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The probability that a purchased light bulb will work is 0.95.
How many light bulbs need to be bought so that with a probability of 0.99, there will be at least five working ones among them?
# | Let's take 6 light bulbs. The probability that at least 5 of them will be working is the sum of the probabilities that exactly 5 of them will be working and that all 6 will be working, which is $6 \cdot 0.95^{5} \cdot 0.05 + 0.95^{6} = 0.9672$.
Let's take 7 light bulbs. The desired probability is $21 \cdot 0.95^{5} \c... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
According to the conditions of the chess match, the winner is the one who outperforms the opponent by two wins. Draws do not count. The probabilities of winning for the opponents are equal. The number of decisive games in such a match is a random variable. Find its mathematical expectation. | Let $X$ be the number of decisive games. At the beginning of the match, the difference in the number of wins between the two participants is zero. Let's list the possible cases of two decisive games, denoting a win by the first participant as 1 and a win by the second participant as 2: 11, 12, 21, 22. Two of the four c... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
At a familiar factory, they cut out metal disks with a diameter of 1 m. It is known that a disk with a diameter of exactly 1 m weighs exactly 100 kg. During manufacturing, there is a measurement error, and therefore the standard deviation of the radius is 10 mm. Engineer Sidorov believes that a stack of 100 disks will ... | Given $\mathrm{E} R=0.5 \mathrm{~m}, \mathrm{D} R=10^{-4} \mathrm{~m}^{2}$. Let's find the expected value of the area of one disk:
$\mathrm{ES}=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(\mathrm{D} R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi$.
Therefore, the expected value of... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The figure shows a track scheme for karting. The start and finish are at point $A$, and the karting driver can make as many laps as they want, returning to the starting point.
The young dr... | Let $M_{n}$ denote the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is either $A B$, $B A$, or a loop $B B$). Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
One can reach point $B$ in... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the Valley of Five Lakes, there are five identical lakes, some of which are connected by streams (on the diagram, dashed lines indicate possible "routes" of the streams). Small carp are born only in lake $S$. While growing up, a carp crosses from one lake to another exactly four times via some stream (the carp choos... | The transition from one lake to another will be called a route of length $n$ if it passes through $n$ streams. Let's prove several statements.
1. There is no route of length 2 from $S$ to any of the lakes $A, C$, and $D$.
Proof. Suppose it is possible to sail from lake $S$ to lake $A$ through one intermediate lake (s... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A fly moves from the origin only to the right or upwards along the lines of an integer grid (monotonic walk). At each node of the grid, the fly randomly chooses the direction of further movement: up or to the right.
a) Prove that sooner or later the fly will reach the point with an abscissa of 2011.
b) Find the expec... | a) A fly will never reach the point with an abscissa of 1 only if it moves up at each step. The probability of this is $0.5 \cdot 0.5 \cdot 0.5 \cdot 0.5 \cdot \ldots = 0$.
Thus, the fly will inevitably reach the point with an abscissa of 1. Reasoning in the same way, we can conclude that sooner or later the fly will ... | 2011 | Combinatorics | proof | Yes | Yes | olympiads | false |
In 2012, the Anchuria School Board decided that three questions were too few. Now, one needs to correctly answer six questions out of 40. The question is, if one knows nothing and simply guesses the answers, in which year is the probability of obtaining the Anchurian certificate higher - in 2011 or in 2012? | If a graduate guesses the answers, the Unified State Exam (EGE) can be considered as a Bernoulli scheme with a success probability $p=$ 0.25 and a failure probability $q=0.75$. In 2011, to pass the exam, one needed to answer at least three questions correctly. It is more convenient to find the probability of the opposi... | 2012 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Every day, the dog Patrick gnaws one slipper from the available supply in the house. With a probability of 0.5, Patrick wants to gnaw a left slipper and with a probability of 0.5 - a right slipper. If the desired slipper is not available, Patrick gets upset. How many pairs of identical slippers need to be bought so tha... | Let in a week Patrick wants to eat $S$ left and $7-S$ right slippers. We need to find such $k$ that the inequality $\mathrm{P}(S \leq k \cap 7-S \leq k) \geq 0.8$ holds. Rewrite the event in parentheses: $P(7-k \leq S \leq k)$. It is clear that $7-k \leq k$, that is, $k \geq 4$.
The probability on the left side of the... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
If one person spends one minute waiting in a queue, we will say that one person-minute is wasted. In a bank queue, there are eight people, five of whom plan to perform simple operations taking 1 minute, while the rest plan to perform long operations taking 5 minutes. Find:
a) the minimum and maximum possible total num... | Let a short operation take $a$, and a long one - $b$ minutes $(a<b)$. A client planning a simple operation will be called a rusher, and one who intends to take a long time - a dawdler. Suppose there are $n$ rushers and $m = -2$ dawdlers.
Obviously, the number of wasted person-minutes depends on the order in which the ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On board an airliner, there are $2 n$ passengers, and the airline has loaded $n$ portions of chicken and $n$ portions of fish for them. It is known that a passenger prefers chicken with a probability of 0.5 and fish with a probability of 0.5. We will call a passenger dissatisfied if they are left with what they do not ... | a) The number of dissatisfied passengers can be any from 0 to $n$. In the case of $n=1$, everything is obvious: there is either no dissatisfied passenger or one, and both cases are equally likely. We will further assume that $n>1$.
