble. After this, the peasant had no money left. How ... | Before coming to the third merchant, the peasant had 1 ruble and as much again - a total of 2 rubles. Before coming to the second merchant, he had $2+2$ = 4 rubles and as much again - a total of 8 rubles. Finally, before coming to the first merchant, the peasant had $8+1=9$ and as much again; so initially, the peasant ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (continued).]
In a bag, there are balls of two different colors: black and white. What is the smallest number of balls that need to be taken out of the bag blindly so that among them there are definitely two balls of the same color?
# | We need to draw three balls in total, so the balls are the "rabbits," and the colors are the "cages." Since there are fewer cages than rabbits, by the Pigeonhole Principle, there will be a cage with at least two rabbits. That is, two balls of the same color. It is easy to notice that by drawing two balls, we might get ... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
a) In the store "Everything for Tea," there are 5 different cups and 3 different saucers. How many ways can you buy a cup and a saucer?
b) The store also has 4 teaspoons. How many ways can you buy a set consisting of a cup, a saucer, and a teaspoon?
c) The store still sells 5 cups, 3 saucers, and 4 teaspoons. How man... | a) Let's choose a cup. To it, we can choose any of the three saucers. Therefore, there are 3 different sets containing the chosen cup. Since there are 5 cups in total, the number of different sets is 5$\cdot$3 = 15.
b) Any of the 15 sets from part a) can be complemented with a spoon in 4 ways. Therefore, the total numb... | 47 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Three merchants' wives - Sosipatra Titovna, Olympiada Karpovna, and Polikseina Uvarovna - sat down to drink tea. Olympiada Karpovna and Sosipatra Titovna drank 11 cups together, Polikseina Uvarovna and Olympiada Karpovna - 15, and Sosipatra Titovna and Polikseina Uvarovna - 14. How many cups of tea did all three mercha... | Notice, the cup drunk by each merchant's wife was mentioned twice in the problem's condition.
## Solution
|If we add up all the cups accounted for, we get twice the sum of the cups drunk. Therefore, we need to divide this sum by two.
## Answer
20 cups. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Sharygin I.F.
A rectangle is composed of six squares (see the right figure). Find the side of the largest square if the side of the smallest one is 1.
The side of the largest square is equ... | Notice that the side of the largest square is equal to the sum of the sides of two squares: the one following it clockwise and the smallest one. Denoting the side of the largest square as \( x \), we can sequentially express the sides of the other squares: \( x-1, x-2, x-3, x-3 \) (see the figure). Now notice that the ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Akulich I.F. }}$.
Avant-garde artist Zmey Kletochkin painted several cells on a $7 \times 7$ board, following the rule: each subsequent painted cell must share a side with the previously painted cell, but must not share a side with any other previously painted cell. He managed to paint 31 cells.
B... | Do not try to completely color the sides, leave the middle of the sides uncolored.
## Solution
a, b) If we can color 33 cells, then 32 cells can be colored by stopping in time. Three examples where 33 cells are colored are shown in the figure (in fact, there are many more such examples). It is impossible to color mor... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Baranov d..V.
The hare bought seven drums of different sizes and seven pairs of sticks of different lengths for her seven baby hares. If a baby hare sees that both its drum is larger and its sticks are longer than those of one of its brothers, it starts to drum loudly. What is the maximum number of baby hares that can... | Not all the bunnies can play the drum, as the baby bunny that gets the smallest drum will not play it. On the other hand, if the same baby bunny is also given the shortest drumsticks, all the other bunnies will play the drum.
## Answer
6 bunnies. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The numbers $1, 2^{1}, 2^{2}, 2^{3}, 2^{4}, 2^{5}$ are written on the board. It is allowed to erase any two numbers and replace them with their difference - a non-negative number.
Can the number 15 be the only number left on the board after several such operations? | The required sequence of operations is evident from the following record: $15=32-16-(8-4-2-1)$.
## Answer
It can. | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (continued).]
