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2. In the notebook, a double leaf (sheet) is torn out. The page numbers on the torn-out leaves are formed by the following digits: $0,1,1,3,3,3,4,5,6,9$. What were the page numbers on the torn-out leaf? Find all possible options and prove that there are no others. | Answer: 59 / 60 / 133 / 134.
Solution:
Of the four numbers on the sheet, two are even, and their sum is 194; therefore, they end in 0 and 4. Then the odd numbers paired with them end in 9 and 3. From the remaining digits 1,1,3,3,5,6, we need to form the beginnings; it is clear that this can be done in only one way. | 59/60/133/134 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,573 |
3. Gosha is looking at the equation: $71 * \ldots+72 * \ldots+73 * \ldots+74 * \ldots=2014$ and wants to fill in the blanks with one digit each to make the equation true. Explain why Gosha will not be able to do this. | Solution:
If the sum of the digits written by Gosha is 27 or less, then the result will not exceed 74*27=1998. If the sum of the digits written by Gosha is 28 or more, then the result will be no less than $(71+72+73) * 9+74 * 1=2018$. Therefore, no number between 1999 and 2017 can be obtained. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,574 |
4. The castle consists of 81 rooms in a square shape $9 * 9$. In some walls between adjacent rooms, there is one door. There are no doors to the outside, but each room has at least two doors. Prove that in some room, there are at least three doors. | Solution:
If all rooms have degree two, then the castle is a union of cycles. Due to the chessboard coloring, each such cycle must have an even length. But the total number of rooms is odd. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,575 |
5. Each of the 777 bankers has several safes. At a meeting, the Chief Oligarch gave each banker N diamonds and ordered them to distribute them among the safes so that each safe contains a different number of diamonds. The Chief Oligarch is certain that this task is feasible. Prove that the bankers can keep no more than... | # Solution:
Obviously, the necessary condition for the fulfillment of the Oligarch's task is that the number of safes is less than N. Also, each banker, except possibly one, has no less than two safes. We will show that after destroying all "extra" safes of each banker, this task can still be accomplished. The banker ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,576 |
6. Let's call a number feerich if it is prime or can be factored into no more than four prime factors. Vasya wrote down in a row 7 consecutive three-digit numbers. Prove that among them, one can choose two adjacent numbers so that the six-digit number formed by them is feerich. | The given six-digit number has the form $1001k + 1$, where $k$ takes on 6 consecutive values. Among the three even values of $k$, the remainders when divided by 3 and 5 do not repeat, so one can choose $k$ such that $1001k + 1$ is not divisible by 2, 3, or 5. Since the chosen number $1001k + 1$ is also coprime with 100... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,577 |
7. 99 people - knights and liars (knights always tell the truth, while liars always lie) - are standing in a row. Each of them said one of two phrases: "To the left of me, there are twice as many knights as liars" or "To the left of me, there are as many knights as liars." In reality, there were more knights than liars... | Answer: 49.
Solution:
There are more knights than half, so:
1) Either they alternate like this: KRK...LK, but this option does not work: the phrase "there are twice as many knights as liars" cannot be said by any knight, but there are more than 50 such phrases.
2) Or some two knights stand next to each other. Two k... | 49 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,578 |
4. Vasya drew a grid figure. It turned out that it can be cut into $2 \times 2$ squares, and it can also be cut into zigzags of 4 cells. How many cells can be in Vasya's figure?
## Solution: | Answer: any number that is a multiple of 8, starting from 16.
Example:
Evaluation: we will prove that the number of cells is a multiple of 8, that is, that the number of figures obtained by... | anythatismultipleof8,startingfrom16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,580 |
5. On a $3 \times 99$ board, a chip is placed in the central cell. Magnus and Sergey take turns moving the chip one or two cells vertically or horizontally. It is forbidden to stop the chip in a cell where it has already been. The player who cannot make a move loses. Who will win with correct play if Magnus starts? | # Solution:
Let's divide the entire board, excluding the central cell, into dominoes. If Magnus places a chip on one of the dominoes, Sergey will move the chip to the second cell of that domino. With this strategy, Magnus will always be forced to move to an unvisited domino, and eventually, they will run out and Serge... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,581 |
7. Someone added several consecutive natural numbers and with shameful satisfaction noted that the result was the sum of two identical powers of two and three. Could he be right? | Solution:
Let's write this equation in a more familiar form
$$
1+\cdots+n=\frac{n(n+1)}{2}=2^{m}+3^{m}
$$
Multiply both sides by 2 and consider the equation modulo 9. The left side can take the values $0,2,3$ and 6, while the right side can take the values $1,2,4,5,7$ and 8. Therefore, both sides are equal to 2 modu... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,582 |
1. Today's date is written as: 22.11.2015. How many other days in this year can be written with the same set of digits? | Solution. The month number cannot start with a two, so it is either 11 or 12. In the first case, it is 22.11, in the second case 12.12 and 21.12. Answer: two. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,583 |
3. Once, two elves Ilsa and Elsa, and two dwarves Bilyn and Dilyn gathered. One of them gave something to their common friend, a human named Vasily, after which each of them spoke:
- The gift was a sword
- I didn't give anything!
- Ilsa gave a necklace.
- Bilyn gave a sword.
It is known that the elves lie when speaki... | Solution. The phrase "I didn't give anything" could not have been said by the elf, so it must have been said by the dwarf. Since this phrase is about a gift, it is a lie, and this dwarf must have given something to Vasily. Then the phrase "Ilse gave the necklace" is a lie, and it could not have been said by the elf. Th... | Dilyngavethesword | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,585 |
4. A group of toddlers in a kindergarten has 90 teeth in total. Any two toddlers together do not have more than 9 teeth. What is the minimum number of toddlers that can be in the group | Solution. If all children have fewer than 5 teeth, then there are no fewer than $90 / 4$, i.e., no fewer than 23 children.
If one child has exactly 5 teeth, then the others have no more than 4, and there are no fewer than $1 + 85 / 4$, i.e., also no fewer than 23 children.