Let us introduce the random variable $\xi$ "Number of dissatisfied passengers". $\xi=0$... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In Anchuria, a checkers championship is taking place in several rounds. The days and cities for the rounds are determined by a draw. According to the championship rules, no two rounds can take place in the same city, and no two rounds can take place on the same day. Among the fans, a lottery is organized: the main priz... | In an $8 \times 8$ table, you need to select $k$ cells such that no more than one cell is selected in any row or column. The number of tours $k$ should be chosen so that the number $N k$ of possible selection options is maximized.
$$
N_{k}=C_{8}^{k} A_{8}^{k}=\frac{8!\cdot 8!}{(8-k)!(8-k)!k!}\left(C_{8}^{k}-\text { th... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Ilya Muromets meets the three-headed Zmei Gorynych. And the battle begins. Every minute Ilya cuts off one of Zmei's heads. With a probability of $1 / 4$, two new heads grow in place of the severed one, with a probability of $1 / 3$ - only one new head, and with a probability of $5 / 12$ - no heads at all. The Zmei is c... | Strikes by Ilya Muromets, in which the number of heads changes, are called successful.
Let's find the probability that at some point there will be a last successful strike. This means that starting from this point, there will be no more successful strikes, that is, all strikes will be unsuccessful. The probability of ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A deck of playing cards is laid out on the table (for example, in a row). On top of each card, a card from another deck is placed. Some cards may have matched. Find:
a) the expected value of the number of matches;
b) the variance of the number of matches.
# | a) Let's number the pairs from 1 to $N$ (we don't know how many there are) in the order they lie on the table. Let
the indicator $I_{k}$ be 1 if the two cards in the $k$-th pair are the same, and 0 if the cards in the $k$-th pair are different.
Obviously, $\mathrm{P}\left(I_{k}=1\right)=1 / N$. Therefore, $\mathrm{E}... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On each of the three axes, there is one rotating gear and one stationary pointer. The gears are connected sequentially. The first gear has 33 teeth, the second has 10, and the third has 7. On each tooth of the first gear, in alphabetical order, a letter of the Russian alphabet is written clockwise:
А Б В Г ДЕ Ё Ж З И ... | Carefully determine the stop moments after the start of encryption. For this, assign the ordinal number to each letter of the Russian alphabet.
## Solution
Determine the stop moments after the start of encryption. Number the letters of the Russian alphabet: А - 0, Б - 1, and so on. The letters of the encrypted word w... | 515355128523864354 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How many integers exist from 1 to 1000000 that are neither perfect squares, nor perfect cubes, nor fourth powers?
# | $1000000=1000^{2}=100^{3}=10^{6}$. Therefore, in the specified range, there are exactly 1000 squares and 100 cubes. 10 numbers among them are sixth powers, meaning they are both squares and cubes. All fourth powers are found among the squares. Therefore, the numbers that satisfy the condition are
$1000000-1000-100+10=... | 998910 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Examples and counterexamples. Constructions] Evaluation + example $\quad]$
On an island, there are 100 knights and 100 liars. Each of them has at least one friend. One day, exactly 100 people said: "All my friends are knights," and exactly 100 people said: "All my friends are liars." What is the smallest possible num... | Consider a resident $A$ of the island, claiming: "All my friends are liars". If $A$ is a knight, then his friend is a liar. If $A$ is a liar, then his friend is a knight. In either case, $A$ is part of a pair of friends of "different" types. Since 100 people made such a statement, the number of such pairs cannot be les... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Poddlisky 0. . Around a round table, 2015 people are sitting, each of them is either a knight or a liar. Knights always tell the truth, and liars always lie. Each of them was given a card with a number on it; all the numbers on the cards are different. After looking at their neighbors' cards, each person said: "My num... | Let $A$ and $B$ be the people who received the cards with the largest and smallest numbers, respectively. Since both of them said the first phrase, $A$ is a knight, and $B$ is a liar. Therefore, neither of them could have said the second phrase. Thus, $k \leq 2013$.
The situation where the remaining 2013 people can sa... | 2013 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
At the New Year's school party in the city of Lzheretsark, 301 students came. Some of them always tell the truth, while the rest always lie. Each of the 200 students said: "If I leave the hall, then among the remaining students, the majority will be liars." Each of the other students stated: "If I leave the hall, then ... | Let's call those who tell the truth knights. We will consider the statement of each of the two hundred students as the first phrase, and the statement of the remaining as the second.
First method. Note that not all students are liars, otherwise there would be no one to say the first phrase. If the second phrase was sp... | 151 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the class, there are 30 students: excellent students, C-grade students, and D-grade students. Excellent students always answer questions correctly, D-grade students always make mistakes, and C-grade students answer the questions given to them strictly in turn, alternating between correct and incorrect answers. All s... | Let $a$ be the number of excellent students, $b$ be the number of poor students, and $c$ be the number of average students who made a mistake on the first question, answered the second question correctly, and made a mistake on the third question.
According to the conditions, $a+b+c=19, b+c=12, c=9$. Therefore, $b=3, a... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shen A.X.
A strip of $1 \times 10$ is divided into unit squares. The numbers $1, 2, \ldots, 10$ are written in the squares. First, the number 1 is written in one of the squares, then the number 2 is written in one of the adjacent squares, then the number 3 is written in one of the adjacent squares to the already occup... | Obviously, the numbers to the left of 1 are in descending order, and those to the right are in ascending order. Therefore, the string is uniquely determined by the choice of a subset of numbers that will stand to the left of one. There are such subsets in the set $\{2, \ldots, 10\}$ of 9 elements, as is known, $2^{9}$.... | 512 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
The game takes place on a $9 \times 9$ grid of squared paper. Two players take turns. The player who starts the game places crosses in free cells, while his partner places noughts. When all cells are filled, the number of rows and columns $K$ in which there are more crosses than noughts, and the numbe... | One possible strategy for the first player: the first move is to the center of the board; thereafter, for each move the second player makes to any cell, the first player responds by moving to the cell symmetric to the center. This is possible because the second player always breaks the symmetry.