A store received 25 boxes of three different types of apples (each box contains apples of only one type). Prove that among them, there are at least 9 boxes of apples of the same type.
# | 25 boxes - "rabbits" will be distributed among 3 cells-sorts. Since $25=3 \cdot 8+1$, we apply the "generalized pigeonhole principle" for $N=3, k=8$ and obtain that in some cell-sort there will be no less than 9 boxes. | 9 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Pigeonhole Principle (continued).]
What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other
# | Answer: 16 kings. Divide the board into 16 squares, in each of which there can be no more than one king. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7,8
What is the maximum number of rooks that can be placed on an 8x8 chessboard so that they do not attack each other
# | Obviously, 8 rooks can be placed, for example, along the diagonal from a1 to h8. Let's prove that it is impossible to place 9 rooks that do not attack each other.
On one horizontal row, there cannot be more than one rook - otherwise, they would attack each other; therefore, the number of rooks that can be placed canno... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[
Several points were marked on a line. After that, a point was added between each pair of adjacent points. This operation was repeated three times, and as a result, there were 65 points on the line. How many points were there initially?
# | If there were $n$ points on a line, then in one operation, $n-1$ points were added.
## Solution
If there were 65 points at the end, then there were 33 (and 32 were added) points before that. Similarly, there were 17 points after the first operation, and 9 before it.
## Answer
9 points. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (continued).]
In a class of 25 students, it is known that any two girls in the class have a different number of boy friends from this class. What is the maximum number of girls that can be in this class?
# | If there are 13 girls in the class, the number of their boy friends from this class can be any integer from 0 to 12 (13 different options), which meets the condition. If there are more than 13 girls (at least 14), then the number of boys in the class will not exceed 11, which means the number of different options for t... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
| |
In the bag, there are 70 balls that differ only in color: 20 red, 20 blue, 20 yellow, and the rest are black and white.
What is the smallest number of balls that need to be drawn from the bag, without seeing them, to ensure that among them there are at least 10 balls of the same color? | By drawing 37 balls, we risk getting 9 red, blue, and yellow balls each, and there will be no ten balls of one color. If we draw 38 balls, however, the total number of red, blue, and yellow balls among them will be no less than 28, and the number of balls of one of these colors will be no less than ten (since $28>3 \cd... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Coordinate method on the plane $]$
Find the equation of the line passing through the intersection point of the lines $3 x+2 y-5=0$ and $x-3 y+2=0$ and parallel to the y-axis. | Solving the system of equations
$$
\left\{\begin{array}{l}
3 x+2 y-5=0 \\
x-3 y+2=0
\end{array}\right.
$$
we find the coordinates of the point $B\left(x_{0} ; y_{0}\right)$ of intersection of these lines: $x_{0}=1, y_{0}=1$.
Since the desired line is parallel to the y-axis and passes through the point $B\left(x_{0} ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$[$ P $[\quad$ Case Analysis $\quad]$
Find all three-digit numbers that are 12 times the sum of their digits.
# | According to the condition, the number is divisible by 3. Therefore, the sum of its digits is also divisible by 3, which means the number itself is divisible by 9.
In addition, it is divisible by 4. Therefore, we need to look for numbers that are divisible by 36. The sum of the digits of a three-digit number does not ... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3 [ Examples and counterexamples. Constructions ]
Can integers be written in the cells of a $4 \times 4$ table so that the sum of all the numbers in the table is positive, while the sum of the numbers in each $3 \times 3$ square is negative? | The central square of size $2 \times 2$ is contained in each square of size $3 \times 3$. If we place the number -9 in one of the cells of the central square, and fill the rest of the cells of this table with ones, then the sum of all numbers in the table is $15+(-9)=6$, and the sum of the numbers inside any $3 \times ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Multiplied several natural numbers and got 224, and the smallest number was exactly half of the largest. How many numbers were multiplied
# | $224=2^{5} \cdot 7$. Consider the two numbers mentioned in the condition: the smallest and the largest. If one of them is divisible by 7, then the other must also be divisible by 7. But 224 is not divisible by 7², so both of these numbers must be powers of two. From the condition, it also follows that these are two con... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Masha has two-ruble and five-ruble coins. If she takes all her two-ruble coins, she will be 60 rubles short to buy four pies. If all five-ruble coins - she will be 60 rubles short for five pies. And in total, she is 60 rubles short to buy six pies. How much does a pie cost? | If Masha takes all her two-ruble and five-ruble coins, she will be short of $60+60=120$ rubles for $4+5=9$ pies. On the other hand, she will be short of 60 rubles for 6 pies. That is, three pies cost 60 rubles.