If any of the children have between 6 and 9 ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,586 |
5. Given nine natural numbers. The first is written only with ones, the second - only with twos, ..., the ninth - only with nines. Can one of these numbers be equal to the sum of all the others? | Solution. No, because in this case the sum of all numbers should be even, but it is odd. Another possibility is to try all options for the last digit. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,587 |
7. At a masquerade ball, there were 20 people. In each dance, two people participated. It turned out that eleven of them danced with three partners, one with five, and the remaining eight with six. Prove that in some dance, people of the same gender participated. | Solution. Let it not be so. Note that the number of dances that everyone except one person has danced is a multiple of three. Suppose this person is a boy. Then the number of dances in which boys participated is divisible by 3, while the number of dances in which girls participated is not. Since these numbers must coin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,589 |
1.3. Prove that no matter how the colonies are initially arranged, migrations can be used to settle the colonies with one on each island. | Solution. Induction on the number of vertices.
The base case is trivial.
Transition. We will prove that a non-zero value can be obtained in any vertex by generalizing the reasoning from point 2. Hang the tree from this vertex. Consider such a semi-invariant: the sum over all vertices of the quantities $x_{i} n^{-d_{i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,592 |
1.4. Let initially each island is inhabited by one colony, and let one of the islands have $d$ neighboring islands. What can the maximum possible number of colonies that can settle on this island be equal to? | Solution. Answer: $d+1$.
Example. We will prove that in a vertex of degree $d$, $d+1$ colonies can gather. Suspend the tree from this vertex as the root and prove that in each vertex from which $e$ edges go down, $e+1$ colonies can gather, conducting only migrations within its subtree. We will prove this by "induction... | +1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,593 |
3.3. Prove that there exists $n$ such that $a\left(n, 10^{100}\right) \leqslant 1+10^{-100}$. | Solution. Consider all possible ways to write the number $1+10^{-100}$ as a sum of positive rational fractions with the denominator $10^{200}$ (not necessarily in lowest terms). Clearly, this number is finite, and we denote it by $n$ - by writing down all corresponding sets on cards, we can verify that such a set of ca... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,599 |
3.4. Prove that there exists $k$ such that $a\left(10^{100}, k\right)>10^{100}$. | Solution. Suppose we have a suitable set of $10^{100}$ cards. We will order the numbers on each card in descending order. For any $l \leqslant k$, we must be able to majorize the set consisting of $l$ numbers equal to $\frac{1}{l}$ and $k-l$ zeros, so for each such $l$, there must be a card where the $l$-th number is n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,600 |
1. Several cells are marked on a chessboard. A rook can pass through these cells but cannot stop on them. Can it happen that the shortest path from A1 to C4 contains 3 rook moves, from C4 to H8 - 2 moves, and from A1 to H8 - 4 moves? | Solution. It can. For example, all cells except A1, A2, C2, C4, C8, H8 are marked. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,601 |
2. From head to tail of the zebra Hippotigris - 360 stripes of the same width. Flea Masha and flea Dasha started crawling from the head of the zebra to its tail. At the same time, flea Sasha started crawling from the tail to the head. Flea Dasha crawls twice as fast as flea Masha. Before meeting flea Sasha, Masha overc... | Solution: 240 stripes.
Mashka crawled half of the zebra. Let her speed be $v$, then the closing speed of Mashka and Sashka is $2v$, and the closing speed of Dashka and Sashka is $3v$. Therefore, Sashka will crawl $3/2$ times fewer stripes before meeting Dashka than before meeting Mashka, i.e., 120 stripes. The remaini... | 240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,602 |
3. In a row, 100 knights and 100 liars are standing (in some order). The first person was asked: "Are you a knight?", and the rest were asked in turn: "Is it true that the previous person answered 'Yes'?" What is the maximum number of people who could have said "Yes"? Knights always tell the truth, liars always lie. | Solution. 150, for example, if there are 100 knights followed by 100 liars. We will now prove that this is the maximum.
Consider any liar, except possibly the first one. Either they said "no," or the previous person said "no." Thus, to each of the 99 liars, we can associate at least one "no" response (either theirs or... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,603 |
4. The country of Dodecahedria has 20 cities and 30 air routes between them. The map of air routes in Dodecahedria is shown in the figure. In one of the cities is Fantomas, whom the police want to catch. Every day, Fantomas flies to another city using exactly one air route. Every evening, the police learn which city Fa... | # Solution.
Method 1, in which it is not even necessary to see Fantomas until he is caught.
The police can catch Fantomas as follows: the police mentally divide the set of cities into two non-intersecting groups A and B, each with 10 cities.
The next actions by the police make all air routes internal for group A. Th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,604 |
5. Kostya replaced even digits in the example with vowels, and odd digits with consonants (the same digits with the same letters, different digits with different letters). Could the puzzle KRONA + UNCIA = TURKA have resulted?
---
Note: The translation maintains the original text's formatting and structure. | Solution. Suppose such an example could have been obtained. Let's make sure that there are no carries in this example. Let's look at the digits, starting with the most significant. $\mathrm{K}+\mathrm{Y}=\mathrm{T}$ (if there is no carry) or $\mathrm{K}+\mathrm{V}=\mathrm{T}-1$ (if there is). By parity, we see that the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,605 |
6. There is a grid table, in which some cells are shaded. "I can shade 5 more cells in each row," said Yakov, "and then in each column there will be as many shaded cells as there are in each row now." "And I can erase 3 cells in each column," Yuri replied, "and then in each row there will be as many shaded cells as the... | Solution. It is clear that if we color 5 more cells in each row, then in all rows there will be an equal number of colored cells; hence, initially, there were an equal number of colored cells in the rows. The same reasoning applies to the columns.
Let there be $k$ rows with $x$ colored cells in each, and at the same t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,606 |
7. There is a strip of 101 cells, on which a chip can move: forward by any even number of cells, and backward by any odd number of cells. Vasya and Petya want to visit all the cells of the board once with their chips: Vasya starting from the first cell, and Petya starting from the fiftieth. Who has more ways to do this... | Solution. The number of ways is the same. Imagine that this is not a strip but a ring (connect the beginning with the end), and you can only move forward and only by an even number of cells. Making a move backward by an odd number of cells is equivalent to making a move forward by an even number of cells along the ring... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,607 |
1. Jerry wrote a palindrome on a sheet of paper (reads the same from left to right and right to left). Tom, running by, tore it into five pieces with his claws. The pieces got mixed up, and the result was: $M S, S U$, US, MUS, UMM. Provide an example of the original palindrome. | Solution. SU MS UMM US MUS (SUM SUMMUS MUS in Latin means "I am the strongest mouse").