In the final position,... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Two players take turns coloring the sides of an $n$-gon. The first player can color a side that borders with zero or two colored sides, the second player - a side that borders with one colored side. The player who cannot make a move loses. For which $n$ can the second player win, regardless of how the ... | For $n=3$, it is obvious that the first player wins, while for $n=4$, the second player wins. Let's show how the first player wins for $n>4$. After the first move of the second player, two adjacent sides are painted. The first player can paint a side "one apart" from them, creating an unpainted "hole" of one side. This... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the zoo, there were 200 parrots. One day, they each made a statement in turn. Starting from the second,
the statements were "Among the previously made statements, more than 70% are false." How many false statements did the parrots make
# | Note that if the first statement was true, then the second will be false, and vice versa, if the first statement was false, then the second will be true. Since the truth of the following statement depends only on the number of false statements among the previous ones, the truth of all remaining statements does not depe... | 140 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.B.
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, all t... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evoookimov M.A.
On one of the cells of an $8 \times 8$ field, a treasure is buried. You are at the center of one of the corner cells of this field with a metal detector and move by transitioning to the centers of adjacent cells along the sides. The metal detector triggers if you are on the cell where the treasure is b... | Author: $\underline{\text { Aviel Boag }}$
Even 25 moves are sufficient. We will move along the path indicated in the figure. When the mine detector first goes off, the treasure can be in no more than three cells. For example, if it went off in the black cell, then the suspicious cells will be the three marked on the ... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Kuznecovv.
On the 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 answe... | 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 |
Bakayev E.v. $\quad$.
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 4 chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order? | Let's number the chips and cells in order from 0 to 99. A free operation does not change the remainder of the cell number when divided by 5.
Evaluation. Let's mentally arrange the piles of chips in a circle. First, the pile of chips with a remainder of 0, then with 1, and so on up to 4. A paid operation swaps a pair o... | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
| :---: | :---: | :---: |
| | Processes and operations | |
| | Evenness and oddness | |
| | Group theory (other) | |
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 satisfied: if a number $m$ is obtain... | 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 sequenc... | 1024 | Combinatorics | proof | 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 to problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.
## Answer
a) Two weights of 67 grams and 66 weights of 1 gram or 66 weights of 3 grams and two of 1 gram.
b) 3 sets.
Author: Fomin D:
G... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Fomin }}$ 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, an... | 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 divis... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Uubanov I.S.
There is a set of weights with the following properties:
a. It contains 5 weights, all of different weights.
b. For any two weights, there are two other weights with the same total weight.
What is the smallest number of weights that can be in this set? | Let $A$ be one of the lightest weights, and $B$ be one of the weights that follow $A$ in weight. Clearly, the pair of weights $\{A, B\}$ can only be balanced by an identical pair.
Therefore, there are at least two weights $A$ and two weights $B$. The pair $\{A, A\}$ can also only be balanced by an identical pair. Ther... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Pooolish $A$.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 222 nuts between two small boxes. After looking at the distribution, Chichikov names any integer $N$ from 1 to 222. Then Nozdryov must, if necessary, move one or several nuts to an empty third box and show Chichikov one or two boxes, where t... | Upper bound. By placing 74 and 148 nuts in the boxes, Nozdryov can, for any $N$, move no more than 37. Indeed, $N$ can be written in the form $74 k+r$, where $k=0,1,2,3$, and $-37 \leq r0$, and the numbers 74, 148, and 222 can be collected with one or two boxes, without moving anything. If $r0,74 k=0,74$ or 148. Taking... | 37 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Polish A.
Chichikov is playing with Nozdryov. First, Nozdryov distributes 1001 nuts among three boxes. After looking at the distribution, Chichikov names any integer \( N \) from 1 to 1001. Then Nozdryov must, if necessary, move one or several nuts to an empty fourth box and show Chichikov one or several boxes where t... | Upper bound. By placing 143, $286=2 \cdot 143$ and $572=4 \cdot 143$ nuts in the boxes, Nosdriv can, for any $N$, move no more than 71. Indeed, $N$ can be represented as $143 k+r$, where $0 \leq k \leq 7$, and $-71 \leq r < 71$. The number $143 k$ can be collected with one or several boxes without moving anything. If $... | 71 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$$
61609
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99 children stand in a circle, each initially having a ball. Every minute, each child with a ball throws their ball to one of the two neighbors; if two balls land with the same child, one of these balls is lost irretrievably. What is the minimum time after which only one ball may remain with the children? | Number the children and balls clockwise from 1 to 99.
Example. Suppose children 1 and 2 toss the first ball to each other. The other balls with odd numbers are always thrown counterclockwise until they reach the second child, who discards them (this happens after an odd number of minutes, and at that moment he also re... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
ЕЕооокимов M.A. Anya calls a date beautiful if all 6 digits in its notation are different. For example, 19.04.23 is a beautiful date, while 19.02.23 and 01.06.23 are not. How many beautiful dates are there in 2023? | The digits 2 and 3 are already part of the year number, so among all the months, only $01, 04, 05, 06, 07$, 08, 09, and 10 need to be considered. Each of these month numbers contains a 0, so in a beautiful date, there will be no day number starting with 0, 2, or 3, and there will also be no days 10, 11, 12, and 13—leav... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The buyer took goods worth 10 rubles from the seller and gave 25 rubles. The seller didn't have change, so he exchanged the money with a neighbor. After they settled and the buyer left, the neighbor discovered that the 25 rubles were counterfeit. The seller returned 25 rubles to the neighbor and began to think. What lo... | Answer: 25 r. The loss consists in the fact that the seller gave 25 r. for fake 25 r; other exchanges can be disregarded. | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
An electrician 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 ... | 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 |
Tomong A.K.