## Answer
20 rubles. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Folklore
To repair the propeller, Karlson needs to buy three blades and one screw. In the store, blades cost 120 tugriks each and screws cost 9 tugriks each. However, after a purchase of at least 250 tugriks, a 20% discount is given on all subsequent purchases. Will Karlson be able to repair the propeller if he has on... | Let the first purchase be two blades and two screws, costing $2(120+9)=258$ tugriks. Since the cost of the purchase is more than 250 tugriks, Carlson can buy the third blade with a $20\%$ discount, spending $120 \cdot 0.8=96$ tugriks. Therefore, in total, Carlson will spend $258+96=354$ tugriks.
## Answer
He can. | 354 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
It is known that ЖЖ + Ж = МЁД. What is the last digit of the product: В $\cdot И \cdot H \cdot H \cdot U \cdot \Pi \cdot У \cdot X$ (different letters represent different digits, the same letters represent the same digits)?
# | Since a two-digit number ЖЖ was added to a one-digit number Ж to get a three-digit number, then Ж $=9$, and МЁД = 108. Four digits have already been used. In the product $\mathrm{B} \cdot \mathrm{U}^{\prime} \cdot \mathrm{H} \cdot \mathrm{H} \cdot И \cdot П \cdot У \cdot \mathrm{X}$, six other digits are used.
Therefo... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Higher Order Roots (Miscellaneous). ] [ Examples and Counterexamples. Constructions ]
Is there a natural number $n$, greater than 1, such that the value of the expression $\sqrt{n \sqrt{n \sqrt{n}}}$ is a natural number? | For example, $n=2^{8}=256$. Indeed, $\sqrt{n \sqrt{n \sqrt{n}}}=n^{7 / 8}$. For $n=2^{8}$, the value of this expression is $2^{7}=$ 128.
## Answer
It exists. | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Irrational Equations $]$ [ Monotonicity and Boundedness
Solve the equation $2017 x^{2017}-2017+x=\sqrt[2017]{2018-2017 x}$.
# | The function $f(x)=2017 x^{2017}-2017+x$ is increasing, while the function $g(x)=\sqrt[2017]{2018-2017 x}$ is decreasing. Therefore, the equation $f(x)=g(x)$ has no more than one root. However, it is obvious that $f(1)=g(1)$.
## Answer
$x=1$.
Author: Volienkov S.G.
A sheet of paper has the shape of a circle. Can fi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
Does there exist a number divisible by 2020 in which all digits $0,1,2, \ldots, 9$ are present in equal amounts? | $2020=20 \cdot 101$, so, for example, the number 10198987676545432320 works.
## Answer
It exists. | 10198987676545432320 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & {\left[\begin{array}{l}\text { Irrational Equations } \\ \text { [ Completing the Square. Sums of Squares }\end{array}\right]}\end{aligned}$
Solve the equation
$$
\left(x^{2}+x\right)^{2}+\sqrt{x^{2}-1}=0
$$ | Since the numbers $\left(x^{2}+x\right)^{2}$ and $\sqrt{x^{2}-1}$ are non-negative, and their sum is zero, then both these numbers are equal to zero. On the other hand, if both these numbers are equal to zero, then their sum is zero. Therefore, the original equation is equivalent to the following system:
$$
\left\{\be... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameter $]
For the quadratic trinomial $f(x)=a x^{2}-a x+1$, it is known that $|f(x)| \leq 1$ for $0 \leq x \leq 1$. Find the greatest possible value of $a$. | Since $f(0)=f(1)=1$, the graph of the quadratic function is a parabola symmetric about the line $x=0.5$. From the condition $|f(x)| \leq 1$ for
$0 \leq x \leq 1$, it follows that the branches of the parabola are directed upwards. The minimum value of $f(x)$ is $f(0.5)=1-\frac{a}{4}$.