| SUMMUS | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,608 |
2. The gnomes went to work, and Snow White is feeling lonely. She laid out a pile of 36 stones on the table. Every minute, Snow White splits one of the existing piles into two, and then adds a new stone to one of them. After some time, she had seven piles, each with an equal number of stones. How many stones ended up i... | Solution. There will be seven piles after six moves. After six moves, there will be $36+6=42$ stones - meaning, 6 stones in each pile. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,609 |
3. Given a paper square of $4 \times 4$ cells. It can be folded along the sides and diagonals of the cells, but not in any other way. Can a 12-sided polygon be obtained using such folds? | Solution. It is possible, the folding pattern is shown in the figure.
 | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,610 |
4. In a certain city, the fare scheme for traveling by metro with a card is as follows: the first trip costs 50 rubles, and each subsequent trip costs either the same as the previous one or one ruble less. Petya spent 345 rubles on several trips, and then on several subsequent trips - another 365 rubles. How many trips... | Solution. A total of 710 rubles was spent. The total number of trips could not have been 14 or less (for 14 trips, a maximum of 700 rubles could be spent), but it also could not have been 17 or more $(50+49+48+\cdots+35+$ $34=714$, and this is the minimum that could be spent). Therefore, the choice is between two optio... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,611 |
5. Seven people stood in a circle, each of whom is either a knight, who always tells the truth, or a liar, who always lies, or a traveler, who alternates between truth and lies.
The first and second said in unison: "Among us there is exactly 1 liar," the second and third: "Among us there are exactly 2 knights," the th... | Solution. If all 7 statements are lies, then everyone lied, meaning everyone is a liar, and then 7 and 1 told the truth. This means that at least one statement is true. On the other hand, there are no more than 2 true statements (one about the liars, the other about the knights). That is, there are either 0 or 1 knight... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,612 |
6. When the child was born, their parents were not yet 40 years old, but they were already adults. When the child turned 2 years old, the age of exactly one of the parents was divisible by 2; when the child turned 3 years old, the age of exactly one of the parents was divisible by 3, and so on. How long could such a pa... | Solution. The condition means that at the moment of the child's birth, the age of exactly one of the parents was divisible by 2, the age of exactly one of the parents - by 3, and so on (as long as this pattern continued).
Let one of the parents be 24 years old, and the other 35 years old. Then this pattern could conti... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,613 |
7. In a company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to triple the salary for those earning up to 500 dollars, and increase the salary by 1000 dollars for the rest, so the total salary will become 24000 dollars. A mean manager proposes to reduce the salary to 500 doll... | Solution. Note that the increase proposed by the kind manager is twice as large as the salary proposed by the evil manager (this is true for both poor and rich employees). The increase according to the kind manager's proposal is 14000, so the salary according to the evil manager's proposal is 7000. | 7000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,614 |
1. In the marked points (see figure), there are 6 burrows. In four of the burrows live hobbits: Frodo, Sam, Merry, and Pippin. The other two burrows are empty, and both are closer to Sam's burrow than Frodo's. Frodo's burrow is closer to the river than Merry's, but farther from the tree line than Pippin's. Who lives wh... | Solution. Let $A, B, C, D, E, F$ be the holes marked in the order "from left to right and from top to bottom". Frodo's hole can be $D, E, F$ from the condition that it is closer to the river than Merry's hole, and $B, C, E$ from the condition that it is further from the forest than Pippin's hole. Therefore, Frodo lives... | FrodoinE,SaminA,PippininF,MerryinC | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,615 |
2. 31 cars started simultaneously from one point on a circular track: the first car at a speed of 61 km/h, the second at 62 km/h, and so on (the 31st at 91 km/h). The track is narrow, and if one car overtakes another by a full lap, they crash into each other, both fly off the track, and are eliminated from the race. In... | Solution. First, the fastest car collides with the slowest, then the second fastest collides with the second slowest, and so on. In the end, the car with the median speed remains, i.e., the 16th. It travels at a speed of $76 \mathrm{Km} /$ h.
Criteria. Full solution - 3 points.
1 point if only the answer 76 is writte... | 76 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,616 |
3. The Elector George has 100 coins, some of which may be counterfeit (possibly all or none). George can show the expert between 10 and 20 coins, and the expert will tell him how many of them are counterfeit. The problem is that the only expert in the area is Baron Munchausen, and he exaggerates: the result given by th... | Solution. It can. We will submit 10 random coins for examination, and then the same 10 and 1 additional coin. If the baron says the same number both times, the added coin is genuine; otherwise, it is counterfeit. Thus, in two questions, we can check any coin for being counterfeit.
Criteria. Full solution - 4 points.
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,617 |
5. Each of the five friends multiplied several consecutive numbers starting from 1. It turned out that one of the products is equal to the sum of the other four. Find all possible values of this product and show that there are no other values. | Solution. Let's call Vasya the friend whose product is equal to the sum of the others. It is clear that Vasya multiplied more numbers than the others. If he had multiplied at least 5 numbers, his product would be at least 5 times greater than the product of any of the remaining friends. But his product is the sum of th... | 1\cdot2\cdot31\cdot2\cdot3\cdot4 | Algebra | proof | Yes | Yes | olympiads | false | 21,619 |
6. Eight dwarfs are sitting around a round table, each with three diamonds. The dwarfs' chairs are numbered in order from 1 to 8. Every minute, the dwarfs simultaneously do the following: they divide all their diamonds into two piles (one or both piles may be empty), then give one pile to their left neighbor and the ot... | Solution. Note that gnomes with even numbers always share with gnomes with odd numbers, and vice versa. Since all distributions happen simultaneously, all diamonds of gnomes with even numbers will go to gnomes with odd numbers and vice versa. That is, the gnomes with even numbers will always have a total of 12 diamonds... | 512 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,620 |
7. Can the equation БАРАНКА + БАРАБАН + КАРАБАС = ПАРАЗИТ be solved by replacing all letters with digits (the same letters with the same digits, and different letters with different digits) to make it true?
Note: The equation uses Russian letters. For the purpose of solving, the equation can be represented as: BАRANKA... | Solution. Note that $3 \cdot \mathrm{APA}+x=\mathrm{APA}+1000 \cdot y$, where $x$ and $y$ are possible values of carry-overs, each from 0 to 2. If we simplify this expression, it turns out that $2 *(\mathrm{APA})+x$ is divisible by 1000. Let's look at the possible values of $x$:
- when $x=0$, it turns out that $\mathr... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,621 |
1. Five brothers were dividing their father's inheritance equally. The inheritance included three houses. Since the houses couldn't be divided, the three older brothers took them, and the younger brothers were given money: each of the three older brothers paid $2000. How much did one house cost?