Petya's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300 dollars or adding 198 dollars.
What is the maximum amount Petya can withdraw from his account if he has no other money?
# | Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars. The maximum number that is a multiple of 6 and does not exceed $500 is 498.
Let's show how to withdraw 498 dollars. Perform the following operations: $500-300=200, 200+198=398, 398-$ $300=98, 98+198=296... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
The 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 number. To open the locker, the correct five-digit number must be entered. What is the minimum ... | 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 |
Twelve chairs are arranged in a row. Sometimes a person sits on one of the free chairs. At this point, exactly one of his neighbors (if they were present) stands up and leaves. What is the maximum number of people that can be sitting at the same time, if initially all the chairs were empty?
# | Evaluation. It is impossible for all chairs to be occupied simultaneously, because at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.
Example. Let's show how to seat 11 people. Number the chairs fro... | 11 | 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? White and black move alternately, one queen per move. | The queen on the squares b1 and b8, which moved earlier, could not move to the opposite rank with that move, meaning these two queens together made at least three moves. Similarly, the pairs of queens on the files c, ..., g each made at least three moves. The first of the four corner queens to move did not land on a co... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Evdokiyov M.A.
A pirate has five bags of coins, each containing 30 coins. He knows that one bag contains gold coins, another contains silver coins, a third contains bronze coins, and each of the two remaining bags contains an equal number of gold, silver, and bronze coins. You can simultaneously take any number of coi... | Example. Let's take one coin from each bag. Among these five coins, there are coins of all three types, so there is only one coin of a certain type. If it is, for example, a gold coin, then it was taken from the bag with gold coins. Indeed, for each coin from the "mixed" bag, there is a matching one from the correspond... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Folkior
There is a set of tickets numbered from 1 to 30 (numbers can repeat). Each student drew one ticket. The teacher can perform the following operation: read out a list of several (possibly one) numbers and ask their owners to raise their hands. How many times does he need to perform this operation to find out the... | We will encode the tickets with binary numbers from 00001 to 11110. On the $k$-th stage, the teacher includes in the list all numbers whose $k$-th bit is one (for example, on the 3rd stage, the numbers are $4,5,6,7,12,13,14,15,20,21,22,23$, $28,29,30$). After the fifth stage, the teacher learns the binary code of the n... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A. I. F. Shark
To a natural number $A$, three digits were appended on the right. The resulting number turned out to be equal to the sum of all natural numbers from 1 to $A$. Find $A$.
# | Let the appended digits form the number $B, 0 \leq B \leq 999$. Then the resulting number is, on the one hand, $1000 A+B$, and on the other hand, $-1+2+\ldots+A=\frac{1}{2} A(A+1)$. The equation $1000 A+B=\frac{1}{2} A(A+1)$ transforms into $A(A-1999)=2 B$, from which $0 \leq A(A-1999) \leq 1998$. Since the left inequa... | 1999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[Algorithm Theory (Miscellaneous).] [ Processes and Operations ]
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 s... | Try to determine in five attempts which of the 6 suitcases the first key fits.
## Solution
olnyaлись equalities in all rows and each
* $\qquad$
##
$\qquad$
![](https://cdn.mathpix.com/... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
This ancient problem was known even in Ancient Rome.
A wealthy senator, on his deathbed, left his wife pregnant. After the senator's death, it was discovered that his estate, worth 210 talents, was left with the following will: "In the event of the birth of a son, give the boy two-thirds of the estate (i.e., 140 talen... | Since it is impossible to fulfill all the testator's requirements, only a part of them will have to be fulfilled. Depending on which specific part we fulfill, one or another method of division will be adopted.
## Solution
Since it is impossible to fulfill all the testator's requirements, we will have to fulfill only ... | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the room, there are 12 people; some of them are honest, meaning they always tell the truth, while the rest always lie. "There is not a single honest person here," said the first. "There is no more than one honest person here," said the second. The third said that there are no more than two honest people, the fourth ... | Note that if someone among those present lied, then all the previous ones lied as well. Such people are in the room; otherwise, the first one told the truth, and according to his words, there are no honest people in the room. For the same reason, there must be honest people in the room.
Let there be $x$ liars in the r... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
A game with three piles of stones. There are three piles of stones: the first has 10, the second has 15, and the third has 20. On a turn, it is allowed to split any pile into two smaller parts; the player who cannot make a move loses.
# | There are 45 stones in total. In the end, we will get 45 piles of one stone each. To break down the first pile into single stones, it takes 9 moves; for the second pile, it will take 14 moves; for the third pile, it will take 19 moves (the number of moves does not depend on whether we separate the stones one by one or ... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ $[\underline{\text { Games-jokes }}]$
On the board, the numbers 25 and 36 are written. In one move, it is allowed to write down another natural number - the difference between any two numbers already on the board, if it has not appeared before. The player who cannot make a move loses.
# | In the process of the game (compare with Euclid's algorithm), the greatest common divisor of the initial numbers will definitely be written out. Consequently, all numbers that are multiples of it, not exceeding the larger of the initial numbers, will also be written out. In our case, the GCD is 1. Therefore, all number... | 34 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Symmetric Strategy ]
There are two piles of stones: one has 30, the other has 20. On a turn, you are allowed to take any number of stones, but only from one pile. The player who cannot make a move loses.