The maximum possible value of $a$... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Tournaments and tournament tables ] Counting in two ways
[ Examples and counterexamples. Constructions ]
In the competition, 10 figure skaters participate. The competition is judged by three judges in the following way: each judge independently distributes places (from first to tenth) among the skaters, after which ... | Evaluation. Since each of the three judges distributed a set of places from first to tenth, the sum of the places awarded by all judges to all participants in the competition is $3 \cdot (1 + 2 + \ldots + 10) = 165$.
On the other hand, if the winner received a sum of no less than 16, then all the others received a sum... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{c}\text { Absolute value of a number }\end{array}\right]$
$\left[\begin{array}{c}\text { Evenness and oddness }\end{array}\right]$
$\left[\begin{array}{l}\text { Partitions into pairs and groups; bijections }\end{array}\right]$
[ Examples and counterexamples. Constructions ]
Find the maximum value... | The modulus of the difference of two non-negative numbers is not greater than their maximum.
## Solution
Estimate. Note that the modulus of the difference of two non-negative numbers is not greater than their maximum. Therefore, $\left|x_{1}-x_{2}\right|$ $\leq \max \left\{x_{1}, x_{2}\right\}$
||$x_{1}-x_{2}\left|-... | 1989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9,10 $[\quad$ The Pigeonhole Principle (etc.).
The sum of ten natural numbers is 1001. What is the greatest value that the GCD (greatest common divisor) of these numbers can take? | Example. Consider nine numbers equal to 91 and the number 182. Their sum is 1001.
Estimate. We will prove that the GCD cannot take a value greater than 91. Note that $1001=7 \cdot 11 \cdot 13$. Since each term in this sum is divisible by the GCD, the GCD is a divisor of the number 1001. On the other hand, the smallest... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shen A.H.
There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions can be performed:
- move a stone from the bag to the first cell of the strip or vice versa;
- if there is a stone in the cell... | Note that for each action, there is an inverse action. Therefore, if we move from situation $A$ to situation $B$ by following the rules, we can move from situation $B$ back to situation $A$ by following the rules.
We will show by induction that if there is a reserve of $n$ stones, then, acting according to the above r... | 197 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9,10
What is the maximum number of self-intersection points that a closed 14-segment broken line can have, passing along the lines of a grid paper so that no line contains more than one segment of the broken line? | Let's first show that any closed 14-segment broken line has 7 horizontal and 7 vertical segments. Indeed, from each vertex of the broken line, one vertical and one horizontal segment emerge. Counting along all 14 vertices, we will thus count 14 horizontal and 14 vertical segments. But in this counting, each segment is ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Processes and Operations] $[$ Motion Problems ]
From point $A$, 100 planes (1 flagship and 99 additional) take off simultaneously. With a full tank of fuel, a plane can fly 1000 km. In flight, planes can transfer fuel to each other. A plane that has transferred fuel to others makes a gliding landing. How should the f... | Let's describe the optimal fuel exchange procedure. Initially, 100 aircraft take off with full tanks. As soon as possible, one of the aircraft distributes its fuel to the others, after which 99 aircraft have full tanks, and the freed-up aircraft lands. By this point, the aircraft will have flown \(1000 \cdot \frac{1}{1... | 5187 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
At a round table, 12 places were prepared for the jury with each place labeled with a name. Nikolai Nikolaevich, who arrived first, absent-mindedly sat not in his own place, but in the next one clockwise. Each subsequent jury member, approaching the table, would take their own place or, if it was alrea... | Let's consider a certain way of seating the jury members. We will call a jury member lucky if they are sitting in their own seat. The first of the unlucky ones (excluding Nikolai Nikolaevich) to approach the table is the one whose seat is taken by Nikolai Nikolaevich (another unlucky one would sit in their still free s... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Examples and counterexamples. Constructions ]
A first-grader has a hundred cards with natural numbers from 1 to 100 written on them, as well as a large supply of "+" and "=" signs. What is the maximum number of correct equations he can form? (Each card can be used no more than once, each equation can contain only on... | In the equality, no less than three numbers are involved, so there cannot be more than 33 equalities. Let's show how to form 33 equalities.