## Solution: | Answer: $5000.
Since each of the brothers who received money got $3000, the total inheritance was estimated at $15000. Therefore, each house was worth a third of this amount. | 5000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,622 |
3. Sasha bought 2 pens, a pencil, and 7 notebooks for exactly 100 rubles at a stationery store. His sister Masha wants to buy a pen, a notebook, and three pencils. She has many 5-ruble coins with her. Will she definitely be able to pay with them without getting change, if each item costs a whole number of rubles?
# | # Solution:
Answer: Yes.
From the condition, it follows that 6 pens, 3 pencils, and 21 notebooks cost 300 rubles, which means that 1 pen, 3 pencils, and 1 notebook differ from 300 by a sum that is a multiple of 5.
Guidance on the problem approach: if a participant gives only the answer without supporting it with rea... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,623 |
4. In the Parliament of the Emerald City, 5 parties are represented, which together developed 100 laws over the year (each law was developed by exactly one of the parties). It is known that any three parties together developed no fewer than 50 laws. What is the maximum number of laws that the Green Lenses party could h... | Solution:
Answer: 33.
Evaluation. Let this quantity be $x$. Since any three parties together have developed no less than 50, any party together with the Green Lenses party has developed no more than 50. Therefore, each of the four other parties has developed no more than $50-x$, and all of them together have develope... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,624 |
2. Alyona erased one digit from a five-digit number divisible by 99, and it turned out that the resulting four-digit number is also divisible by 99. Which digit in order could have been erased? (Neither the original nor the resulting number can start with zero.) | Solution. Answer: any, except the first and second Examples:
$$
\begin{aligned}
99099 & \rightarrow 9999 \\
49500 & \rightarrow 4950 \\
99990 & \rightarrow 9999
\end{aligned}
$$
First, note that according to the divisibility rule for 9, the digit 0 or 9 was removed.
The first digit cannot be removed: since it is not... | any,exceptthefirst | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,626 |
4. The distance between two palaces in the Kingdom of Zenithia is 4 km. The king ordered to build a circular Arena with a diameter of 2 km between them. Prove that no matter where the king orders to build the Arena, the traffic police will be able to build a road between the palaces so that it is no longer than 6 km. | Solution. Let the segment between the palaces run from west to east. Consider the center of the arena; if, for example, it is not north of this segment, then any point of the arena north of this segment is no more than 1 km away. Therefore, we can first walk 1 km north, then 4 km parallel to the segment, and then 1 km ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,627 |
7. Two players take turns writing down ones or twos. The player who, after their move, makes the sum of the last few digits equal to (a) 533; (b) 1000, loses. Who will win if both players play optimally? | Solution. (a) The first player wins. His first move is to write the number 1, and then he complements the opponent's move to 3 (responding to a one with a two, and to a two with a one). After 147 such complements, before the second player's move, the sum of the numbers on the board will be 532, which means that the sec... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,628 |
1. In the class, there are 10 people, and each person has three friends among their classmates. They split into 5 teams of 2 people, with each team consisting of friends. The teacher said they should reorganize so that Vasya and Petya are not on the same team. Vasya noticed: "But then, one of the teams will definitely ... | Solution. Let's provide an example. In this picture, the vertices (points) are students, and
the edges are friendships between them. In this case, they can be paired up. For example, $\mathr... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,629 |
2. Five patricians each have five statues, each costing a natural number of sesterces. First, the patricians compared who had the greater total cost of their statues, then each of them smashed their cheapest statue. Then they compared again, smashed their cheapest remaining statues, and so on, three more times, until a... | Solution. This could happen. For example, with such sets of statue values:
| I: | $10^{10}+1$, | $10^{9}$, | $10^{8}$, | $10^{7}$, | $10^{6}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| II: | $10^{10}$, | $10^{9}+10$, | $10^{8}$, | $10^{7}$, | $10^{6}$ |
| III: | $10^{10}$, | $10^{9}$, | $10^{8}+100$, | $10^... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,630 |
3. Andrey placed chips of a hundred different colors in the cells of a $10 \times 10$ board. Each minute, one of the chips changes color, and only a chip that was unique (i.e., differed in color from all others) in its row or column before this operation can change color. After $N$ minutes, it turned out that no chip c... | Solution. Let's look at the moment when there are no moves left. Suppose we have $k$ colors left, then we have repainted at least $100-k$ cells. Note that if nothing can be repainted, then there are at least 4 cells of each color. In total, there are at least 25 colors left after this process. Therefore, we will repain... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,631 |
4. Given three real numbers greater than 1. The product of any two of them is greater than the sum of these two. Prove that the product of all three numbers is greater than their sum, increased by 2. | Solution. Let $a \leqslant b \leqslant c$ be our numbers. Since all numbers are positive, multiplying both sides by $a, b$, or $c$ does not change the inequality sign. Given that $a b>a+b$, then
$$
a b c>(a+b) c=a c+b c>a+c+b+c=a+b+2 c .
$$
Notice that $a c>a+c \geqslant 2 a$. Therefore, $c>2$. Hence,
$$
a b c>a+b+c... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 21,632 |
6. In triangle $A B C$, angle $A$ is equal to $50^{\circ}, B H$ is the altitude. Point $M$ on $B C$ is such that $B M=B H$. The perpendicular bisector of segment $M C$ intersects $A C$ at point $K$. It turns out that $A C=2 \cdot H K$. Find the angles of triangle $A B C$. | Solution. Draw a perpendicular to $BC$ from point $M$. Let $K^{\prime}$ be the point of its intersection with $AC$. Then the segment $MK$ is the midline in $\triangle K^{\prime}MC$, that is, $K^{\prime}K = KC$. From the fact that $HK = KC + AH$, we get $AH = HK^{\prime}$. Right triangles $BHA$ and $BHK^{\prime}$ are eq... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,634 |
7. Points are placed at equal intervals on a circle, forming a 33-sided polygon by connecting adjacent points with segments. Each side of this 33-gon is painted in one of three colors, and it turns out that there are an equal number of segments of each color. Prove that the 33-gon can be divided into triangles by non-i... | Solution. Let's find two adjacent sides of different colors on the perimeter and cut off the corresponding triangle with a diagonal (painting the third side in the third color), reducing the problem to a polygon with fewer vertices. We will continue to act this way - we just need to make sure that we do not end up with... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,635 |
8. Two players take turns writing down ones or twos. The player who, after their move, makes the sum of the last few digits equal to (a) 533; (b) 1000, loses. Who will win if both players play optimally? | Solution. (a) The first player wins. His first move is to write the number 1, and then he complements the opponent's move to 3 (responding to a one with a two, and to a two with a one). After 147 such complements, before the second player's move, the sum of the numbers on the board will be 532, which means that the sec... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,636 |
1. Show how to cut an $8 \times 8$ square along the grid lines into four parts, each with the same area but pairwise different perimeters. | Solution. One of the possible solutions is shown in the figure.
 | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,637 |
2. In front of Misha, there are four girls. One girl has one cat at home, another has two, the third has three, and the fourth has four, but Misha doesn't know which girl has how many cats. Misha can point to one or several girls and ask how many cats they have in total. How can he find two girls who have exactly 5 cat... | Solution. Ask the first two girls how many cats they have. Ask the first and third girls how many cats they have. If at least one of the answers is "Five," then the problem is solved. If not, then the first and fourth girls will have five cats, since for any girl, there will be another girl with whom the total number o... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,638 |
3. We have two cards with the digit 5, two with the digit 6, two with the digit 7, two with the digit 8, and two with the digit 9. From these cards, two five-digit numbers were formed and the resulting numbers were added. The first digit of the result, as expected, is 1. Can it be that each of the remaining digits is n... | Solution. Answer: It cannot.
Method 1. Notice that in each of the five digits, there is a carry. Let's calculate the sum of the digits of the result. The sum of the digits on the card is $10+12+14+16+18=70$. Each carry reduces the sum of the digits of the result by 9 compared to the sum of the digits of both addends. ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,639 |
4. Masha bought 2021 weights of pairwise distinct masses. Now Masha places one weight on each pan of a two-pan balance (weights placed on the balance previously are not removed). Each time the balance is in equilibrium, Masha rejoices. What is the maximum number of times she can find a reason to be happy | Solution. Answer: 673 times.
Example. Let Masha have bought 673 triples of the form $x, y, x+y$ (we will choose the weights of the new triples so that they do not duplicate the old ones). The last two weights are any. Masha puts $x$ on the left pan, then $y$ on the same left pan, and finally $x+y$ on the right pan and... | 673 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,640 |
5. An Indian engineer thought of a natural number, listed all its proper natural divisors, and then increased each of the listed numbers by 1. It turned out that the new numbers are all the proper divisors of some other natural number. What number could the engineer have thought of? Provide all options and prove that t... | Solution. Answer: 4 or 8.
Let the original number be $n$, and "some other number" be $m$. The number $m$ is odd because an even number has a divisor of 2, which could not have resulted. Therefore, all its divisors are odd. This means all divisors of $n$ are even. Thus, $n$ is a power of two. The options $n=4$ and $n=8... | 4or8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,641 |
6. There is an unpainted board $101 \times 201$. The first player has a bucket of yellow paint, the second player has a bucket of blue paint. On each turn, each player can paint a row (horizontal or vertical) with their color. When yellow and blue paints mix, they produce green. Further mixing of green with yellow or b... | Solution. Answer: The second player wins. Let's divide the rows into pairs: we will match the first row with the first column, and the remaining rows and columns will also be paired. If the first player paints a certain row, then the second player paints the paired row.
We will prove that with this strategy, the secon... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,642 |
7. Seven fishermen stand in a circle. The fishermen have a professional habit of exaggerating numbers. Each fisherman has a measure of lying (each has their own, an integer) - how many times the number mentioned by the fisherman is greater than the true value. For example, if a fisherman with a lying measure of 3 catch... | Solution. Note that the product of all the numbers named by the fishermen in each of the surveys is the product of the number of fish caught, multiplied by the product of all the measures of lying of these fishermen. Therefore, the seventh answered $\frac{12 \cdot 12 \cdot 20 \cdot 24 \cdot 32 \cdot 42 \cdot 56}{12 \cd... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,643 |
1.2. Prove that no matter how the colonies are initially arranged, migrations can be used to settle the colonies with one on each island. | Solution. Induction on the number of vertices.
The base case is trivial.
Transition. We will prove that in any vertex, it is possible to obtain a non-zero value. Hang the tree from this vertex. Consider such a semi-invariant: the sum over all vertices of the quantities $x_{i} n^{-d_{i}}$, where $x_{i}$ is the number ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,645 |
1.3. Prove that the number of colonies on a given island will never exceed the number of adjacent islands by more than 1. | Solution. We will show that any possible distribution of numbers on the tree can be achieved by organizing migrations so that each colony moves no further than to a neighboring island. This immediately implies the required statement.
We prove this by induction on the number of migrations. Base case: zero migrations, w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,646 |
1.4. A certain number of migrations have occurred. After this, an ornithologist landed on each island. Each ornithologist can fly to another island in a personal helicopter along the same air corridors. However, for safety reasons, it is prohibited to take off from neighboring islands and fly over the same corridor or ... | Solution. By the considerations from the previous solution, we can assume that each colony has not moved further than to the neighboring island. The island to which each colony will migrate will be chosen as in the inductive assumption of the previous point. We will draw an arrow on the edge of the tree if the colony h... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,647 |
2.4. Let $\angle R=135^{\circ}$. The perpendicular from $A$ to the nearest side of the angle intersects the smaller circle at point $P$, the perpendicular from $A$ to the second side intersects $B P$ at point $Q$. Finally, let $O_{1}$ and $O_{2}$ be the centers of the original circles, and $O$ be the center of the circ... | Solution. Let's perform the same inversion as in the previous step, and we will again obtain a right angle and a pair of circles inscribed in it, intersecting at an angle of $135^{\circ}$. The perpendicular $AP$ will transform into the diameter of the larger of the circles, and the point $i(P)$ will be its intersection... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,648 |
3.2. Let $a=4, b=3$. Prove that there will be a desired pair containing one of the extreme numbers. | Solution. If $p=3$, then everything is clear.