# | The first one wins. With the first move, he equalizes the number of stones in the piles, after which he plays as in problem 10. | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[$ Problems with constraints $]$
A row of soldiers is called incorrect if no three consecutive soldiers stand in order of height (neither in ascending nor in descending order). How many incorrect rows can be formed from $n$ soldiers of different heights if
a) $n=4$;
b) $n=5$? | Let's place a "+" between neighboring soldiers if the right one is taller than the left one, and a "-" otherwise. According to the condition, the plus and minus signs in the incorrect rows should alternate. It is clear that the number of rows starting with a plus is equal to the number of rows starting with a minus.
a... | 32 | 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 |
[Algorithm Theory (Miscellaneous).]
Given two fuse cords, each of which burns for exactly one minute if lit from one end (but may burn unevenly).
How can you use these cords to measure 45 seconds? (You can light the cord from either of the two ends.)
# | One wick can be lit at the moment when the other one is completely burned out.
## Solution
We will light the first wick from both ends simultaneously, and the second wick from one end. The wick lit from both ends will burn twice as fast, that is, in 30 seconds. At the moment the first wick is completely burned, we li... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
We are considering all possible triangles with integer sides and a perimeter of 2000, as well as all possible triangles with integer sides and a perimeter of 2003. Which set of triangles is larger? | ## Solution
Let the sides of the triangle be integers $a, b, c$ and its perimeter be 2000. We will associate this triangle with a triangle with sides $a+1, b+1, c+1$, whose perimeter is 2003 (it is easy to verify that the triangle inequality holds for these sides). Under this correspondence, different triangles corres... | 2003 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
On Uncle Fyodor's birthday, Postman Pechkin wants to find out how old he is. Sharik says that Uncle Fyodor is more than 11 years old, and cat Matroskin claims that he is more than 10 years old. How old is Uncle Fyodor, given that exactly one of them is wrong? Justify your answer. | Note that if Sharik did not make a mistake, then Matroskin did not make a mistake either, which contradicts the condition. Therefore, Sharik must have lied, while Matroskin told the truth. Thus, Uncle Fyodor is more than 10 years old, but not more than 11. Therefore, Uncle Fyodor is 11 years old.
## Problem | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Old cobbler Karl made boots and sent his son Hans to the market to sell them for 25 talers. At the market, two disabled men (one without a left leg, the other without a right leg) approached the boy and asked him to sell them each a boot. Hans agreed and sold each boot for 12.5 talers.
When the boy came home and told ... | Think about how much money Karl should have received, how much he actually received, and why.
## Solution
The disabled people paid 23 talers for the boots, but Karl received only 20, as Hans spent the remaining 3 talers on candies. Sitting in the pantry, Hans added the income (23 talers) to the expense (3 talers). Th... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
There are five chain links, each with 3 rings. What is the minimum number of rings that need to be unlinked and relinked to connect these links into one chain?
# | For connecting two links, one ring is required.
## Solution
Example: we unlock 3 rings from one link. The remaining 4 links are connected using the three unlocked rings.
Evaluation: if fewer than 3 rings are unlocked, at least 5 separate links will remain, which would require at least 4 unlocked rings to connect - a... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
|
There are 85 balloons in the room - red and blue. It is known that: 1) at least one of the balloons is red; 2) in every arbitrarily chosen pair of balloons, at least one is blue. How many red balloons are in the room?
# | Think about whether there can be two red balls in the room.
## Solution
Since among any two balls, one is blue, there cannot be two red balls in the room. Therefore, there are 84 blue balloons and 1 red balloon in the room.
## Answer
1 ball. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The director of the plant, reviewing a list of phone numbers and surnames of his employees, noticed a certain relationship between the surnames and phone numbers. Here are some surnames and phone numbers from the list:
| Achinsky | 8111 |
| :--- | :--- |
| Butenko | 7216 |
| Galich | 5425 |
| Lapina | 6131 |
| Martyan... | Think about what the first digit of the phone number might mean? What two numbers are derived from the remaining three digits?
## Solution
The first digit of the phone number is equal to the number of letters in the surname, and the three remaining digits are the alphabetical positions of the first and last letters o... | 5163 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
How can you use a balance scale without weights to divide 24 kg of nails into two parts - 9 and 15 kg?
# | Try to weigh out 12 kg first, then -6 kg, and then -3 kg.
## Solution
Weigh out 12 kg of nails and set them aside. From the remaining 12 kg, weigh out 6 kg and set them aside in a different place. From the remaining 6 kg, weigh out 3 kg and combine them with the 6 kg that were set aside. We get the desired 9 kg of na... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Five first-graders stood in a line and held 37 flags. All those to the right of Tanya had 14 flags, to the right of Yasha - 32, to the right of Vera - 20, and to the right of Maksim - 8. How many flags does Dasha have?
# | ## Solution
Obviously, the more flags to the right of a first-grader, the "further left" their place in the line. Someone is standing to the right of Maksim (otherwise there would be no flags to his right). But everyone except Dasha is definitely standing to the left of Maksim. Therefore, Dasha is standing to the righ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & {[\text { Mathematical Logic (other) }]} \\ & {[\text { Arithmetic. Mental calculation, etc. }]}\end{aligned}$
Rabbits are sawing a log. They made 10 cuts. How many chunks did they get? | Into how many parts is a log divided by the first cut? How does the number of pieces change after each subsequent cut?
## Solution
The number of chunks is always one more than the number of cuts, since the first cut divides the log into two parts, and each subsequent cut adds one more chunk.
## Answer
11 chunks. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (miscellaneous).]
In the wallet, there are 2 coins with a total value of 15 kopecks. One of them is not a five-kopeck coin. What are these coins?