25 equalities: $49+51=100, 47+52=99, \ldots, 1+75=76$. The remaining even numbers are from 2 to 50. Another 5 equalities: $18+30=48, 14+32=46, \ldots, 2+38=40$. The remaining numb... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8,9
The areas of the projections of a certain triangle onto the coordinate planes Oxy and Oyz are $\sqrt{6}$ and $\sqrt{7}$, respectively, and the area of the projection onto the plane $O x z$ is an integer. Find the area of the triangle itself, given that it is also an integer. | Let the vector perpendicular to the plane of the original triangle form angles $\alpha, \beta$, and $\gamma$ with the coordinate axes $O x, O y$, and $O z$ respectively. Then
$$
\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=1
$$
Let the area of the original triangle be denoted by $S$, and the areas of the projections on t... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the plane, $n$ points are given. It is known that among any three of them, there are two whose distance does not exceed 1. Prove that it is possible to cover all these points with two circles of radius 1 on the plane. | Since we have a finite number of points, we can choose two of them, the distance between which is the greatest. From these points, as centers, we will draw circles of radius 1. Any third point will fall into one of these circles. This happens because if the distance between the chosen points is greater than 1, then the... | 99 | Geometry | proof | Yes | Yes | olympiads | false |
Berlov S.L.
At the alumni meeting, 45 people attended. It turned out that any two of them who had the same number of acquaintances among those present were not acquainted with each other. What is the maximum number of pairs of acquaintances that could have been among those who attended the meeting? | Since $45=1+2+3+\ldots+9$, we can divide 45 people into groups of 1, 2, ... 9 people. Let people in the same group not know each other, while people in different groups know each other. Then each person in the $k$-th group has $45-k$ acquaintances.
Under this condition, the problem is solved, and the total number of p... | 870 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Berrov S.L.
On some cells of a $10 \times 10$ board, $k$ rooks were placed, and then all cells that are attacked by at least one rook were marked (it is assumed that a rook attacks the cell it stands on). For what largest $k$ can it happen that after removing any rook from the board, at least one marked cell will no l... | Let's consider the placement of $k$ rooks satisfying the condition. There are two possible cases.
1. Suppose there is at least one rook in each column. Then the entire board is under attack, and a rook can be removed from any column that has at least two rooks. Therefore, in this case, there is exactly one rook in eac... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
At the beginning of the school year, Andrey started recording his math grades. Upon receiving a new grade (2, 3, 4, or 5), he called it unexpected if, up to that point, it had occurred less frequently than each of the other possible grades. (For example, if he had received the grades 3, 4, 2, 5, 5, 5, 2, ... | The first unexpected grade will be the last one received for the first time. The second unexpected grade will be the last one received for the second time, and so on. Therefore, there will be a total of 10 unexpected grades.
## Answer
It can be. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).] Case analysis $\quad]$
After a hockey game, Anton said that he scored 3 goals, and Ilya only one. Ilya said that he scored 4 goals, and Seryozha as many as 5. Seryozha said that he scored 6 goals, and Anton only two. Could it be that together they scored 10 goals, given that each ... | There are two cases.
1) Anton told the truth about himself, that is, he scored 3 goals. Then Seryozha lied about Anton, so he told the truth about himself, that is, he scored 6 goals. Therefore, Ilya lied about Seryozha and told the truth about himself, that is, he scored 4 goals. In this case, the boys scored a total... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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