If $3^{(p-1) / 2} \equiv 1(\bmod p)$, then by Fermat's Little Theorem
$$
3^{\frac{p-1}{2}}+4^{\frac{p-1}{2}}=3^{\frac{p-1}{2}}+2^{p-1} \equiv 1+1 \equiv 3^{p-1}+4^{p-1} .
$$
Otherwise, $3^{(p-1) / 2} \equiv -1$. For example, because from Fermat's Little Theorem $3^{p-1} ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,650 |
3.3. Prove that the desired pair will be found if $a=4, b=7$. | Solution. Cases $p \leqslant 7$ are checked manually.
- $7^{(p-1) / 2} \equiv 1(\bmod p)$, this case is handled exactly the same as in $\mathbf{3 . 2}$;
- $7^{(p-1) / 2} \equiv-1(\bmod p)$.
Then $4^{\frac{p-1}{2}}+7^{\frac{p-1}{2}} \equiv 0(\bmod p)$, so if all $p-1$ residues are distinct, their sum is not equal to 0... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,651 |
3.4. Prove that the desired pair will be found if $\frac{p-1}{2}$ is prime, $a=2$ and $b=3$.
# | # Solution.
Lemma 1. Let $q>2$ be a prime, and $u k<q$ such that for any $x=1,2, \ldots, q-1$, the number $x u$ and the remainder of $k x$ modulo $q$ have different parities. Then $k=q-1$.
Proof of Lemma 1. Substituting $x=1$ gives that $k$ is even, and thus $k x$ is always even, meaning the remainder of $k x$ has th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,652 |
4.1. Let \( f(x) = x^2 - x + 1 \), \( g(x) = -x^2 + x + 1 \). Find a non-constant polynomial \( h \) such that \( h(f(x)) = h(g(x)) \). | Solution. For example, such a polynomial is $h(x)=(x-1)^{2}$. | (x-1)^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,653 |
4.2. Prove that there do not exist quadratic trinomials \( f \) and \( g \) such that \( f(g(x)) = x^4 - 3x^3 + 3x^2 - x \). | Solution. Notice that $x^{4}-3 x^{3}+3 x^{2}-x=x(x-1)^{3}$. Therefore, $f$ has two roots (possibly coinciding). Let $f(s)=f(t)=0$, and let $g(a)=$ $g(b)=s, g(c)=g(d)=t$ (possibly, some of these numbers coincide, then they are considered as multiple roots). But we know that three of the numbers $a, b, c, d$ are equal to... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,654 |
4.3. Let $h$ be a non-constant polynomial, $f \neq g$, and let $h(f(x)) = h(g(x))$. Prove that $f + g$ is a constant polynomial. | Solution. The degrees of $f$ and $g$ coincide, and the leading coefficients are equal or differ in sign (if the degree of $h$ is even). In any case, they can be represented as $k+l$ and $\pm(k-l)$, where $\operatorname{deg} k=\operatorname{deg} f=\operatorname{deg} g>\operatorname{deg} l$. Now let's expand the brackets... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,655 |
4.4. Let non-constant distinct polynomials $f$ and $g$ with positive leading coefficients be such that
\[
\begin{aligned}
& f(f(x) g(x)) + f(g(x)) \cdot g(f(x)) + f(f(x)) \cdot g(g(x)) = \\
& \quad = g(f(x) g(x)) + f(f(x)) \cdot f(g(x)) + g(f(x)) \cdot g(g(x))
\end{aligned}
\]
Prove that $f$ and $g$ differ by only on... | Solution. Let $h(x)=f(x)-g(x)$. Then the scary condition
$$
\begin{aligned}
& f(f(x) g(x))+f(g(x)) \cdot g(f(x))+f(f(x)) \cdot g(g(x))= \\
& \quad=g(f(x) g(x))+f(f(x)) \cdot f(g(x))+g(f(x)) \cdot g(g(x))
\end{aligned}
$$
collapses into a simpler form $h(f(x) g(x))=h(f(x)) h(g(x))$. It remains to prove that only $h(x)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,656 |
6.1. Let $k=4, n=3$. Can Pasha always ensure that one of the chips reaches the last cell? | Solution. Yes. Move 2 chips to the second cell, without loss of generality, Roma will remove the moved chips from the second strip. Now move one of the just moved chips and one from the second strip, Roma must remove the first one, otherwise it will reach the end on the next move. Now we have one chip in the second cel... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,657 |
6.2. Let $k=4, n=100$ and Pasha moves chips from only two cells (one in each strip) on each turn. Prove that Roma can ensure that no more than 50 chips (including those removed) end up in the last cells of their strips. | Solution. Roma needs to ensure that no more than 50 chips, upon reaching the third cell, are immediately removed. Strategy: If Pasha moves all chips from the 1st cell to the 2nd or from the 2nd to the 3rd, we remove the group in the strip where more chips were moved. For example, if in one strip $k$ chips are moved fro... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,658 |
6.3. Let $n<2^{k-3}$. Prove that Roma can make it so that no chip reaches the end. | Solution. Let the rating of a chip standing on cell $k$ be $2^{k}$, and the rating of the chip arrangement be the sum of the ratings of all chips in the arrangement. When Pasha makes a move, the rating of some group of chips doubles; by removing one of the parts of this group, Roma can ensure that the rating of the gro... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,659 |
6.4. Let $n > k \cdot 2^{k}$. Prove that Pasha can ensure that at least one of the chips will reach the end. | Solution. Let's divide everything into blocks of $2^{k}$ chips. In one move, we shift the rightmost block in each strip, or one of them, and split it into blocks of half the size. Two of these four new blocks are removed by Roma, so the total number of blocks remains unchanged. Note that we can always split in half unt... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,660 |
2. The gnomes have gone to work, and Snow White is feeling lonely. She laid out a pile of fifteen stones on the table. Every minute, Snow White splits one pile into two non-empty piles and adds a stone to one of them. How can Snow White use these actions to get seven identical piles? | Solution. First, let's understand how many stones are in the piles. With each action, the number of piles increases by one, as does the number of stones. Therefore, 7 piles will arise after 6 actions, with the total number of stones being $15+6=21$. This means each pile should have three stones. From these consideratio... | 3333333 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,661 |
4. In the company, several employees have a total monthly salary of 10000 dollars. A kind manager proposes to double the salary for everyone earning up to 500 dollars, and increase the salary by 500 dollars for the rest, so the total salary will become 17000 dollars. A mean manager proposes to reduce the salary to 500 ... | Solution. Answer: 7000
Let's look at the difference between the total salary from the kind manager and the current one, and we will understand that this is the total salary from the evil manager:
up to 500 - from the doubled we subtract the current, we get the current
over 500 - from the increased by 500 we subtract... | 7000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,663 |
5. Red Martians always tell the truth, while blue Martians lie and then turn red. In a company of 2018 Martians, each in turn answered the question of how many red ones there were among them. The answers were the numbers $1,2,3, \ldots, 2018$ (in that exact order). How many red ones could there have been initially? | Solution. Answer: 0 or 1
Let's prove that there cannot be two or more red individuals. Suppose they do exist, and let's look at the first two red individuals. The difference in their answers is the number of people who spoke between them, +1. This means that the number of red individuals should have changed by this nu... | 0or1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,664 |
6. Dima wrote a ten-digit number in which even and odd digits alternate. Then he swapped the digits, so that even and odd digits alternate again. He added both numbers and found that the sum also has alternating even and odd digits. Can the sum be an eleven-digit number? | Solution. Answer: No, it cannot.