# | Two coins can only be 5 kopecks and 10 kopecks. If one of them is not a five-kopeck coin, then it is a ten-kopeck coin, and the other one is a five-kopeck coin. | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Mathematical logic (miscellaneous).]
We are looking for the correct statement. In the notebook, there are one hundred statements written:
1) In this notebook, there is exactly one false statement.
2) In this notebook, there are exactly two false statements.
..
100) In this notebook, there are exactly one hundred ... | In this notebook, there are 99 false statements ( $100-1=99$ ). | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[
There are pan scales without weights and 3 visually identical coins, one of which is counterfeit: it is lighter than the genuine ones (the genuine coins weigh the same). How many weighings are needed to determine the counterfeit coin?
# | When searching for a counterfeit coin among three coins, try placing one coin on each pan of the balance.
## Solution
We will need only 1 weighing. Place one coin on each pan of the balance. If one of the pans is lighter, the counterfeit coin is on it. If the balance is even, the counterfeit coin is the one that was ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a bookshelf, four volumes of Astrid Lindgren's collected works stand in order, with 200 pages in each volume. A worm living in this collection gnawed a path from the first page of the first volume to the last page of the fourth volume. How many pages did the worm gnaw through?
# | Try to recall how the volumes of a collected works stand on a bookshelf.
## Solution
Note that when the volumes stand on the shelf in order, the first page of the 1st volume touches the last page of the 2nd volume, and the last page of the 4th volume touches the first page of the 3rd volume. Thus, the worm only bored... | 400 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical logic (other)]
In the room, there are 85 balloons - red and blue. It is known that: 1) at least one of the balloons is red, 2) in every arbitrarily chosen pair of balloons, at least one is blue. How many red balloons are in the room?
# | Think about whether there can be two red balls in the room.
## Solution
Since among any two balls, one is blue, there cannot be two red balls in the room. Therefore, there are 84 blue balloons and 1 red balloon in the room. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In the garden of Anya and Vitya, there were 2006 rose bushes. Vitya watered half of all the bushes, and Anya watered half of all the bushes. It turned out that exactly three bushes, the most beautiful ones, were watered by both Anya and Vitya. How many rose bushes remained unwatered? | Vitya watered 1003 bushes, of which 1000 he watered alone, and three - together with Anya. Similarly, Anya watered 1003 bushes, of which 1000 she watered alone, and three - with Vitya. Therefore, together they watered $1000+1000+3=2003$ bushes. Thus, 2006 - 2003 = 3 rose bushes were left unwatered.
## Answer
3 bushes... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Golenischeva-Kumuzova T.I.
Yura has a calculator that allows multiplying a number by 3, adding 3 to a number, or (if the number is divisible by 3) dividing the number by 3. How can Yura use this calculator to get from the number 1 to the number 11? | Comment. Note that on Yura's calculator, any number can be increased by $1: (x \cdot 3+3): 3=x+1$. Therefore, in principle, from one, any natural number can be obtained on it.
## Answer
For example, $((1 \cdot 3 \cdot 3 \cdot 3)+3+3): 3=11$ or $(1 \cdot 3+3): 3+3+3+3=11$. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Zsspasii A.A.
The number of the current Olympiad (70) is formed by the last digits of the year of its holding, written in reverse order.
How many more times will such a situation occur in this millennium?
# | Let in some year the described coincidence occurred.
If the number of the Olympics is two-digit, then the sum of this number and the number formed by the last two digits of the year is divisible by 11 (the sum of two numbers consisting of digits $a$ and $b$ is $11(a+b)$). Since each year this sum increases by 2, the e... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Consider $C$.
$N$ digits - ones and twos - are arranged in a circle. An "image" is a number formed by several consecutive digits (either clockwise or counterclockwise). For what smallest value of $N$ can all four-digit numbers, whose representation contains only the digits 1 and 2, be among the images? | Note that the images of the numbers 1111, 2112, and 2122 cannot have common units, and the images of the numbers 2222, 1221, and 1211 cannot have common twos. Therefore, if all these numbers are among the images, then there must be at least 14 digits around the circle - 7 ones and 7 twos. The equality $N=14$ is possibl... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
On each square of a chessboard, there is a rook initially. 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 rooks betwee... | None of the rooks standing in the corner squares 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 either attacked by two rooks (if it stood on the edge of the board), or by ... | 59 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 two selected numbers,
3) find the roots of the equation $x^{2}+a x+b=0$ for selected numbers $a$ and $b$, and if there are no roots, display a message about it.
The res... | 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 $x=1$.
Send a comment | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Diding M.
There are 100 piles, each with 400 stones. In one move, Petya selects two piles, removes one stone from each, and earns as many points as the absolute difference in the number of stones in these two piles. Petya must remove all the stones. What is the maximum total number of points he can earn? | Evaluation. Let's assume that the stones in the piles are stacked on top of each other, and Petya takes the top (at the moment) stones from the selected piles. Let's number the stones in each pile from bottom to top with numbers from 1 to 400. Then the number of points Petya gets on each move is the difference in the n... | 3920000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Anzhans A.
There are 100 silver coins, ordered by weight, and 101 gold coins, also ordered by weight. It is known that all coins have different weights. We have a two-pan balance that allows us to determine which of any two coins is heavier. How can we find the coin that ranks 101st in weight among all the coins with ... | Let's prove that if from $n$ silver and $n$ gold coins, the $n$-th by weight coin can be found in $k$ weighings, then the $2n$-th by weight coin can be found in $k+1$ weighings from $2n$ silver and $2n$ gold coins.
Indeed, suppose the $n$-th silver coin is heavier than the $n$-th gold coin. Then the $n$ first silver c... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Koschei the Deathless has kidnapped the three daughters of the tsar. Ivan the Tsarevich set out to rescue them. He comes to Koschei, and Koschei says to him: "Tomorrow morning, you will see f... | a) Let all the princesses call Koshchei's daughters princesses (see the left image). Then Koshchei's daughters will be called at least three times, while princesses will be called no more than twice. This is how Ivan will distinguish them.
 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 cle... | 33 | Logic and Puzzles | 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 from 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?" O... | 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 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 |
7,8 |
Petya plans to spend all 90 days of his vacation in the village and during this time, he will go swimming in the lake every second day (i.e., every other day), go to the store for groceries every third day, and solve math problems every fifth day. (On the first day, Petya did the first, second, and third activit... | Let's number the days from 0 to 89. Petya swims on days with even numbers, goes to the store on days with numbers divisible by 3, and solves problems on days with numbers divisible by 5.
Thus, we need to answer two questions:
1) How many even numbers from 0 to 89 are not divisible by 3 or 5?
2) How many odd numbers f... | 24 | Other | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
| :---: | :---: | :---: |
| | Division with remainder | |
| | Product rule | |
| | Cooperative algorithms Evaluation + example | |
Authors: Knyaz K.A., Leontyeva O.
A magician and an assistant are going to perform the following trick. A spectator writes a sequence of $N$ digits on a board. The assist... | Suppose that for some value of $N$ the trick is possible. Then for each variant of the sequence with two closed digits (let their quantity be $k_{1}$), the magician can restore the original; hence, to each sequence with two closed digits, the magician can uniquely correspond the restored sequence of $N$ digits (let the... | 101 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Authors: Bogdanov I.Y., Knop K.A.
King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are 1, 2, ..., 11 kg. He also has a bag that will tear if more than 11 kg is placed in it. Archimedes has learned the weights of all the ingots and wants to prove ... | Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then put ingots weighing 1, 4, and 6 kg. In both cases, the bag will not tear.
We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_1, \ldots, w_6 \) kg instead of ingots ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Algorithm Theory (miscellaneous).] Evaluation + example $\quad]$
Authors: Blinkov A.D., Blinkov Y.A., Gorska E.S.
A casino offers a game with the following rules. The player bets any whole number of dollars (but no more than they currently have) on either heads or tails. Then a coin is flipped. If the player guesse... | The main observation in this problem is this: as soon as the player guesses correctly how the coin will land once, they can guess incorrectly in all remaining games. Therefore, all bets after a win should be $1.
1. The first time, the player has to bet at least $3, because they could win this game and lose the next fo... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10,11
There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be broken so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 150 g can be formed (a broken link also weighs 1 g)? | Answer: 4 links. According to the solution of problem 5 for grades $7-8$, for a chain consisting of $n$ links, where $64 \leq n \leq 159$, it is sufficient to unfasten 4 links. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10,11
In the corners of a 3x3 chessboard, knights are placed: white knights in the top corners, black knights in the bottom corners. Prove that to swap their positions, it will take no fewer than 16 moves. (The knights do not necessarily move in the order of white, then black. A move is defined as one knight moving.) | Let's number the fields of the board (except the central one) in the order of their traversal by a knight (Fig. 40, a); white knights stand on fields 1 and 3, black ones - on fields 5 and 7. Consider the auxiliary scheme shown in Fig. 40, b. The positions of the knights are denoted by symbols: o - white knight, $x$ - b... | 16 | Logic and Puzzles | proof | Yes | Yes | olympiads | false |
A tanker of milk was delivered to the store. The seller has balance scales without weights (fляги can be placed on the pans of the scales), and three identical flasks, two of which are empty, and the third one contains 1 liter of milk. How can you pour exactly 85 liters of milk into one flask, making no more than eight... | First, let's place two empty flasks on one scale pan and on the other pan - a flask containing 1 liter of milk, and pour milk into one of the empty flasks until the scales balance. Then, this flask will contain (1 - a) liters of milk, where $a$ is the number of liters of milk that balances one empty flask. Next, let's ... | 85 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
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 receives nine handfuls, after which the other takes all the remaining coins (the division may... | 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 |
36 tons of cargo are packed in bags with a capacity of no more than 1 ton. Prove that a four-ton truck can transport this cargo in 11 trips.
# | Each time we load the car until the weight of the cargo exceeds 4 tons. After that, we remove one bag and set it aside. After eight trips, we will set aside 8 bags, and less than 4 tons of cargo will remain on the warehouse, which we will transport on the ninth trip. On the tenth and eleventh trips, we will transport t... | 1993 | Combinatorics | proof | Yes | Yes | olympiads | false |
Tokarev S.i.
Among 2000 indistinguishable balls, half are aluminum with a mass of 10 g, and the rest are duralumin with a mass of 9.9 g. It is required to separate the balls into two piles such that the masses of the piles are different, but the number of balls in them is the same. What is the smallest number of weigh... | Let's compare the mass of 667 balls with the mass of another 667 balls. If the masses of these two piles are not equal, the required condition is met.
Suppose the specified masses are equal. Then the mass of 666 balls that did not participate in the weighing is not equal to the mass of any 666 balls lying on one of th... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The Cantor Set. A segment of the number line from 0 to 1 is painted green. Then its middle part, the interval $(1 / 3 ; 2 / 3)$, is repainted red. Next, the middle part of each of the remaining green segments is also repainted red, and the same operation is performed on the remaining green segments, and so on to infini... | Numbers from 0 to 1 can conveniently be considered as infinite ternary fractions of the digits 0, 1, and 2. The numbers mentioned in point v) are those numbers whose ternary representation contains no 1.
| | Game Theory (Other) |
| :---: | :---: |
| | Similarity Criteria |
| | [ Auxiliary Similar Triangles |
| Prob... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Franklin B.P.