Suppose it can. When adding two 10-digit numbers, a new digit can only arise from a carry, i.e., the first digit of the 11-digit number can only be 1. By the alternation of parity, we have: $1 O E O E O E O E O E$
At the end, there is an odd digit, which means the last digits of the a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,665 |
7. Vova cut out a shape without holes from a grid paper (the side of each cell is 1 cm) along the grid lines, and Nikita made straight cuts (along the grid lines) totaling 2017 cm in Vova's shape, and the shape fell apart into individual cells. Prove that Vova's shape had a straight segment of the boundary with a lengt... | Solution. We will prove by contradiction. Suppose there are no such plots, i.e., there are no two adjacent cells on the boundary. Then, for any cell on the boundary, its neighbors are corner neighbors. We will color the cells of the figure in a checkerboard pattern. Choose a cell on the boundary and walk around the bou... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,666 |
6. A stork, a cormorant, a sparrow, and a pigeon decided to weigh themselves. The weight of each of them turned out to be an integer number of parrots, and the total weight of all four was 32 parrots. Moreover,
- the sparrow is lighter than the pigeon;
- the sparrow and the pigeon together are lighter than the cormora... | Solution:
Let's denote the weights of the birds by A, B, V, and G respectively.
Since $\Gamma +$ B $< \mathrm{A} + \mathrm{B}$, then $\Gamma +$ B is no more than 15 parrots, while $\mathrm{A} + \mathrm{B}$ is no less than 17 parrots.
If a sparrow could weigh at least 5 parrots, then a pigeon must weigh no less than ... | A=13,B=4,\Gamma=5,V=10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,668 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,670 |
1. Konstantin pronounced the names of all natural numbers from 180 to 220 inclusive, while Mikhail - from 191 to 231 inclusive. Who pronounced more words and by how many? | Solution. Let's remove the numbers that both have: 191-220. Then each will have 11 numbers left: Konstantin has 180-190, and Mikhail has 221-231. Note that in the names of the numbers 181-189, 221-229, and 231, there are three words each, while in 180, 190, and 230, there are two words each. This means that Mikhail sai... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,674 |
2. Numbers from 1 to 9 are arranged in a row. It is known that any two numbers standing one apart differ by 1. Can the number 4 be at the end of this row? | Solution. Note that the number immediately following 1 can only be 2. Therefore, the number 1 is either at the end or next to the end. Let's start counting the positions from this end. If 1 is in the first position, then 2 is in the third, 3 can only be in the fifth, and 4 ends up in the seventh position. If 1 is in th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,675 |
4. A girl is standing on the first floor of a 24-story building, on the 13th, $16th$ and 24th floors of which her friends live. There are no stairs in the building, but there is an elevator that can only move 7 or 10 floors up or down. Can the girl visit all her friends by making no more than 10 elevator trips? | Solution. Yes, it can. For example, like this: 1 - 11 - 21 - 14 - 24 - 17 - 10 $20-13-6-16$ | 1-11-21-14-24-17-10-20-13-6-16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,677 |
5. At the festival "Hobbits - for cultural diversity!", more than 20 participants arrived. A correspondent found out that among any 15 participants of the festival, there are at least 4 humans and at least 5 elves. How many hobbits participated in the festival? Provide all possible answers and prove that there are no o... | Solution. Suppose there is at least one hobbit. If there are 10 people among the participants, then in their company with the hobbit and 4 other participants, there will not be 5 elves, which contradicts the condition. Therefore, there are no more than 9 people. Since the total number of participants is more than 20, t... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,678 |
6. Cheburashka bought as many mirrors from Galina in the store as Gen bought from Shapoklyak. If Gen had bought from Galina, he would have 27 mirrors, and if Cheburashka had bought from Shapoklyak, he would have 3 mirrors. How many mirrors would Gen and Cheburashka buy together if Galina and Shapoklyak agreed and set t... | Solution. Let Gena have $x$ times more money than Cheburashka. If we swap Gena's and Cheburashka's money, then with the second method of purchase, the number of mirrors should be equal. Therefore, $3 x=\frac{27}{x}$, from which $3 x^{2}=27$, and $x=3$. So, Gena initially had three times more money than Cheburashka, and... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,679 |
7. The set contained weights of 5, 24, and 43 grams, with an equal number of each type. All the available weights were weighed, and the total mass was found to be $606060 \ldots 60$ grams. Prove that a) at least one weight is lost; b) more than 10 weights are lost. | Solution. a) Note that the mass of one complete set is 72 grams. This number is divisible by 8, but the total mass is not, which means something has definitely been lost. b) The mass of one complete set is divisible by 24, but the total mass is not, although it is divisible by 12. Since the mass of the weights of the s... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,680 |
1. Can we obtain the triplet of numbers $2,6,9$ in some order from the triplet with numbers $2,4,7$? | Solution:
If $f$ is a polynomial with integer coefficients, then $f(x)-f(y)$ is divisible by $x-y$ for any integers $x, y$. Now notice that $7-2=5$, and among the numbers $2, 6, 9$ there are no such numbers whose difference is divisible by 5. (Here, either apply the above statement to the polynomial that is the compos... | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,681 |
2. Can we obtain $(1,10,7)$ (numbers in exactly this order) from the triplet $(1,4,7)$?
Solution: | Answer: No.
Let some sequence of trinomials transform $(1,4,7)$ into $(1,10,7)$. It is easy to see that such a single trinomial does not exist (if this trinomial is $a x^{2}+b x+c$, then by substituting we get, for example, the system: $a+b+c=1,16 a+4 b+c=10,49 a+7 b+c=7$, from which, for example, $a=-\frac{2}{3}$).
... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,682 |
3. Let $N=400$. What is the maximum number of balls that the magician can guarantee are not in their correct vessels?
# | # Solution:
Answer: 533.
Let's imagine that inside the vessels, the balls are divided into cells, and each cell contains one ball. Then we can imagine that (regardless of whether they are in the same vessel or different ones) balls $i$ and $j$ simply swap cells upon the command ( $i j$ ). Thus, any sequence of comman... | 533 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,683 |
1. In the marked points (see figure) there are 4 burrows. In them live hobbits: Frodo, Sam, Merry, and Pippin. Frodo's burrow is closer to Merry's burrow than to Pippin's. And Sam's burrow is closer to the river than Merry's, but farther from the tree line than Pippin's. Who lives where? Justify your answer. | Solution. Let $A, B, C, D$ be the holes marked in the order "from left to right and from top to bottom". Sam's hole is $C$ (it is not the farthest from the river, i.e., not $A$ and not $B$, but also not the closest to the forest strip, i.e., not $D$). From the conditions, it follows that Frodo does not live in $A$ (oth... | FrodoinB,MerryinA,PippininD,SaminC | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,685 |
2. The Elector George has 100 coins, some of which may be counterfeit (possibly all or none). George can show one or several coins to an expert, who will tell him how many of them are counterfeit. The problem is that the only expert in the area is Baron Munchausen, and he exaggerates: the result given by the Baron is a... | Solution. Yes, there are many ways to do this. For example, we can bring the expert one coin at a time, 100 times. If not all of his answers are the same, then the fake coins are those on which the baron named a larger number. If all the answers are the same (let's say they are $x$), then either all the coins are genui... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,686 |
3. 31 cars started simultaneously from one point on a circular track: the first car at a speed of 61 km/h, the second at 62 km/h, and so on (the 31st at 91 km/h). The track is narrow, and if one car overtakes another by a full lap, they crash into each other, both fly off the track, and are eliminated from the race. In... | Solution. First, the fastest car collides with the slowest, then the second fastest collides with the second slowest, and so on. In the end, the car with the median speed remains, i.e., the 16th. It travels at a speed of $76 \mathrm{Km} /$ h.
Criteria. Full solution - 7 points.
Solution that correctly describes the o... | 76 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,687 |
6. Eight dwarfs are sitting around a round table, each with three diamonds. The dwarfs' chairs are numbered in order from 1 to 8. Every minute, the dwarfs simultaneously do the following: they divide all their diamonds into two piles (one or both piles may be empty), then give one pile to their left neighbor and the ot... | Solution. Note that gnomes sitting on even-numbered stools always share with gnomes on odd-numbered stools, and vice versa. Since all sharing happens simultaneously, at any moment in time, the total number of diamonds with gnomes on even-numbered stools will be the same as the total number of diamonds with gnomes on od... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,689 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,693 |
1. In a football match between the teams "Zubilo" and "Shayba," bets on the victory of "Zubilo" were accepted at 1 to 2 (meaning the bookmaker would pay out twice as much as the bet if "Zubilo" won), and on the victory of "Shayba" at 1 to 3. Volka managed to place a bet, knowing that he would definitely get back exactl... | Solution. Let Volya bet $2x$ on "Shayba", then his bet on "Zubilo" is $3x$. Since he will receive $6x$ in any case, his bet on a draw is $x$, thus the coefficient for a draw is 1 to 6. | 1to6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,697 |
2. Vasya likes natural numbers that are divisible by each of their non-zero digits, for example, 10 or 122. What is the maximum number of consecutive numbers that Vasya can like? | Solution. Answer: 13.
For example, numbers from 111111111111111111000 to 111111111111111111012 work.
Note that if Vasya likes numbers $x$ and $x+10$, then $x$ cannot have the last digit 3, 6, or 9 (since $x$ and $x+10$ cannot both be divisible by 3). Therefore, more than 13 consecutive numbers cannot exist. | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,698 |
3. On the diagonal $BD$ of the isosceles trapezoid $ABCD$, there is a point $E$ such that the base $BC$ and the segment $CE$ are the legs of a right isosceles triangle. Prove that the lines $AE$ and $CD$ are perpendicular. | Solution. Since the right-angled triangle is isosceles, and the trapezoid is isosceles, the angles between the diagonals and the bases are $45^{\circ}$. Then the diagonals intersect at a right angle. Then in triangle $A C D$, the altitudes intersect at point $E$, from which it follows that line $A E$ intersects $C D$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,699 |
5. On the side $A B$ of an isosceles triangle $A B C$, a point $D$ is marked, on the side $A C$ - a point $E$, and on the extension of the base $B C$ beyond point $B$ - a point $F$, such that $C D = D F$. On the line $D E$, a point $P$ is chosen, and on the segment $B D$ - a point $Q$ such that $P F \| A C$ and $P Q \|... | Solution. We will prove that triangles $D F Q$ and $C D E$ are equal by two angles and the side between them, from which the required equality of segments will follow.
- $C D=D F$ by the cond... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,700 |
6. Each cell of a $10 \times 10$ board is painted black or white. A cell is said to be out of place if it has at least seven neighbors of a different color than itself. (Neighbors are cells that share a common side or corner.) What is the maximum number of cells on the board that can be out of place at the same time? | # Solution.
Answer 26. The example consists of 13 dominoes, which do not share any cells and do not touch the border of the board (see right).
Estimation. Let's call a cell that is not in its place an NVCT-cell. Obviously, NVCT-cells cannot be adjacent to the border. Let's examine what form a connected component of N... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,701 |
7. Vasya has $10 n^{2}$ triples from an $n$-element set. Prove that in this set there exist elements $a, b, c, d, e, f$ such that Vasya has the triples $\{a, b, d\},\{b, c, e\},\{c, a, f\}$. | Solution. Let's call a pair of elements bad if it is contained in more than 0 but fewer than 4 triples. We will remove such triples until there are no bad pairs left. Note that each pair of elements was considered no more than once, meaning that we removed no more than $4 C_{n}^{2}=4 \cdot \frac{n(n-1)}{2}<2 n^{2}$ tri... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,702 |
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