A traveler visited a village where every person either always tells the truth or always lies. The villagers stood in a circle, and each one told the traveler whether the person to their right was truthful. Based on these statements, the traveler was able to uniquely determine what fraction of the village... | Let $x$ be the fraction of liars. Imagine that all truthful residents became liars, and all liars "reformed." Then the traveler would hear the same thing! Indeed, the truthfulness of any resident has changed, but the truthfulness of the neighbor they are talking about has also changed. But the fraction of truthful peop... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Mathematical Logic (Miscellaneous).] $[$ Evenness and Oddness $]$
Author: Khaitumuren A.v.
13 children sat around a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their... | It is clear that there were both boys and girls at the table. A group of boys sitting next to each other is followed by a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so their number is even. Incorrect statements were made at the tr... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$:$ Folkiore
It is known that among 63 coins there are 7 counterfeit ones. All counterfeit coins weigh the same, all genuine coins also weigh the same, and a counterfeit coin is lighter than a genuine one. How can you determine 7 genuine coins in three weighings using a balance scale without weights? | 1) Let's set aside one coin and put 31 coins on each pan of the scales. If the pans balance, then we have set aside the counterfeit coin, and there are 3 counterfeit coins on each pan. If one of the pans is heavier, then there are no more than three counterfeit coins on it. Thus, after the first weighing, we will be ab... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On an island, there live three tribes: knights, who always tell the truth, liars, who always lie, and tricksters, who sometimes tell the truth and sometimes lie. At a round table, 100 representatives of these tribes are seated.
Each person at the table made two statements: 1) “To my left sits a liar”; 2) “To my right ... | Since knights tell the truth, a liar sits to the left of each knight. Let's now prove that a knight sits to the right of each liar. Indeed, since liars lie, a trickster does not sit to the right of each liar. Moreover, a liar cannot sit to the right of another liar, because then the right liar would have told the truth... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Around a large round table, 60 people are sitting, each of whom is either a knight or a liar. Each of them said: “Of the five people sitting in a row to my right, at least two are liars.” How many knights can sit at this table? | Let's divide 60 people sitting at a table into 10 groups of 6 people each and prove that in each group there are exactly two liars. Let's consider two cases.
1) Suppose the first person in such a group is a knight. Then he told the truth, and among the five people in this group sitting to his right, at least two are l... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a large box, 10 smaller boxes were placed. In each of the nested boxes, either 10 even smaller ones were placed, or nothing was placed. In each of the smaller ones, either 10 or none were placed again, and so on. After this, it turned out that there were exactly 2006 boxes with contents. How many are empty? # | Each box A, except for the very first one, is associated with a box B in which it lies (if the box B itself lies in some other box, we do not associate it with box A). Thus, each box, except for the first one, is associated with exactly one box, and each non-empty box is associated with exactly ten boxes. Let S denote ... | 18055 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Winning and losing positions ]
There are 300 matches in a box. On a turn, it is allowed to take no more than half of the matches in the box. The player who cannot make a move loses. | The first player wins. Winning positions are those in which there are $2^{n}$ - 1 matches left in the box.
The first move is to leave 255 matches. | 255 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kazitsyna T.v.
Four mice: White, Gray, Fat, and Thin were dividing a piece of cheese. They cut it into 4 visually identical slices. Some slices had more holes, so Thin's slice weighed 20 grams less than Fat's, and White's slice weighed 8 grams less than Gray's. However, White was not upset because his slice weighed ex... | Notice that now the slice of Gray weighs as much as the slice of White, i.e., exactly a quarter of the total cheese mass. This means that both White and Gray have already received their share, and all 28 grams should be divided between Fat and Thin. Currently, they have an equal amount of cheese, so they should receive... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}{\left[\begin{array}{l}\text { Riddles }\end{array}\right]} \\ \text { [ CaseAnalysis }\end{array}\right]$
Authors: Tokoreve S.I. Khachatryan A.B.
Find the smallest four-digit number СЕЕМ for which there is a solution to the riddle МЫ + РОЖЬ = СЕЕМ. (Identical letters correspond to identical di... | With $C > P$ and in particular, $C > 1$.
## Solution
Since $C > P$, then $C > 1$. As we are looking for the smallest number, let's try $P=1, C=2$, and $E=0$. Then $M=3$. The case $CEEM = 2003$ is possible: $35 + 1968 = 2003$ or $38 + 1965 = 2003$.
In addition to the solutions mentioned, the puzzle has 38 more soluti... | 2003 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
It is known that among the members of the government of Lemonia (and there are 20 members in total) there is at least one honest member, and that among any two members, at least one is a bribe-taker. How many bribe-takers are there in the government?
# | Note that in the government of Lemonia, there is exactly one honest official. Indeed, according to the condition, there is one honest official; but there cannot be two honest officials—otherwise, none of them would be a bribe-taker, which contradicts the condition. Therefore, there is exactly one honest official in the... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
On the Island of Misfortune, only knights, who always tell the truth, and liars, who always lie, live. In
the Island's Council - there are 101 deputies. To reduce the budget, it was decided to reduce the Council by one deputy. But each of the deputies stated that if they were removed from the Council, the majority amo... | Write two inequalities corresponding to the statements of the deputy liar and the deputy knight.
## Solution
Let $P$ be the number of knights in the Duma, and $L$ be the number of liars ($P+L=101$). According to the statement of the deputy knight, $P-1<50$. Therefore, $P<51$ (from this, in particular, it follows that... | 51 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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